Dauda Yakwo

- • ^NDORSED BY: ENGR. M.D. ABDULLAHI, BSc (Physics), H S c ^

FOREWORD BY:DR. JACOB TSADO, B.Ei.g„ MEng., P?»FI ET, FNSE, FAEng, MFR

H n iB e n )

SE N IO R

ALL-INCLUSIVE

CALCULATIONS IN PHYSICSFOR

SENIOR SECONDARY SCHOOLS

Solomon Dauda Yakwo B.Eng., p g d e

REESE MAU CONCEPT LIMITED

© SO LO M O N Y. DAUDA 2 0 0 9First Published 2010

All rights reserved. No part of this book may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the written permission of the author and

copyright owner.

ISBN 978-978-49081-3-9

Published byReese Mau Concept Ltd.BEFS Electronic Villa,Plot 485, Utako District, Abuja.

For enquiries, criticism or bulk purchase of this book, contact the author on any of these addresses;

Mr. Solomon Dauda Yakwo,P.O.BOX 4157, Minna, Niger State.

Email: Solomon.dauda@yahoo.comGSM: 08057779257, 08065039265, 08084143149, 08095112354

ii

f o r e w o r d

It has been ascertained that many students fail physics as a result of poor

mathematical background. It is also true that 70% of physics is mathematics. If you are

writing or preparing for physics in any of these examinations: WAEC, NECO, NABTEB,

NDA, JAMB or Unified Tertiary Matriculation Examination, this book - ALL-

INCLUSIVE CALCULATIONS IN PHYSICS - is your sure companion to success.

This book has been greatly simplified to the most basic level and the examples

arranged in proper ascending order. It has not only covered a lot o f topics but treated

them to the highest satisfactory level. The content is suitable for senior secondary school

level, as well as students preparing for pre-degree and preliminary courses in universities,

polytechnics and colleges of education. The exercises are numerous and widespread,

covering examinations of many years.

This book seems to be the best of in its class. Thus, I strongly advise that you

should have this book and at least one theory physics textbook. However, if you cant

afford many but one, let that one be THIS ONE: ALL-INCLUSIVE CALCULATIONS

IN PHYSICS.

The author is an injector and generator. I commend the author, Solomon Dauda

Yakwo, for writing this book. It is a masterpiece. Above all, I congratulate him for the

formulae he has produced in radioactivity. Through this book, the author has “valentined”

mathematics and physics. Those who buy and study this book to the fullest can be sure of passing any physics calculation examination.

A tree cannot make a forest. The best of this book will include your constructive suggestions sent to the author.

Dr. Jacob Tsado B.Eng., M.Eng., Ph.D(Uniben) Electrical and Computer Engineering Department, Federal University of Technology, Minna, Nigeria.

BRIEF CONTENTS

BOOK ONEChapter 1SPEED VELOCITY AND ACCELERATION............................................................................ ..

Chapter 2FRICTION.......................................................................................................................22

Chapter 3ELASTIC PROPERTIES OF SOLIDS...................................................................................33

Chapter 4LINEAR, AREA AND CUBIC EXPANSIVITIES.................................................................... 41

Chapter 5WORK, ENERGY AND POWER....................................................................................... 50

Chapter 6ELECTRIC FIELD............................................................................................................. 59

Chapter 7ELECTRICAL ENERGY AND POWER.................................................................................84

BOOK T W OChapter 1SCALAR AND VECTOR.................................................................................................... 95

Chapter 2PROJECTILES................................................................................................................... I l l

Chapter 3EQULIBRIUM OF FORCES............................................................................................... 121

Chapter 4EQULIBRIUM OF BODIES IN LIQUID................................................................................ ..

ChaptersCIRCULAR & SIMPLE HARMONIC MOTION........................ ............................................ 14g

Chapter 6LINEAR MOMENTUM: NEWTON'S LAW OF MOTION................................................... 162

Chapter 7MACHINES........................................................................................................................ ..Chapter 8TEMPERATURE AND ITS MEASUREMENT........................................................................ ..

Chapter 9MEASUREMENT OF HEAT ENERGY...

.200

Chapter 10 GAS LAWS.

Chapter 11 PRESSURE IN FLUIDS

Chapter 12 WAVES.....

Chapter 13 mRELECTION OF LIGHT WAVES........................................................................................

Chapter 14REFRACTION OF LIGHT WAVES.................................................................................... 284

Chapter 15SOUND WAVES............................................................................................................312

BOOK THREEChapter 1GRAVITATIONAL FIELD...... ..................................................

Chapter 2ELECTRIC FIELD: STATIC ELECTRICITY...................................

Chapter 3ELECTRIC FIELD: CURRENT ELECTRICITY................................

Chapter 4

ELECTROLYSIS, MAGNETIC AND ELECTROMAGNETIC FIELDS

Chapter 5SIMPLE A.C. CIRCUIT................................

Chapter 6

ENERGY QUANTIZATION AND WAVE-PARTICLE PARADOX

Chapter 7RADIOACTIVITY AND NUCLEAR REACTION

APPENDIX.......................................

.334

...344

...365

...378

...389

.409

• 424

■ 452

DETAILED CONTENTS

Preface.....................................................

To The Physics Student..........................

Acknowledgement.................................

....... xii

......xiv

....... xii

BOOK 1

Chapter 1SPEED VELOCITY AND ACCELERATION..................................Spee i ..............................................................................................Velocity...........................................................................................Rectilinear Acceleration..............................................................Equations of Uniformly Accelerated Motion..........................Problem Solving Using Equations of Motion...........................Instantaneous Velocity and Instantaneous Acceleration.....Graphical Solution To Problems In Rectilinear AccelerationMotion under Gravity: Gravitational Acceleration................Exercise 1........................................................................................

.1

...1

...1

...1

...2...3...6..8.13..15

Chapter 2 FRICTION.Exercise 2.

.2230

Chapter 3ELASTIC PROPERTIES OF SOLIDS.....................................................................................................33Young M odulus....................................................................................................................................33W ork Done In Springs and Elastic Strings.......................................................................................33Exercise 3............................................................................................................................................... 38

Chapter 4LINEAR, AREA AND CUBIC EXPANSIVITIES.....................................................................................41Linear Expansivity..................................................................................................................................41

Area and Cubic Expansivities.............................................................................................................44

Real and Apparent Expansivities....................................................................................................... 44

Exercise 4 .................................................................................................................................................. 47

Chapter 5WORK, ENERGY AND POWER........................................................................................... ..Work.....................................................................................................................................Energy.................................................................................................................................50Power................................................................................................................................. ..EXERCISE 5.......................................................................................................................... ..

vi

Chapter 6ELECTRIC FIELD: CURRENT ELECTRICITY.................Electric Current........................................................Ohm's law................................................ ............... , r nsSeries and Parallel Arrangement of Resistors an E.m.f, Internal Resistance, Terminal p.d and Lost Vo ...

EXERCISE 6............................................................................

.by

..59

..59..59..66..73

Chapter 7 84ELECTRICAL ENERGY AND POWER.................... 84Electrical Energy.................................................................................. 84Electrical Power..................................................................................... 85Electrical Heat Energy............................................................................... 90EXERCISE 7........................................................................................................................

BOOK 2

Chapter 1SCALAR AND VECTOR..........................................................................................................95

Vector Addition............................................................................................................... ^5Two Parallel Forces Acting in the Same Direction........................................................95Two Forces Acting In Opposite Direction in a Straight Line.......................................95Two Forces Acting Perpendicularly............................................................................... 95Two Forces Acting at Acute Angle..... ............................................................................96More Than Two Forces Acting at a Point...................................................................... 99Vector Resolution...........................................................................................................Exercise 1......................................................................v ................. 1

Chapter 2PROJECTILES........................................................................

Objects Projected Horizontally From a Height........Objects Projected at an Angle, 0, to the Horizontal Exercise 2...........................................................

Chapter 3EQUUBRIUM OF FORCES..............................

Moment of a Force....................Moment of a Couple..............Triangle of Forces..................Exercise 3..............................

I l l..111..111119

121-121-.127-1 2 8132

EQULIBRIUM OF BODIES IN LIQUID...Archimedes Principle..............Density and Relative Density- Exercise 4..................................

Chapter 4 . .140.140.140...145

Chapter 5CIRCULAR & SIMPLE HARMONIC MOTION...

Circular Motion......................................Simple Harmonic Motion.....................Simple Pendulum..................................Energy of Simple Harmonic Motion .... Exercise 5...............................................

...149

....149

....151

....153....157.....158

Chapter 6LINEAR MOMENTUM: NEWTON'S LAW OF MOTION

Force, Impulse and Momentum.......................Conservation of Linear Momentum.................Explosion: Recoil of a Gun.................................Jet and Rocket Propulsion.................................Weight of a Body in a Lift..................................Exercise 6.............................................................

...162

......152......166......169...... 169...... 170.......171

Chapter 7MACHINES.................................................................................

Mechanical Advantage, Velocity Ratio and Efficiency.Types of Machines...........................................................The Pulley or Block and Tackle System.........................The Inclined Plane............................................................The Screw...........................................................................The Wheel and Axle.........................................................Gear W heels......................................................................Exercise 7...........................................................................

175..175177..177..179...182..184...185..185

Chapter 8TEM PERATURE AND ITS MEASUREMENT........................................................................... 189

Temperature Scales.............................................................................................. 189Conversion Between Temperature Scales............................................................1 8 9

Thermometric Calculations................................................................................... ..Liquid-in-glass Thermometers............................................................................. 190Resistance Thermometers.................................................................................. ....193Gas Thermometers..................................................................................................196Exercise 8................................................................................................................. 197

viii

Chapter 9MEASUREMENT OF HEAT ENERGY....................

Specific Heat Capacity and Heat Capacity.Methods of Mixtures................................Electrical method......................................Temperature Difference in a Water Fall-Latent Heat...............................................Latent Heat of Fusion...............................Latent Heat of Vapourization...................Relative Humidity.....................................Exercise 9..................................................

.200...200..202...207..210..211...211...212..217..218

Chapter 10GAS LAWS.........................................................................................................

Gas Pressure Measurement.........................................................................Boyle's Law..................................................................................................Application of Boyles Law to Gas Trapped in a Tube by Mercury ColumnApplication of Boyle's Law to Air Bubbles in a Water Body........................Charles Law..................................................................................................Pressure Law................................................................................................General Gas Law...........................................................................................Work Done by an Expanding at Constant Pressure.....................................Cubic and Pressure Expansivity of a Gas.................1...................................Exercise 10...................................................................................................

.224...224...226...227..233...235..237..238..240.241..244

Chapter 11PRESSURE IN FLUIDS................................................... .................

Fluid Pressure.........................................................................Atmospheric pressure............................................................Pascal Principle......................................................................Hare's Apparatus...................................................................Exercise 11..............................................................................

Chapter 12WAVES.........................................................................................

General Characteristics of Waves..........................................Stationary Waves...................................................................Mathematical Representation of Progressive Wave Motion Exercise 12............................. ................ ...............................

Chapter 13RELECTION OF LIGHT WAVES...................................

The Pinhole Camera: Magnification.......................Reflection of Light at Plane Surface..................Images formed by Inclined Mirrors.................Rotation of Plane Mirrors................................Spherical Mirrors: Concave and Convex.............

.250

.251

.251252253 255

.257-257-2 6 2-2 6 3...268

.271-271.271

...275.275-.277

ix

Exercise 13lOU

Chapter 14REFRACTION OF LIGHT WAVES..................................................

Laws of Refraction and Refractive Index ..........................Real and Apparent Depth...................................................Critical Angle and Refractive Index....................................Refraction of light Through Triangular Prisms..................Refraction of light Through Concave and Convex Lenses.Application of Light Waves................................................Microscopes and Telescopes and Magnifying Glass........Exercise 14...........................................................................

.284

.284...290...294..294...297....303....304....305

Chapter 15SOUND WAVES....................

Velocity and Reflection of Sound Waves...................................Echoes..........................Vibration of Strings................Fundamental Frequency. Harmonics. Overtones.....................Sonometer: Factors Affecting Frequency of Vibrating Strings.Vibration of Air Column in Pipes................................................Vibration in Closed Pipes...........................................................Vibration in Open Pipes............................................................The Resonance Tube...................................................................Beat Frequency...........................................................................Exercise 15...................................................................................

.312

..312.313.317„317.320„323„323. 325...326...329...329

BOOK 3

Chapter 1

GRAVITATIONAL FIELD...........................................Gravitational Force........................................Gravitational Potential..................................Gravitational Field Intensity.........................Weight of Objects in Space...........................Escape Velocity...............................................Exercise 1........................................................

J Chapter 2ELECTRIC FIELD: STATIC ELECTRICITY....................

Coulomb's Law................................................Electric Field Intensity...................................Electric Potential and Potential Difference

.334334.336.336.338„340.342

.344344.345.347

X

Electron Volt..........................................................Capacitors and Capacitance................................Series and Parallel Arrangement of Capacitors.

Parallel Connection...............................................Series Connection.................................................Combined Series and Parallel Connection........Energy Stored In Capacitor..................................Exercise 2................................................................

...348....349...350..350...351...353...355..359

Chapter 3'^^tECTRIC FIELD: CURRENT ELECTRICITY.

Resistivity and Conductivity......................................Galvanometer Conversion: Shunt and Multiplier...Conversion of Galvanometer to Ammeter..............Conversion of Galvanometer to Voltmeter.............Potentiometer. Wheatstone Bridge. Meter Bridge.Potentiometer.............................................................Wheatstone Bridge.....................................................Meter Bridge................................................................Exercises.......................................................................

...365

...365,...367..367..369..370..370..371..372..373

r£bapter 4^ELEC T R O LY SIS , MAGNETIC AND ELECTROMAGNETIC FIELDS

ELECTROLYSIS...................................................................Faradays First Law of Electrolysis.....Faraday's Second Law of Electrolysis. Electrochemical Equivalent...............Exercise 4A..........MAGNETIC FIELDExercise 4B..........................ELECTROMAGNETIC FIELDTransformers......................Exercise 4C.........................

/

.378

.378

.378

.378

.378380,382383.384.384.387

Chapter 5SIM PLE A.C. CIRCUIT....................................................................

Peak and Root Mean Square Values..................................Reactance and Impedance.............................................— •A.C. Circuit Containing Only Resistor.................................A.C. Circuit Containing Only Inductor................................A.C. Circuit Containing Only Capacitor..............................A.C. Circuit Containing Only Inductor and resistor..........A.C. Circuit Containing Only Resistor and Capacitor........A.C. Circuit Containing Only Inductor and Capacitor.......A C. Circuit Containing Resistor, Inductor and Capacitor.

.....389

..... 389

..... 391

..... 391

..... 392..... 393.....394..... 396.....399.....400

xi

Resonance in A.C. Circuit............................................. 403Power in A.C. Circuit...................................................... 405Exercise 5..............................................................................

Chapter 6 409^ E N E R G Y QUANTIZATION AND WAVE-PARTICLE PARADOX........................... 4 0 g

Energy Quantization................................................................... 411Electron Volt and Kinetic Energy................................................ 413Photoelectric Effect: Einstein Equation........................................... 410Stopping Potential................................................................................ 47gWave-Particle Duality............................................................................. 4^9Heisenberg Uncertainty Principle.......................................................... 420Exercise 6..........................................................................................................

H/Chap,er7 424^ R A D IO A C T IV IT Y AND NUCLEAR REACTION...............................................................

I 424Radioactivity...................................................................................................... ^

Half-life: Equations of Radioactive Disintegration...............................................Decay Constant...........................................................................................................

1 n 4-38Relation Between Decay Constant and Disintegrating Ratio................................440Disintegrating Ratio and Graphical Solutions...........................................................

Transformation of Elements Through Radioactivity................................................ 442443Nuclear Reaction:.......................................................................................................444Binding Energy...............................................................................................................

c 448

APPENDIX........................................................................................................................... 452

..............402

PREFACEin physics in an entirely different,

This book deals specifically with calculations tQ ensure that a student inhighly simplified and very comprehensive manner. ^ skills o f physics calculations Senior Secondary School acquires the basic Prl" clP JAMB> n a BTEB, and NDA and prepares adequately in advance for WAE ,examinations in such a way that nothing is left to j “ "“ . programmes in universities,

Also, students undergoing pre-degree or p ^ ALL-INCLUSIVE

polytechnics and colleges of education wi J understanding of basicCALCULATION IN PHYSICS extremely useful to th d d Levelcalculations in physics as a prelude to the more complex calculations ,n AdvancePhysics courses.

For every topic, concept or chapter, a brief theoretical explanation is given and the relevant equations are stated. Based on similarity, degree of difficulty and frequency of appearance, various typical examples are taken from a pool of ALL past JAMB (1978 2009), WAEC (1988 - 2009) and NECO (2000 - 2009) questions and solved in a detailed and simplified manner that will ensure that students are fully equipped to handle A N Ycalculation problems, no matter how tricky or difficult the question may be.

At the end of each chapter is a question and answer section that includes ALL other calculation questions (except those used as examples) in JAMB (1978 - 2009), WAEC (1988 - 2009) and NECO (2000 - 2009) related to the concepts or topics treated in that chapter. The author is bold and sincere to declare that if a student invests quality time and effort in going through all the solved examples and in solving all the exercises at the end of each chapter, then there is assurance that the student will confidently and correctly solve any calculation question in WAEC, NECO, JAMB, NABTEB, NDA examinations.

In fact, a renowned Nigerian physics author and prolific writer of a great many science textbooks, G.O. Ewelukwa, affirm that, “...over 70% o f the physics questions are mathematical problems...” In other words, it is almost impossible for any student to pass any physics examination if he or she does not have a good grasp of calculations in physics. Therefore, ALL-INCLUSIVE CALCULATIONS IN PHYSICS is highly recommended to all physics students and teachers of physics in all senior secondary schools.

The author accepts responsibility for any error and encourages both students and teachers to write to the author if any mistake is detected. Suggestions for improvement of this book are welcomed and will be duly acknowledged in subsequent editions or reprints

To all who contributed in making this book a success, I say, thank you and Godbless you abundantly.

The author would like duly acknowledge and profoundly thank the

West African Examinations Council (WAEC), the National Examinations Council

(NECO) and the Joint Adm.ss.ons and Matricu.ation Board (JAMB) for using their pas.physics questions.

Solomon Dauda Yakwo

TO THE PHYSICS STUDENTI wrote this book for you with the understanding that you may be facing one

more of the following challenges as a physics student. nll1- While you accept the fact that physics is very relevant to your future c , y

just don’t like or understand the too many calculations involved in almost every cone p in physics.2- You are among the thousands of Nigerian physics student who have no physics teacher in their school.3. Due to laziness or incompetence, your physics teacher only concentrates on the theoretical aspects of physics and partially or completely ignores the calculation problems.4- After reading or been taught a topic or concept, you are always not sure if you have adequately prepare to solve any calculation question on that topic/concept that may come up in the future in any of these exams; WAEC, NECO, JAMB, NABTEB or NDA.5. You are preparing for post-JAMB test or undergoing a preliminary or pre-degree physics course in a university, polytechnic or college of education, and know that you still do not have the competence and confidence in solving physics calculation problems.

Congratulations! The book you are holding will help you overcome one or more of the challenges you are facing. Read the preface if you have not done so. If you adhere strictly to the following suggestions it will go a long way in increasing your ability, confidence and speed when solving calculation problems in physics.(i) Because this book deals with calculations, I strongly advise you to have and use a scientific calculator, preferably the Porpo* scientific calculator. Please read the manual again and again until you can efficiently and effectively use the calculator to solve problems without delay or mistakes. From my experience as a teacher, the PorpoR scientific calculator is the cheapest, most advanced and easy to use calculator best suited for senior secondary school students and even tertiary students.

(ii) Most questions and examples in this book are objective questions. The only difference between objective and essay question is that in objective questions, you are NOT required to show how you obtained your answers. It is absolutely unnecessary and will not fetch you any mark, hi fact, you are allowed to use any shortcut method (and should devise various shortcut methods peculiar to you) because the examiner is ONLY interested in your answers. On the other hand, an essay question requires you to show in every detail how you arrived at your answers. This involves you stating the formula or equation to be used and showing how you carried out the substitution and calculations. It is very important for you to show ALL your workings or else you will lose a lot o f marks.

(iii) This book deals mainly with calculations in physics and should be used in combination with other major textbooks that concentrate more on the theory o f physics. Apart from M. Nelkon’s Principle o f Physics and A.F. Abbott’s Physics, I recommend any o f these textbooks written by great Nigerian authors: Senior Secondly Physics SSS 1 2&3 by B.L.N. Ndupu, P.N. Okeke and O.A. Ladipo; New School Physics fo r Senior Secondary Schools by M. W. Anyakoha and Senior Secondary Physics by P.N. Okeke and M.W. Anyakoha.

(iv) If you find out any mistake in this book or have a suggestion on how any calculation method can be improved, please feel free and write to me. I will be extremely glad to read from you and will duly acknowledge you in subsequent edition or reprint.

xiv

SPEED, VELOCITY AND ACCELERATIONSPEEDSpeed is defined as the rate of change of distance with time.

Average speed = distance = change in distance = s time change in time t

_ final distance - initial distance _ s final time-initial time t

1 he S.l. unit for speed is m/s.

Example 1A student walks a distance of 3km in 20 minutes. Calculate his average speed.Solution

Average speed = — stance _ 3km _ 3 x 1000m _ 2 5 ^ time 20min 20 x 60s

Example 2A driver traveling at a speed of 115km/hr received a text message on his mobile phone. How far is he, in kilometers, 20s later from when he received the text?Solutionspeed = 115km/hr; time,t = 20s, (20s = -^5hr) distance^ = ?

nicio . 115km 20hr 2300 .Distance = speed x time = -------- x ------ = -------= 0.639kmhr 3600 3600

1

VELOCITYVelocity is defined as the rate of change of distance moved with time in a specified direction. It is also defined as the rate of change of displacement with time. Displacement means distance traveled in a specified direction.

displacement change in displacementvelocity = ------ :--------- = ------ ;-------:— :---------time change in time

final displacement - initial displacement final time - initial time

However speed is used in place of velocity and vice versa. The S.l unit for velocity is m/s.

RECTILINEAR ACCELERATIONThe term rectilinear acceleration means the rate of increase of velocity along a straight- line path in a unit time. When the velocity of an object changes it could be said to accelerate or decelerate. Acceleration is defined as the increasing rate of change of velocity with time. Deceleration on the other hand is defined as the decreasing rate of change of velocity with time.

Deceleration is also called retardation or negative acceleration.Change in velocity

Acceleration (Deceleration) - Time taken for change

_ Final velocity - Initial velocity Final time - Initial time

1

E quations o f Uniform ly Accelerated M otionEquations of motion for a body traveling a ong a acceleration are derived as follows.

straight line with uniform

FimT?LZ7‘Z ora,ready nroving w„h an initial ve.ocUy n begins to accelerate at a m/s2, after a particular time, t sec, it will obtain a final ve oci y v The acceleration, a, will be defined by

_ Change in velocity Time interval

_ Final velocity - Initial velocity Time interval

Substituting letters for words we have,_ v-u

tCross-multiplying we have,v - u = atRearranging,v = u + at ..........................................(1)

This is the first equation of motion.

Second equation o f motion:The second equation of motion can be derived if we consider that a body moving with uniform or constant acceleration must have had an initial velocity u, before attaining a final velocity, v.

Therefore the average velocity is equal to the sum of the initial velocity, u and the final velocity, v divided by two.

U + VThus, Average velocity = —-—

Substituting the first equation of motion, v = u + at into the above we obtain, u + u + atAverage velocity = -----------

_ 2 u + at ~ 2 _ 2 u at~ T ~2= u + / 2 at

, displacement (distance) sAverage velocity = -------------;---------------= -

time t t

- = u + y2 at

Cross multiplying,s = t(u + a t)s = ut + y2 at2 ......................^

This is second equation of motion.

Third equation o f motion:A combination of first and second equation of motion yields the third equation of motion as follows:Square both sides of first equation of motion, v = u + at

2

(v)2 = (u + at)2 v2 = (u + at)(u + at) v2 = u2 + uat + uat + (at)2 v2 = u1 + 2uat + a2t2

Factor out 2a from the last two terms of the right hand side of the above equation.

v2 =u2 + ^ -(2 uat + a2t2)

V= 2a 2a J

v2 = u2 +2a(ut +y2 at2)Substituting the second equation of motion s = ut + y2at2 into the above, we ob^tn

v2 = u2 + 2 as ......................................... (3)This is the third equation of motion.

The three equations of motion with uniform acceleration are:1. v = u + at2. s = ut + !/2at23. v2 = u2 + 2as

Where u = initial velocity in m/s or ms-1 v = final velocity in m/s or ms-1 t = time in sec s = distance in ma = uniform acceleration/deceleration in m/s2 or ms-2

ALW AYS REMEMBER:1. When an object moves or accelerates from rest, its initial velocity, u = 0.2. When a body comes to rest or stops, its final velocity, v = 0.3. When a body’s velocity is constant or not changing, its acceleration, a = 0.

Problem Solving Using Equation Of Motion

Example 3A particle accelerates uniformly from rest at 6.0m/s for 8s and then decelerates uniformly to rest in the next 5s. Determine the magnitude of the deceleration. WAEC 2005 SolutionYou are to calculate the particle’s deceleration. It decelerated to rest from a particular velocity which was attained during acceleration. So, you are to calculate the velocity attained as follows:Initial velocity, u = 0; acceleration, a = 6m/s ; t = 8s; velocity attained, v = ? Substituting into the most appropriate equation, v = u + at

v = 0 + 6x8. v = 48/wIs x

Next the particle decelerated to rest from a velocity of 48m/s, therefore, the deceleration is calculated thus:Initial velocity, u = 48m/s; final velocity, v = 0; time, t = 5s; deceleration, a = ? Substituting into the most appropriate equation,

v = u + at 0 = 4 8 + a x 50 = 48 + 5o

- 5a - 48

3

a = 9.6 ms- 5a _ 48 5 ~ 5

A*body'accelerates »„tfotmly from res. a, the rate o f 3ms'! for 8s. C . l c u l a ^ d i s m c e covered by the body during acceleration.SolutionUsing appropriate symbols deduce, the known and unknown from the question. Acceleration, a = 3m/s2; initial velocity, u = 0; time, t = 8s; distance, s „„notionTherefore the most appropriate equation that includes a, u, t and s is t e secon of motion:

s = ut + X a t2 Substituting:

s = 0x8 + j^x 3 x 8 2 = 0 + ^ x 3 x 6 4= 96 m

Exam ple 5Starting from rest, a Formular One car accelerates uniformly at 25m/s" for 30s. What distance does it cover in the(i) last one second of motion.(ii) seventeenth second of motion.SolutionThe distance in the last one second of motion is the distance it covered between time interval o f the 29th and 30lh second. It is calculated by subtracting the distance in the 29' sec from that of the 30th sec.Distance, s = S30 - S29

But, S30 = ut + l/ 2 at30 and S29 = ut + / 2 at 29

Starting from rest, initial velocity, u = 0. Therefore, ut = 0 in each case s - /K a t30 - X at29

' X a 0-3O _ ^29)

From the question, acceleration, a = 25m/s2; t30 = 30s; t29 = 29s. So substituting into above equation:

s = Vi X 25(302 - 292)= 12.5 (900-841)= 12.5(59)= 737.50m.

(ii) From question acceleration, a = 25m/s2; initial velocity, u = 0; time, t is not equal to 17s because the 17' second is actually the time interval between the 16th and 17th second, therefore time, t = tn - ti6- Substituting into s = X a(t?7 ~ t26)

s = >/2 X 25(172-162)= 12.5 (289 -256 )= 12.5X 33 = 412.5m.

Note: In calculating distance between time intervals, the second equation o f motion,s = ut + Xat2 is modified to s-X a(t2 -tf), where initial velocity u = 0, t2 and ti are the time intervals.

4

! l aT ' e 6 f , , , ft .„„ .g m /s* In how much time will the body travel aA body uniformly accelerates from rest at om/s . udistance of 2.5km?SolutionAcceleration, a = 8m/s2; initial velocity, u = 0; distance, s = 2.5km = 2.5X1000m = 2500m; time, t = ?Substituting into the most appropriate equation, s = ut + at

2500 =0 + } ^ x 8 x t 2 2500 = 4t2

. ’ = ^ 0 0 = 625

t = V625 =25s

Example 7A particle starts from rest and moves with a uniform acceleration of 4m/s . What is its velocity after covering a distance o f 8m.SolutionInitial velocity, u = 0; acceleration, a = 4m/s2; distance, s = 8m; final velocity, v = ?The equation containing u, a, s, and v is the 3rd equation o f motion, v2 = u2 + 2as. Substituting in to most appropriate equation,

v 2 = 0 + 2 x 4 x 8 v 2 = 64

v = Vó4 = 8m/s

Example 8A motorist, travelling at 120km/hr sees a broken down truck at the middle of the road and immediately applies his brakes and comes to a stop with uniform retardation in 20s. What distance does the car travel after the brakes were applied?SolutionConvert 120km/hr to m/s: 120km/hr = -------- — = 33.33m/s

60 x 60 sInitial velocity, u = 33.33m/s; final velocity, v = 0; time, t = 20s.

The retardation, a, must v - u _

Reterdation, a = —------

first be found using v = u + at0 —33.33 ,2—*----------= 1.67m/s

20

Then any o f the other two equations of motion can be used to calculate distance(i) s = ut + X a t2

= 33.33 x 20 + '/2( - 1.67) X 202

= 6 66 .6 -54 X 1.67 X 400

= 666.6 - 334

GO

= 332.60m or

v2 = u2 + 2as s = 2c

0 -(3 3 .3 3 )2 _ -1110 .89 s = ——, , - 7— — - 332.60m

2 x (-1 .6 7 ) -3 .34

5

Example 9 ^ n f lO mA body which is uniformly retarded comes to rest in 5s after travelling aWhat is its initial velocity?Solution _ ?Final velocity, v = 0; time, t = 5s; distance, s = 1 Om; initial velocity, u

Did you notice that acceleration or deceleration is not given? Therefore, . ^asequation of motion that can be applied because all the 3 equations of motion as deceleration/acceleration. A combination of two equations will do.

The deceleration, a, is not given and must first be determined in order to ca cu a e initial velocity. Substituting into v = u + at

v— u 0 - ua = ------ = -------t 5

Substituting this deceleration,—— into the third equation, v2 — u2 + 2as

u 2 = v2 -2as

u 2 = 0 -2 x [ — |x l0

- 2 x - u x l0

u 2 20u „= -----= 4u5

u 2 =4uu 2 4u— = — dividing by u u uu = 4m/s

However, this type of question can be solved more easily using graphical method. See example 15 (Graphical solutions).

Instantaneous Speed/Velocity and Instantaneous Acceleration.Apart from being average, uniform or non-uniform, the speed, velocity and

acceleration of objects could also be instantaneous, that is, at particular point or instant in time.

Instantaneous speed/velocity. This is defined as the actual speed/velocity of an object at any particular point in time. The speedometer of a moving car gives the instantaneous speed/velocity of that car at any moment in time.

Instantaneous velocity at a given instant t is the final, limiting value approached by the average velocity when the average velocity is calculated for a smaller and smaller range of time intervals that includes the instant t. Mathematically, instantaneous speed or velocity is written as;

.. As dsV = l im — = —

At->o At dt

The instantaneous speed at a particular instant has the same value as the average speed provided two conditions are met:

6-

1. The particular instant must be included in At. curve that is nearly a2. The ratio ^ must cover a very small part of the distance timstraight line segment. ^

Instantaneous acceleration at a given instant t is d include the desiredvelocity with time during an infinitesimally small interva o instant t. Mathematically, this is written as;

a =Av

Iim —At-*o At

d vd t

E> imple 10 . g uniformly acceleratedCalculate the instantaneous velocity at 7s of a rocket un ergoin seconds.motion if the position is given by s = 5t + 12t2, where s is in m

Solution . obtained whenFrom differential calculus, we know that speed/ve ocitydistance/displacement is differentiated with respect to time.

If

Then

y = x n dy dx = n xn~x

Also, the derivative of a constant is zero.

Therefore, differentiating s = 5t + 1212

We obtain instantaneous velocity,

— = 5 t1-1 + 2(12t2-1) = 5t° + 24tx d t = 5 + 241

ror t = 7s, —dt ■ o -f- 44 X 7

= 173ms“1Example 11

The velocity v of a space shuttle in a time t is given by v = 25 + 3 t2 Find th instantaneous acceleration at t = 9s. ' na tfleSolutionWhen velocity is differentiated with respect to time, acceleration is obtained

Differentiating v = 25 + 3 t2

We obtain instantaneous acceleration,^ = 0 + 2(3 t2-1)

••• for t = 9s,

= 61dvdt = 6 x 9 = 54m /s

7

raphical Solution To Problems In Rectilinear Accelerationition>Clt tlmC ~ ^ are commonly used to solve problems in Rectilinear

°n and could be in different shapes depending on whether the velocity isaccelerationincreasing, decreasing or constant.The following are various v - t graphs:

Velocity ' m/s A

Time (s)Pig. 1.1: Velocity—Time Graph For An Object Moving With Constant Velocity.

The graph AB is drawn parallel to the time axis.

Fig.1.2 Velocity-Time Graph For An Object Accelerating From Rest.The graph OA begins from the origin. The slope of line OA gives the acceleration of the object.

3.

Fig 1.3 Velocity-Time graph for an object decelerating to rest.The slope OR gives the deceleration or retardation of the object

Example 12A body moving with uniform acceleration, a, has two points (5, 15) and (20, 60) on the velocity - time graph of its motion. Calculate a. WAEC 1991SolutionOn a velocity - time graph, the uniform acceleration, a, is equal to the gradient (slope) of the graph.From question, first point is (5, 15), equivalent to (xi, yi)

Second point is (20, 60), equivalent to (X2, y2)

The two points are plotted below

8

The slope of AB = a = = y?~ ZiAx x 2 — x,

60 -152 0 -5 — = 3.00ms 2

15

Example 13A car runs at a constant speed of 15m/s for 300s and then accelerates uniformly to a speed of 25m/s over a period of 20s. This speed is maintained for 300s before the car is brought to rest with uniform deceleration in 30s. Draw a velocity - time graph to represent the journey described above. From the graph find(*) The acceleration while the velocity changes from 15m/s to 25m/s.(ii) The total distance travelled in the time described.(iii) The average speed over the time described. A.F. ABBOT (J.M.B.)Solution“ ...Constant speed of 15 m/s... ", is represented by AB drawn parallel to the time axis.

“.. fo r 300s... ”, is represented by OH.

.accelerates uniformly to a speed o f 25m/s... " is shown by BC. over a period o f

20s... ”, by HG.“...Speed (25m/s) is maintained... ", signifies constant or uniform speed/velocity and is

therefore represented by CD drawn parallel to the time axis; .maintained for 300s... ”

is presented by GF.“ brought to rest with uniform deceleration... ", is represented by downward sloping DE

and FE represents the time (30s) for deceleration.

Fig. 1.5(i) The acceleration, a is equal to the slope or gradient of BC.

9

. , CJ 2 5 -1 5 __IO = l/ = o.5m/s2AcceleratDn,a = sIopeBC = — = 20--300 20 7

(ii) In a v-t graph, the total distance is equal to the aiea under th &

Total distance travelled is equal to area OABCD Area OABCDE = ABHO + BCGH + CDFG + DEFArea ABHO = Distance covered during constant speed of 15m/s or s Area BCGH = Distance covered during acceleration or velocity change rom

to 25m/s. - ,nnArea CDFG = Distance covered during when speed (25m/s) is maintame Area DEF = Distance covered during uniform deceleration in 30 seconds

Total distance = OABCDE = ABHO + BCGH + CDFG + DEF

= AB x BH + '/2 (BH + CG)HG + (CD x DF) + 54(FE X DF)

= 300 X 15 + 1/2(15 + 25)20 + 300 X 25 + ‘/2(30 X 25)

= 4500 + 400 + 7500 + 375

= 12775m

. , Total distancetravelled 12775 ,(111) Averagespeed=------------------------------= ---------= 19.65m/sTotal time taken 650

Note: Time taken is obtained from the graph, represented on the time axis by OE.

Example 14A car starts from rest at a check point A and comes to rest at the next check point B, 6km away, in 3 minutes. It has first, a uniform acceleration for 40s, then a constant speed and is brought to rest with a uniform retardation after 20s. Sketch a velocity — time graph o f the motion. Determine(i) the maximum speed (ii) the retardation NECO 2003SolutionConvert all units to S.I. units.

6km = 6 X 1000 = 6000m 3mins = 3 X 60s = 180s

The v - t graph is drawn as shown below.

parallel sides of the trapezium; which of course is equivalent to DF or CE.

Total distance, 6000m = Area of trapezium ADCB

6000 = ‘/2(AB + DC) S

6000 = ‘/2(180 + 120) S

6000 = '/2(300) S

10

6000 = 150S

S 6000150

= 40/ms 1

(ii) Retardation, a, is equal to the slope of CBCE 4 0 -0 40 _ . »a = ----= ------------ = ------= - 2m/sEB 160-180 -2 0

. Retardation = 2m/s2

jLxampie i s ,s2 for 30s after which thea) A body at rest is given an initial uniform acceleration ot 8. . e speed attainedacceleration is reduced to 5.0m/s2 for the next 20s. The body main 1 graph of thefor 60s after which it is brought to rest in 20s. Draw the ve ocimotion using the information given above.b) Using the graph, calculate the:

i. maximum speed attained during the motion;ii. average retardation as the body is being brought to rest,

iii. total distance travelled during the first 50s;iv. average speed during the same interval as in (iii).

Solutiona) Graph z frir m sl is calculated as

The velocity attained during acceleration (8.0m/follows: initial velocity, u = 0; acceleration, a=8m/s2; time, t=30s

Substituting into v = u + atv = 0-t- 30x8 = 240m/s

started from rest.“...After which the acceleration is reduced to 5m/s2 fo r the next 20s...”. This signifies deceleration/retardation of 5m/s , therefore the graph BC slopes downward.

The velocity attained after this retardation is found as follows:

The retardation begins from a speed of 240m/s Initial velocity, u = 240m/s

Retardation is negative acceleration Retardation, a = 5m/s2

11

Time, t = 20sTime for deceleration is 20s So, final velocity (speed) is v = u + at

v = 240 + (-5X20)

= 240-100= 140m/s.

Next the graph (CD) is drawn parallel to the time axis to represent the period (60s) du b which the body maintained the speed of 140m/s.

Finally, the graph (DE) slopes downward as the body retards to rest.

b) i. As shown in the graph the maximum speed attained is 240m/s.

ii. Average retardation = slope of DE = = 7m/s'H 110-130 20

iii. Total distance during first 50s = Area of OBCG = Area of triangle OBHArea of trapezium BCGH.

Distance = '/2 O H X B H + >/2 (BH + CG) GH

= '/2 X 30 X 240 + '/2[(240 + 140)] 20

= 3600 + 3800

= 7400m

/ ’ \ a , total distance covered in 50s 7400 , .( i v ) Average speed = --------------------------------------------------- = ------- = 148m/s

total time taken 50

Example 16If a car starts from rest and moves with a uniform acceleration of 10m/s2 for ten seconds, the distance it covers in the last one second of its motion is JAMB 1985SolutionInitial velocity, u = 0; acceleration, a = 10m/s2; time, t=10s

Final velocity, v = u+ atv = 0+ 10x l0 = lOOm/s

The graph is drawn as shown below.

Fig. 1.8

The last one second o f motion is the time between the 9th and the 10th second. Therefore the distance covered in the last one second of motion is:

=Distance in the 10th sec - Distance in the 9th sec.

12

=Area of triangle OBC - Area of triangle OED

='/2 X OC X BC - Vt X OD X DE

=1/2 X 10 X 100 - ‘/ 2 X 9 X 90

= 5 0 0 -4 0 5 = 95m.

Example 17A body which is uniformly retarded comes to rest in 5s after travelling a distance of 10m. What is its initial velocity? NECO 2002Solution

The graph is drawn as shown above. The area of the graph is equal to the distance, 10m. Initial velocity is equal to the height of triangle OAB, which is length OA.

A reaofO A B = 10

Z i X OB X O A = 10

'/2 X 5 X O A = 10

20O A = — = 4/775’_1

5

M otion Under Gravity: G ravitational AccelerationWhen an object is thrown upwards or released from a height, its motion is

governed or affected by gravity. An object thrown upwards experiences a negative acceleration (-g) because its motion is in opposite direction to the gravitational pull of the earth. On the other hand, when an object falls downwards or is released from a height, it experiences positive acceleration (+g) because its motion is in the same direction as that o f the gravitational attraction of the earth.

Equations of motion for gravitational acceleration are derived from the equations of motion for rectilinear acceleration:

v = u + at

s = ut + Xat2v 2 = u 2-t-2as

1. For a body thrown upwards, the following equations apply.v = u - g t

h = u t - j ^ g t2

v2 = u 2 -2 g h2. For a body falling downwards or released from a height the following equations apply.

v = u + g t

h = ut + j^g t2

v2 = u 2 + 2gh

13

Where v = final velocity in m/su = initial velocity in m/s h = height in mt = time in s . 2g = acceleration due to gravity m s

Always remember: „ ., , • . _et v = 01. When an object is thrown upwards, its final velocity, a 1 g . vgl0city u =02. When an object falls downward or is released from a eig , 1 come down3 The time it takes an object to travel upwards is the same t,me it takes to come

Example 18 , tueAt exactly 2hr:00min:00sec a pendulum bob is thrown vertically upwar s ground with an initial velocity of 75m/s. At what time will the bob return to t e grou (Take g=10m/s2)SolutionInitial velocity, u = 75m/s; g=10m/s2; final velocity, v = 0; time, t=?

The equation containing all the known and the unknown items is v = u - gt Substituting:

0 = 75- lO xtlOt = 75

75 c t = — = 7.5s 10

The time, t = 7.5s is the time taken to go upwards which is the same time it takes to come down. Therefore time taken to return to the ground from the moment of throw is,

(7.5 + 7.5)s = 15s.Time of return will be 2hr:00min: 15sec

Example 19Calculate the maximum height a ball of mass 1.2kg will attain if projected vertically upward with an initial velocity of 17m/s.SolutionInitial velocity, u = 17m/s; final velocity, v = 0; g=10m/s2; h=?The mass of the ball is not required.

Substitute into v2 = u - 2gh0 = 172 - 2 x l 0 x h

20h = 172, 289 . . . .h = -----= 14.45m

20

Example 20An object falls freely from a height of 25m onto the roof of a building 5m high. Calculate the velocity with which the object strikes the roof. WAEC 1997SolutionInitial velocity, u = 0; g=10m/s2; h=20 i.e.(25 - 5), because the object does not reach the ground instead it fell on a 5m high roof.Substituting into v2 = u2 + 2gh

v2 = 0 2 + 2 x10x20

v2 =400

v = -J400 = 20m/s

14

Example 21 fA carpenter working at a construction site mistakenly drops his hammer from a eig o 125m. How long does it take to reach the ground?SolutionInitial velocity, u = 0; g=10m /s2; height, h= 125m; time, t=?

Substituting into h = ut + / 2 g t2

125 = 0 + ^ x l 0 x t 2

125 = 5t2

t = V25 = 5s

EXERCISE 1

1. A boy cycles continuously through a distance of 1.0km in 5 minutes. Calculate hisaverage speed. WAEC 1995Ans: 3.33m/s2. An air force jet flying with a speed of 335m/s went past an anti aircraft gun. How far is the aircraft 5s later when the gun was fired.A. 838m B. 3350m C. 670m D. 1675m E. 67m JAMB 1978 Ans: 1675m3. A car travels with a constant velocity of 45 km/hr for 20s. What distance does it coverin this time? NECO 2005 Ans: 250m4. A body starts from rest and accelerates uniformly at 5m/s2 until it attains a velocity of 25m/s. Calculate the time taken to attain this velocity. WAEC 2005 Ans: 5s5. A train has an initial velocity of 44m/s and an acceleration of -4m/s2. Its velocity after 10 sec is A. 2m/s B. 4m/s C. 8m/s D. 12m/s E. 16m/s

JAMB 1983 Ans: 4m/s6. If a car starts from rest and moves with a uniform acceleration of 10m/s2 for ten seconds, the distance it covers in the last one second of motion isA. 95m B. 100m C. 500m D. 905m E. 1000m JAMB 1985 Ans: 95m7. A body starts from rest and moves with uniform acceleration of 6m/s2. What distance does it cover in the third second?A.15m B. 18m C. 27m D. 30m JAMB 1992 Ans: 15m8. A body accelerates uniformly from rest at the rate of 3m/s2 for 8s. Calculate thedistance covered by the body during acceleration. WAEC 1992 Ans: 96m9. A particle starts from rest and moves with a constant acceleration of 0.5m/s2. Calculate the time taken by the particle to cover a distance of 25m. WAEC 1993 Ans: lOsec10. How far will a body move in 4 seconds if uniformly accelerated from rest at the rateof2m s'2. WAEC 1993 Ans: 16m11. A car takes off from rest and covers a distance of 80m on a straight road in 10s.Calculate the magnitude of its acceleration. WAEC 2002 Ans: 1.6m/s212^ A bus travelling at 15m/s accelerates uniformly at 4m/s2. What is the distance

"edvered in 10s? WAEC 2000 Ans: 350m13. A body starts from rest and accelerates uniformly at 5ms'2. Calculate the time takenby the body to cover a distance of 1 km. WAEC 2005 Ans: 20 secs14. A particle starts from rest and moves with a constant acceleration of 0.5m/s2. The distance covered by the particle in 10s isA. 2.5m B. 5.0m C. 25.0m D. 50.0m JAMB 1990 Ans: 25m

15

_ / 2 pairiilate its velocity after travelling15. A body accelerates uniformly from rest at 2m . WA.EC 1988 Ans: 6m/s

/ i .rti f»ni in rest bv a constant force after16. A boy moving with a velocity of 3m/s is bi g ^ 2000 Ans: 0.3m/s2travelling 15m. Calculate the retardation b “ the appi1Cation of the17. A car moving with a speed of 90km/hr was bio g „brakes in 1 Os. How far did the car travel after the brakes werJ *PP 1 ' Ans; x 25mA. 125m B. 150m C. 250m D. 15km ^ . . . rnil(jht to a st0p18. An aeroplane lands on a runway at a speed of 180 m r an . coming touniformly in 30 seconds. What distance does it cover on the runway 750mrest? A 360m B. 540m C. 750m D. 957m JAMB 1993 750m19. Two points on a velocity time graph have coordinates (5s, 1 ms ) anCalculate the mean acceleration between the two points.A. 0.67ms'2 B. 0.83ms'2 C. 1.50ms'2 D.2.00ms'2 JAMB 1989 Ans:0.67ms"20. A motor vehicle is brought to rest from a speed of 15ms 1 in 20 seconds. Calculate the retardation.A. 0.75ms'2 B. 1.33ms'2 C. 5.00ms'2 D. 7.50ms'2 JAMB 1994 Ans:0.75m/s221. A particle moving in a straight line with uniform deceleration has a velocity of 40m/s at a point P, 20m/s at a point Q and comes to rest at a point R where QR=50m.Calculate the: (i) distance PQ; (ii) time taken to cover PQ; (iii) time taken to cover PR.

WAEC1990 Ans: (i) 150m; (ii) 5s (iii) 10s22.

V(ms ')

The diagram above (Fig.1.10) represents the velocity - time graph of a body in motion.The total distance travelled by the body is 195m. Calculate the acceleration of the body in section OP of the graph. WAEC 1994 Ans: 2.5m/s223. The diagram below(Fig. 1.11) illustrates the velocity -time graph of the motion o f a body. Calculate the total distance covered by the body. WAEC 1996 Ans: 225m

V(m/s)

24. The diagram below (Fig. i.i2 ) renresentQ ndistance covered in the first 20s °ct y - time graph. Determine theWAEC 2005 Ans: 60m

16

V(m/s ')

25. The figure below (Fig. 1.13) shows the velocity - time graph of a car which starts from rest and is accelerated uniformly at the rate of 3m/s2 for 5 seconds. It attains a vc »city which is maintained for 1 minute. The car is then brought to rest by a uniform lelardation after another 3 seconds. Calculate the total distance covered.A. 900meters B.1920meters C.1020meters D.960meters E.860meters

JAMB 1979 Ans: 960mV(m/s)

26. A ball is dropped from a height of 45m above the ground. Calculate the velocity o fthe ball just before it strikes the ground. WAEC 1994 Ans:30ms''27. An orange is dropped from a height of 100m above the ground level. Calculate thevelocity o f the orange just before it strikes the ground. IVAEC 2005Ans 44.7m s'128. The graph below(Fig. 1.14) describes the motion of a particle. The acceleration o f the particle during the motion is A.O.OOms'2 B.0.25ms'2 C.4.00ms'2 D.8.00ms'2 E. 10.00ms'2

JAMB 1985 Ans: 4.00m s'2V ( m s 1)

30. A ball is thrown vertically upwards lrom the ground with an initial velocity of 50m/s. What is the total time spent by the ball in the air? W AEC 1991 Ans: 10s31. A stone is projected vertically upward with a velocity of 20ms'1. Two seconds later a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10ms'1. Neglecting air resistance, calculate the (i)

17

length of time the second stone is in motion before they meet, (ii) Vel « ,stone when they meet. (Take g-lOm/s’) W A E C Ans: (0 Is32. A ball bearing is projected vertically upwards from the groun 2-.15ms \ Calculate the time taken by the ball to return to the ground, [g ms

W A E C2004 Ans: 3.0s ^33. A stone of mass 0.7kg is projected vertically upwards with a speed of 5 Calculate the maximum height reached. [Take g as 10ms'2 and neglect air resistance)

\ i v r n in n s Ans:1.25m

V(ms'')

34. What is the average velocity of the sprinter whose velocity - time graph is shown in the figure Fig. 1.16 above?A. 85.0ms'1 B. 17.0ms'1 C. 8.5ms'1 D. 1.7ms'1 M M B 1990 Ans: 8m/s35.Vms'1

Fig. 1.17The diagram above (Fig. 1.17) shows a velocity- time graph representing the motion of a car. Find the total distance covered during the acceleration and retardation periods o f the motion. A. 75m B. 150m C. 300m D. 375m JAMB 1993 Ans: 75m36. A stone is released from a height of 80m above the ground. Calculate its velocity justbefore it strikes the ground. NECO 2003 Ans: 40m/s37. A body falls freely under gravity (g=9.8m/s2) from a height of 40m on to the top o f a platform 0.8m above the ground. It’s velocity on reaching the platform isA. 784ms'1 B. 80ms'1 C. 78.4ms'1 D. 39.2ms'1 E. 27.7ms'1

JAMB 1981 Ans: 27.7ms 138. The diagram below (Fig. 1.18) shows the velocity-time graph of a vehicle. Its acceleration and retardation respectively are.A. 8.0 ms'2,4.0 ms'2 B. 4.0 ms'2,8.0 ms'2 C. 4.0 ms'2,2.0 ms'2 D. 2.0 ms'2, 4.0 ms'2

JAMB 1999 Ans: 4ms'2, 2ms'2

18

39. A mango fruit drops from a branch 10m above the ground. Just before hitting the

ground its velocity is A. 10>/2ms 1 B. -^Lms"1 C. 100ms"1 D. 5V2ms 1V 2

E. 200ms'1 JAMB 1982Ans: lO-v/^ms'140. An orange drops to the ground from the top of a tree 45m tall. How long does it taketo reach the ground? WAEC1991 Ans: 3s41. An object is released from rest at a height of 25m. Calculate the time it takes to fallto the ground. WAEC 2002Ans:2.24s42. A palm fruit dropped to the ground from the top of a tree 45m tall. How long does it take to reach the ground? [g=10ms"2]A. 9s B. 4.5s C. 6s D. 7.5s E. 3s JAMB 1978 Ans: 3s43. A stone is dropped from the top of a tower of height 11.25m. Calculate the time itwill take to reach the ground. NECO 2004 Ans: 1.50s44. A small metal ball is thrown vertically upwards from the top of a tower with an initial velocity of 20m s1. If the ball took a total of 6 seconds to reach the ground level, determine the height of the tower.(g=10ms"2A. 60m B. 80m C. 100ms D. 120m JAMB 1991 Ans: 20m45. A piece of stone is projected vertically upwards with a speed of 25ms"1. Determine

€peed at its highest point reached. WAEC 2005 Ans:0.0m/sIn free fall, a body of mass 1kg drops from a height of 125m from rest in 5s. How

j will it take another body of mass 2kg to fall from rest from the same height? [g=10ms"2] A. 5s B. 10s C. 12s D. 15s JAMB 1998 Ans:5s47. A ball thrown vertically upward reaches a maximum height of 50m above the level o f projection. Calculate the : (i) time taken to reach the maximum height.(ii) speed of the throw. WAEC 2001 Ans: (i) 3.16s (ii)31.6m/s48. A car accelerates uniformly from rest at 5ms"2. Determine its speed after 10s.

WAEC 2006 (Ans: 50.0ms"1)

The diagram above (Fig. 1.19) shows the speed-time graph o f a car. If the car covered a total distance of 600m in 25s, calculate its maximum speed. WAEC 2006 Ans:30ms"‘

50. Two particles X and Y starting from rest cover the same distance. The acceleration of X is twice that o f Y. The ratio of the time taken by X to that taken by Y is?A.'A B. 2 C. y ri D.V2 E. 4 JAMB 1981 Ans:

19

Si- A bullet fired vertically upward from a gun held 2m above the ground, rea _ maximum height in 4s. Calculate the (i) magnitude of the initial velocity o f the bu ,00 total distance the bullet travelled by the time it hits the ground. [g-lOms J

NECO 2006 Ans: (i) 40m/s (n) 162mWhat is the total time of flight of an object projected vertically upwards with a >f 30ms'1? [g=10ms'2] NECO 2007 Ans: 6.0sA car starts from rest and accelerates uniformly at a rate of 5.0ms . Calculate the

— » w r ’ o 1) r n A n s : 31,62ms

52.speed of 30ms'1? [g=10ms"53. A car starts from rest ana accciciaiw--------- -- \ ip r n 7007magnitude of its velocity after moving through 100m. i f 5 j ng a straight54. An object of m ,ss 2kg moves with uniform speed oflOms for 5s along^ ^ ,path. Determine the magnitude o f its acceleration. vplncitv of v55. A body moving with an initial velocity u accelerates until it attams a veloctywithin time t. The distance, s covered by the body is given by the expression.

' 72v - u '

A .s = B.v + 2 m C.s =

v+u2

D.s =2 1 " 1 2 , ,

WAEC 2007 Ans:C

56. A body accelerates uniformly from rest at 2ms 2. Calculate the magnitude o f ivelocity after traveling 9m. WAEC 2007 Ans. - ms57. A body moving with a velocity of 50ms 1 is brought to rest in 30s by a cons an retarding force. Calculate the distance covered by the body. NECO 2008 Ans. 750m58. A ball is thrown vertically upwards with a speed of 20ms 1. Calculate the maximumheight reached. (g= 10ms'2) NECO 2008 Ans: 20.0m59. A bullet fired vertically upward from a gun held 2.0m above the ground reaches its maximum height in 4.0s. Calculate its initial velocity, (g = 10 ms )A. 10 ms’1 60.

Vms'

B. 8 ms'1 C. 40 ms'1 D. 20 ms'1 JAMB 20098 Ans: C

The velocity-time graph above represents the motion of a car. Calculate the total distancetravelled by the car. NECO 2008 Ans: 24061. What is the acceleration between two points on a velocity-time graph which hascoordinates (10 s, 15 ms ) and (20 s, 35 ms )?

-2 ^ i A A ___ -L —ms’2 B. 3.50 ms’2 C. 1.00 ms'^ D. 2.00 ms JAMB 2009 Ans: D A car accelerates uniformly from rest at 4 ms'2. How far will it travel in the fifth

JAMB 2009s Ans: D

62.complete second?A. 100m B. 50m C. 32m D. 18m63. A car moves with a speed of 30ms'1. Calculate the distance travelled in 30s.

WAEC 20097 Ans: 900m64. A body at rest is given an initial acceleration of 6.0ms'2 for 20s after which the acceleration is reduced to 4.0ms'2 for the next 10s. The body maintains the speed attained

for 30s.Draw the velocity-time graph of the motion using the information given above. From the graph, calculate:

(i) Maximum speed attained during the motion;(ii) Total distance travelled during the first 30s;(iii) Average speed during the same time interval as in (ii) above

20

WAEC 2009e" Ans: (i) 120ms'1 (ii) 2200m (iii) 73.33ms'1 65. An object o f mass 2kg moves for 5s with a uniform velocity o f 10ms . What is the magnitude of its acceleration in ms'2? NECO 20 094 Ans: 0.067. A car starts from rest and accelerates uniformly for 5s until it attains a velocity of 30ms *. It then travels with uniform velocity for 15s before decelerating uniformly to rest in 10s.(i) Sketch a graph of the motion.(ii) Using the graph above, calculate the

I. acceleration during the first 5s,II. deceleration during the last 10s,III. total distance covered throughout the motion.

NECO 2009E11 Ans: I. 6ms'2 II. 3ms'2 ID. 675m

21

2f r i c t i o n

Friction is force which opposes relative motion between two !"„„ frictional forceother. Coefficient of static friction, p is defined as the ratio of the limiting factional(F) to the normal reaction (R).

iBlock of wood

.. f- Direction of motion(1Horizontal surface _

Fig. 2.1f

Coefficient of static friction = „ = L =R Normal reaction

The normal reaction, R is equal to the weight, W of the block i.e. R = W = mg, where m is the mass of the block in kg and g is the acceleration due to gravity.

However, for a block of wood (or any object) sliding down an inclined plane, AB as shown below the coefficient of friction is obtained as follows:

The weight, W, of the block Q acts vertically downwards and can be resolved into two components:

(i) Componenty, parallel to the plane: sin6 = — y = W sin6W

ii) Component x, perpendicular to the plane: Cos6 = — x = IT cos 6*W

To obtain the coefficient of static friction, fi, the angle of inclination of the plane, 0, is slowly increased until the block just begins to slide down the plane. At this point,

22

the parallel component of the weight, Wsin 0 acting downwards along the plane is q to the limiting frictional force, F(=7xR) acting upwards along the plane.

That is, F = Wsin 0 .............................. (1)

or fiR = Wsin 0

Also, on an inclined plane, the normal reaction R is not equal to the weight, W,

but to the perpendicular component of the weight, Wcos 0.

That is R = Wcos 0 ...............................(2)

Combining equations (1) and (2),

The coefficient of friction, ju = S'n ^R W cos 6

, . , Odd y sin 6Alternatively, u = Tan# = = — = ----------

Adj x W cosO

The angle o f inclination, 6, is called the angle o f friction.

Example 1A wooden block of mass 1,6kg rests on a rough horizontal surface. If the limiting frictional force between the block and the surface is 8N, calculate the coefficient of

friction (g=10m/s2). WAEC 2000Solution

R = 16N

F = 8N«-

Fig. 2.3 W

Frictional force, F = 8N; normal reaction, R = W = mg = 1.6 X 10 = 16N

F 87VCoefficient of friction, u - — = ----- = 0.5

R 167V

Example 2Calculate the magnitude of the force required to just move a 20kg object along a horizontal surface if the coefficient of friction is 0.2.A.400.0N B.40.0N C.4.0N D. 0.4N (g=10ms‘2) JAMB 1994

R = 200N 4

F

Fig. 2.4 W

SolutionCoefficient of friction, n = 0.2; normal reaction, R = W = mg = 20 X 10 = 200N

FFrom f = — , frictional force, F = /rR = 0.2 X 200 = 40N

R

23

Example 3If an object just begins to slide on a surface inclined at

R i 1coefficient of friction is A. V3 B. — C. D- ^

30° to the horizontal, the

JAMB 2003

Solution , .The coefficient of friction for an inclined plane is the tangent of the angle of the plane when an object on the plane just begins to slide.Therefore, coefficient of friction = Tan 30

From trigonometry, Tan 30 = 1

Example 4A horizontal force of 45N applied to a crate of mass 9kg is just sufficient to move it. If the crate is now pulled at an angle of 50° to the horizontal, find the force required to move the crate over the horizontal surface. (g=10m/s2)SolutionThe coefficient of friction, g is found from the first sentence of the question as follows.

F = gR +

Fig. 2.6

R =a90N

W

* 45N

Normal reaction, R = W = mg = 9X10 = 90N Frictional force, F = 45N; 0 = 50°

F 45 NCoefficient of friction, u = — - ----- = 0.5R 90 N

The second part of the question is represented below. The pulling force, P has been resolved into its vertical component (Psin 0) and horizontal component (Pcos 0).

In such a case as this, the normal reaction, R is not equal to the weight, W. The vertical component.must also be considered. Therefore,

W = R + Psin 0 ................................. (1)

At the point in which the crate moves forward, the horizontal component of the pulling force, Pcos 0 is equal to the limiting frictional force, F. Therefore,

F = Pcos0

or gR = Pcos 0 ......................................(2)

Equation 1 and 2 are solved simultaneously to obtain the value of the pulling force, P Substitute the given and calculated values (0 = 50°, W = 90N, F = 45N, g = 0.5) into equation 1 and 2.

24

90 R + l\sin 50

0.5R I’cos 50

90 = R + 0.766P

0.5R = 0.643P

From (2b), 0.5R=0.643P

0.643 P 0.5

1.286P

Substitute R = 1.286P into ( 1 b)

90 = R + 0.766P

90 = 1.286P + 0.766P

90 = 2.052P

:. P = — =43.86N 2.052

( 1 )

..........(2)

,..(lb)

.........(2b)

Example 5An object of mass 15kg is at the point of sliding down a plane inclined at 36° to the horizontal. Find the least force parallel to the plane required to make the object to begin to move up the plane. (g=10m/s2)Solution

From the question; 0 = 36°; m = 15kg; g = lOm/s2; W = mg = 15 x 10 = 150N.

For the object to move up the plane, it must overcome the opposing limiting frictional force, F and the parallel component o f the weight, WsinO. That is, the upward pulling force, P is

P = Parallel component of weight + Limiting frictional force

P = Wsin 0 + F

= Wsin 0 + /xR [F = /xR]

P = Wsin 0 + ¿iWcos 0 [R = Wcos 0|

Coefficient of friction, n is required to find P

25

Wsin 6 _ 150xsin 36 _ 150 x 0.588 -73Wcos 6 ~ 150xcos 36 _ 150 x 0.809

, the fact that the angle o f Alternatively, the coefficient of friction, ¡1 is calcu ate 0 fh e tangent o f thisinclination, 0 at the point o f sliding is called the angle of friction, angle o f friction gives the coefficient of friction.

[à = tan 6 = tan 36 = 0.73

Substitute into P = Wsin 0 + /xWcos 0

= 150 sin36 + 0.73 X 150 X Cos36

= 150 X 0.588 + 0.73 X 150 X 0.809

= 88.20 + 88.59

= 176.79N

Example 6 , , , o tnA concrete block of mass 25 kg is placed on a wooden plank inclined at an angle othe horizontal. Calculate the force parallel to the inclined plane that will keep the bloc atrest if the coefficient of friction between the block and the plank is 0.45.SolutionR = W = mg = 25 X 10 = 250N; 0 = 32°; n = 0.45As shown in the figure below, the block is under the influence of 2 parallel and opposite forces:

The diagram above can be redrawn as follows:(i) Wsin 0, the parallel component of the weight of the block acting dowr^yards along the plane.

Wsin 0 = 250 X sin32 = 250 X 0.53 = 132.48N

(ii) F (=/rR), the limiting frictional force acting upwards along the plane;

F = /iR = /xWcos 0

= 0.45 X 250 X cos32 = 112.5 X 0.848

= 95.41N

26

It becomes obvious that the force required to keep the block at rest is the difference between the downward sloping force, Wsin0 (132.48N) and the upward sloping frictional force, /xWcosG (95.4IN);

132.48 - 95.41 = 37.07N

Example 7A body of mass 25kg, moving at 3ms'1 on a rough horizontal floor is brought to rest after sliding through a distance of 2.50m on the floor. Calculate the coefficient o f sliding friction. WAEC 2005

Solution

u =3m/s

Direction of motion -------- ►

at restV = Om/s

S = 2.50m

Fig. 2.10

The acceleration (in actual sense, deceleration) is calculated first. Initial velocity, u - 3m/s; final velocity, v = 0; distance, s = 2.50m.

Substitute into v2 = u2 + 2as

02 = 32 + 2 X a X 2.50 0 = 9 +5a

5a = -9

a = —- = -1 .8 ms~25 :* , aivooH

:. Deceleration y - i .8jn/s-

Frictional force,2

Normal reaction, R = V» '= ]h '|^v2v X

- ma; v he®*' m is mass in kg and a is acceleration

(deceleration) in m/s

:. F = ma = 25 X 1.8 —45N,-

27

Coefficient of sliding friction, =R

Example 8 . , , ne i f the b o d y slides downThe diagram above shows a body resting on an inc me WAEC 1988the plane, what will be its acceleration? (g=T0m/s )

Fig. 2.11

Solution , object to slideAs shown in the diagram below, mgsinG is the force that ma es downward the plane, though it is opposed by the 25N frictional force.

The resultant force, F = mgsinG - 25N

where mass, m = 15kg; G = 30; g = lOm/s':. F = 15 X 10 X s in 3 0 -2 5

= 75 - 25 = 50NF

From F = ma, the acceleration a = —m

5015

3.33 m is2

---Two bodies X and Y of masses 5.0kg and 7.5kg, respectively, are connected by » fight inextensible string as illustrated in the diagram above. If X is placed on a rough sm£*c?afcoefficient of friction 0.5, calculate the magnitude of the:

i. Normal reaction on Xii. Frictional force between X and the surface,

iii- Tension in the string (g = 10ms'2) NECO 2907e11

Solutioni. Normal reaction on X is, R = mg

= 5 X 10 = 50N

ii. Frictional Force, F = /xR= 0.5 X 50 = 25N

iii. For 5kg mass, T -/xR = ma Substituting, T - 0.5 X 50 = 5a

T - 25 = 5a ...................................... (1)For 7.5kg mass, mg - T = ma Substituting, (m = 7.5kg), 7.5 X 10 - T = 7.5a

75 - T = 7.5a

Rearranging - T + 75 = 7 .5 a ...............(2)

Equation (1) and (2) are simultaneous equations and can be solved as follows: Add equation (1) and (2)

T - 25 = 5a

- T + 75 = 7.5a

0 + 50 = 12.5a . 50 = 12.5a

acceleration, a = = 4m s'112.5

The tension T is found by substituting a = 4ms'2 into any of equations (1) and (2). Substituting into (1), T - 25 = 5a, we obtain

T - 25 = 5 X 4. T = 20 + 25 = 45N

The acceleration of the system shown above isa. 2ms'2 b. 4ms'2 c. 6ms'2 d. 8ms'2 JAMB 1995SolutionThe equation for the mass on the inclined plane takes into consideration the angle of inclination. The parallel component of the mass is mg sin 0.

For 20kg mass, T - mg sin 0 = ma

Substituting, T - 20 X 10 X Sin30 = 20a

T - 100 = 20a ............... (1)

For 30kg mass, mg - T = ma

29

30 X 10 - T = 30a

300 - T = 30a

Rearranging - T +300 = 30a

Add equations (1) and (2)

T - 100 = 20a

- T + 300 = 30a

0 + 200 = 50a

200 = 50a

, 200 „ -2acceleration, a = ---- = 4ms50

Example 11 i fA motorcycle of mass 100kg moves round in a circle of radius 10m with a ve ocity o 5ms'1. Find the coefficient of friction between the road and the tyres.A. 25.00 B. 2.50 C. 0.50 D. 0.25 (g=10m /s ) JAMB 2uUoSolutionMass m = 100kg; radius r = 10m;

velocity v = 5ms'1, coefficient of friction ¡1 = 1

When an object of mass m moves with a velocity v, the force at work is the inward actingm v2centripetal force (F) given by F = -----

r100x52

10

100x25ÏÔ

= 250W

R = m g= 100 X 10= 1000N

Coefficient of friction, /u ■F 250

10000.25

EXERCISE 2.1. A force of 20N applied parallel to the surface of a horizontal table is just sufficientto make a block of mass 4kg move on the table. Calculate the coefficient of friction between the block and the table (g=10m/s2). Ans: 0.5 WAEC 19922. A metal block of mass 5kg lies on a rough horizontal platform. If a horizontal force of 8N applied to the block through its center of mass just slides the block on the platform, then the coefficient of limiting friction between the block and the platform isA. 0.16 B. 0.63 C. 0.80 D. 1.60 E. 2.00 Ans:0.16 JAMB 19853. A lOOKg box is pushed along a road with a force of 500N. If the box moves with a uniform velocity, the coefficient of friction between the box and the road is

30

A. 0.5 B. 0.4 C.1.0 D. 0.8 Ans: 0.5 JAMB 20044. What is the coefficient of static friction between a load °horizontal surface, if the limiting frictional force is ION. Ans: 0.5 NE ^ .5. A block weighing 15N rests on a flat surface and a horizonta orce oexerted on it. Determine the frictional force on the block. Ans:3N WA . , f6. A concrete block of mass 35kg is pulled along a horizontal floor wit t e ai 0 rope inclined at an angle of 30° to the horizontal. If the coefficient of nction iscalculate the force required to move the block over the floor. Ans. 211.52

A body of mass 10kg rests on a rough inclined plane whose ang e o ti t isvariable. 0 is gradually increased until the body starts to slide down the p ane at The coefficient o f limiting friction between the body and the plane is?A. 0.30 B. 0.50 C. 0.58 D. 0.87 Ans :0.58: JAMB J99U8. A force, 10N drags a mass 10kg on a horizontal table with an acceleration o 0.2m/s2. If the acceleration due to gravity is 10m/s2, the coefficient o f friction etween the moving mass and the table is? A. 0.02 B. 0.08 C. 0.20 D. 0.80

Ans:0.08 JAMB 19889. The coefficient of static friction between a 40kg crate and a concrete surface is 0.25. Find the magnitude of the minimum force needed to keep the crate stationary on t e concrete base inclined at 45° to the horizontal.A. 400N B. 300N C. 283N D. 212N (g = 10ms'1) Ans :212 N JAMB 198710. When a box of mass 45kg is given an initial speed of 5m/s, it slides along a horizontal floor a distance of 3m before coming to rest.What is the coefficient of the kinetic friction between the box and the floor?

A .5/ 6 B .^ 2 c X D A Ans: 0.417 or JAMB 198611.

Fig. 2.15

A body o f mass 10kg on a smooth inclined plane is connected over a smooth pulley to a mass o f 15kg as shown (Fig. 1.13). The acceleration o f the system is

A . j '/ g B V l 5 g C- Y l g D g E‘% 8 Ans;2/5& JAMB 1982 12.

Fig. 2.16

A mass o f 2kg on a surface ( f i = / 2) is connected to a second mass o f 4kg over a frictionless pulley as shown above. If the acceleration due to gravity is 9.8ms'2- the masses will

31

13.

Accelerate at 4.9ms'2 Remain stationary Accelerate at 9.8ms'2 Accelerate at 19.6ms"Accelerate at 39.2ms'i An object of mass 80kg is pulled on a

Ans: A JAMB 1981/\IK>. r \ ******— -- ___ ,JWW ................. o „ r ___ horizontal rough ground by a force of

500N. Find the coefficient of static friction. (g.= 10 m s2)A -0.8 B. 0.4 C. 1.0 D. 0.6

JAMB 2009" Ans: D

3ELASTIC PROPERTIES OF SOLIDS

Hooke s law states that the change in length or extension, e, of an elastic body is directly proportional to the applied force or load, F, provided that the elastic limit of the body is not exceeded.

Force a Extension F a e

F = Ke

where K is the force constant or stiffness constant of the material in Nm’1.F is the force or weight in N. F = mg (m is mass in kg and g is

acceleration due to gravity in m/s2) e is extension or compression of the elastic material in m.

Young ModulusYoung Modulus, y is defined as the ratio of tensile stress to tensile strain.

Tensile stressY - -----------------Tensile strain

Tensile strain is the ratio of the extension of a material to its original length.

Strain - Change in lenght _ Extension/Compression _ e Original lenght Original lenght /

Tensile strain has no unit.Tensile stress is the ratio of the force acting on a material to the cross sectional area of the material.

0 Force F FStress = --------=— =— -Area A n r

Therefore Young Modulus, y = = /A . _ X Lstrain y A e

The S.I unit for y is N /m 2

Work Done In Springs and Elastic StringsIf an applied force, F caused an elastic spring of original length, /, to undergo an

extension or compression, e, then the average force is;

= £ ± £ = l F 2 2

Work done = force X Distance

= average force X Extension

= F/2 X e

W= 'AFe or 'AKe2 [F=Ke]

This is the work done by a spring when compressed or extended by a force.The energy stored in a spring or string is also given by W = 'AFe or 'AKe2 The unit o f work or energy is Joules, J.

33

Example 1 An object weighing 0.6N is hung on a spiral spring and causes it to exten y 6.0cm. The object is removed and a block of wood replacing it causes the spring o extend by 10.0cm. What is the weight of the block of wood? N bU J ¿UUJSolutionApplied force, F = 0.60N; Extension, e = 6.0cm = 0.06m

P F 0.60• • Force constant, K = — = ----- = lON/me 0.06

For e = 10.0cm = 0.1m; and K = lON/m

Weight of wood, F = Ke = 10 X 0.1 = 1.0N

Example 2A load of 5N gives an extension of 0.56cm in a wire which obeys Hooke’s law. What is the extension caused by a load of 20N?A. 1.12cm B. 2.14cm C. 2.24cm D. 2.52cm JAMB 1987SolutionApplied load, F = 5N; Extension, e = 0.56cm

Force constant, K = — = 5N = 8.93N/cm e 0.56cm

For F = 20N and K = 8.93N/cm

_ . F 20NExtension e = — = -------------= 2.24cmK 8.93N/cm

Example 3Use the following data to determine the length, L of a wire when a force of 30N is applied, assuming Hooke’s law is obeyed.

Force applied(N) 0 5 10 30

Length of wire (mm) 500.0 500.5 501.0 L

SolutionChange in force applied, F = 5 - 0 = 5N Change in length, e = 500.5 - 500.0 = 0.5mm

FFrom, F = Ke, force constant, K = — =e

For F = 30N;FExtension, e = —

3 ONlON/mm

= 3mm

5N0.5mm

= lON/mm

L = Original length + Extension

= 500.0mm + 3mm = 503.0mm

Example 4A spiral spring balance is 25.0cm long when 5N hangs on it and 30.0cm'when the weight is 10N. What is the length of the spring if the weight is 2N, assuming Hooke’s law is obeyed? NECO 2000SolutionLet the original length of the spiral spring be L

34

Extension, e = New length resulting from applied force - Original length First case: extension, e = 25.0cm - L

applied force, F = 5N 2nd case: extension, e = 30.0cm - L

applied force, F = IONSubstitute into F = Ke for 1st and 2nd cases.

5 = (25-L )K ..................... (1)10 = (30 - L)K ...............

1st2nd

■ (2)

From (1), K =

From (2), K =

Equate (3) and (4):

25 - L

1030 - L

5

(3)

1025 - L 30 - L

..(4)

Cross multiplying, 5(30-L) = 10(25-L)150-5 L = 250- 10L 10L -5L = 250- 150

5L = 100, 100 ™L = ---- = 20c»j

5Considering the Cease;

Extension, e = 25.0 - L = 25 - 20 = 5.0cm = 0.05m

Applied force, F = 5NF 5

From, F = Ke, force constant, K - — = ----- = 100Nm 1e 0.05

For F = 2N and K = 1 OON/m

F 2Extension, e = — = — = 0.02m - 2cm K 1OU

New length of spring = Original length (L) + Extension (e)= 20cm + 2cm

= 22cm or 0.22m

Example 5A spiral spring of natural length 20.00cm has a scale pan hanging freely in its lower end. When an object of mass of 40g is placed in the pan, its length becomes 21.80cm. When another object of mass 60g is placed in the pan, the length becomes 22.05cm. Calculate the mass of the scale pan. WAEC 2001SolutionThe extension of the spring is caused by both the mass of the object and the mass of the scale pan. Let M be the mass of the scale pan.

Is'case: extension, <? = 21.80 - 20.00 = 1.8cm

mass, m = M + m = M + 40

35

2nd case: extension, e = 22.05 - 20.00 = 2.05cm

mass, m = M + m = M + 60

F = Ke or mg = Ke [F=mg]

Substitute into mg = K e for 1st and 2nd cases

1SI (M + 40)g = 1.8K ..................................

2nd (M + 60)g = 2.05K........................................F rom (l) F8K_ ............................... (3)

M + 40

From (2) g = -2-05K ...................................... (4)M + 60

Equate (3) and (4): =M + 40 M t 60

_ M + 40 1.8KRCarrangmg: m T w = I0 S K

M + 40 _ 1.8 M + 60 ~ 2.05

- (1) (2)

Cross multiplying, 1.8(M + 60) = 2.05(M + 40) 1.8M + 108 = 2.05M + 82 2 .0 5 - 1.8M= 108-82

0.25M = 26

M = — = 104g 0.25

Example 6A force o f 40N applied at the end of a wire of length 4m and diameter 2.00mm process and extension of 0.24mm. Calculate the;(a) Stress on the wire (b) strain in the wire(c) Young’s Modulus for the material of the wire (7t =3.14). WAEC 2003Solution

„ Force(a) Stress = -------- =

Arearadius, r = diameter/2 :.

Fnr1r = 2.00mm/2 = 1mm = 0.001m

p or p --- 40N and r = 0.001m 40

StrCSS_ 3.14x (0.001 )212.74xl06N/m2

(b) Strain =extension _ e

original lenght / 0.24

/ = 4m; e = 0.24mm = —— = 2.4x10 m

Strain = -2.4x10"

1000

= 6.0x10"

stress 12 .74x l06 ,, ,(c) Yount' Modulus, / = - =2.12 X 10 N/m2v0-' 1 c strain 6.0x10

36

Example 7 i > of force constantA force o f magnitude 500N is applied to the free end of a spira sPn "k 20051.0 x 104Nm '. Calculate the energy stored in the stretched spring-Solution _ or ,/ j ie 2The energy stored in the spring is equal to the work done, W - 2 LForce, F = 500N; force constant, K = 1.0 x 104Nm

From F - Ke, Extension, e -

500~ l.OxlO4

0.05m

K

Energy stored, W = V2Fe = '/2 X 500 X 0.05 = 12.5J

or W = YzKe2 = '/2 X 1.0 X 104 X 0.052 = 12.5J

Exam ple 8 _A catapult used to hold a stone o f mass 500g is extended by 20cm with an app i orceF. If the stone leaves with a velocity o f 40ms'1, the value of F isA.4.0 x 10*N B A O x lO ^ C .2 .0x l0*N D .4 .0 x l0 2N JAMB 2UUUSolution „1 t, rtieThe work done in stretching the catapult is equal to the kinetic energy gam ystone, i.e. Yz Fe = */2 mv2Extension, e = 20cm = 0.2m; velocity, v = 40m/s; mass, m = 500g — 0.5kg.Substituting,

l/ 2 X F X 0.2 = '/z X 0.5 X 4020.2F = 0.5 XI600

800 _ 4 o o c N o r4 x l0 3N0.2

EiAttlllplC 7

On top o f a spiral spring o f force constant 500Nm~' is placed a mass o f 5 x 10T3kg. If the spring is compressed downwards by a length o f 0.02m and then released, calculate the height to which the mass is projected. (g = 10m'2)A. 8m B. 4m C. 2m D. lm Ans: 2m JAMB 2003SolutionThe work done in compressing the spring is equal to the potential energy stored in thespring.

YzKe2 = mgh

Force constant, K = 500Nm ' ; extension, e = 0.02m;

Mass, m= 5 x 10'3kg; g = lOm/s

SubstitutingX x500x(0 .02 )2 = 5 x lO '3xlO xh0.1 = 0.05h

EXERCISES 3 e ;ii the wire bv I 25cm1. A force of i 8N extends a wire by 0.4cm. What ^ru. " ^iTthc elastic limit is not exceeded? Ans ’ ' , . , mounl 0 f force2. If a force of 50N stretches a wire from 20m to 20.01m. whatrequired to stretch the same material from 20m to 20.05m in/)JA. IO0N B. SON C.250N I>. 200N Atts: 250N J A U B 2 0 0 4 Hooke's low ¡,3. The extension in a spring when 5g w. was hung from it was 0.56cm. If I looks law i.obeyed, what is the extension caused by a load of 20g wt. lllQIA. 1.12cm B. 2.14cm C. 2.52cm D. 2.80cm Ans: 2.24cm JAMB 1J814. A lOg mass placed on the pan of a spring balance causes an extension o 15g mass is placed on the pan of the same spring balance the extension is .A. 3.3cm B. 6.5cm C. 7.5cm D. 10.8cm E .15.0cm

Ans: 7.5 cm JAMB 19825. A spiral spring extends from a length o f 10.00cm to 10.01cm when a force o f 20N is applied on it. Calculate the force constant of the spring. -I r>r r-*X-»

6. The total length o f a spring when a mass of 20g is hung from its end is 14cm, while its total length is 16cm when a mass of 30g is hung from the same end. Calculate the unstretched length of the spring assuming Hooke’s law is obeyed.A.9.33cm B.10.00cm C.10.66cm D.12.00cm E.15.00

Ans: 10.00cm JAMB 19837. A force of 100N stretches an elastic string to a total length of 20cm. If an additional force o f 100N stretches the string 5cm further, find the natural length of the spring.A.15cm B.12cm C.lOcm D.8cm E.5cm Ans:15cm JAMB 19858. The spiral spring o f a spring balance is 25.0cm long when 5N hangs on it and 30.0cm long when the weight is 10N. What is the length o f the spring if the weight is 3N assuming Hooke’s law is obeyed? A. 15.0cm B. 17.0cm C. 20.0cmD. 23.0cm Ans:23.0cm JAMB 19919. A force o f 15N stretches a spring to a total length of 30cm. An additional force o f 1 ON stretches the spring 5cm further. Find the natural length of the spring.A. 25.0cm B. 22.5cm C. 20.0cm D. 15.0cm Ans: 22.5cm JAMB 199410. A piece o f rubber 10cm long stretches 6mm when a load of 100N is hung from it.What is the strain? A.60 B.6 C .6 x l0 ’2 D .6 x l0 '3 E. 0.6

Ans:6 x \0 2 JAMB 197811. A load o f 20N on a wire o f cross-sectional area 8 x 10'7m2, produces an extension o f 10^m. Calculate Young’s Modulus for the material o f the wire if its length is 3m.A. 7.0 x 10n Nm'2 B. 7.5 x 10u Nm'2 C. 8.5 x 10nNm'2 D. 9.0 x 10n N tn 2

Ans: 7.5 x 10“ Nm'2 JAMB 199712. The tendon in a man’s leg is 0.01m long. If a force o f 5N stretches the tendon by 2.0 x 10’5m, calculate the strain on the muscle.A. 5 x 106 B. 5 x 102 C. 2 x l 0 '3 D. 2 x 10'7 Ans: 2 x 10'3 JAMB 199813. I f the stress on a wire is 107Nm'2 and the wire is stretched from its original length o f 10.0cm to 10.05cm. The Young’s Modulus of the wire isA. 5.0 x 104Nm'2 B. 5 .0xl05Nm'2 C. 2.0 x 108Nm'2 D. 2.0 x l O 'W 2

Ans: 2 x 109N/m2JTA/B 199914. A spiral spring o f natural length 30.0cm and force constant o f 20Nm 1 is compressed to 20.0cm. Calculate the energy stored in the spring. Ans: 0.1Ji.WAEC^002,-15. An elastic string o f force constant 200Nm ‘ is stretched through 0.8m within itselastic limit. Calculate the energy stored in the string. Ans: 64J WAEC 200416. A spring o f force constant 1500Nm'' is acted upon by a constant force o f 75N. Calculate the potential energy stored in the spring. A. 1,9J B. 3.2J C. 3.8J D. 5.0J

Ans: 1.9J JAM B 199117. When a force o f 50N is applied to the free end o f an elastic cord, an extension o f 4cm is produced in the cord. Calculate the work done on the cord.

AnsM.OJ WAEC 2004

rn j . A r " 8 0^ length 25cm 1S extended to 30cm by a load of 1 SON attached to one of its a ^ , 'S thC Cnergy Stored in the spring?1Q3 ™°J B ' 2500J C 3 -75J D.2.50J Ans: 3.75J JAMB 1994

e energy contained in a wire when it is extended by 0.02m by a force of 500N is A .5J B. 10J C. 103J D. 104J Ans: 5J JAMB 1995

20. A spiral spring is compressed by 0.03m. Calculate the energy stored in the spring ifits force constant is 300Nm''. Ans: 0.135J WAEC 200421. Calculate the work done to stretch an elastic string by 50cm, if a force of 12.5Nproduces an extension of 5cm in it. Ans: 31.250J NECO 200722. A spring of force constant 500Nrn' is compressed such that its length shortens by 5cm. The energy stored in the spring isA. 0.625J B. 6.250J C. 62.500J D. 625.000J JAMB 2008 Ans: 0.625J23. Use the data in the table below to determine the length X of a wire asummmg Hooke’s law is obeyed

Force applied/N 0.0 5.0 10.0 50.0

Length of wire/mm 800.0 801.0 802.0vicrn

Xlons^AnsSlO.Omml tLCVi.vuo ' *------

24. A wire of length 5.0m and diameter 2.0mm extends by 0.25mm when a force of 50N was used to stretch it from its end. Calculate the(a) Stress on the wire(b) Strain in the wire (7t=3.142) . - . U O . 1 n7Nm-2 h 5 X 10‘5 WAEC 2007

25.

10m

Fig. 3.17 7 /7 > 7 /7

A 50.0kg block is dropped on a spring from a point 10m above (above figure). If the force constant o f the spring is 4.0 x 104N m ', find the maximum compression of the

spring [g=10m/s].A. 1.25m B. 0.50m C. 0.25m D. 0.05m Ans: 0.5m JAMB 1987 26. A string o f length 4m is extended by 0.02m when a load of 0.4kg is suspended at its end. What will be the length of the string when the applied force is 15N?(g = 10ms'2) A. 6.08m B. 5.05m C. 4.08m D. 4.05m Ans: 4.08m JAMB 200727. The diagram below shows the force extension curve o f a piece of wire. The energy stored when the wire is stretched from E to F is? A. 1.5 x 10 2J B. 7.5 x 1 0 ‘j C. 7.5 x* T ” r * » *** ss\n -»

10'3J D. 2.5 x 10' JAns: 2.5 x 10 J JAMB 2002

39

Force (N)R

28. If a load o f mass ION stretches a cord by 1.2cm, what is the total work done?A. 6.0 x 10“2J B. 7.6 x 10~2J C. 1.8 x 10~2J D 6.6 X 10_2J

JAMB 2009,6Ans: A29. The diagram below represents the graph of the force applied in stretching a spiral

against the corresponding extension (X). The force constant o f the spring is

A. 20 Nm 1 B. 40 Nm 1 C. 30 Nm'1 D. 10 Nm'1JAMB 200915 Ans: A

30. The work done in extending a spring by 40 mm is 1.52J. Calculate the elastic constant o f the spring. NECO 200953 Ans:1900.0Nm 1

40

4l i n e a r , a r e a a n d c u b i c e x p a n s i v i t i e s

LINEAR EXPANSIVITYI lie lincai expansivity or coeffieient of linear expansion a of a substance is defined as raclional increase in length of a piece of that substance per degree rise in temperature.I length

Linear expansivity, a = ______ Increase in length______Original length x Temperature rise

h z i ■ _ A/iA d

where a = linear expansivity/1 = original length before expansion

l2 = final length after expansion 0, = initial temperature before expansion 02 = final temperature after expansion A1 = increase in length or change in length (l2 - /,)A0 = temperature rise or change in temperature, (02 - 0i)

The S.l. unit of linear expansivity is per Kelvin (K 1).

Example 1Steel bars, each of length 3m at 29°C are to be used for constructing a rail line. If the linear expansivity of steel is, 1.0 x 10'5K'‘, calculate the safety gap that must be left between successive bars if the highest temperature expected is 41°C. WAEC 1989 SolutionThe safety gap to be left is found by calculating the change in length.

MFrom a = ------------ Change in length, A/ = a /,(02 - 0i)l\(Q2~d\)

Given: linear expansivity, a = 1.0 X lO^K"1 Original length, /, = 3mInitial temperature, 0| = 29°C Final temperature, 02 = 41°C

M = 1.0 X 10'5 X 3(41-29)= 0.00003(12)= 0.00036m or 3.6 X loAn

Example 2A metal rod of length 50cm'is heated from 40 C to 80 °C. If the linear expansivity of the material is «, calculate the increase in length of the rod (in meters) in terms of a.

WAEC 2008'9SolutionGiven: Original length, /, = 50cm = 0.5m Initial temperature, 0| = 40°C

Final temperature, 02 = 80°C Linear expansivity, a = a

From cc =M

/,(02-* .)Increase in length, A/ = a x l ^ -(9,) = ax0.5(80-40)= arx0.5x40= 20a

41

*Xamp,e3 J u u ,oo0r calculate the change in length ofA metal rod of length 100cm, is heated through 100 ’ lO^K.'1 ) WAEC 2003the rod. (Linear expansivity of the material of the ro is

G^ven.° Original length, /, = 100m; ? , temperature change ,A0, = 100°C;Linear expansivity, a = 3.0 X 10 K

tr A/From a =-----/, A0

Change in length, A/ = a X /) X A0= 3.0 x 10'5 X 100 X 100 = 0.3cm

Example 4A bridge made of steel is 600m long. What is the daily variation in its length if the night­time and day-time temperature are 10°C and 35°C respectively. The linear expansivity o f steel is 0.000012°C'‘A. 0.18cm B. 1.80cm C. 18.0cm D. 1800cm JAMB 1992 Ans: 18.00cmSolutionDaily variation is the same as change in length or increase in length.

Given: Original length, /,= 600m; final temperature, 02 = 35°CInitial temperature, 0, = 10°C; linear expansivity, a = 0.000012°C"1

/.(02-0 ,)

0.000012=-------—------ A/ =0.000012x600x25 = 0.18m = 18cm600(35-10)

Example 5A metal rod of length 40.00cm at 20°C is heated to a temperature of 45°C. If the new length of the rod is 40.05cm, calculate its linear expansivity. WAEC 1994 Solutioninitial length,/, = 40.00cm final length,/, = 40.05cminitial temperature, 0j = 20°C final temperature, 02 = 45°C

w ,Linear expansivity, ^ = 77^— 77*1 yyj ~ J

40.05- 40.00 40.0(45-20)

0.0540x25

= 0.00005 = 5xlO~3k-'

Example 6A brass rod is 2m long at a certain temperature. What is the length for a temperature rise o f 100K, if the expansivity o f brass is 18 x 10‘6K''A. 2.0036m B. 2.0018m C. 2.1800m D. 2.0360m

JAMB 1991SolutionInitial length, /, = 2m; temperature rise, AG = 100K; a = 18 x 1 O'6 K '1

Substituteinto a - h z k/,A0

18xl0~61 - 2

2x100/2 - 2 = 18x10^x200 = 36x10^

42

/2 =2 + 36x10^ = 2.0036m

Example 71 he linear expansivity of a metal P is twice that of another metal 0- When these material ate lealed through the same temperature change, their increase in length is the same.C alcu,atc ,hc rat'(> oi'tlie original length of P to that of Q. WAEC1996 SolutionLinear expansivity of Q is a Linear expansivity of P is 2a

Linear expansivity, a ~ All,A0 ( 1)

Where A / = change in length./, = original length

A8 = change in temperatureSubstitute into equation (1) for Q where lQ is original length of Q;

AllQAG

Substitute into equation (1) for P where lp is original length of P; Al

(2)

2a = -LAO

(3)

Note that Al and A0 are the same in both cases (eqn 2 & 3) because temperature and length increase are the same.

, AlFrom eqn (2), original length of Q, lQ = -

From eqn (3), original length of P, lp =

Al

aAOAl

2aA0

AlRatio of L : L =p 0 2a AO aA 9

AO

1

Multiply both sides by 2Ratio of original length of P to that of Q = 1 : 2

Example 8The ratio of the coefficient of linear expansion of two metals — is 3:4. If, when heated

through the same temperature change, the ratio of the increase in lengths of the two

metals, — is 1:2, the ratio of the original lengths -j- is

B yg C. X D. X JAMB 2007A. X

43

SolutionMetal 1

Metal 2

Coefficient of expansion a Tempt change A0 ( ao is the same) Increase in length A/

From OC —----- ( / — original length)/ A0 °

Original length, / =A /

orA0

a{ = 3 A0 = e A/ = f , = 1

30

«2= 4A0 = 0A/ = f 2 =2

Ratio of original length, A.4

j_^_2_ 30 ‘ 40

1 46 230X 2 3

AREA AND CUBIC EXPANSIVITIESArea or superficial expansivity (p) of a substance is defined as the fractional increase in area of a piece of that substance per degree rise in temperature.

increase in areaArea expansivity, ß =original area x temperature rise

A,- A, _ AAß- 4 ( 0 2 - 0 ,) 4 (A 0 )

Also, area expansivity, (ß) = 2 X linear expansivity (a) That is, ß = 2a

Volume or cubic expansivity (y) of a substance is defined as the fractional increase in volume of a piece of that substance per degree rise in temperature._ , . increase in volumeCubic expansivity, y = ----------------------------------------

original volume x temperature rise

V2- V, _ AVr ~ v x(e2- e x) vt(a0)

Also, cubic expansivity (y) = 3 X linear expansivity (a)That is, y = 3a

Where A, = original area or area at temperature 0,A2 = final area or area at temperature 02AA = change in area or increase in area (A2 -A,)V, = original volume or volume at initial temperature 0, V2 = final volume or volume at final temperature 0:AV = change in volume or increase in volume (V2 -V,)

Both area and volume expansivity are measured in K'1

REAL AND APPARENT EXPANSIVITIESApparent cubic expansivity (ya) of a liquid is the increase in volume per unit

volume per degree rise in temperature when the expansion of the vessel is not considered

44

Apparent volume expansivity, y -

Also.

apparent increase in volume

Or

yu =

r„ =

original volume x rise in temperature

mass of liquid expelledMass of liquid remainingx rise in temperature

_________ volume of liquid expelledvolume of liquid remainingx rise in temperature

Real or absolute cubic expansivity of a liquid (yr) is defined as the actual increase in volume per unit volume per degree rise in temperature, when the expansion of the vessel is taken into consideration.

Real cubic expansivity (yr) of the liquid

Yr =

Apparent cubic expansivity of liquid, ya

Ya

+ Cubic expansivity of the container

or vessel y

+

Example 9A solid metal cube of side 10cm, is heated from 10°C to 60°C. If the linear expansivity of the metal is 1.2 x 10’5K"', calculate the increase in its volume. WAEC 1993 SolutionOriginal length, /,= 10cm, original volume, V] =/, x/, x/, = 10x10x10= 1000cm3 Initial temperature, 0, = 10°C; final temperature, 02 = 60°C; a = 1.2 X lO^K'1

Cubic expansivity, y = 3a = 3(1.2 X 10‘5)

AVr=

v f a - e x)Increase in volume, AV = y x V,(#, - 0,)

= 3.6 X IO'5 X 1000 X (6 0 - 10) = 3.6 X IO’2 X 50

= 1.8 cm3Exam ple 10A piece of brass of mass 170kg has its temperature raised from 0°C to 30°C. Calculate its increase in volume, given the density of brass at 0°C as 8.5 x l03kgm'3 and its cubic expansivity as 5.7 x 10 5K 1 WAEC 1998Solution ' „Original temperature, 0, = 0°C; $ Final temperature, 02 = 30°C;Cubic expansivity, y = 5.7 x 10 K

The original volume, Vi is calculated from this equation

Density = — /. volume, (Vt )= ------—De'1Slty volume dens,‘y

Given: mass = 170kg; dens.ty = 8.5 x 103kgm-3

V, =■170

AV

F, ( & - 3 )

8.53x10

Increase in volume, A V = y x Vt(02 - 6>,)

= 0.02m3

45

= 5.7 X IO'5 X 0.02 X (3 0 -0 ) = 3.42 X 10'5m3

Example 11The linear expansivity of brass is 2 x 10'5C"'. If the volume of a piece o f brass is15.00cm at 0°C, what is the volume at 100°C? JAMB 1998A-16.03cm B. 16.00cm3 C. 15.09cm3 D. 15.03cm3SolutionGiven: a = 2 X lO^C'1; Cubic expansivity, y = 3a = 3(2 X 10'5) = 6 X 10'5 °C

Original volume, V| = 15.00cm3; Initial temperature, 0| = 0°C;final temperature, 02 = 100°C; final volume V2 = ?

Y =V - V

6x10-5 K - 15

’l5(l00-0)V2 -15 = 6xlO~sxl5xlOO

V2- 15 = 0.09 V2 =15 + 0.09 = 15.09cm3

Example 12A rectangular metal block of volume lO^m3 at 273K is heated to 573K. If its coefficient of linear expansion is 1.2 x 10'5K'', the percentage change of its volume is? JAMB 1994 Solutionoriginal volume,V! = lO'Vi3; initial temperature,© 1 = 273K; final temperature, 0; 573K; a = 1.2 X 10‘5K_1 Cubic expansivity, y = 3a; Change in volume, AV = ?

From Y =AV

change in volume,V i fa - 4 )

AV = yxV i(02- 0 l)= 3.6 X 10'5 X 1 O'6 (573-273) = 3.6 X IO'11 X 300

= 1.08 X 10'8m3

Percentage change in volume = change m volume ___— T-T------------- xlOOoriginal volume

= ---- xlOOK

1.08xl0~810‘6

xlOO

= 1.08 =1.1%

Example 13A density glass bottle contains 44.25g of a liquid at 0°C and 42.02 at 50°C. Calculate th real cubic expansivity of the liquid (linear expansivity of glass = 1.0 x 10'sK ''l C & SolutionThe linear expansivity, a = 1.2 X 10'5K '‘

The cubic expansivity, y = 3a = 3 X(1.0 X lO^K'1) = 3 X 10'5K '‘

Next the apparent expansivity, ya, is found first.

__________mass of liquid expelled________mass of liquid remaining x temperature rise

46

Mass of liquid expelled = mass of liquid at initial tempt - mass of liquid at final tempt. Mass of liquid remaining is the mass of liquid at the final temperature.

44.25-42.0242.02x(50-0)

7 77— ?- = 1.062x1 O'3 2101

Real expansivity of the liquid, yr = y + ya= 3 X 10'5 + 1.062 X 10 = 1.0913 X 10'3K‘‘

-3

Example 14A relative density bottle of volume 50cm3 is completely filled with a liquid at 30°C. It is then heated to 80°C such that 0.75cm3 of the liquid is expelled. Calculate the apparent cubic expansivity of the liquid. WAEC 200723SolutionApparent cubic expansivity, = ________ volume of liquid expelled_________

volume of liquid remaining x rise in temperature

Yn______ 0/75______(50-0.75)x(80-30)

Ya0.75

49.25x50= 0.0003/T'

Example 15A metal cube of linear expansivity a is heated through a temperature rise of t. If the initial volume of the cube is v, what is the increase in the volume of the cube? NECO 200719 SolutionLinear expansivity, a = a; temperature rise, A0 = t; initial volume, Vt = v Cubic expansivity, y = 3a

AVr ~Vx(A0)

Increase in volume, AV = yxV t Ad Substituting above values, we obtain, AV = 3a vt

EXERCISES 4.1. A brass rod is 2m long at a certain temperature. Calculate the linear expansion of the rod for a temperature change of 100K [Take the linear expansivity of brass as1.8 x 10 5K"‘] q WAEC 1995 Ans: 3.6 x 10'3m2. Steel bars each of length 3m at 28 C are to be used for constructing a rail line. If the linear expansivity of steel is 1.0 x 10 K , what is the safety gap that must be left between successive bars, if the highest temperature expected is 40°C?

WAEC 1997 Ans: 3.6 x lO-'m3. An iron rod of length 30cm is heated through 50 Kelvin. Calculate its increase in length °C [Linear expansivity of iron = 1.2 x 105K''j WAEC 2000 Ans: 1.8 x 10'2cm4. Metal rods of length 20m each are laid end to end to form a bridge at 25°C. What gap will be provided between consecutive rails for the bridge to withstand 75°C?

47

A. 0.22r B. 0.25m C. 0.02m D. 0.20m

Ans: 0.052mF1 210'5K '. what is its

Ans: 1.9 x 10 "K 1

. . . i _ -> n v lO ^K 'l JAMB 2004 Ans: 0.02m[Linear expansivity of the material - 2.U x rv l „ jf the change in length5. A wire. 20m long is heated from a temperature of 5 C to 5 5 ^ , ^ ^ x ^is 0.020m, calculate the linear expansivity of the wire. • - lemith of the6. The length of a ™.e rod a. 23"C « 200m. Calculate lite mercase ,n M g fco * frod when its temperature rises to 33°C. | Linear expansivity o t ■ -

WAEC. 2UUo7. If the cubic expansivity of brass between 27nC and 100 C is 5.7 x 1linear expansivity? WAEC 1998 Ans.8. A metal rod 800mm long is heated from 10°C to 95°C. If it expands y j ™ 111,linear expansivity of the metal is? WAEC 1993 Ans. 2 x 109. An iron rod of length 50m and at a temperature of 60°C is heated to 70 C. C a cu ate its new length. [Linear expansivity of iron = 1.2 x 10‘5K '] WAEC 1999 Ans. 50.006m10. An iron rod is 2.5Sm long at 0°C. Calculate the length of a brass rod at 0lC if the difference between the lengths of the two rods must remain the same at all temjjeratui es [Linear expansivity of iron = 1.2 x 10‘5K '', linear expansivity of brass = 1.9 x 10 K ].

NECO 2006 Ans: 1 629m11. On a fairly cool rainy day when the temperature is 20°C, the length of a steel rail road track is 20m. What will be its length on a hot dry day when the temperature is 40 C?A. 20.13m B. 20.009m C. 20.004m D.20.002m[Coefficient of linear expansion of steel = 11 x lO^’K 1] JAMB 2003 Ans: 20.0044m12. A steel plug has a diameter of 5cm at 30°C. At what temperature will it fit exactly into a hole of constant diameter 4.997cm? [Coefficient of linear expansion o f steel is 1 1 x 10'6C '‘) A. 27.3°C B. -27.3°C C .-2 .T C D. 7.32°C E. 2.7°C

No correct option JAMB 1978 Ans:-24.58°C13. The ratio of (he linear expansivity of copper to that of iron is approximately 1.5. \specimen of iron and a specimen of copper expand by the same amount per unit rise in temperature. The ratio of their length is A. 3 B. 1.5 C. 1.32 D. 1 E. 0.67

JAMB 1980 Ans: 0.6714. A solid material of volume 100cm' is heated through a temperature difference o f40°C. Calculate (he increase in the volume of the material if its linear expansivity is 2.0 x 10‘V . WAEC 1997 Ans: 2.4 x 10’W15. When a metal ball is heated through 30°C, its volume becomes 1.0018cm’. If the linear expansivity of the material of the ball is 2.0 x 1 O^K1, calculate its original volume.

WAEC 2004 Ans: 1 .00cm316. The linear expansivity of a substance is 1.2 x lO^K'1. A cube of this substance has a volume of 8.0 x 103cm3 at 30°C. Calculate the increase in its volume at 80°C.

WAEC 2005 Ans: 1.44 x 102cm317. A brass cube of side 10cm is heated through 30°C. If the linear expansivity o f brass is 2.0 x 10"5K_I, what is the increase in its volume? NECO 2002 Ans: 1.80cm318. The linear expansivity of brass is 2 x 10'5 ° C ‘. If the volume of a piece o f brass is10cm3 at 0°C, what will be its volume at 100°C? A. 10.02cm3 B. 10 04cm3C. I0 .06cm 3 D. 10.20cm3 E. 102.00cm3

JAMB 1983 Ans: 10.06cm319. The length of a side of metallic cube at 20°C is 5.0cm. Given that the linear expansivity of the metal is 4.0 x 10 SKL1, find the volume of the cube at 120°CA. 126.50 cm3 B. 126.25 cm3 C. 126.00 cm3 D. 125.00 cm3

JAMB 1987 Ans: 126.5cm320. If L, S and V arc the linear, area and volume expansivities of a given metalrespectively, which of the following equation is correct? A. L - S = 0 B. V - 2S = 0 C. S - 2 L = 0 D. 2 S - L = 0 E. 3 V - L = 0 WAEC 1998 Ans: C21. The temperature of glass vessel containing 100cm3 of mercury is raised from 10°C to 100"C. Calculate thejipparent cubic expansion of the mercury. [Real cubic expansivity o f mercury = 1-8 x 10 K J [Cubic expansivity o f glass = 2.4 x 10‘5K '']

WAEC 1995 Ans: 1.422cm3

48

, , .... . . . f io^ K 1 and the linear expansivity of glass22. I he cubic expansivity of mercury is l.o x m rv ulass container,is 8.0 x lO'^K1, calculate the apparent expansivity of mercury 1 lO^K'1

WAEC2000 Ans: l o x i o r- o{-(23. A cube made of a metal of linear expansivity is wanned thr g cube? A. '/3aIf the initial volume of the cube is V, what is the increase w vo u ^Vt B. ‘/2 aVt C. aVt D. 2 aVt E. 3aVt WAEL u ^ ' what wi,i be its24. A thin aluminum plate has a surface area of 1.500m a aluminum to besurface area when it is cooled to -20°C? [Take the linear expansivi m32.5 x 10'SK'‘] A. 1.503 m3 B.1.500m3 C.1.498 nr D .M 97m Ans; 1.497m2

JAMo 1 so q If the25. A cube made of metal of linear expansivity a is heated through a temp initial volume of the cube isF0, the correct expression for the increase ^cube is A j a Vd6 B A ,av.6 C.2aV,6 D. to V .6 WABC2007 ^ ^26. A blacksmith heated a metal whose cubic expansivity is • x 10’6K''exp.nsivity is A. 4.2 x lO^K.’1 B. 2.0 x 10'6K 1 C. 6.3 x 10 K • • in^’K"1

JAMB 2007 2 Ans: 4.2 x 10f -,« r is heated to 80°C. Calculate the27. A rod of initial length 2m at a temperature o f-5 L is _ 1

increase in length of the rod if its linear expansivity is 4.0 2oo918 Ans: 0.44m

49

5WORK, ENERGY AND POWER

W ORKWork is defined as the product of a force and the distance in the direction

force.Work done, W = force X distance (displacement)

W = F X s or W = mg X s

However, if the force applied is at an angle 0 to the horizontal as shown below, the work done depends on what direction the objects moves.

Vertical direction

Horizontal directional

If the box moves in the horizontal direction, the work done(W) is;

W = F cos0 X s

Or W = mg cos0 X s

If the box moves in the vertical direction, the work done(W) is;

W = F sin0 X s

Or W = m g sin0 X s

Where F = Force in Newton (N)s = distance or displacement in meter, m

m = mass in kilogram, kg g = acceleration due to gravity, m/s~

Work is measured in Joules.

ENERGYEnergy is the capacity to do work. Kinetic Energy (KE) is the energy possessed by a bodyin motion.

KE = ‘Arnv2

Potential energy (PE) is the energy possessed by a body at rest or at a height

PE = nigh

Where m = mass in kilogram, kgv = velocity in meter per second, m/s g = acceleration due to gravity, m/s2 h = height in meter (m)

50

POW ERPower is defined as the rate o f doing work or the rate o f energy transfer.

n Work done (W)Power(P) = ---------------- -— -time taken (t)

_ Forcex Distancetime taken

p = W = F x s = n tg x s = p x v t t t

Power is measured in Joules per second (J/s) or watts (W)1 horse power (hp) = 746W = 0.746kW.

Exai pie 1A boy drags a bag o f rice along a smooth horizontal floor with a force o f 2N applied at an angle 60° to the floor. The work done after a distance o f 3m is A. 6J B. 3J C .4 J D. 5J JAMB 2007SolutionForce F = 2N; 0 = 60°; Distance s = 3m

Work done, W = F co s^ x s = 2 cos60x3

= 2 x 0 .5 x 3 = 3J

_ Distance „ , , , ..= Force x -------------= Force x Velocitytime

Example 2A constant force o f 40N acting on a body initially at rest gives an acceleration of 0.1 m/s2for 4s. Calculate the work done by the force.A. 8J B. 10J C. 32J D. 160J JAMB 1987SolutionGiven: Force, F = 40N; initial velocity, u =0; acceleration, a = 0.1m/s2; time, t = 4s.

Distance, s is obtained from 2nd equation o f motion, s = ut + '/-at2

s = 0 X 4 + '/a x o . l X 42 s = 0 + 0.05 X 16 s = 0.8m

Work done = F X s = 40 x 0.8 = 32J

Example 3

Force / N

Using the force-displacement diagram shown above, calculate the work done.WAEC 2000 SolutionGenerally, in a force - displacement (F-d) or force distance graph, the work done is equal to the area o f the shape formed by the F-d graph. If the area under the F-d graph is a

S t

trapezium or semicircle, then tho work done ls or area o f a semicircle ‘/a(jrr‘). , ,|, is a rectangle thoa ° ,c *

In the above case, the area mulct the -l b

Work done 100 x 20 = 2000J

xample 4 . - , n in 20s. What is the power ofn engine raises 100kg o f water through a height o 1 If'A ECte engine? Take g=10m/s‘

Example 4An theSolution • j = 20sGiven: mass, m = 100kg; g = lOm/s*: distance, s - b( in.

Power - work done (w) _ force x Distance _ mg x s time taken (s) time taken t

^ m g ^ l O O x l O x Wt ?n

Example 5 .A machine o f efficiency 80% is used to raise a body of mass 75kg through a vertica height of 3m in 30s. Calculate the power input. [g=10m/s‘ | IIAEC -000 SolutionGiven: efficiency, E = 80%; mass, m = 75kg;

distance (height), s = 3m; time, t =30

. Power output, P innEfficiency, E = ------------- --------^xIOOPower input, P,

Power output.m gxs

t

75x10x330

225030

= 75JF

Power outputPower input, Pj = — — ---------

efficiency

P, =75x100

8093.75W

Example 6The engine o f a train produces a force of 3000N when moving at 30m/s r 1.power o f the engine. ‘ f^ £ r cu u,c thcSolutionGiven: force = 3000N; speed = 30m/s

Force x DistancePower = ----- ;------- ;--------

tune taken' Speed = djs*an^e , time

P = force X speed (velocity)

P = 3000N X 30m/s = 90000W = 9.0 x 104W

Example 7Mow long will it take a 60kg man to climb a height o f 2',ni if In- rate o f 0.25kW? [g=10m/s2] A. 5.3s 13. 34.5s C. 41.6s D. S2xpended

52.8s ./.energy at theIA//Î /9SS

52

SolutionGiven: mass, m=60kg; height, s = 22m; g=10m/s2;

Power, P = 0.25kW = 250W

F x s _ mg x sPower, P =

. m g x s. . time, t = —- — = P

t t6 0 x 1 0 x 2 2

250= 52.8s

Example 8A car o f mass 800kg attains a speed o f 25ms"1 in 20 seconds. The power developed in the engine is A. 1.25 x 104W B. 2.50 x 104W C. 1.25 x 106 W D. 2.50 x 106W JAMB 1999SolutionGive.i: mass, m=800kg; speed or velocity, v = 25ms"1; time, t = 20s.

Acceleratbn, a = X e'oclty - H = i.25m /s2 Time 20

Force, F = m a = 800 X 1.25 = 1000N

Power, P = Force X Velocity

= 1000 X 25

= 25000W or 25KW

Exam ple 9A body o f mass 5kg falls from a height o f 10m above the ground. What is the kinetic energy o f the body just before it strikes the ground? [Neglect energy losses and take g as 10m/s2] W AEC1992SolutionThe question is solved using the equation, KE = ‘Amv2. However, in this and many other cases the velocity is usually not given and has to be calculated using any of the three equations o f motion and their variations for gravitational acceleration or motion under gravity. (See the chapter on speed, velocity and acceleration).

The final velocity, v before it strikes the ground is found using the 2nd equation of motion, v2 = u2 + 2gh, for a body falling downwards or released from height.Given: initial velocity, u=0; height, h= 10m; g=10m/s2; mass, m = 5kg

v2 = u2 + 2gh

v2 = 02 + 2 X 10 X 10

v2 = 200

v = >/200 = 14.14 m/s

KE = W = Yi X 5 X 14.142 = 500J

Example 10A body o f mass 4kg is acted on by a constant force o f 12N for 3 seconds. The kinetic energy gained by the body at the end o f the time isA. 162 J B. 144 J C. 72 J D. 81 J JAMB 2004SolutionMass, m=4kg; Force, F = 1 2 N ; time, t=3s

F = ma acceleration, a = F/m =12/4 = 3m/s2

53

Also, acceleration. vclocih(v)a = -------- -----time (()

Velocity, v - a X t

= 3m/s2 X 3s - 9m/s

t- KE = Vi mv’ = '/i X 4 X 9: = 162J

A ball of mass 200g falls from a height of 5m on to a hard floor and rebounds to t of 3m. What energy is lost by the ball as a result ot the impact on the oor^ ^ 2006

SolutionGiven: mass, m=200g = 0.2kg; g=10m/s‘

Potential energy at height, h = 5m. PE, = mgh= 0.2 X 10 X 5 = 10J

Potential energy at height, h = 3m, PE, = mgh= 0.2 X 10 X 3

= 6J

Energy lost = PE, - PE, = 10 - 6 = 4J

Example 12A body rolls down a slope from a height of 100m. Its velocity at the toot of the slope is 20m/s. What percentage of its potential energy is converted into kinetic energy/ [g=10m/s2]. A. 40% B. 35% C. 20% D. 15%

JAMB 19S7SolutionGiven: height, h= 100m; velocity, v=20m/s; g=10m/s~

PE at height, h=100m, mgh = m X 10 X 100= 1000m J

KE at foot of slope, ¡/¡mv2 = 'A X m X 202 = 200m J

Percentage of PE converted to KE =KE x 100

PE200m Jx 100

1000 m J= 20%

Example 13

A box of mass 40kg is being dragged along the floor by a rope inclined at 60° to the horizontal. The frictional force between the box and the floor is 100N and the tension on the rope is 300N. How much work is done in dragging the box through a distance of 4m‘> A. 680 J B. 400 J C. 200 J D.100 J JAMB 1995SolutionWork done = Force x distance = F x s

Effective horizontal component of tension force, T = TcosO = 300cos60 150N

Effective forward dragging force, F = I ension force - Frictional force.

54

F = 150 - 100 = 50NW = F X s = 50N X 4m = 200J

Example 14A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is [g=10m/s2] JAMB ¡999

Height, h = 102 - 0 2 100 2x10 ~~ 20

Total height from ground level = h = 10 + 5 = 15m

PE at maximum height = KE just before hitting the ground.

PE = mgh = 0.1 X 10 X 15 = 15J

Example 15If a body of mass 5kg is thrown vertically upwards with velocity u, at what height will the potential energy equal to the kinetic energy?

, u2 u2 „ . 2u2 , u2A. h = — B. h = — C. h = ----- D. h = —

g 4g g 2gJAMB 2008

SolutionEquate potential energy (mgh) with kinetic energy ( X mu )

mgh = —mu22

^ _ 1 mu2 2 mg

55

EXERCISE 5. 1.

Force(N)

Distance (m)

A force varying linearly with the distance acts on a body as shown above. The wo on the body by the force during the first 10 meters of motion is A. 100J B. 150J C. 200J D. 300J E. 600J JAMB 1984 Ans: 100J 2.

A body is under the action of a force F such that the force-displacement graph o f the body is semicircle as shown above. The work done on the body by the force in moving through 24 meters is A. 367tJ B. 72ti J C. 144tiJ D. 288tiJ

JAMB 1991 Ans: 72k J ot 226.3,J3. A load of mass 120kg is raised vertically through a height of 2m in 30s by a machine whose efficiency is 100%. Calculate the power generated by the machine [g=10m/s2].

WAEC 2002 Ans: 80W4. A girl whose mass is 20kg climbs up 25 steps each of height 15cm in 10 seconds.Calculate the power expended [g=10m/s2] NECO 2005 Ans: 75 W5. A man of mass 50kg ascends a flight of stairs 5m high in 5 seconds. If acceleration due to gravity is 10ms'2, the power expended isA. 100W B. 200W C. 250W D. 400W E. 500W JAMB 1983Ans: 500W6. If a water pump at Kainji dam is capable of lifting 1000kg o f water through a verticalheight o f 10m in 10s, the power o f the pump is A. 1 .OkW B. 10.OkW C. 12.5kW D. 15.0kW E. 20.0kW [g=10m/s2] JAMB 1984 Ans: 10..0 kW7. A man weighing 800N climbs up a flight of stairs to a height o f 15m in 12.5secs. What is the man’s average power output?A. 667W B. 8 10W C. 960W D. 15000W JAMB 1992 Ans: 960W8. An electric water pump rated 1.5kW lifts 200kg o f water through a vertical height o f 6m in 1 Osees. What is the efficiency of the pump?A. 90.0% B. 85.0% C. 80.0% D. 65.0% JAMB 1997 Ans: 80%9. A man whose mass is 80kg climbs a staircase in 20s and expends a power o f 120W Find the height of the staircase [g=10m/s2]A. 1.8m B. 2.0m C. 2.5m D. 3.0m JAMB 1998 Ans: 3m10. A stone o f mass 2.0kg is thrown vertically upward with a velocity of 20.0ms'1.Calculate the initial kinetic energy of the stone. WAEC 2006 Ans: 400J

56

11. A 500kg car which was initially at rest traveled with an acceleration of 5m sec,2 its kinetic energy after 4 second was A. 105J B. 2.5 x 103J C. 2 x 103JD. 5 x 103J E. 5 x 105J JAMB1979 Ans: 105J12. An object of mass 50kg is released from a height of 2m. Find the kinetic energy just before it strikes the ground.A. 250J B. 1000J C. 10,000J D. 100,000J JAMB 1994 Ans: 1000J13. An object of mass lOOg projected vertically upwards from the ground level has a velocity of 20ms'1 at a height of 10m. Calculate its initial kinetic energy at the ground level. [g=10m/s2, neglect air resistance],A. 10J B. 20J C. 30J D. 50J JAMB 1997 Ans: 30J14. An object of mass 1000kg is dropped from a height of 10m. Calculate its energy when it strikes the ground. [g=10ms'2) WAEC 2005 Ans: 100000J or 102kJ15. The efficiency of a machine is 80%. Calculate the work done by a person using the machine to raise a load of 300kg through a height o f 4m [g=10ms’2]

WAEC 2006Ans: 15000J16. A bead travelling on a straight wire is brought to rest at 0.2m by friction. If the mass of the bead is 0.01kg and the coefficient of friction between the bead and the wire is 0.1, determine the work done by the friction. [g=10ms 2] Hint: w = Fxs = (iRxsA. 2 x lO^J B. 2 x 10"3J C. 2 x lO 'j D. 2 x l 0 2J JAMB 2003 Ans:2 x 10 3J17.

A force, F is applied to a body, P as shown in the diagram above. If the body P moves through a distance x, which of the following represents the work done?

A. Fx B. ^ C° S C. FxtanO D. FxsinO E. FxcosG WAEC 1990 Ans:E x

18. A box is pulled a distance, s along a smooth horizontal floor by a force of magnitude F, inclined at an angle, 0 to the horizontal. The work done is? NECO 2005Ans: FscosG19. A force F is applied to a body P as shown in the diagram above. If the body moves through a distance, a, which of the following represents the work done?

P

Fig- 5.8

A Fa B FacosG C. Fa— D. FasinG E. NECO 2000Ans: DCosG SinG

20. A car travelling with uniform velocity of 30ms"1 along a horizontal road overcomes a constant frictional force of 600N. Calculate the power o f the engine of the car.

WAEC 20086 Ans: 18000W or 18kW21. An object o f mass 0.25kg moves at a height h above the ground with a speed o f 4ms"1.

If its mechanical energy at this height is 12J, determine the value of h.[g = 10ms"2] WAEC 200816 Ans: 4.8m

22.1f a cage containing a truck of coal weighing 750kg is raised to a height o f 90m in 1 minute, what is the total power expended? (g = 10 ms"2)

A. 11.50kW B. 12.60kW C. 11.25kW D. 12.10kW JAMB 200910 Ans: C

57

23. A stone of mass 0.5kg is dropped from a height of 1,2m. ialculatc .ts maximumkinetic energy. (» = ] om s>) WAKC 2009 Ans: 60.0J24. A ball of mass lOOg falls from a height of 5m onto a concrete floor and rcboum sto a height of 3m. Calculate the energy lost, (g = 10ms ) NfcCO 20 ns f25. A steam engine of efficiency 70% bums 20g of coal to produce )7a ° C . . ^ it bums 200g of coal per second, calculate its output power. NIX O 200 ns.

6CURRENT ELECTRICITY

ELECTRIC CURRENTElectric current is defined as the rate of flow of charge along a conductor.

Electric curreni (1) = Quanlilyof char8°(Q)Time (t)

That is, I = — t

The unit o f electric current is ampere (A).

Example 1Calculate the quantity of charge flowing through a conductor if a current of 10A passes through a conductor for 10s. WAEC 1995Solution

I = 10 A; t = 10s

Cm Tent(l)=Quantityofcharee = S Time t

Quantity of charge, Q = lt = 1 0 X l 0 = 100C

O H M ’S LAW: Ohm’s law states that the current flowing through a metallic conductoris directly proportional to the potential difference across its ends, provided temperature and other physical conditions of the conductor are kept constant That is, V a I

V VV = IR or I = — or R = —R I

Where I = current in Amperes, AV = potential difference in Volts, V R = resistance in Ohm, Q

SERIES AND PARALLEL ARRANGEMENT OF RESISTORS AND CELLS

Resistors in Series:I Ri R2 R3

The combined, total, equivalent or effective resistance for resistors connected in series as shown above is R = R| + R2 + R3 Generally, in series connection of resistors;1. The same current passes through each resistor.2. The potential difference across each resistor depends on the value of its resistance.3. The potential difference, V, across the whole scries connection is equal to the sum of

59

. v== Vi + V2 + v 3the individual potential differences across ^aC^ r resistance; R = + R2 +

4. The total resistance R is the sum of the individual

Resistors in Parallel

The combined, total, equivalent or effective resistance for resistors connected para

R1 1

R , + R 2

If there are two resistors, Ri and R2, the combined resistatw»,_ ^1^2

r , + r 2Generally, in parallel connection o f resistors;1. The potential difference across each resistor is the same and is equal to the potential difference across the whole connection, V = V i = V 2 = V3

2. The total current is equal to the sum of the currents flowing through each resistor,I = Ii + I2 + I3

3. The combined resistance is always less than the least individual resistance.

Cells in Series

E2,r2 £3, riFor cells connected in series the total e.m.f, E = E| + E2 + E3

The total internal resistance, r = ri + r2 + r3

Cells in Parallel

____

____1

l____

1?'2’ r2

Fig. 6.4 1------------

For cells connected in parallel, the total e.m.f, E = E; = E2 = E1 1 1 1

The total internal resistance, - = — H— H—' r r, r, r,

60

For two internal resistors, r, and r2 in parallel, effective resistance, r = •

Example 2What is the effective resistance of the circuit below.

20.

50

Fig. 6.5

SolutionTlu 512 and 1012 are in series, therefore the combined resistance, R = Ri +R2 = 5 + 10 = 1512.The circuit is redrawn as follows.

20

The

R = -

212 and 412 are in parallel, therefore

R, + R : 2 + 4 6The circuit is drawn again as follows.

the combined resistance,

is n i.33nFig. 6.7 • --------- --------- -------- •

The 1512 and 1.3312 are in series, therefore the combined resistance, R = Ri + R2R = 15 + 1.33 = 16.3312

Example 3lfi 2f2

Fig. 6.8 in 20

The total resistance measured at PQ in the diagram above is?

A. 18.0 Q B. 11.012 C. 4.0 12 D. 2.0 Q JAMB 1998

SolutionThe three 2Q resistors are in series, therefore the total resistance,

r = r ( + r ,+ r , ;. R= 2 + 2 + 2 = 612. The circuit is redrawn as follows.

61

Fig. 6.9

The 3Q and 6i2 resistors are in parallel, therefore the combined resistance,

R,R, _ 3x 6 _" R, + R, _ 3 + 6

The circuit is redrawn once again as follows:in

The three resistors (li2, IQ, and 2i2) are in series, therefore equivalent resistance, R = Ri + Ri + R3

= 1 + 1 + 2 = 4QThe total resistance measured at PQ is 4

Example 4A car fuse is marked 15A and operates normally on a 12V battery. Calculate the resistance of the fuse w ire. WAEC 1994SolutionCurrent, I = 15A; p.d, V = 12V

From V - 1R. R = — = — 0.8Q

Example 5The sum o f the current I| and T in the figure below is

Fig. 6.11 Mi ionA. 1.2A B. 0.6A C .0.8A f>. 2A E. 1A SolutionGiven: V - 1 2 V ; R=10£2

V 12 1 7AFrom V = IR, K = — = — = 1 2A

JAMB 1979

62

I or V 12V and R -5 + 1 0 - 1 5£i (series connection)

. V 12!. = — = — = 0.8A I, + I2 = 1.2 + 0.8 = 2.0A

R 13

Alternatively, this could be solved as follows.(i) rhc 5£2 and 10£2 resistor through which I| flows are in series. Therefore, effective resistance, R = 5 + 10 = 15Q.(ii) I he 15£ 2 resistor is connected in parallel with the 10£2 resistor. Therefore the

effective resistance,R R

R = ~ ' p vvherc R, = 15£2; R: = 10£2K. + K,

R = 10x15 10+ 15

(iii) Total circuit current (I, + 1, ), Potentialdifference total circuit resistance

Calculate the current in the 3Q resistor shown in the diagram above. WAEC 1990 Solution_ . , 1 1 1 1 1 1 1Equivalent resistance, — - — + ------ 1----- = — + — + —

R R, R , R 3 2 3 4

1 _ 6 + 4 + 3 _ 13 12QR _ 12 _ 12 13

Voltage or p.d across the parallel connection, V = IR

Current in the 3Q resistor,

V = 1 3 x — = 12V 13

12

T = 4A

Example 7An electrical circuit is connected up as shown below. What is the value o f the current

63

JAMB 1979*• / 1 3 15

Solutionu .

Fie. 6.1412V

(i) R, and R2 are in series Combined resistance, R i,2 = 3 + 3 = 6D(ii) Rj and R„ are in series Combined resistance, Rj.* = 3+3 = 6D(iii) The circuit can be redrawn as follows.

Fie. 6.1512V

(iv) R| 2 and R3 4 are connected in parallel

Effective résistance, R6 = ---—— = —X^ = — = 3Î2R ,, + R 34 6 + 6 12

(v) The circuit now becomes

Rô=3n R,= m

Fig. 6.16 12V

R,, and R5 are in scries :. Total resistance, R = R? + Rc = 3 + 1 = 4QThe current through the ID resistor is the same as the current flowing through the wholecircuit.

VFrom V = IR, 1 = — Substitute V=12V; R=4D

R

, I = — = 3A 4

Example 8In the figure below, a voltage V is applied across the terminals P and Q pjle voj across the ID resistor is a~e

64

-1Í2

SolutionThe effective resistance across the parallel combination,

R — — R| R - = 4 x 4 = 26 = 2 fl R, + R, 4 + 4 8

The circuit becomes

in 2Í2p 1------------1 —Fig. 6.18

Total circuit resistance, R = 1 + 2 = 3QTotal circuit voltage, V = V

V VTotal circuit current, 1= — = —R 3

From V = IR

Voltage across the l i i resistor is (where R =1)

JAMB 1987

Example 9If a resistance is halved in value and the p.d across it is tripled, then the ratio o f the newcurrent to the old is A. 1:6 B. 1:3 C. 2:1 D. 6:1 JAMB ¡995SolutionI f a resistance is halved in value, R = '/>R; p.d. across it is tripled, V = 3 V.

V _ 3V 3V 6V1 = — The new current, I = ----- = ------- ■ = —R y i R 0.5R R

VThe old current, I = —

R

So, the ratio o f the new to old current is

ELECTROMOTIVE FORCE, INTERNAL RESISTANCE, TERMINAL POTENTIAL DIFFERENCE AND LOST VOLT

The electromotive fo rce (e.m.f.), E of a cell is defined as the total work done in driving one coulomb of electricity round a circuit. The e.m.f of a cell is also defined as the potential difference between the terminals of a cell when it is an open circuit, i.e. not delivering current to an external resistance.

6V V R R

= 6 :1

65

The internal resistance, r, o f a cell is the opposition to current How olfeicd by 1*,<J when it is discharging current to a circuit.

Terminal potential difference, V, is defined as the potential difference betwee terminals o f a cell when it is delivering current to an external resistance, R.

The lost volt or voltage, v is the p.d across the internal resistance, r, ol a cell.

E .m .f = p.d across external resistance + p.d across internal resistance.E = V + v ..................................................... ( ' )

Applying ohm ’s law to eqn (1) we get

E = IR + Ir (V= IR, v = Ir)E = I(R + r)

Example 10A battery o f e.m .f 24V and internal resistance 4Q is connected to a resistor o f 32i2. What is the terminal p.d o f the battery? WAEC 1998SolutionW hether you are asked to or not, always draw a circuit diagram before solving any question on current electricity.

Fig. 6.19

E=24V ■

II ¡0

—►-

------

-

R = 32il

First, calculate the circuit current, I, E 24 24 2 ,I = ------- ----------- = — = — A or 0.67A

R + r 32 + 4 36 3

Terminal p.d, V = IR = 0.67 X 32 = 21.3V

Alternatively, the answer is gotten from the equation, E = V + v or E = V + Ir

Terminal pd, V = E - IrV = 2 4 - 0 . 6 7 X 4

= 2 4 - 2 . 6 7 = 21.3V

A ^ eH ^ o f e m .f 1.5V is connected in series with a resistor o f resistance 3Q. A highresistance voltm eter connected across the cell registers only 0.9V. Calculate the internal

• .. tVi/= WAEC 1996resistance of the cell.Solution

Q V=0-9Vr = ?

_________ 1 ._______ E = 1.5V1 '

r = in

E = 1.5V

i

V = 0.9V R = 3Q

L r = ?

Fig. 6.20

66

Current flowing in the circuit, I = —R + /• 3 + r

p.d across cell, v = E - V = 1.5 - 0.9 = 0.6V v = 0.6

Substitute v = lr in the above Ir = 0.6

1.5 1.5 _ n rSubstitute I = ------ into above equation to obtain, —---- * r - u 0

3 + r 3 + r

Cross multiply; 1.5r = 0.6(3+r)1.5r = 1.8 + 0.6r

1 .5 r- 0.6r = 1.8

— — o n0.9r = 1 .8 :. Internal resistance, r - — -

Example 12An electric bell takes a current o f 0.2A from a battery of two dry cells connected in series. Each cell has an e.m.f of 1,5V and an internal resistance of 1 .0i2.(i) Calculate the effective resistance of the bell.(ii) What current would the bells take if the cells were arranged in parallel?Solution(i)

E,=1.5V ,= 1.5V

r,= 1.0Ì2 ry= i .oni

BellFig. 6.21

Let R be the resistance of the bell

Total e.m.f, E = Ei + Ej = 1.5 + 1.5 = 3VTotal internal resistance, r = ri + r2 = 1 + 1 = 2Q ; Current, I = 0.2A

ESubstitute the above into I = -------

R + r

R + 2Cross multiplying we have

0.2(R+2) = 3 0.2R + 0.4 = 3

0.2R = 3 - 0.49

0.2R = 2.6 R = — = 13Q0.2

( i i )E, = I.SV

67

E.m.f, E = E| = E: = 1.5V

r = - ^ = — = 1 = 0 .5« and R = 13Q r, + r, 1 + 1 2

Current, 1 = —- — = — — — = = 0.11AR + r 13 + 0.5 13.5

Example 13 ThTwo cells, each o f e.m.f. 2V and internal resistance 0.5Q, are connected in series, are made to supply current to a combination ol three resistors; one o f the resistances ' connected in series to a parallel combination of two other resistors each o f resistance Draw the circuit diagram and calculate: r(i) Current in the circuit (ii) Potential difference across the parallel com “tabon o

IVA EC 1998the resistor, (iii) Lost volts o f the battery.Solution

Fig. 6.23

2V 2V

(i) Total e.m.f, E = 2 + 2 = 4V

Effective internal resistance, r = 0.5 + 0.5 = 1.0f2

3x3 9Effective external resistance, R = 2 + -— - = 2 + — = 2 +1.5 = 3.5Î2

3 + 3 6

.-. I = = 0.89AR + r 3.5 + 1 4.5

(ii) p.d across parallel connection, V = 1R

3x3(Total parallel resistance, R = - — - - 1.5£2)

V = 0.89 X 1.5 = 1.3V

(iii) Lost volt = p.d across the battery

= Ir = 0.89 X 1 = 0.9V

Example 14An electric cell with nominal voltage, E has a resistance o f 3fi connected across it. If the voltage falls to 0.6E, the internal resistance o f the cell is? JAltfB 2002Solution

I

Fig. 6.24

E,r_________ 11______________

rR=3fi

68

The nominal voltage, E is the e.m.f of the cell.The terminal p.d, V = 0.6E

The resistance, R = 3Ì2

The e.m .f o f a cell, E, is equal to the sum of the terminal p.d, V, and the lost volt, v

i.e E = V + v ........................................................................ (1)or E = IR + ir

Terminal p.d, V = IR

0.6E = 31 (2)From Eqn( 1), E = V + v

The lost voltage, v = E - V

v = E - 0.6E

v = 0.4E

Ir = 0.4E .......

From eqn (2), I =3

j- y 0.4EFrom eqn (3), I = ------

(3)

(4)

(5>

Equate eqns (4) and (5)0.6E _ 0.4E

3 ~ r

„ 0 .4Ex3 1.2E „Re-airangmg, r = - ^ = — = 2 Internal resistance, r = 2Ì2

Example 15A cell can supply currents o f 1.2A and 0.4A through a 4Q and 14fi resistors respectively. Calculate the internal resistance of the cell.SolutionThis problem involves two instances (6.25:a & b) as shown below) and two unknown quantities, that is, the e.m.f E and the internal resistance, r.

E E

---------------11---------------

Jl 1=1.2 A

---------------1 1---------------I=0.4A

Fig. 6.25 R=4i2 R=14i2(a) (b)

ESubstitute each o f the above case into I =

R + r

1st case: 1.2 = E 4 + r

E = 1.2 (4 + r) ........... ......... (1)

2nd case: 0.4 = —- — :. 1 4 + r

E = 0 .4 (14 + r) .......... ......... (2)

69

Equations 1 and 2 are equated and then solved simultaneously.

1.2 (4 + r) = 0.4 (14 + r)

4.8 + 1.2r = 5.6 + 0.4r

5 .6 -4 .8 = 1.2r - 0.4r

0.8 = 0.8r

Worthy o f note is the fact that the c.m.f, E can be calculated by substituting the internal resistance, r = 1 .Ofi into eqn 1 or 2 as follows

(i) E = 1.2(4+r)E = 1.2(4+1) = 1.2(5) = 6V

(ii) E = 0 .4(14+ r)E = 0.4(14+1) = 0.4(15) = 6V

Example 16

R

In the circuit diagram above, the ammeter reads a current of 3A when R is 5E2 and reads 6A when R is 2Q. The value of the unknown resistance X is A. 1 B. 2 C. 3 D. 4 JAMB 1988Solution

EFrom I = ------- we modify and apply it to the diagram above, making r = X.

R + r

I= -R+X (1)

I s1 instance: I = 3A, R = 5fi substituting in (1), we obtain E

3 = -5 + X

£ = 3(5 + 2Q =15 + 32f (2)

2nd instance: I = 6A, R=2f2 substituting in (1), we obtain

6 =E

2 + XE = 6(2 + X) = 12 + 6X (3)

Equate equation (2) and (3)

15 + 3X = 12 + 6X

1 5 - 1 2 = 6 X -3 X

3 = 3X

, x = 3/3 = \ n

70

Example 17

Fï p . I

X

— G X — 1--------------

In the circuit diagram above, the ammeter reads a current o f 3 A, when R is 5 when R is 20 . Determine the value of X.A. 8Q B. 2Q C. 10Q D. 4Q Solution JAMB 2001

n X are resistors in parallel. The effective resistance is obtained from the equation of two resistors in parallel.

p -R, + R,

t Ri be the given resistance and R2 the unknown resistance, X

First instance: R = —5 -EX for R| = 50, R2 =X, I = 3A

Second instance: R = ~ ~ for r , = 2Sì, R2 =X, I = 6A

From Ohm’s law, V = IR Substitute each case into I = V/R

Is1 instance

5X. 3 _ V(5+X)

5X5+X

M.^ke V the subject of formula, 15XV =-

5 + X ( 1)

2nd instance

Os II l< Os II V (2+X)

2X2+X

Making V the subject o f formula,

v - 12X

2X

2 + XEquate (1) and (2)

15X 12X 5 + X 2 + X

Cross multiply30X + 15X2 = 60X + 12X2

60X - 3OX =

3 OX = 3X2

30 = 3X

15X2 - 12X2

(2)

71

30X = — = 10ft 3

Example 18A cell of internal resistance, r supplies current to a 6.0ft resistor and its efficiency is 75 /o. Find the value of r. A. 4.5ft B. 1 .Oft C. 8.0ft D. 2.0ft JAMB 2001Solution

Efficiency of a cell, E . = Powcr QutPut x l00 power input

I2RE / =

Ef =

I2(R + r) R

-xlOO

(R + r)xlOO

Also, Ef = P.d across external resistor e.m.f of the cell

xlOO

R is external resistance(s) and r is internal resistance of the cell. Depending on the question, any o f the above formula can be used to calculate efficiency o f a cell.

From the question, efficiency, E f = 75%; External resistance, R = 6.0ft.

E f =

75 =

<R + r)6

(6 + r) 75 _ 6

100 “ (6 + r)

0.75 = — — (6 + r)

0.75(6 + r) = 6 4.5 + 0.75r = 6

0.75r = 6 - 4 .5

0.75r = 1.5 1.5

■v

vl#0

xlOO

r = ■0.75

= 2 0

Alternatively, we make r the subject of equation before substituting. r _ R(l00 —E f )

Ef_ 6(100-75)

756x25r - --------

75r = 2i?

Example 19A resistor o f resistance, R is connected across a cell. If the terminal p.d of the cell is reduced to one-quarter o f its e.m.f., express the internal resistance, r of the cell in terms of R. NEC0 2004Solution

e.m.f, E = V + v or E = IR + Ir

E - I RInternal resistance r = ---- -—

but terminal p.d, V — ’/>E (“ . ..terminal p.d... is... one quarter o f e.m.f. . .”)

. £ = 4V

4 V -IRSubstituting, r = ----------- (V = IR)

72

r _ 4IR- ir I

EXERCISE 6.1* A current o f 100mA electricity transported is?A.

passes through a conductor for 2 200C B.50C C.12C D.0.02C

minutes. The quantity of

JAMB 1993 Ans:12C

Fig. 6.28

Calculate the effective resistance between P and Q in the diagram shown above.WAEC2007 Ans: 4.00Q.

' parallel combination of 3i2 and 4i2 resistors is connected in series with a resistor o f and a battery of negligible resistance. Calculate the effective resistance in the circuit.

WAEC2000 Ans: 5.7Q 4.

4Í2

Fie. 6.29

4Û

4Í2 4QFind the effective resistance in the diagram above. A. 6 fi B. 12Î2 C. 18Q D. 24Q 5.

JAMB 2008 Ans: 24 Q

Fig. 6 JO

Calculate the effective resistance between point X and Y in the diagram above.WAEC 1994 Ans: 2Q.

6. Two resistors o f resistance 6Q and 4i2 are connected in parallel. A third resistor o f 8 ii is then connected in series with the parallel connection. Calculate the effective resistance o f the arrangement. WAEC 2005 Ans: 10.4 £1

73

7.in

Fig. 6.31

2n

The effective resistance o f the circuit shown in the figure above is ?NECO 2006Ans:3Q

8. What is the resistance in the diagram below?ton

RjJAMB 1980 Ans: 25Q

9. Which o f the circuits illustrated below will give a total resistance o f 1H20 30 60

A. ----------- -------- --------

60

JAMB 1982 Ans: C

74

10.

20

W hat is the resultant resistance of the circuit given above? . s oA im B .8 0 C .4 0 D. 3.6a E. 4 .30 Ans. 8011. W hat is the resistance o f the circuit shown below?

20

30

Fie. 635 _ ___ ___________________________t

A. 4 fl B. l l f i C. '%Q D. E. 8Q JA M B 1984 Ans:12. Which o f the following arrangement will produce an equivalent resistance of 1.5Q from three IQ resistors.

lO lO lOA.

B.

lO

JAMB 1986 Ans: D

75

13.

Fig. 6.37 40

Calculate the current I in the diagram shown above (neglect the internal resistance o f the cell). WAEC 1993 (Ans: 1A)

14. Using the data on the diagram below. Calculate the p.d across the 2 0 0 resistor. (Neglect the internal resistance of the cell)

WAEC 1999 Ans: 60V15. A car fuse marked 3A operates optimally on a 12V battery, calculate the resistanceof the fuse. WAEC 2004 Ans: 4 .00016. A 12V battery supplies a current of 0.4A to a lamp. Calculate the resistance o f thelamp. NECO 2005 Ans: 30 .00017. Three 5 ohm resistors connected in parallel have a potential difference o f 60V applied across the combination. The current in each resistor isA. 4A B. 36A C. 12A D. 24A E. 10A JAMB 1978 Ans: 12A18.

6VI---------- --------------------------------

The effect o f closing the key K in the circuit A. increase the current by 0.6A C. reduce the current by 0.4AE. keep the current unchanged.

shown above would be toB. reduce the current by 0.6A D. increase the current by 0.4A

JAMB 1981 Ans: D

76

19.

<C)

The diagrams above show three circuits. The internal resistances o f the batteries are negligible. W hich o f the currents is the largest?A. I, B. 12 C. I3 D. I4 E. I5 JAM B 1983 Ans: I520. A 24V potential difference is applied across a parallel combination o f four 6 resistors. The current in each resistor is?A. 1A B. 4A C. 16A D. 18A E. 36A JAMB 1983 Ans: 16A21.

3n

Calculate the current I in the diagram above (neglect the internal resistance o f the cell).WAEC 1994 Ans: I = 1.0A

22.

Fig. 6.42

5Q

E

In fig 6 42 E is an accumulator with negligible internal resistance. If the e.m .f is 9.0V,

i ' S f r B o l T c i . O A D. 1.8A JAMB 1986 Ans: 1.0A ' Th resistors with resistances 250i2, 500Q and lki2 are connected in series. A 6V

re ctej to either end o f the combination. Calculate the potential difference

bb: " : l ,S.b °"n d s o f .be A. 0.20V B. 0.86V C. 1.71V D. 3.43V JAMB 1989 Ans: 0.86V

77

24.2i2

Fig. 6.4312V

The current I in the figure above isA. 4.00A B. 1.30A C. 0.80A D. 0.75A JAMB 1993 Ans: 4.0A25.

3£2

Fig. 6.446V

In the diagram above, the current isA. | B. £ C. ^ D. f JAMB 2001 Ans: 9/n or 0.82A26.

Using the data in the circuit illustrated above, calculate the value of R.W AEC1991 Ans: 5Q

27. A battery o f e.m.f E and negligible internal resistance supplies a current I to the combination o f two resistances R| and R2 as shown in the diagram below. Calculate the current through R2.

Fig. 6.46

WAEC 1994 Ans: ^R \ + & L

79

28. A cell o f e .m .f 1.5V and internal resistance o f 2 .5 ft is connected in scries w ith anam m eter o f resistance 0 .5 ft and a resistor o f resistance 7 .0ft. C alculate current in the circuit. W AEC 1990 Ans: 0 .15A29. A battery o f e .m .f 10V and internal resistance 2 ft is connected to a resistance of 6 ft.Calculate the p.d across the terminals. W AEC 1994 Ans: 7.5V30. What is the potential difference between X and Y in the diagram below if the battery is o f negligible internal resistance? Calculate the current in the 3 ft resistor.

W AEC 1995 Ans: 3V 1.0A

2V

0

31. Calculate the terminal p.d across 20ft resistor connected to a battery o f e .m .f 15Vand internal resistance 5 ft WAEC 1997 Ans: ! 2V32. Two identical cells each o f e.m .f 2V and internal resistance 1.0ft are connected inparallel. The combination is connected to an external load o f 1.5ft, calculate the current in the circuit. WAEC 20(03 Ans: 1.0A33. Four identical cells each o f e.m .f and internal resistance r are connected in series with a resistance R. W rite down the formula for (he current in the circuit.

34.

WAEC 1989 Ans: 4ER \ 4 r

3il

Fig. 6.48

In the diagram above, the current passing through the 6ft resistor is 1.5A. Calculate the (i) current in the 3 ft resistor (ii) terminal p.d. o f the battery.

WAEC 2000 Ans: (i) 0.6A (ii) 10.8V35. In the circuit shown below, the cell P has an e.m .f 1.5V and an unknown internal resistance r while the cell Q has an e.m .f 2.0V and an internal resistance 1ft. If the ammeter reads 50mA, then r is equal to? A. 0.5 ohms B. 0.62 ohms C. 1.0 ohms D. 2.0 ohms E. 3.0 ohms JAM B 1979 Ans: 2ft

P 1.5V Q 2.0V

36. An electric cell has an internal resistance o f 2ft. A current of'0.5A is found to flow when a resistor o f 5ft is connected across it. What is the e.m.f. o f the cell?A. 5 volts B. 3.5 volts C. 2.5 volts D. 1 volts E. 10 volts

79

JAM B 1979 Ans: 3.5V37. A cell o f c.m .f 2V and internal resistance lO supplies a current ol 0.5A to a resistance whose value isA. 0 ,5 0 B. 1 O C. 2 O D. 2.5 O E. 3 O JAM B 1981 Ans: 3 038. A high-resistance voltmeter reads 3.0V when connected across the term inals of abattery on open circuit and 2.6V when the battery supplies a current o f 0.2A through a lamp. The resistance o f the lamp is A. 2.00 0 B. 13.00 OC .0 .5 2 O D. 0.13 O E. 1 .5 0 0 JAM B 1982 Ans: 13 .000

39. The difference o f potential between the terminals o f a cell is 2.2 volts. W hen a 4 0resistor is connected across the terminals o f this cell, the p.d is 2 volts. W hat is the internal resistance o f the cell? A. 0.10 ohms B. 0.25 ohms C . 0.40ohms D. 2.50 ohms E. 4.00 ohms JAM B 1983 Ans: 0 .4 040. Two cells, each o f e.m .f 1.5V and an internal resistance 2 0 are connected in parallel. Calculate the current flowing when the cells are connected to a 10 resistor.A. 0.75 0 B. 1.5 0 C. 0.5 0 D. 1.0 0 E. 0.6 O JAM B 1984 Ans: 0 .7 5 041. Three cells o f e.m .f 1,5V and an internal resistance o f 1 .00 are connected in parallel across a load resistance o f 2 .670. Calculate the current in the load.A. 0.26A B. 0.41 A C. 0.50A D. 0.79A JAM B 1990 Ans: 0.5A42.

The internal resistance o f each o f the cells E| and E2 shown in the figure above is 2 0 . Calculate the total current in the circuit.A. 0.80A B. 0.50A C. 0.40A D. 0.004A JAM B 1991 Ans: 0.4A43.

12fl

In the circuit diagram above, calculate the current in the 120 resistor if the cell has an e.m .f o f 12V and an internal resistance o f lO.A. 0.8A B. 1.0A C. 1.6A D. 2.4A JAM B 1994 Ans: 0.8A44. A 12V battery has an internal resistance o f 0.50. If a cable o f 1.00 resistance is connected across the two terminal o f the battery, the current drawn from the batters is

A. 16.0A B. 8.0A C. 0.8A D. 0.4A JAM B 1995 Ans: 8.0A45. Three electric cells each o f e.m .f 1.5 V and internal resistance 1.00 are connected in parallel across an external resistance o f 2/30 . Calculate the value o f the current in the resistor. A. 0.5A B. 0.9A C. 1.5A D. 4.5A

JAM B 1997 Ans: 0.9A46. Four cells each of e.m .f 1.5V and internal resistance o f 4 0 are connected in parallel. What is the effective e.m .f and internal resistance o f the combination?

80

A. 6.0V, 16 O B. 6.0V, lO C. 1.5V, 4 f i D. 1.5V, lOJAM B 1999 Ans; 1.5V, 1.00

47.

Fig.

In the diagiam above, the values o f V| and V2 are respectively?JAM B 2 0 0 0 Ans: IV, 2/ iV

48. A radio is operated by eight cells each o f e.m .f 2.0V connected in series. I f two o f the cell are wrongly connected, the net e.m.f o f the radio isA. 16V B. 12V C. 10V D. 8V JA M B 2 0 0 0 Ans: 12V49. A cell whose internal resistance is 0 .50 delivers a current o f 4A to an external resistor. The lost voltage o f the cell isA. 1.250V B. 8.000V C. 0.125V D. 2.000V JA M B 2 0 0 4 Ans: 2.0V50. A resistance R is connected across the terminal o f an electric cell ct internalresistance 2 0 and the voltage was reduced to 4 of its nominal value. The value o f R is A. 3 O B. 2 O C. 1 O D. 6 O JAM B 2001 Ans: 3 051. A cell supply current o f 0.8A and 0.4A through a 2 0 and 5 0 resistor respectively. Calculate the internal resistance of the cell. WAEC 1988 Ans: 1.0052. A cell supplies current of 0.6A and 0.2A through 1.00 and 4 .0 0 resistors respectively. Calculate the internal resistance o f the cell. W AEC 2 0 0 5 Ans: 0 .5 053. A cell gives a current o f 0.15A through a resistance o f 8 0 and 0.3A when the resistance is changed to 3 0 the internal resistance of the cell isA. 0.05 0 B. 1.00 0 C. 1.50 0 D. 2.00 O E. 2.50 0 JAM B 1985 Ans: 2 054. The terminal voltage o f a battery is 4.0V when supplying a current o f 2.0A and 2.0V when supplying a current o f 3.0A. The internal resistance o f the battery isA. 0.5 0 B .l.O O C. 2.0 0 D. 4.0 O JA M B 1993 Ans: 2 055. A cell can supply currents o f 0.4A and 0.2A through a 4 .0 0 and 10.00 resistors respectively. The internal resistance of the cell is A. 2.0 0 B . l .O O C. 2.5 0 D. 1.5 056.

JA M B 2001 Ans: 2 0

Fig. 6.53

What is the value o f R in the diagram above?A. 3 0 B. 4 0 C. 5 0 D. 6 O57.

JA M B 1 9 8 6 Ans: 6 0

81

In the circuit above, the ammeter reads a current o f 5.0A when R = 5Q. The value o f the unknown resistance X is A. 10. O fi 13. 7.5 i2 C .S .O fi D. 2.5 D 58.

when R—8f2 and reads

JAMB 1999 Ans: 2.5Q

7.0A

Fig. 6.55

3 £2

12VFrom the diagram above, determine the value of the resistance X.A . 9 Q B. 12 i2 C. 15 Q D. 6 Q. JAMB 2002 Ans: l SCI59. A cell o f internal resistance l i i supplies current to an external resistor o f 3f2. The efficiency of the cell is? A. 75% B. 50% C. 33% D. 25% JAMB 2000 Ans: 75%60. A cell o f internal resistance 2Q supplies current to a 6fi resistor. The efficiency o f the cell is A. 12.0% B. 25.0% C. 33.3 % D. 75.0 % JAMB ¡988 Ans: 75%61. A chemical cell o f internal resistance 1£1 supplies electric current to an externalresistance 3Q, calculate the efficiency o f the cell. WAEC 2005 Ans: 75%62. A cell o f e.m.f 1.5V and internal resistance 1.0Q is connected to two resistors o f resistance 2.0Q and 3.0Q in series. Calculate the current through the resistors.

WAEC 2006 Ans: 0.25 A63. Three identical cells each o f e.m.f 1.5 V and internal resistance 1.0i2 are connected in parallel across an internal load of resistance 2.67Q. Calculate the current in the load.

N E C 0 2007 Ans: 0.50A64. A cell supplies a current o f 0.6A through a 2 £2-resistor and a current o f 0.2 A through a 7 Q-resistor. Calculate the internal resistance o f the cell.

NECO 2008 Ans: 0.5 Q65.

The cells each o f e.m .f 1.5V and internal resistance of 2 .5il are connected as shown in the diagram above. Find the net e.m.f and the internal resistance.A. 4.5V, 0.83D B. 4.5V, 7.50 Cl C. 1.5V, 0.83 Q D. 1.5V, 7.50 Q

JAMB 2007 Ans. C66. A resistor o f resistance R is connected to a battery of negligible internal resistance I f a similar resistor is connected in series with it the

A. effective resistance o f the circuit is halved.B. total power dissipated is doubled.C. total current in the circuit is halved.D. terminal voltage is halved WAEC 200938 Ans: C

82

67.

Fig. 6.57

Calculate the e.m.f o f the cell in the above circuit if its internal resistance is negligible.A. 12V B. 8V C. 2V D. 36V JAMB 20094n Ans: B

68. A cell o f e.m .f 1,5 V is connected in series with a resistor o f resistance 3.Oil. Avoltmeter connected across the cell registers 0.9V. Calculate the internal resistance of the cell. WAEC 2009'W Ans: 2il

69. A battery of e.m .f 12.0 V and internal resistance 0.5 il is connected to 1.5il and 4.0 il series resistors. Calculate the terminal voltage o f the battery.

WAEC 20094'A ns: 11V70. When a resistor of resistance R is connected across a cell, the terminal potential

difference of the cell is reduced to the three-quarters o f its e.m.f. The cell internal resistance in terms of R isA. R/ a B. R/ 3 C. R/ 2 D. 2 r / 3 E. R

NECO 20094’ Ans:B71. Determine the p.d across the load in the diagram below, if the ammeter is o f

negligible internal resistance. NECO 200944 Ans: 10.7V

72. A cell of electromotive force E and internal resistance r is connected in series with an ammeter and an external load of resistance R. Derive an expression for the power dissipated by the external load in terms of the parameters.

NECO 2009E7 Ans:— -(R+r)2

83

7ELECTRICAL ENERGY AND POWER

ELECTRICAL ENERGYElectrical energy (W) = quantity of charge (Q) X Potential difference (V)

W = QV................................................... (1)

Substituting Q = It into (1) we have

W = IV t................................................... (2)

Substituting V = IR into (2) we have

W = I2Rt ................................................. (3)

Substituting I = V/R into (3) we have V 2tW = — .................................................(4)R

So, Electrical Energy can be calculated using any of these equations.

W = QV = IVt = l2Rt = —R

ELECTRICAL POWERElectrical energy transferred (W)

Electrical Power (P) = ------------------—--------------------Time taken (t)

WP = - ................................................ (5)

tSubstituting W = QV into (5) we have

P °tV .................................. (6)Substituting W = IVt into (5) we have

P “ = IV ........................................... (7)

Substituting W = I2Rt into (5) we have„ I2Rt l2„p = —¡— = I R ......................................(8)

Substituting I = V/R into (8) we have „ V2R V 2

R 2 “ R (9)So, Electrical Power can be calculated using any of these equations.

W _ QV t t

Where, W = electrical energy in Joules (J)P = electrical power in watts (W)I = current in amperes (A)V = potential difference in volts (V)

R = resistance in ohms (Q)Q = quantity of charge in coulomb (C)

t = time in seconds (s)

IV = i2rR

84

ELECTRICAL HEAT ENERGYElectrical Heat energy, H = I2Rt. This is in agreement with Joule’s law of electrical heating which states that the heat energy produced in a wire as a result of electnca current passing through it is proportional to(i) the resistance of the wire, R(ii) the square of the current, I2(iii) the time (t)So, electrical heat energy can be calculated using any of the three equations

, V2tH = l2Rt = IVt = —— = Pt

R

Example 1A work of 30 Joules is done in transferring 5 milli-coulombs of charge from a point B to a point A in an electric field. What is the potential difference between B and A? WAEC1990 SolutionEnergy or work done in an electric field, W = QV

Given: work done, W = 30J; charge, Q = 5mC = 5 X 10"3C

Potential difference, V = — = — — -r = 6.0 x 103 V Q 5xl0~3

Example 2Calculate the resistance of the filament of a lamp rated 240V, 40W. WAEC 1990SolutionGiven; potential difference, V = 240V Power, P = 40W

P V2Resistance, R = — P

R = 24040

5760040

= 1440S2

Example 3The maximum power dissipated by a 100Q resistor in a circuit is 4W. Calculate thevoltage across the resistor. __ __ WAEC 2003SolutionGiven: Power, P = 4W; resistance, R = 100Î2

V2 2P = — V2 = PxRR

V = V P x R = V 4 x l0 0 = V4ÔÔ = 20V

Example 4An electric iron is rated at lOOOwad^.^SOV. The corresponding maximum resistance and accompanying current is A. 62.5A, 4.0Q B. 16.0A, 62.5 i2C. 62.5A, 16.0 Q D. 4.0A, 62-^Pvy & 4.0A, 250 Q JAMB 1982 Ans. DSolutionGiven: Power, P = 1000W, voltage, R = 250V.

. P 1000 . . 0 0 ‘P = IV . current,/ = — = ------' - 4 A ~

V 250

, n P 1000 1000P = I2R . resistance, R = -— = — — = — = 62.5Q I 4 16

Ans :62.5£2,4A

Example 5 .When connected to a mains of 250V, the fuse rating in the plug of an electric device ol lkW isA. 4A B. 2A C. 5A D. 3A JAMB 2002 Ans: 4ASolutionThe fuse rating is the maximum safe current a fuse can take before breaking.Given: Power, P = lkW = 1000W, p.d., V = 250V

P = IV current, / = — = ------= 4AV 250

Example 640

Calculate (i) the rate of energy consumption; (ii) the total heat energy developed in 15 minutes; in the circuit illustrated by the diagram above.Solution(i) Rate of energy consumption = PoweT used

Power, P = IV = I2R4 x 4Current, I = 2.5A, Effective resistance, R =------= 2Q,4- 4

P = I2R = 2.52 X 2 = 12.5W

(ii) Given, I = 2.5A, R= 2£2, time, t = 15 X 60 = 900sec.

Heat energy, H = I2Rt

H = 2.52 X 2 X 900= 11250J

Example 7

In the diagram above, three identical lamps each of 100W are connected in parallel across a p.d of 250V. Calculate the reading of the ammeter. fVAEC 1993SolutionPotential difference, V = 250V; Power, P = 100W each across each lamp.

From P = IV

P 100Current through each lamp, I = — = = 0.4A

The ammeter reads the total current from the 3 lamps, therefore total current is3X0.4A = 1.2A

86

Example 8In the diagram below, X and Y arc resistance 4£i and 6£2 respectively. If power dissipation in X is 10W, then what is the power dissipation in Y?

Fig. 7.2

A. 2.4W B. 4.0W C. 6.0W D. 6.7W E. 15.0W Solution

Voltage2 V2Power = -------- -— !.c. P = —Resistance R

JAMB1984 Ans:6.7W

For resistor X, P = 1OW and R = 4i2

V2From P = ----, the p.d across the 4£2 resistor isR

V1 = P x R = 10x4V2 =40

K = V40 =6.32KThe p.d across Y is also 6.32V because the resistors are connected in parallel. For resistor Y, V = 6.32V and R = 6«

V2From P = ----, Power dissipated in Y is

Example 9ion i5n

In the diagram above, X and Y are resistance 10£2 and 15£2 respectively. If the power dissipation in X is 40W, then what is the power dissipation in Y?SolutionFrom P = I2R, the current through the circuit is, I2 = */R For resistor X; P = 40 W and R = 10f2

I2 = — = 4 10

Circuit current, 1 = >/4 = 2A

87

For resistor Y, R = 150 and I=2APower dissipation in Y is P= I2R — 2 X 15

= 60W

Example 106Q

In the diagram above, the ratio of the electric power dissipated in the 60 and 30 resistors respectively is A. 2:3 B. 1:2 C. 1:3 D. 2:1

JAMB 2004 Ans: 1:2Solution

. . . . . _ /?,/?, 3x6Effective circuit resistance, R = — L — = ----- = 20R,+R2 3 + 6

V 12Current through 30 resistor, I = — = — = 4AR 3

Power dissipated in 30 resistor, P = I2 R= 42 X 3= 48W

Current through 60 resistor, I - — = — = 2AR 6

Power dissipated in 60 resistor, P = I2 R= 22 X 6 = 24W

Ratio of power dissipation in 60 and 30 is 24: 48 or 1: 2

Example 11A portable generator is connected to six 100W lamps and a 600W amplifying system. How much energy is consumed if the generator runs for 6 hours? WAEC 1989Solution

_ Energyconsumed(W)Power, P = -----—------ -----—-----

Time taken (t)

Energy consumed, W = P X t

Power consumed by lamps = 6 x 100W = 600W Power consumed by amplifier = 600W

Total power consumed, P = 600 + 600 = 1200W

Time taken, t = 1 hr

8 8

W = P X t = 1200 X 1 = 1200 Wh = 1.2kWh Example 12An electric kettle, connected to a 240V mains produces 6.0 x 10 J of heat energy to boil a quantity of water in 5 minutes. Find the resistance of the kettle.A. 14.4Q B. 28.8D C. 144i2 D. 288i2 E. 2880Q JAMB 1985 Ans. 28.8Q SolutionGiven: H = 6.0 X 105J; V = 240V; t = 5min = 5 X 60 = 300sec.

VztElectrical heat energy, H - IVt - ----R

V2tResistance, R - ----H

2402 x300 6.0x10s

1.728 x107 6.0x10s

28.8ß

Example 13Calculate the time in which 4.8kJ of energy would be expended when an electric heater of resistance 1.8 x 103 il is used on a 240V mains supply. WAEC 1992SolutionEnergy, W = 4.8kJ = 4800J; resistance, R = 1.8 x 103 Q; Potential difference, V = 240V

_ _ w . w wFrom P = — , time, t, = — = —- t P j v l

R

t = -WR 4800x1.8x l032402

= 150i

V /Alternatively, we could use H =

time, t = HR 4800x1.8xl03V2 2402

= 1505

Example 14A working electric motor takes a current of 1.5A when the p.d across it is 250V. If its efficiency is 80%, the power output isA.300.0W B. 469.0W C. 133.0W D. 4.8W JAMB 2001 Ans: 300W SolutionGiven: E = 80%, current, I = 1.5A, voltage, V = 250V

Pj = IV= 1.5 x 250V = 375W

E//icie»cy=pOWerOMpU' x i m power input

E = — xlOOP,

„ ExP 80x375Power output, Pn = ------ =----------- = 300W0 100 100

Example ISAn electric lamp is rated 240V, 40W. What is the cost of running the lamp for 72hr if the electricity authority charges N2.50 per kWh? WAEC 2005SolutionGiven: Power, P = 40W; time, t = 72hr

W = 40 X 72 = 2880Wh = 2.880kWh

89

P = — Energy consumed, W - P X t

Cost of electrical energy (C f) = cost per kWh X electrical energy consumed in k

Cost of running lamp = N2.50/kWh X 2.880kWh

= N7.20

Example 16An electric heater takes 4A when operated on a 250V supply. What is the cost o t e electricity consumed at 10k per kWh when the heater is used for 5 hrs.? NECO 2002 SolutionGiven: I = 4A; V = 250V; time,t = 5hr

P = IV Power, P = 250 X 4 = 1000W

Energy consumed, W = P X t = 1000 X 5 = 5000Wh = 5kWh

10&Cost of electricity = -----x5kWh = 50kkWh

Example 17A land lord has eight 40W electric light bulbs, four 60W bulb and two 100W bulb in his house. If he has all the points on for five hours daily and if NEPA charges 5k per unit, his bill for 30 days isA. £15.70 B. N7.25 C. 143.65 D. N8.05 E. W.50 JAMB 1979 Ans: N5.70 SolutionTotal power consumed = (8 X 40W) + (4 X 60W) + (2 X 100W)

= (320 + 240 + 200)W

= 760W

Total time taken = 5hr/day X 30 days = 150hrs

Electrical energy used (kWh) = Power consumed X time taken

= 760W X 150hr

= 114000Wh

= 114kWh

Cost of electricity = 5k/kWh X 114kWh = - ^ - x l 14kWhkWh

= 570k = N5.70k

EXERCISE 71. Find the work done in moving a 2C charge between two point X and Y in an electric field if the potential difference is lOOVolts.A. 50J B. 100J C. 200J D. 400J JAMB ¡998 Ans: 200J2. Calculate the power delivered by a 3-phase line if its voltage and current are 132kVand 60A respectively. WAEC 1996 Ans: 7.92 x 106W

90

3. An electric bulb is rated 60W, 220V. Calculate the resistance of its filament when itis operating normally. WAEC 2001 A ns: 806.67«4. A lamp is rated 240V, 60W, calculate the resistance of its filament.

WAEC 2003 Ans: 960«5. Calculate the resistance of the filament rated 240V, 60W.

WAEC 1998 Ans: 960«6. An electric kettle contains a 720W heating unit, calculate the current it takes from a240V mams. NECO 2004 Ans: 3A7. Two lamps rated 40W and 220V each are connected in series. The total power dissipated in both lamps is A.10W B. 20W C. 40W D. 80W E. none of the above

JAMB 1978 Ans: 80W8. The bulb of a motor cycle head lamp is marked 40W, 6V. The resistance of the filament when it is switched on is?A. 6 2/402 ohms B. 40/62 ohms C. 40/6 ohms D. 6x40 ohms E. 62/40 ohms

JAMB 1979 Ans: 0.9« or ohms

9. The resistance of a 240V, 60watts electric filament bulb isA. 0.25« B. 480« C. 60« D. 240« E. 960« JAMB 1981 Ans: 960«10. Which of the following is most suitable for protecting the circuit of a 2000W electric iron connected to a 250V mains?A. 13A B. 8A C. 5A D. 3A JAMB 1990 Ans: 8A11. A lamp is rated 240V, 60W. The resistance of the filament is?A. 960« B. 16« C. 15« D. 4 « JAMB 1990 Ans: 960«12. An equipment whose power is 1500W and resistance is 375 ohms would draw a current of A. 0.10A B. 2.00A C. 4.00A D. 77.5A JAMB 1991 Ans: 2A13. Which of the following apparatus will require the smallest fuse rating for its protection? A. 60W, 240V B. 60W, 40V C. 40W, 12V D. 40W, 5V

JAMB 1992 Ans: A14. A 40W instrument has a resistance of 90«. On what voltage should it be operated normally? A. 60V B. 150V C. 225V D. 3600 V

JAMB 1993 Ans: 60V15. If the maximum voltage across a 100 ohm resistor is 20V, then the maximum power it can dissipate is A. 5.00W B. 4.00W C. 2.00W D. 0.25W

JAMB 1995 Ans: 4W16. A 3000W electric cooker is to be used on a 200V mains circuit. Which of the fusesbelow can be used safely with the cooker? A. 2A B. 5A C. 10A D. 20A

JAMB 1999 Ans: 20A17. An electric generator with a power output of 3.0kW at a voltage of 1,5kV distributes power along cables of total resistance 20.0«. The power in loss the cable is A. 0.1 W B. 10.0W C. 40.0W D. 80.0W

JAMB 2000 Ans: 80.0W18. An electric iron is rated 1000W, 230V. What is the resistance of its element?A. 57.6« B. 55.9« C. 52.9« D. 51.9«. JAMB 2002 Ans: 52.9«19. Calculate the total heat energy developed in 5 minutes by the system below

40Q

91

20.2 il

2A

Fig. 7.72Q

Calculate the rate at which energy is used up in the circuit illustrated by the diagram above. WAEC 1992 Ans: 2W21. Three identical lamps each of power 100W are connected in parallel across a potential difference of 250V. Calculate the current in the circuit.

WAEC 2001 Ans: 1.2A22.

10Ì1

Fig. 7.8

I

5fiIn the diagram above, current (I) passes through the parallel combination. If the power dissipated in the 5i2 resistor is 40W, then the power dissipated in the 10Q resistor isA. 10W B. 20W C.40W D. 80W E. 100W JAMB 1985 Ans: 20W23.

20

In the diagram above, if each of the resistors can dissipate a maximum of 18W without becoming excessively heated, what is the maximum power that circuit can dissipate?A. 27W B. 18W C.9W D. 5W JAMB 2003 Ans: 27W24.

Fig. 7.9

40

12V

The total power drawn from the cell in the circuit diagram above isA. 12W B. 24W C. 32W D. 40W JAMB 1989 Ans: 24W25. Two resistors, R| = 4Q and R2 = 5Q are connected in parallel across a potential difference. If Pi and P2 represent the power dissipated in R| and R2 respectively, then the ratio P,:P2 is A. 4:5 B. 5:4 C. 16:25 D. 25:16

JAMB 1994 Ans: 5 : 426. A lamp is marked 220V 60W. Calculate the energy it would consume whenconnected to a 220V source for 1 hour. WAEC 1988 Ans: 216000J27. A potential difference of 6V is used to produce a current of 5A, for 200s thtough aheating coil. The heat produced is A. 4800 cal B. 6000 calC. 2400J D. 240k cal E. 6000J JAMB 1978 Ans: 6000.1

92

28. A stand by generator is connected to fifteen 40W lamps and a musician’s 600W amplifying system. How much energy is used if the generator runs for 6 hours ? A. 3.84kwh B. 7.20kwh C. 8.40kwh D. 12.00kwh E. 14.56kwh

JAMB 1980 Ans: 7200Wh =7.2kWh29. A current of 0.5A flows through a resistor when connected to a 40V battery. How much energy is dissipated in 2 minutes?A. 1200J B. 1500 J C. 2400 J D. 96000 J JAMB 1987 Ans: 2400J30. An electric current of 2 amperes flows in a heating coil of resistance 50 ohms for 3 minutes 20 seconds. Determine the heat produced.A. 0.5kJ B. 8.OkJ C. 20.0kJ D. 40.0kJ JAMB 1995 Ans: 40.0kJ31. What is the total electrical energy consumed by using an electric cooker rated 1000Wfor 5hrs? A. 5.3 x 103J B. 6.5 x 103J C. 1.8 x 107J D. 2.3 x 107J

JAMB 1997 Ans: 1.8 x 107J32. Electricity is supplied to a school along a cable of total resistance 0.5i2 with the maximum current drawn from the mains as 100A. The maximum energy dissipated as heat for lhr is A. 3.6 x 103J B. 5 .0 x l0 3J C. 3 .0 x l0 5J D. 1 .8 x l0 7J

JAMB 1998 Ans: 1.8 x 107J33. A bread toaster uses a current of 4A plugged in a 240 volts line. It takes one minute to toast slices of bread. What is the energy consumed by the toaster?A. 5.76 x 104J B. 1.60 x 104J C. 3 .6 0 x l0 3J D. 1 .60x l02J

JAMB 2001 Ans: 5.76 . 04J34. An electric iron is rated 500W,200V calculate the :

(i) current required to operate the iron,(ii) heat generated in 30 minutes. 'NECO 2006 Ans: (i) 2.5A (ii) 9.0 x 105J

35. A lamp marked 100W, 250V is lit for 10 hours. If it operates normally and lkWh of electrical energy cost 2k, what is the cost of lighting the lamp?

WAEC 1994 Ans: 2k36. A man uses a 900W electric iron to press his cloth for an average of 4 hours a weekfor 5 weeks. If the cost of electrical energy is 30 kobo per unit, calculate the cost of energy for pressing. WAEC 2005 Ans: 540k37. An electric iron rated 250V has a coil of resistance 125f2. What is the cost of using itfor 1 hour at 10k per kWh. NECO 2000 Ans: 5.00k38. Find the cost of running a 60W lamp for 24hrs, if 1 kWh cost 5 Naira

WAEC2004 Ans: N7.2039. An electric heater takes 4A when operated on a 250V supply. What is the cost of the electricity consumed at 10k per kWh when the heater is used for 5 hours.

NECO 2002 Ans: 50k40. If NEPA charges 5k per kWh, what is the cost of operating for 24hours a lamp requiring 1A on a 200V line?A. 24k B. 55k C. 40k D. 26k E. 32k JAMB 1978 Ans:24k41. What is the cost of running five 50W lamps and four 100W lamps for 10 hours if electrical energy costs 2 kobo per kWh? A. N0.65 B.N0.13 C. N .90D. N39.00 E. N234.0 JAMB 1984 Ans: NO. 13 or 13k42. A household refrigerator is rated 200 watts. If electricity cost 5k per kWh, what is the cost of operating it for 20 days.A. N4.80 B. N48.00 C. N480.00 D. N4800.00

JAMB 1988 Ans: $44.8043. Which of the following instruments consumes the highest current?

Instruments Voltage Rating Power RatingA Electric iron 250V lkWB Television set 220V n o wC Torch light 6 V 30WD Immersion heater 110V 500W

JAMB 1988 Ans: C

93

44. The cost of running five 60 W lamps and four 100W lamps for 20 hours if electrical energy cost N10.00 per kWh is A. *4280.00 B. *4160.00 C. *4120.00 D. J l140 000 JAMB 2001 Ans: N 140.00

electric lamp rated 120V is used on a 240 Vrm„ calculate the resistance of its filament. WAEC2006. Ans: 4800.

^ steady current of 2A flows in a coil of e.m.f. 12V for 0.4s. A back e.m.f was induced during this period. The stored energy in the loop that can be utilized

,S: A' 12 3 B 12 0 J C. 2.4 J D. 9.6 J JAMB 2004 42 Ans. 7.2 J41. * u i. Hint: E = l(V,-VOtomoT t a U b mar^ed 240V, 40W is used for 30 minutes. Calculate the heatge"erated' WAEC 200743 Ans. 72000J48. Five 80-W and three 100-W lamps are run for 8 hours. If the cost of energy is N5.00 per unit, calculate the cost of running the lamps. [ 1 unit = kWh]

WAEC 2007 Ans. U28.0049. An electric generator has an e.m.f of 240V and an internal resistance of lii. If the current supplied by the generator is 20A when the terminal voltage is 220V, find the ratio of the power supplied to the power dissipated.A. 11:1 B. 1:11 C. 12:11 D. 11:12 JAMB 2008 Ans: 12:11 Hin«.E:V50. A generator is on daily use and in the process ten 60W and five 40W tungsten bulbs are on for the same time interval. The energy consumed daily isA. 0.9kWh B. 1.92 kWh C. 9.60 kWh D. 19.20 kWh

JAMB 2008 Ans: 19.20 kWh51. A man has five 60W bulbs and a 240W water heater in his apartment. If the bulbs and the water heater are switched on for four hours daily and the cost of electricity is N 1.20 per kWh, calculate his bill for 30 days. NECO 2008 Ans: # 77.7652. An electric lamp marked 240V, 60W is left to operate for an hour. How much energy is generated by the filament? A. 3.86 x 105J B. 3.56 x 105JC. 1.80 x 104J D. 2.16 x 10SJ JAMB 200941 Ans: D

1SCALARS AND VECTORS

Scalar quantities arc those quantities which arc described by their size or magnitude only. They have no direction. Examples are temperature, speed, distance, time, mass etc.

On the other hand, vector quantities are those quantities which can be described by their size or magnitude and direction. Examples are force, velocity, acceleration, pressure, displacement etc.

VECTOR ADDITIONWhile scalars can be added or subtracted easily (e.g. 40kg + 50kg — 90kg or 3nt — 2m = lm), vector addition takes into consideration not only the magnitude but the direction ol the vectors. The method of calculation also depends on whether the vectors are acting in a straight line or at an angle to each other, illustrated by the following typical cases.

A. (i) Two Parallel Forces Acting in the Same Direction.

------------------------► 100N_______ *■ SON

The resultant force, R = 100 + 80 = 180NFig 1.1

(ii) Two Forces Acting In Opposite Direction in a Straight Line

SON > 100N

The resultant force, R = 100 80 = 20NFig 1.2

li. Two Forces Acting Perpendicularly (At Right Angle Or 90°) To Each Other

Fig. 1.3The resultant (R) is the diagonal of the rectangle (If F, and FT each other) or the square (I f /q = F2) as shown below.

are not equal to

Fig. 1.4

9 5

1 lie resultant R, is found using Pythagoras theorem as follows.

R2 = F2 + Fj Or R = + F2

flic direction (to the horizontal) at which the resultant act is angle 0 and is found using trigonometrical formular as follows'

Fitan 9 = —

F, 0 = tan'

Angle Between the Two Acting Forces is Acute (Between 0 And 90 ) or Obtuse (Between 90° and 180°).

Fig. 1.5 (a)

Assume that F| and F2 in Fig. 1,5a acts at 55° to each other and F3 and F4 in Fig. 1.5b act at 135° to each other. Therefore, to find the resultant for each case, a parallelogram is drawn as shown below.

Always remember the following characteristic of a parallelogram,1. The opposite sides are equal.2. The opposite sides are parallel.3. The opposite angles are equal.

On applying the three characteristics of a parallelogram, Fig. 1.6a and Fig. 1.6b becomes;

The sum of all the angles in a parallelogram is 360°.From Fig. 1.7a; ADC + ABC = 360° - (55° + 55°)

= 360°-110°= 250°

250°ABC and ADC are equal. =. ABC = ADC = - y - = 125°

96

Similarly in Fig. 1.7b;EFG +EHG = 360°-(135°+135°) = 360°-270°= 90°

90°EFG and EHG are equal. EFG = EHG = — = 45°

With the above information, Fig. 1.7a and Fig. 1.7b are redrawn as follows.

The resultant, R of the forces Fi and F2 in Fig. 1.8a is the diagonal of the parallelogram ABCD. Note, the diagonal is drawn from the point of intersection of forces Fi and F2. Similarly, the resultant, R of the forces Fj and F4 in Fig. 1.8b is the diagonal of the parallelogram EFGH. Again, note that the diagonal is drawn from the point of intersection of forces F3 and F4.

The resultant (R) and the angle (a) it forms can be calculated using cosine rule and sine rule respectively. We can apply cosine rule to triangle ACD in Fig. 1.9a to find the resultant. Assume Fi = 10N and F2 = 12N.

Fig. 1.10

R2 = F2 + /r | - 2F1F2 cos Dr 2 = 102 + 122 — 2 x 10 x 12 cos 125R2 = 100 + 144 - 240 x (-0.5736)

= 244-(-137 .66)= 244+ 137.66 = 381.658

R = V381.658 R = 19.53NThe direction of the resultant vector (19.53N) is the angle a it forms with the F2 vector as shown in Fig. 1.10. Applying sine rule, we obtain,

97

AC CD ADsin D sin A ~ sin C

R F1 Fzsin 125 sin a sin C

19.53 10 12sin 125 sin a sin C

Wc pick the first two and ignore — — because it is not relevant, except if we are asked to find angle C. S'"C

19.53 10sin 125 sin a

Cross multiplying, sin a x 19.53 = Sinl25 x 10

sin 125 x 10 8.19sin a = —------------ = --------19.53 19.53 0.419

sina= 0.419a = sin-1 0.419 = 24.8°

If we substitute R = 19.53N and a = 24.8° into Fig. 1.9a, we obtain the following

Note, BAD = 55°, CAD = 24.80° BAC = 55° - 24.80° = 30.20°

The magnitude and direction of the resultant is 19.53N at 24.80° to the 12N force or 19.53N at 30.20° to the 10N force.

The above approach or procedures above can be applied to Fig. 1,9b as follows assuming F3 = 10N and F4 = 12N.

R2 = Ff + F} - 2F3F4 cos Hr 2 = 102 + 122 - 2 x 10 x 12 cos 45°

98

R* 244 240 x 0.707R: - 244 160.70 74.29R v'74.20 R 8.62/9

We use sine rule to calculate the tingle uR /’a

sin// sin E

H.62 10■ --------- = ■ — ■

sin 45 sin a

Cross multiplying, sin a x 8.62 10 x Sin45

10 x sin 45sin a = ----- —------= 0.8203

8.62

n Sin'10.8203 55.12°

Substitute R 8.62N and a 55.12° into Fig. 1,9b, to got the following;

F h = 1 2 N G

Fig. 1.13Note, FE1I = 135°, GEM = 55.12° FEG 135° 55.12°

= 79.88°The magnitude and direction of the resultant of vector Fi (ION) and vector F4

(12N) is 8.62N at 55.12° to the 12N force or 8.62N at 79.88° to the 10N force.

D. More Than Two Forces Acting at a Point

99

I rCSl'l,ant ^ ant l*1c anyl (0) it makes with the horizontal is found as follows.Resolve all the forces in the x (horizontal) and y (vertical) directions taking note 0‘ t,1c P°s‘t've and negative x and y axes.Horizontal x components: Vertical y components:F .cosG, v ■ a_ FiSinG,F;COS0: —F2sin02-FjeosGj c • n-F,stn0j

2.

F4cos04 F4sin04The x and y components arc added separately and algebraically.X = F,cos0, + F:cos0: — F,cos0, — F4cos04Y = F,sin0, — F2sin0: - F,sin0j + F4sin04 . .

If a vector is directly aligned with the x axis, its angle of inclination, 0 -0 . In contrast, a vector is directly aligned with the y axis, its angle of inclination, 0 - 90 .

The following trigonometric ratios can be very useful when solving problem vectors resolution.

tan 60° = V3 = 1.73 tan 30° = 4 = 0.58 V3

sin 60° = cos 30° cos 60° = sin 30°

Example 1An object is acted upon by two forces of 5N and 12N. Calculate the résultant of the two forces if

(i) the forces act perpendicular to each other.(ii) The two forces act at an angle of 40° to each other.(iii) The two forces act at an angle of 125° to each other.

Solution(i)

Fig 1.15Because the angle between the two forces is 90°, we use the pythagoras theorem as

follows.R2 = 52 + 122 = 25 + 144 = 169

R = \f l69 = 13 N

(Ü)

Fig 1.16

1 0 0

Applying cosine rule, R2 = 52 + 122 2 x 5 x 12 x cos 140R2 = 1 6 9 - 120(-0.766)R2 = 169 + 91.93 = 260.93 R = V260.93 = 16.15 N

Using sine rule, we have;R CD

sin 140 sin016.15 5

••• sinö =

sin 140 sin0 5 x sin 140 5 x 0.643

16.15 16.156 = sin-1 0.199 = 11.48°

The resultant is 16.15N at 11.48° to the 12N force.

0.199

= 169 - 120 x 0.574 = 100.1

R = V 100.12 = ION We apply sine rule to find the direction

R CDsin 55 sin0

10 _ 5sin 55 sin0

5 x sin 55 5 x 0.819■■■ sin 9 = ------ ----- = -------------

10 109 = sin-1 0.4095 = 24.17°

The resultant is 10N at 24.17° to the 12N force.

0.4095

Example 2Two forces, whose resultant is 100N are perpendicular to each other. If one of them makes an angle of 60° with resultant, calculate its magnitude (Sin60° = 0.8660, Cos60° = 0.5000). WAEC1991Solution

Let X and Y be the two perpendicular forces

101

It is the force X that makes an angle of 60° with the resultant as shown above and is calculated as follows:

Xcos 60 = ——

100X = 100 x cos 60 = 100 x 0.5000 = 50N

Example 3A body of mass 20kg is set in motion by two forces 3N and 4N acting at right angles to each other. Determine the magnitude of its acceleration. WAEC 1997SolutionMass, m = 20kg; acceleration, a = ?The resultant, F, of the two forces, 3N and 4N is found as follows:

F 2 = 32 + 42 = 9 + 16 = 25

F = V25 = 5 NF 5 _2

Force - mass x acceleration ••• acceleration, a = — = — = 0.25msm 20

Example 4A motorcyclist, passing a road junction, moves due west for 8s at a uniform speed of 5m/s. He then moves due north for another 6s with the same speed. At the end of the 6s his displacement from the road junction is 50m in what direction? WAEC 1999SolutionDistance covered due west is 5m/s x 8s = 40m Distance covered due north is 5m/s x 6s = 30m

tan 0 = — = — = 0.75 0 = t a n '1 0.75 = 37°ta 40 4Therefore, the direction is, 9 0 - 37 = 53

i l

w ----------

s

E

102

Because the displacement is in the North-West direction, the direction is stated thus, N53°W or 53° West of North.

Example 5A boy travels 12km eastwards to a point B and then 5km southwards to another point ( Calculate the difference between the magnitude of displacement of the boy and the distance travelled by him. NECO 2006Solution

The displacement is the distance AC and is found using Pythagoras theorem. AC2 = AB2 = BC2 AC2 = 122 + 52 = 144 + 25 = 169 AC = a/169 = 13 km

The distance covered = AB + BC = 12km + 5km = 17km Difference between distance and displacement = 1 7 -1 3 = 4km.

Example 6A tug boat is travelling from Asaba to Onitsha across the River Niger with a resultant velocity of 20 knots. If the river flows at 12 knots, the direction of motion o f the boat relative to the direction of water flow isA. 36.87° B. 53.13° C.90° D. 136.87° £.143.13° JAMB 198(Jr7Solution

Fig 1.22Direction of river flow

Angle 0 is the direction of motion of the boat relative to the direction of water as show n above.

12Applying trigonometric ratio, we have; cos 0 = — = 0.6

6 = cos-1 0.6 = 53.13°

Example 7An aircraft travelled from Calabar to Kano as follows: It flew first to florin covering adistance of 300km, 30° West of North, and then flew 400km, 60^ East of North to Kano. What is the resultant displacement?A. 567km B. 410km C. 594km D.500km E. 600km JAMB 1981

I103

SolutionK

V

C, L, K represent Calabar, llorín and Kano respectively.

Fig 1.23

The above figure is a right angle triangle (angle L is 90°). Therefore the resultant displacement is found using Pythagoras theorem CK2 = CL2 + KL2CK2 = 3002 + 4002 = 90000+ 160000CK2 = 250000CK = V250000 = 500km

Example 8A stream is flowing at 0.75m/s and a boat heading perpendicular for the stream landed at the opposite bank at an angle 30°. Calculate the velocity of the boat.A. 0.65ms-1 B. 0.86ms-1 C. 1.00ms-1 D. 1.50 m s-1 JAMB 2000Solution

'Final position of boat

Direction of boat

Direction of streamFig 1.24

V is the velocity of the boat. Using trigonometric ratio, we have

Example 93V3N

▲10N

>3N

Fig 1.26 6N

104

The figure above shows four forces 3N, 1 ON, 3 V3N and 6N acting on a particle F. ITie resultant of the four forces isA. ION B. 10V3N C. 5N D. 5V3 N JAMB 2007uSolution9, = 90°; 02 = 30°; 03 = 0; 04 = 60"

F3 cos 03 = 3>/3 x cos 90 = 3V3 x 0 = 0

V3 rF2cos62 = 10 x cos 30 = 10 — = 5v3

F3 cos 93 = 3 x cos 0 = 3

—F4 cos 04 = —6 x cos 60 = —6 x 0.5 = —3

Sum of x components, X = F3 cos 0X + F2 cos 02 + F3 cos 03 — F4 cos 04

= 0 + 5V3 + 3 - 3 = 5V3Horizontal y components:

Fj sin 03 = 3V3 x sin 90 = 3V3 x 1 = 3V3

F2 sin02 = 10 x sin 30 = 10 x - = 5i. 1. 2

F3 sin 03 = 3 x sin 0 = 3 x 0 = 0

V3—F4 sin 04 = - 6 x sin 60 = - 6 x — = -3V3

Sum of y components, Y = Fx sin 0X + F2 sin 02 + F3 sin 03 - F4 sin 04

= 3V3 + 5 + 0 - 3V3

= 3V3 - 3V3 + 5

= 5

The resultant of X and Y is found as follows.

Fig 1.25

105

Apply Pythagoras theorem,

R2 = 52 + (5V3)2 = 25 + [52(V3)2j R2 = 25 + 25 x 3 = 25 + 75 R2 = 100 R = VlOO = IONThe direction of the resultant, 0 = tan-1(|)

fl = tan‘ 1( i 7 f ) =tan“ ( ^ ) = tan" 0S77 =30°

VECTOR RESOLUTIONA single ector V, inclined at angle 0 to the horizontal can be resolved into two perpendicular components. The horizontal (Vx) and vertical (Vy) components are shown below.

(b)

Vertical component of V is, Vy = V sin 0 Horizontal component of V is, Vx = V cos 6

In some cases, the vector (V) might be inclined at angle 9 to the vertical as shown below.

*■ x

Fig 1.29 (a)Vertical component of V is, Vy = cos 6 Horizontal component of V is, Vx — V sin 6

(b)

Example 10A wheel barrow inclined at 30° to the horizontal is pushed with a force of 150N.

(i) What is the vertical component of the applied force?(ii) What is the horizontal component of the applied force?(iii> h what direction will the wheel barrow move and why?

106

Solution

Fig 1.30

Fx

(i) Vertical component, Fy = Fsin 0 = 150 x sin 30 = 150 x 0.5 = 75N(ii) Horizontal component, Fx = Fcos 0 = 150 x cos 30 = 150 x 0.87 = 130N(iii) It will move in the horizontal direction because the horizontal component of the

force is greater than the vertical component.

Example 11A force of 10N is applied to a block of wood as illustrated below.

What is the vertical component of the force? NECO 20076SolutionForce, F = 10N; Angle of inclination 0 = 30°Vertical component of force = Fsin0 = 10 x sin30 = 10 x 0.5 = 5.0N

Example 12

Fig 1.32A body on the ground is acted on by a force of 10N at% point P as shown in the diagram above. What force is needed to stop the body from moving eastward?A. 5N in the direction of East B. 5N in the direction of WestC. 5V3N in the direction of West D. 10N in the Southwest direction JAMB 1998SolutionThe horizontal (eastward) component of the 10N force is = 10 cos60 = 10 x 0.5 •= 5N. The equilibrant force needed to stop the body moving eastward must act in exactly the opposite direction.Therefore, the needed force is 5N in the direction o f West as shown below.' noii».-.

.»ION

Fig 1.33 West «

107

EXERCISES 1.1 The dnvcr ° f a car moving with uniform speed of 40ms observes a truckapproaching in the opposite direction with a speed of 20m/s. Calculate the speed ot the car relative to that of the truck. WAEC 2000 Ans: 60m s

T""' r------ox, • - • 1S due North and due LastTwo lorces j N and 4N act on a body in directions respectively. Calculate their équilibrant.

B. 5N 53° West of South E. 7N 37° South of West.

v. 5N 53° East of North.D. 7N 37° West of North 3. What is the TV'cnit'ir

C. 5N 37° North of East.WAEC 1997 Ans: B

Fig 1.34

4ns: 91.65S5. What is the magnitude of the resultant of two forces 20N and 15N inclined at 90to each other? NF.CO 2006 Ans: 25S6. A man walks 1km due East and then 1km due North. What is his displacement ?

WAEC 1978 Ans: \l2 S45°E7. A body of mass 5kg initially at rest is acted upon by two mutually perpendicular lorces 12N and 5N as shown in the figure below. If the particle moves in the direction ot QA, calculate the magnitude of the acceleration.A. 0.40ms-2 B. 1.40ms-2 C. 0.26ms-2 D. 2.60m s 2 E. 3.40ms-2

Fig 1.35

JAMB 19S4 Ans: 2.6m s'8. A block of mass 2.0kg resting on a smooth horizontal plane is acted uponsimultaneously by two forces, 10N due North and 10N due East. The magnitude of the acceleration produced by the forces on the block is A. 0.10m s-2 B. 7.07ms- :C. 10.00ms-2 D. 14.10ms-2 E. 20.00ms-2 JAMB 1985 Ans: 7.07m/s:9. A man walks 8km North and then 5km in a direction 60° East of North. Find the distance from his starting point. A. 11.36km B. 12.36km C. 13.00kmD. 14.36km JAMB 19S7 A/is: 1156km10. Two forces each of 10N act on a body, one towards the North and the other towards the East. What is the magnitude and direction of the resultant force?A. 10 V2 N, 45°E B. 10 V2 N,45°W C. 20N. 45°W D. 20N. 45°E

J.4MB 2002 Ans: ¡O sfl N45°E11 A lorry travels 10km northwards, 4km eastwards, 6km southwards and 4kmwestwards to arrive at a point T. What is the total displacement? A. 6km southB 4km north C. 6km north D. 4km east JAMB 198S Ans: 4km North12 Two forces whose resultant is 100N are at right angles to each other. If one of them makes an angle of 30° with the resultant, determine its magnitude. A. S.66NB 50 ON C. 57.7N D. 86.60N JAMB 19SS Ans: 86.60S13 Two forces of magnitude 7N and 3N act at right angles to each other. The angle 0between the resultant and the 7N force is given by: A. cos 0 = 3/7 B. sin 0 = \C. tan 0 = h D. cot 0 = h JAMB 1995 Ans: C14 A boat travels due East at a speed of 40nt/s across a river flowing due South at 30m/s. What is the resultant speed of the boat? A. 1.3 m s-1 B. 10.0ms-1

108

C. 50.0m s'1 D. 70.0ms"1 JAMB 1994 Ans: 50m/s15. A ball is moving at 18m/s in a direction inclined at 60° to the horizontal. Thehorizontal component of its velocity is A. 9V3ms_1 B. 6-\/3ms"1C. 6V3ms_1 D. 9 m s"1 JAMB 1998 Ans: 9m/s16.

résultant?A. 2V5 N in a direction N56°E B. 2V5 N in a direction N56°WC. 2Vl3 N in a direction N56°E D. 2VÏ3 N in a direction N56°WE. -v/ÏÔN in a direction N56°E. JAMB 1982 Ans: C17. Calculate the distance between thc points P(2,4) and Q(-5,3).

NECO 2007 Ans: 8.00 units18.

Fig 1.37

The figure above illustrates three vectors P, Q and R. Which of the following is correct about the vectors? A. P+Q+R=0 B. R = P - Q C. R = P + Q D. R = P.QE. Q = P. R NECO 2007 Ans: A19. A ship moving northward with speed F* is acted upon by a wind blowing due west at a speed Vw. Which of the following diagrams correctly indicates the velocity VR of the wind relative to the ship?

Fig 1.38 WAEC 200810 Ans: A

20. Thc resultant of two forces 12N and 5N is 13N. What is the angle between the twoforces? A. 0 B. 45° C. 90° D. 180° JAMB 20083 Ans: 90°21. An aircraft attempts to fly due north at 100 kmh'1. If the wind blows against it from east to west at 60kmh ‘, its resultant velocity isA. 117 kmh'1, N 31° E B. 127 km h1, N 31° E

D. 127 kmh'1, N 31° WC. 117 kmh' , N 31° W JAMB 2009 Ans: A

109

y _ r n jJ* _ (¿\ :-^CCt0'; agramsjiclow represents the following position vectors? i - 2 - l J , V3 = (_2) and V4 = ( J2) WAEC 2009'° Ans:C

Fig 1.39

23. Two forces 6N and 8N, act eastwards and northwards respectively on a body.Calculate the magnitude of their équilibrant. WAEC 2009" Ans: 1 ON24. Two forces of magnitudes 8N and 4N act at right angle to each other. What is the angle between the resultant and the 8N force? NECO 20095 Ans: 26.6°25. Two forces A and B act a point. If their resultant is given by (B-A) in the direction of B, thenA. A and B are equalB. A is greater than BC. The angle between A and B is 0°D. The angle between A and B is 90°E. The angle between A and B is 180° NECO 20096 Ans: E26. Two forces of magnitudes 4N and 5N act on a body in the directions due North and East respectively. Calculate their équilibrant. NECO 200910 Ans: 6.4N, S51.3°W

110

2PROJECTILES

A projectile is any object thrown or released into space that travels along a curved or parabolic path. At any point in time a projectile motion is characterized by;

(i) horizontal motion at constant velocity where linear acceleration, a = Om/s"(ii) vertical motion (up or down) with constant acceleration where gravitational

acceleration, g = 9.8m/s2Calculations in projectile motion are basically in two forms viz;

(a) Objects Projected Horizontally From a Height.

Note; The initial horizontal velocity, U does not change. It remains constant during the duration of motion.

The initial vertical velocity, V = 0. However, the subsequent vertical velocities, Vi, V2.......Vn changes with time.

Equations of uniformly accelerated motion (v = u + at, v" = u2 + 2as, s = ut + ‘Aat~) are used in such cases as shown in the diagram above.

You are advised to review the topic motion under gravity in Chapter 1 of BookOne.

(b) Objects Projected at an Angle, 0, to the Horizontal.

Ill

At the point of projection (t=0), the initial velocity U, can be resolved into the component, Vy - Usin 0, and the horizontal component, Vx = Ucos 0.

ie horizontal component of the initial velocity, Ucos 0 remains the same throughout the time of flight (from t=0 to t=4s).changes Vert'ca* component of the initial velocity (Usin 0) after projection

—2 ~ time. After time t the velocity becomes Vy = Usin0 - gt, during upwardmotion or V, = Usin 0 +■ gt. during downward motion.

spent in theS S °WS va*ue °f Vy depends on how much time (t) the projectile hasfl ^ ^ ^ owever’ at maximum height or exactly midway through the time of

fe » e oeity of the projectile has no vertical component, therefore, Vy=0.Aft C vel°city or speed of a projectile at its maximum height is UcosO.

result it (V\ ^r°^eCt'0n mto sPace, the motion of the projectile is governed by the below the vertical and horizontal components of the initial velocity as illustrated

Fig 2.3 (i) During upward motion (ii) During downward motion

In whatever case, the resultant velocity, V is found using the Pythagoras theorem.V2 = V2 + Vy2

:. Resultant velocity, V = Jvx2 + Vf

The resultant velocity forms an angle 0 with the horizontal plane and is calculated as follows.

t a n 0 = ! ••• « = « " - ' ( £ )

For a projectile in motion as shown in fig 2.2, the following applies.U sin0

Time taken to reach maximum height, t = --------9

2U sin 9Time of flight, T = ----------

Maximum height, H

Horizontal Range, R

U2s'\n202 9

(/2 sin 20

or

9

i/2( sin 9)2

2 9

112

An arrow is horizontally projected from the top of a tower wit a n ' 13 °15m/s. If it reaches the ground 7 seconds later, calculate t e eig 0 ower.(g=10m/s2)Solution 2Given: initial velocity, u=0; time, t = 7s; g=10m/s .Equation of motion under gravity is used to find the height; h u t + 2fltNote that the 15m/s given in the question was not used as the initial velocity because15m/s represents the initial horizontal velocity and not the initial vertical ve ocity.

Height of tower, h = ut + ±gt2= 0 x 7 + i x l 0 x 72 = 245m.

Example 2A ball is projected horizontally from a height of 20m above the ground with an initial velocity of 0.4m/s. Calculate the horizontal distance moved by the ball before hitting the ground. (g= 1 Om/s2) WAEC 2005SolutionGiven: height, h = 20m; initial horizontal velocity, u = 0.4m/s;

initial vertical velocity, u = 0The time taken to reach the ground is governed by gravity, therefore, equation of motion under gravity is used and also initial velocity, u = 0.

Substituting into ,h — ut + ±gt220 = 0 x t + i x 10 x t 2 20 = 5 t2

t 2 = 4 ••• t = V4 = 2s

After calculating the time taken to reach the ground, the horizontal distance, s, can then be calculated using equation of uniformly accelerated motion and remembering that acceleration, a = 0 and u = 0.4m/s

Substituting into s = ut + \at2= 0.4 x 2 + \ x 0 x 22 = 0.8 + 0 = 0.8 m

Example 3An object is projected horizontally from the top of a cliff with a velocity of 7m s-1 and lands on the ground level at a point 56m from the base of the cliff. Calculate the heHit of cliff SolutionHorizontal velocity V = 7m s '1; horizontal distance d = 56m

distanceFrom velocity = tm e' - time taken by object to land on the ground is

horizontal distance 56horizontal velocity 7

Equation of motion under gravity is used to find the height; h = ut + ' q t 2 Substitute initial velocity u - 0m s“1 (not 7ms-1); g - lOnis-1- time f - \

113

••• h = 0 x 8 + - + 10 x 82 = 0.5 x 64 = 32m2

Example 4A stone is projected upwards at an angle of 30° to the horizontal from the top o a

tower of height 100m and it hits the ground at a point Q. If the initial velocity of projection is 100ms"1, calculate the;

(i) maximum height of the stone above the ground.(ii) Time it takes to reach this height;(lii) Time of flight.(iv) Horizontal distance from the foot of the tower to the point Q.

(Neglect air resistance and take g as 10ms'2) fVAEC 1995Solution

Fig. 2.4

(i) The maximum height of the stone above the ground, S, is equal to the maximum height, H from the point of projection and the height of the tower, 100m ie. S = 100 + H.

Given ', initial velocity, u = 100ms"1; angle of projection, 0 = 30°

Maximum height, H =U2sin29 1002 x sin230

2 g 2 x 10

10000 x 0.52H = --------—------- = 500 x 0.25 = 125m

S = 100 + H = 100 + 125 = 225m

(ii) Time taken to reach highest height, t =U sin 6 100 X sin 30

9 10100 x 0.5

ÜT ■ = 5s

2U sin 6 2 x 100 x sin 30(iii) Time of flight, T ------------- = ------------------------= 10s

9 9

(iv) The total time the projectile travels from A to Q (see the diagram above) must be found before the horizontal distance EQ can be calculated.

114

Applying equation of motion under gravity we can find the time taken to travel from the highest point, B, to the ground, Q.

At the highest height, BD or h = 225m, and considering the vertical motion of the projectile, the initial velocity, u=0. substituting into h = ut + 1gt2 we have

225 = 0 x t + 2 x l 0 x t 2

225 = 5 t299C _

t 2 = — = 45 t = V45 = 6.71s

6.71 s is the time taken to travel from the highest point B, to the ground, Q.From (ii) and (iii) above, the time taken to travel from A to B is 5 sec; the time

taken to travel from B to C is also 5 s.Therefore the time taken to travel from C to Q is 6.71 - 5 = 1.71 sec and the total

time taken to travel from the point of projection, A, to the ground Q is 5 + 6.71 = 11.71 s.

The diagram below shows time duration for each section of the flight.

Horizontal distance (EQ) = horizontal component of initial velocity x total time taken to reach the ground from point of projection.

= u cos 6 x 11.71= 100 cos 30 x 11.71 = 100 x 0.866 x 11.71 = 1014.12m

Alternatively, the horizontal distance can be calculated as follows. The distance EP (see above diagrams) is the range, R of the projectile.

£/z sin20 1002 sin 2 x 3 0R = ----------- = ---------77--------= 866.025m

9 10

The time taken to travel the horizontal distance PQ is 1.7Is. The horizontal component of the velocity is u cos Q = 100xcos30 = 86.60m/s.

distance ,Velocity = ---------- ••• distance = velocity x time

time

Distance, PQ = 86.60 X 1.71 = 148.1m Horizontal distance EQ = Range(EP) + PQ

115

= 866.025 + 148.1 = 1014.12m

Example 5Two bodies X and Y are projected on the same horizontal plane, with same initial speed but at angles 30° and 60 respectively to the horizontal. Neglecting air resistance, the ratio of the range of X to that of Y isA. 1:1 B. 1:2 C. V3:l D. 1: V3 JAMB 1997Solution

Range, RU2 sin 26

g

For X, 0 = 30°,U2 sin 2 x 30 U2 sin 60

Range, Rx = ----------------- = -------------g g

For Y, 0 = 60°, Range, RY =U2 sin 2 x 6 0 U2 sin 120

g gRatio of range X(Rx) to range Y(Ry) is

Rx U2 sin 60 U2 sin 120g

U2 sin 60g

gU2 sin 120

sin 60 0.866sin 120 ~ 0.866

Rx : Ry = 1:1

Example 6A body is projected from the ground at an angle 0 to the horizontal with a velocity of 30m/s. it reached a maximum height of 11.25m. Calculate

(i) The value of 0.(ii) The time taken to strike the ground.(iii) The range.(iv) Its velocity 2s after projection. NECO 2004

Solution(i) Initial velocity of projection, U = 30m/s; maximum height, H = 11 25m-

g - 1 Om/s2 't/2sin20

Substituting into, H = — ------2 g

we have,

rearranging.

302(sin 9)2 900 x (sin 9)2~ 2 x 10 ~ 20

(sin#)2 x 900 = 11.25 x 20

(sin 9)211.25 x 20

900 0.25

sin 9 — V 0.25 sin0 = 0.5

9 — sin-1 0.5 9 = ,30^

116

(ii) The time taken to strike the ground is the time of flight, T

2 x 30 x sin 30 60 x 0.5 _ ^

2 U sin# 9

T = 10

, „ U2 sin 29 302 x sin 2 x 3 0 900 x sin 60 _ 77 qAm(ni) The Range, R = ---------- = ----------^ ----------- iq

(iv) The velocity of the body after 2 seconds of projection is found by calculating the resultant of the horizontal (V*) and vertical (Vy) components of the initia ve ocity a eT 2s.

Initial horizontal component of the velocity,Vx = U cos 0 . Substitute U = 30m/s and 0 = 30°Vx = 30 cos 30 = 25.98m/s.Vx does NOT CHANGE during the time of flight.

Initial vertical component of the velocity,Vy = U sin 8Vy changes with time, therefore after time t,Vy = U sin 9 — gtSubstitute U = 30m/s, 9 = 30° and t = 2s Vy = 30 x sin 30 — 10 x 2

= 1 5 -2 0 = - 5m/s

The vertical, horizontal and resultant velocities can be illustrated below.

Fig. 2.6

The resultant velocity, V? + Vy

= V(25.98)2 + ( - 5 ) z

= V674.96 + 25

= V 699.96 = 26.46m/s

This resultant velocity makes an angle, 0 with the horizontal given by

tan 9 = - f = ■- 5

= -0.1925Vx 25.98

9 = tan-1 (—0.1925) = -10.9°

117

Example 7A missile is projected with a velocity of 250ms-1 at an angle of 9 to the vertical. If the total time of flight of the missile is 25s, the value of 9 is A. 30° B. 60° C. 45 D. 50Solution

Fig 2.7

2 x 250 x sin#

25 = 50 x sin025

sin0 = — = 0.5 50G = sin-1 0.5 — 30°

u = 250ms-1 T = 25 s g = 10ms- 1

From the given options, your answer will be option A, isn’t it? YOU ARE ABSOLUTELY WRONG!!Read the question again and look at the diagram once more. "... at an angle o f 6 to the vertical..In all the formulae in projectile motion, 6 is defined as the angle of projection to the horizontal. Therefore when the given angle of projection 9 is to the vertical, two approaches can be applied.

(i) The angle of projection 9 to the horizontal is obtained by subtracting the given or calculated angle of projection to the vertical from 90°. That is, angle of projection to thehorizontal, 9 = 90 — 30 = 60°

(ii) Alternatively, change sin 9 to cos 9 in the formula to be used

Therefore, T = ■2usm 9

9becomes T =

2u cos 99

Substitute u — 250sm 1; T = 25s g = 10ms-2

2 x 250 x cos025 = -------- —--------

1025 = 50 x cos 9

25c°s 9 = — = 0.5 50

9 = cos 1 0.5 = 60°

118

EXERCISES 21.

R

An object R leaves a platform XY with a horizontal velocity of 7m s and lands at Q ^ takes the same object 0.3sec to fall freely from Y to P, calculate the istance g=10m/s2). WAEC1990 Ans: 2.1m2. A ball is projected horizontally from the top of a hill with a velocity of 20m/s. Ifit reaches the ground 4 seconds later, what is the height of the hill? (g-lOm/s ).

WAEC 1991 Ans: 80m3. An object is projected with a velocity of lOOm/s from the ground level at an angle 0 to the vertical. If the total time of flight of the projectile is 1 Os, calculate 0. (g-lO m /s)

WAEC 1993 Ans: 60°4. An object is projected with a velocity of lOOm/s at an angle of 60 to the vertical. Calculate the time taken by the object to reach the highest point. (Take g=10m/s ).

WAEC 1998 Ans: 5s.5. A stone is projected horizontally from the top of a tower with a speed of 5m/s. Itlands on the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the height of the tower. (g=10m/s2). WAEC 1999 Ans: 80m6. In his first attempt, a long jumper took off from the spring board with a speed of 8m/s at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is U2sin20/g, where all the symbols have their usual meanings, show that he gains a distance of 0.8576m in his second attempt. (g=10m/ss).

WAEC2004 Zlns: ?z sinl2x45°> _ 8f_sin(2*3o°) 857510 10 U.OO/JIII7. A stone thrown horizontally from the top of a vertical wall with a velocity of15m/s, hits the horizontal ground at a point 45m from the base of the wall. Calculate the height of the wall. (g=10m/s2). WAEC 2005 Ans: 45m8. A stone is projected horizontally with a velocity of lOm/s from the top of a cliff ofheight 45m. Calculate the horizontal distance covered by the stone from the foot of the cliff. (g=10m/s2) NEC0 2003 Ans: 30m9. An object is projected with a velocity of 50m/s from ground level at an angle 0 to the vertical. If the total time of flight of the projectile is 5s, what is the value of 0?(g=10m/s2). NECO 2002 Ans: 60°10. A ball is kicked with a velocity of 8m/s at an angle of 30° to the horizontalCalculate the time of flight of the ball. (g=l Orn/s'). NECO 2000 Ans 0 8s11. A stone, Q is thrown with velocity U at angle of 75° to the horizontal. Another stone R is thrown with the same velocity U but at an angle of 15° to the horizontal The ranges covered by the stones will beA. greater for Q B. greater for R C. the same for Q and RD. greater for heavier of two stones. JAMB 1992 Ans -C12. An object is projected with a velocity of 80m/s at an angle of 30° to the horizontal. The maximum height reached is A. 20m B. 80m C. 160m D 320m

JAMB 1994 Ans: 80m

119

13. A particle is projected horizontally at lOni/s from a height of 45nt. Calculate the horizontal distance covered by the particle before hitting the ground. (g=10m/s2).

WAEC 2006 Ans: 30m14. An object is projected from a height of 80m above the ground with a velocity of 40m/s at an angle of 30° to the horizontal. What is the time of flight?A. 4s B. 16s C. 10s D. 8s JAMB 2005 Ans: 4s15. A shooter wants to fire a bullet in such a way that its horizontal range is equal to three times its maximum height. At what angle should he fire the bullet to achieve this?A. 53° B. 68° C. 30° D. 45° JAMB 2006 Ans:53°16. A particle is projected at an angle of 30° to the horizontal with a speed of 250ms '. Calculate the

(a) total time of flight of the particle(b) speed of the particle at its maximum height [g=10ms'2]

WAEC 2007 Ans: (a) 25s (b) 216.5ms'17. A stone projected horizontally from the top of a tower with a speed of 4ms 1 landson the level ground at a horizontal distance 25m from the foot of the tower. Calculate the height of the tower. [g=10ms'2] WAEC 2007 Ans: 195.3m18. A tennis ball is projected horizontally from a height of 30m above the ground withan initial speed of 20ms’1. Calculate the horizontal distance travelled by the ball when it hits the ground. [g=l 0ms'2] NECO 2007 Ans: 48. f 8m19. What is the total time of flight of an object projected vertically upwards with aspeed of 30ms'1? [g=10ms'2] NECO 2007 Ans: 6 Os20. The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50m s-1 . If the acceleration of free full due to gravity is 10ms-2, determine its (a) initial speed

(b) speed at the maximum height reachedWAEC 2008£3 Ans: (a) 57.74ms-1 (b) 50ms-1 HintT = jVJ + VJ

21. A particle is projected horizontally at 15ms'1 from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting the ground.

(g = 10ms'2) WAEC 2009E2 Ans: 30m22. A projectile attains its maximum height when the angle of projection isA. 30° B. 45° C. 60° D. 15° D. 90°

NECO 20097 Ans: E23. A stone is thrown with a speed of 10ms'1 at an angle of 30° to the horizontal.Calculate the time of flight, (g = 10ms'2) NECO 20098 Ans: 1.0s

120

EQUILIBRIUM OF FORCESMOMENT OF A FORCE

The moment of a force about a point is the product of the force and the perpendicular distance of its line of action from the point.

Moment of a force = Force (N) x Perpendicular distance(m)

The principle of moments states that when a body is in equilibrium, the sum of the anticlockwise moments about any point is equal to the sum of the clockwise moments about the same point. In addition, the sum of the forces in one direction is equal to the sum of the forces in the opposite direction.

Example 1A body of mass 58g is suspended at the 20cm mark of a uniform meter rule. The meter rule is adjusted on a pivot until it settles horizontally at the 40cm mark. Determine the mass of the meter rule. WAEC 1992Solution

40cm 10cm.

20cm

" 5 ^Fig 3.1 58g

Because the meter mle is uniform the centre of gravity, C of g is at the 50cm mark or 10cm from the pivot. Let the mass of the meter rule be m.Taking moment about the pivot

m x 10cm = 58g (40 - 20)cm m x 10cm = 58g x 20cm

m = 58g x 20cm 116010cm 10 = 116g

Example 2Using the diagram below, calculate the moment of the force of 10N about

point P.

SolutionThe distance OP has to be resolved into its perpendicular distance, OQ as follows

Fig 3.3 ION

y/3OQ = 12 cos 30 = 12 x — = 6V3

moment of force ION = forcc(lON) X perpendicular distance OQ

= 10 x 6V3 = 60V3 or 104 NmExample 3What is the moment of the force F about point P in the diagram shown below?

Fig 3.4 WAEC 1994

SolutionThe perpendicular distance of y is found by resolving the distance y into its horizontal component as follows:

Fig 3.5

but cos(90 — 9) = sin 9

-------- ...-----OQ-= ysinflMoment of F about P = Fy sin 9

Example 4A uniform bar 15m long is balanced on a pivot placed at its midpoint. A boy of mass 55kg sits on one arm of the bar at a point 5m away from the pivot. What mass can be placed 2m away from the other end of the bar to keep the bar horizontally? WAEC 1997 Solution

7.5m■4 — ------------------------------------------------------- ►

« 5m_______^ ___5.5m____ M 2m ^

Fig 3.6 55 kg▼m

122

Sum of anticlockwise moment = sum of clockwise moment. Taking moment about the pivot where m is the placed mass.

m x 5.5 = 55 x 5

m =55 x 5

5.5275T5

= 50 kg

Example 5 , fA meter rule is found to balance horizontally at the 48cm mark. When a o y o mass 60g is suspended at the 6cm mark, the balance point is found to be at t le cm mar Calculate the

(i) mass of the meter rule. ,(ii) distance of the balance point from the zero end, if the body were mo^e e

13cm mark. IVAEC 2007Solution(i)

Fig 3.7 60g M

Let M be the mass of meter rule.Taking moments about the fulcrum,

sum o f clockwise moments = sum of anticlockwise moments M x (48 - 30) = 60 x (30 - 6)

18M = 60 x 24

M =60 X 24

18 = 80g

(ü)■ 48cm

-13 cm

77 j 100 cm

Fig 3.8 60g 80g

Taking moment about the fulcrum,Sum o f clockwise moments = sum o f anticlockwise moments

80 x [4 8 - (1 3 + * ) ]= 60 x *

80 x [ 4 8 - 1 3 - * ] = 60 x *

80 x [3 5 -* ]= 60 x *

2 8 0 0 - 80* = 60*

/123

2800 = 80a- + 60a 2800= 140a

2800x - 140

= 20cm

Therefore, the distance frormthe zero end is 13 + a = 13 + 20 — 33cm.

Example 6A beam PQ pivoted.at P carries a load of 80N as shown above. Calculate the

0.5 mt 2.5 m ^

\ f

Fig 3.9SON

WAEG 1001

SolutionTaking moments about i

E X (2.5 + 0.5) = 80 x 0.5

E x 3 = 40 40

E = — = 13.3N

Example 7

Fig 3.10Determine the magnitude of P in the diagram above. WAEC 2006SolutionThe force P has to be resolved into its vertical component, Psin30

Fig 3.11

fPsin30

Taking moment about the pivot,

P sin30 X 0.5m = 40 X 0.2m (30+20 = 50 cm = 0.5 m; 20 cm = 0.2 m)

124

0.5P x 0.5 = 8 0.251* =8

P = —— = 32N 0.25

8

Example 8 ■ ! i • ■A uniform beam PQ of mass 20kg and length 5m is balanced as shown below. A man of weight SOON stands at M such that QM = 2m. The reactions at P and Q respectively are

........ - - ..................... C. 220N and 380N(g=I0m/s2)

A. 300N and 500N D. 580N and 420N

B. 420N and 580N E. 380N and 220N

2m—H

M5m

0

JAMB 1982■ f ... •;Fig 3.12

SolutionBecause the bar is uniform its weight (w = mg = 20kg x 10 = 200N) act at the middle of the bar.

800NL______^ m _________^ _ 0.5m „ 2m |i T

------------------------ r------------- n

* iJ

<\\

200NFig 3.13

Taking moment about P ------ ------- .

Q x 5 = (200 x 2.5) + 800 x (2.5 + 0.5)

5Q = 500 + 800 x 3 = 500 + 2400

5Q = 2900

2900 ; tQ = —- — = 580N

Sum of downward force = sum of upw ard forces

200 + 800 = P + Q

1000= P + 580

P = 1000 - 580 « 420N

The reaction at P and Q respectively are 420N and 580N.

Example 9In the figure 3.14 below, MN is a light uniform meter rule pivoted at O, the 80cm mark A load o f mass 3.0kg is suspended on the rule at l., the 10cm mark. If the rule is kept in equilibrium by a string RP. fixed at P and attached to the rule at K. the 20cm mark, w hat is the tension. T in the string?A. 3.0 N B. 3.4 N C. 3.5 N D. 6.0 N E. 7.0 N

125

p

m L1 10 20 801 ^$30° i’ 1

3kgJAMB 1985

Fig 3.14 ' :• ;SolutionThe tension force T is resolved into its perpendicular (vertical) component, TsinG

Taking moment about the pivot O I -----Tsin30 x (0.80 - 0.20) = 3 x (0.8 - 0.10)

0.5T x 0.6 = 3 x 0.70 0.3T = 2.1

2;lT = —— = 7 N 0.3

Example 10A 90cm uniform lever has a load of 30N suspended at 15cm from one of its ends. If the fulcrum is at the center of gravity, what is the force that must be applied at its other end to keep it in horizontal equilibrium? JAMB 2003Solution ’ ’ ' . \ u 1 t . .j.;,'-,) .. , ;.i(

30N

90cm

▼F

Fig 3,16

126

I aking moment about the fulcrum,30 X (0.45 - 0.15) = F x (0.90 0.45)

30 x 0.30 = F x 0.459 = 0.45F

F = ----- = 20 N0.45

MOMENT OF A COUPLEThe moment of a couple is defined as the product of one of the forces perpendicular distance between the lines of action of the two forces.

F

Jx >

Pig 3.17Moment of couple = F x x

Example H

0.75m

Fig 3.18

15N

1SNCalculate the magnitude of the couple m the figure above.SolutionMoment of the couple = One force x perpendicular distance between the forces.

= F x x= 15 X 0.75 = 11.25Nm

Example 12Two forces, each of magnitude 25N, act in opposite directions at the opposite ends of a circular ring. If the diameter of the ring is 65cm, calculate the magnitude Df couple acting on the ring.Solution

127

Moment of a couple = Force x perpendicular distance = 25N x 0.65m = 16.25Nm.

THE TRIANGLE OF FORCESThe principle of triangle of forces is applied to solve problems in

forces act on a body to keep it in equilibrium. It states that when a o y is ' ^ ^under the action of three forces, the three forces can be represente in madirection by the three sides of a triangle taken in order. ,0ive

The following examples illustrate how the tnangle of forces is applied problem in equilibrium.

Exampl- 13 f , ^ j y ¿sThe body P shown in the diagram below is in equilibrium. If the mass o 10kg, calculate the tension T in the string (g=10m/s ).

Fig 3.20

WAEC 1990

ouiuuun ,The weight of the body P acts downward and F acts in the horizontal direction as shown

Next a triangle is formed by taking the magnitude and direction of each force in sequence.

The triangle formed is a

right angle triangle and

trigonometric ratio can be

applied to find T.

Fig 3.22

100cos 60 = ~~j~

100 cos 60

10005

= 200N

128

^\\N

VsN

N

"X.Pythagoras theorem can be applied to find T

T2 = 122 + 52 = 144 + 25 = 169 \

T = a/169 = 13N

Example 15The diagram below illustrates three forces T|, T2,' and 100N in equilibrium. Determine the magnitude of Ti.

I l!i 1

WAEC20Ö0*

p/; 129

Solution

Fig 3.26. VUsing trigonometric ratio, Ti can be found as follows:

T-itan 30° =

/ \)\ >t )' I *

100; lTx = 100 tan 30 = 57.74 N or. r 'Ji i

Example 16

Fig 3.27What’s the value of T in the figure above? A. 10.0 N D. 40.0 N

Note that it is the angle the string makes with the vertical draw the mangle of forces and used for the calculation, not a right angle triangle, we use Sine rule to find T.

B. 11.8 N C. 20.0NJAMB 1995

(90° - 30° = 60°) that is used to Because the triangle formed is

20 _ T _ __ 20 x sin 60sin 60 sin 60 sin60 =

Alternatively, we could find T by resolving the two Ts into their vertical component and applying one of the condmons for equilibnum: Sum of upward forces = sum ofdownward forces.

130

Therefore, Tsin30 + Tsin30 — 2 ON 0.5T + 0.5T = 20N

T = 20N

Example 17A mirror of weight 75N is hung by a cord from a hook on the wall. If the cor ma es an angle of 30° with the horizontal, what is the tension of the cord? A. 150N 'C.1500N D.250N JAMB 2006Solution

Fig 3.29

The vertical component of T = Tsin30

Upward vertical forces = Downward vertical forces T sin 30 = 75

0.5T = 75

T750.5 = 150 N

Example 18

Fig 3.30

The diagram above illustrates three forces 7\, T2 and 250N in equilibrium. Calculate the magnitude of T2. NECO 20087Solution

Fig 3.31

131

fan 30" = ~ t = tan 30 x 250 = 0.577 x 250 250= 144.34A/

I Ising trigonometric ratio, T2 can be fourni as follows:

EXERCISE 3.!• A meter rule is found to balance at the 48cm mark. When a body of mass 60g is suspended at the 6cm mark the balance point is found to be at the 30cm mark. Calculate, i) the mass of the meter rule ii) the distance of the balance point from the zero end, if the body were moved to the 13cin mark. IVA EC 1988 Ans (i) 80g (ii) 33cm2- uniform meter rule AB is balanced on a knife edge which is 55cm from B. If amass of 10g is hung at P, which is 10cm from A. Calculate the mass of the meter rule.

WAEC 1994 Ans: 70g3. A uniform meter rule of mass 90g is pivoted at the 40cm mark. If the rule is inequilibrium with an unknown mass m placed at the 10cm mark and a 72g mass at the 70cm mark, determine m. WAEC 2000 Ans: 102g4.

A 40cm P

Fig 3.32 7N " A "

In the diagram above, AB represents a uniform rod of length 1.50m which is in equilibrium on a pivot at P. If AP = 40cm, calculate the mass of the rod. (g=10m/s2)

WAEC 2001 Ans :0.8kg5. Masses mi and m2 at the 20cm and 65cm marks respectively of a uniform meterrule freely suspended at its centre of gravity. If the meter rule balances horizontally, determine the ratio m2:m|. WAEC 2001 Ans: 2:16. In the diagram below, what is the value of W.

/ / / / / /T = 20N

lm lm ->G<-

W

lm >“ X

Fig 3.33 7.

Fig 3.34

JAMB 1979 Ans: 10N

A uniform beam HK of length 10m and weighing 200N is supported at both ends as shown in the figure above. A man weighing 1000N stands at a point P on the beam. If the reactions at H and K are respectively 800N and 400N, what is the distance HP?

JAMB 1979 Ans: 3m8. The figure 3.35 below shows a see-saw which is exactly in balance. The weights o f the people sitting on it are 50kg, X kg and 15kg as shown. They sit at distances of 1.5, 1.5 and 2m from the pivot respectively. What is the value of X?

132

B ■ JAMB 1979Ans:26.67kg9. In the figure below, PT is a uniform meter rule pivoted at R, the 70cm mark. Two forces 0. IN and 0.4N are applied at Q, the 60cm mark and S, the 85cm mark. If the meter rule is kept in equilibrium by the forces and its weight, calculate the weight o f the meter rule. A. 0.25N B. 0.30N C. 0.35N D. 0.50N E. 0.56N JAMB 1984 Ans: 0.25N

P

Fig 3.36

Q

▼0.1 N 0.4N

10. Two masses 40g and 60g respectively are attached firmly to the ends o f a light meter rule. The centre of gravity of the system is A. At the midpoint o f the meter ruleB. 40cm from the lighter mass C. 40cm from the heavier mass D. 60cm from the heavier mass E. indeterminate because the meter-rule is light. JAMB 1985 Ans: C11.

h»10cm HQC : r

Fig 3.37 10g

A uniform meter rule QR is balanced on a knife edge which is 55cm from R. When a mass o f lOg is hung at P as shown above, what is the mass of the meter rule?A.550g B. 350g C. 70g D. 35g JAMB 1986 Ans: 70g12.

A force o f 5N acts at a point Y on a rod XYZ as shown in the diagram above. If XY is 2m, what is the moment of the force about point X? A. 0 Nm B. 3 Nm C. 7 NmD. 10 Nm JAMB 1990 Ans: 013. ________________________ ___________

♦ — a ------------------------ yV— c~*\i V - n — tF, F2

i i d _ l ir f , f

Tip 3.39

133

A uniform light rod is kept in horizontal equilibrium under the influence of four forces as shown below. Which of the following equations correctly represents the condition of equilibrium for the rod?A) F, + F2 = F, + F, B) F, + F2 — F3 + F4 = 0 C) (F, + F2)ab = (F, + F4)cdD) F,a + F2b - F,c - F4d = 0 JAMB 1992 Ans: D14.

Fig. 3.40The diagram above shows three forces Ft , F2 and F3 which keep the bar AB in horizontal equilibrium. Which of the following equation is correct?A. F3 = F1+ F2 B. F2 = Fj + F3 C. F1 = F2 - F3 D. Fx = F2 + F3

WAEC 200811 Ans: D15.

Fig. 3.41

R O s

The diagram above shows a plank RS pivoted at its centre of gravity O and is in equilibrium with the weights P and Q. If a weight 2P is added to P, the plank will be in equilibrium again byA. moving Q nearer to OB. moving P nearer to OC. adding a weight Q to QD. moving P further away form O JAMB 2008s Ans' B16.

a rFig. 3.42 200N

The diagram above shows a uniform wood of weight 200N and length 50m. It is pivoted at one end and suspended by a cord at the other end at an angle of 30° to the wood. Calculate the tension in the cord if the wood is horizontal. JAMB 1994 Ans: 200N17. A uniform rod PQ of length 1 m and mass 2kg is pivoted at the end P. If a load of 14N is placed at the centre of the rod, find the force that should be applied vertically upwards at Q to maintain the rod in equilibrium horizontally. A. 68 N B. 28N C. 17 N D. 7N (g=10m/s2). JAMB 1995 Ans: 17N

134

18.

Fig. 3.43A uniform light beam XZ is hinged at X and kept in equilibrium by the forces T and I as shown in the diagram above. If XO = 20cm and OY = 30cm, express T in terms of F.A.T = — F B. T = 2F C .T = — F D. T = SF JAMB 1997 Ans: T = 2F

3 319. A uniform meter rule weighing 0.5N is to be pivoted on a knife-edge at the 30cm mark. Where will a force of 2N be placed from the pivot to balance the meter rule?A. 95cm B. 25cm C. 20cm D. 5cm

JAMB 1998 Ans: 25cm mark or 5cm from the pivot20. Two forces each of 4N act on the opposite sides of a rectangular plate as shown in the diagram below. Calculate the magnitude of the couple acting on the plate.

4N "*■

Fig. 3.44

Eo

♦ 4N WAEC1990 Ans: 1.6Nm

21. Two forces, each of magnitude 4N, act in opposite directions at the ends of a rod. If the length of the rod is 40cm, calculate the magnitude of the couple acting on the rod.

NECO 2006 Ans: 1.6Nm22.

Fig. 3.45

In the figure above, calculate the magnitude of the couple. NECO 2005 Ans: 2.50Nm23. Two forces forming a couple are separated by a distance of 25cm. If one of theforces equals 40N, what is the moment of the couple? NECO 2000 Ans: 1 ONm24. In the diagram below, MN is perpendicular to ON and MP. What is the difference between the moment about N of the force of 20N applied along MP and its moment about O?

Fig. 3.46

o

135

A. Zero J3 q txj25. A boy pui]s , 7 r C ° '4Nm D. 0.8Nm E. 0.6Nm JAMB 1980 Ans.O at an angle of 60°V* T ' r°m wa^ a string tied to the nail. The string is inclined list'd in ™ ir , ° 1 lc W£,H- K the tension in the string is 4N, what is the effective force used in pulling the nail? WAEC m<> An,: U W

A 151 mass suspended from a ceiling is pulled aside with a horizontal force, F, as shown in the diagram above. Calculate the value of the tension, T. (g=10m/s )

WAEC1999 Ans: 173.2N27.

An object A, is held in the diagram, determine28.

equilibrium as illustrated in the diagram the magnitude, W, of the weight of A.

Fig. 3.49

above. Using the data on WAEC 2002 Ans: 17N

In the given diagram above, what are the values of the vertically suspended weight W and the angle X? A. 2kgf, 30.9° B. 4kgf, 36.9° C. 6kgf, 53.1°D. 8kgf, 73.8° E. 12kgf, 106.2° JAMB 1982 Ans: 4kgf, 36.9°29. Find the tension Ti in the diagram below if the system is in equilibrium.

136

b . ^ n C. % N D. 100 (g=10ms 2)JAMB 1991 Arts: 57.74Nor ^ N

A. 20 0

VT

30.

Consider the three forces acting at O and in equilibrium as shown above. Which o e following equations is/are correct?(i) P,cos 0, = P,cos 02 (ii) P3 = Picos 0, + P2cos 02(iii) P,sin 0, = P2sin 02A. (i) only B. (ii) only C. (iii) only D. (ii) and (iii) only E. (i) and (iii) only

JAMB 1983 Arts: D31.

Fig. 3.52In the figure above, the three forces Fi, F2, F3 acting at O are in equilibrium. If the magnitude of Fi is 10.0N and the magnitude of F2 is 5.0N, find the magnitude of F3.A. 26.4N B. 15.ON C. 13.2N D. 10.0N JAMB 1986 Arts: 13.23N32. The diagram below shows forces 4N, 6N, ION and 8N which act at a point O in the directions indicated. What is the horizontal force? A. 7>/3 N B. 17 N C. V3N D. 13 N JAMB 2004 Arts: 13N

33 Two horizontal forces, ION and 8N and other force F, inclined at 30° to the vertical acting as shown in the diagram below, keep the body P in equilibrium. What is the weight of the body? A .^ N B. V3N C .^ N D. 2V3N

JAMB 1990 Ans:3.46N or 2\i3

137

ION'*

Fig. 3.54

34. What is the value of F in the figure below when it is in equilibrium?

A. ION B. 12 N C. 27 N D .20N JAMB 2005 Ans: 20N35. For what value of 0 are the forces in the diagram below in equilibrium?

JAMB 1997 Ans: 60°36. A hand bag containing some load weighing 162N is carried by two students each holding the handle of the bag next to him. If each handle is pulled 60® to the vertical, find the force on each student’s arm. A. 324N B. 162N C. 121N D. 81N

JAMB 2000 Ans: 162N37.

Fig. 3.57

JAMB 2005 Ans: 9N38. A particle is under the action of three forces in equilibrium. Two of the forces are as shown below. What is the third force?

p T

Fig.. 3.58

138

A. 80N along P B. 160N at 60° to P C. 40V F N at 45° to either P or QD 0N E. 40\/2 N at 30° to Q JAMB 1981 Ans: C39- In the diagram below, a rod 50cm long of uniform cross-section is suspended orizontally on a fulcrum, F, by the action of two forces. What is the weight of the rod?

Fig. 3.59

A. 200N B. 120N

40.

15

TF

200N

C. 80N

45

D. 40N

50

d

40N

JAMB 2007 Ans: 80N

Fig. 3.60

The diagram above illustrates three forces acting on an object at point O. If the object is in equilibrium, determine the magnitude of the force P. WAEC 2007 Ans: 20. ON

41.

Using the diagram above, calculate the magnitude of the about the point Q. moment of the force of 250N

NtCO 2008£2 Ans: 21.21Nm

139

4equilibrium of bodies

IN LIQUIDSARCHIMEDES PRINCIPLE:n r in r in lc bodies in liquid is governed by Archimedes principle andobjects and fluids tl0n' ^*ose*y related to these, are the densities and relative densities of

immersed in^fl^H ^ r‘"c'^ e states that when an object is completely or partiallydisplaced by the o b je c ! .^ ™ 065 “ Upthrust which is eclual t0 the vvei8ht of the fluid 11 non it °f.floatat‘on states that an object will float when the upthrust exertedupon it by the fluid in which it floats is equal to the weight of the body.

Upthrust of object in fluid, = Weight of object in air - Weight of object in fluid, Up = WA — Wp

Note that the fluid could be water, gas or any other liquid, therefore Uf and Wf could be Uw, Ww and Ul, Wl-Upthrust of object in fluid Weight of displaced fluid

Mass of displaced fluid X g Density of fluid x volume of displaced fluid x g Density of fluid x volume of object X g

Up = p x V x gDensity, p, is defined as the mass per unit volume of a substance, measured in kg/m3. Relative density, R.d, of a substance is the ratio of the mass or weight of any volume of itto the mass or weight of an equal volume of water. •- -----

mass of substanceMass of equal volume of water

weight of substanceweight of equal volume of waterdensity of substance

density of waterweight of object in air

weight of equal volume of waterweight of object in air

upthrust of object in waterupthrust of object in liquid

R. d of liquids Upthrust 0f object in water

volume of liquid displaced by object volume of water displaced by same object

140

apparent loss of weight of a solid in liquid apparent loss of weight of same solid in water

Example 1An object weighs 10.ON in air and 7.ON in water. What is its weight when immersed in a liquid of relative density 1.5? WAEC 1990SolutionGiven: Relative density of liquid = 1.5 Weight of object in air, WA = 10 N

Weight of object in water, Ww = 7N Problem. Weight of object in liquid, WL••• Upthrust on object by water, Uw = WA — Ww = 10 — 7 = 3N The appropriate formula is

R. d of liquid =upthrust of object in liquid (UL) upthrust of object in water (Uw)

Substituting, UL = 3 x 1.5 = 4.5N

Generally, upthrust = weight in air — weight in fluid (liquid)Weight of object in liquid = weight in air — upthrust in liquid

WL = WA- UL WL = 10 — 4.5 = 5.5 N

Example 2A block of material of volume 2 x 10'5m3 and density 2.5 x 103kgm'3 is suspended from a spring balance with half the volume of the block immersed in water. What is the reading of the spring balance? (Density of water = 1.0 x 103kgrn3, g=10m/s2) WAEC 1993 SolutionGiven: Volume of object V = 2 x 10~5m3 ; Density of object, p = 2.5 x 103kgm ~3

Volume of object immersed = i V = (2 x 10~5) = 1 x 10_5m3

Density of water = 1.0 x 103 kgm ~3

Problem: Spring balance reading = weight of object in water, Ww

MassDensity (p) = Vo[ume^ " Mass of object = p V = 2 x 10~5 x 2.5 x 103

= 0.05 kg

Weight of object in air, WA = mass of object x g = 0.05 x 10= 0.5N

Upthrust of object in water, Uw = Density of water x Volume of object immersed x g

= p x V x g = l x 10~s x 1.0 x 103 x 10 = 0.1 N

Uw - WA - Ww weight of object in water, Ww = WA - Uw = 0.5 - 0.1 = 0.4N

Example 3A rectangular block of wood floats in water with two-third of its volume immersed. When placed in another liquid, it floats with half of its volume immersed. Calculate the relative density of the liquid. WAEC 1998

141

Solution

Given : Volume of block immersed in water = Vw = - V

Volume of block immersed in liquid = VL = j V

Problem. Relative density of liquid

R d of liquid - volume of liquid displaced by object (VL)volume of water displaced by same object (Vw)

Note. The volume of fluid (liquid or water) displaced is equal to the volume of object immersed in fluid.

& 1R.d = ^ ± = £- = l v + - V = ^ V x ^ = - = 0.75Vw fv 2

2 1 3 3V x3 2 2V 4

Example 4A solid plastic cube of side 0.2m is submerged in a liquid of density 0.8kgm . Calculate the upthrust of the liquid on the cube (g=10m/s2). WAEC 1997SolutionGiven: Length of cube, l = 0.2m Density of liquid, p = 0.80kgm 3Problem: Upthrust on the cube

Volume of cube = V = l3= (0.2)3 = 0.008m3

Upthrust = density of liquid X volume of object X g

= 0.8 X 0.008 x 10 = 0.064N

Example SThe density of water is lg/cm3 while that of ice is 0.9g/cm3, calculate the change in volume when 90g of ice is completely melted. WAEC 1994SolutionGiven: density of water, pw = lgcm-3 density of ice, p, = 0.9pcm~3

mass of ice or water, m = 90g

massDensity = —;-----volume Volume =

massdensity

mVolume of ice = V; = —

Pi90(L9 100cm3

mVolume of water = Kv = —

Pw— = 90cm3

Change in volume = Vj- -V w = 100cm3 - 90cm3 - 10 cm3

Example 6What volume of alcohol with a density of 8.4 x 10'kgnT will have the same mass as 4.2m3 of petrol whose density is 7.2 x 102kg/m3. NECO 2006SolutionGiven: density of alcohol, pa = 8.4 x kgm~3\ Density of petrol, pp= 7.2 xlO2 kgm~3

Volume of petrol, Vp = 4.2m3

142

Problem: Volume of alcohol, Va

mass m, « = density x volumeDe"sit>, = ^ w " M

u — 7 ? x 102 X 4.2 = 3024kg * mass of petrol, Mp — Pp x Vp

Therefore, mass of alcohol, ma — 3024kg

ma _ ___3024kg____ = 3 6m 3 Volume of alcohol, l/a - — - g4 x jo 2 kgm 3

Example 7 y Calculate the relativeA solid weighs 5.0N in air, 4.2N in water and 4.4N in liq NECO 2000density of liquid Y.Solution f _ AGiven: Weight of solid in air, Wa = 5.ON Weight of solid in water, w

Weight of solid in liquid Y, Wl = 4.4N Problem: Relative density of liquid Y.Upthrust of solid in liquid, UL = WA — WL = 5.0 — 4.4 — 0.6 N Upthrust of solid in water, Uw = WA — Ww = 5.0 — 4.2 = 0.8 N

upthrust of object in liquid (UL) WA — WL __ 0.6 NR. d of liquid Upthrust of object in water (Uw) WA — Ww 0.8 N

Example 840m3 of liquid is mixed with 60m3 of another liquid Q. If the density of P and Q are 1.00kg/m3 and 1.6kg/m3 respectively. What is the density of the mixture? A. 0.05 k g m ~3B. 1.25 kgm~3 C .130kgm ~3 D. 1.36 kgm~3 JAMB 1990SolutionGiven: Volume of liquid P, VP = 40m3 Volume of liquid Q, VQ = 60m 3

Density of liquid P, pP = 1.0kgm~3 Density of liquid Q, pQ =1.6 kgm .-2 Problem: Density of mixture.

massDensity = vo]ume ••• Mass = density x volume

Mass of liquid P, MP = pP x VP = 1.0 x 40 = 40kg

Mass of liquid Q, MQ = pQ x VQ = 1.6 x 60 = 96kg

Total mass of mixture, M = MP + MQ = 40 + 96 = 136kg

Total volume of mixture, V - V P + Vq = 40 + 60 = 100m3

Density of mixture, p = — = _ 1 ,V 100 ~ L36li5 m

Example 9The mass of a stone is 15.0g when completely immersed completely immersed in a liquid of relative density 2.0 what air? A. 5.0g B. 12.0g C. 20.0g D.25.0gSolution

in water and 10.0g when is the mass of the stone in

JAMR 1991Given: mass of stone in water, Mw = 15.0g =0.015kg

143

Mass of stone in liquid, ML - 10.Og = 0.010kg Relative density of liquid, r.d = 2.0

Problem: mass of stone in air.

•j r s upthrust in liquid Weight in air - Weight in liquid R. d of liquid (p) - upt|irust ¡n water - Weight in air - Weight in water

From W = mg, weight of stone in water, Ww = 0.015kg X 10 = 0.15N Weight of stone in liquid, WL = 0.010 X 10 = 0.1N Weight of stone in air = WA

R.d of liquid = — — ^ WA - Ww

WA - 0.1 WA - 0.15

Cross multiplying, 2(WA - 0.15) = WA - 0.12WA - 0.3 = WA — 0.1 2Wa - WA =0.3 - 0.1

WA = 0.2NWA

W = mg mass of stone in air, m = —a

0.2— = 0.02 kg = 20 g

Example 10The mass of a specific gravity bottle is 15.2g when it is empty. It is 24.8g when filled with kerosene and 21.2% when filled with distilled water. Calculate the relative density of kerosene. A. 1.25 B. 1.10 C. 0.90 D. 0.80 JAMB 1994 SolutionGiven-, mass of empty bottle,m = 15.2g

Mass of empty bottle with kerosene, MK = 24.8g Mass of empty bottle with water, Mw = 27.2g

Problem: Relative density of kerosene.

R. d =mass of liquid

mass of equal volume of waterMk — m Mw — m

24.8 - 15.2 9.627.2 - 15.2 “ I T

Example 11If a spherical metal bob of radius 3cm is fully immersed in a cylinder containing water and the water rises by 1cm. What is the radius of the cylinder?A. 12 cm B. 1 cm C. 3 cm D. 6cm JAMB 2001SolutionGiven: Radius of bob, r = 3cm

Height by which water rises in cylinder, h = 1cm Problem: Radius of cylinder

4 4tt 4ttVolume of metal bob = -Ttr3 = — x 3 3 = — x 2 7 = 36tt

In general, volume of liquid displaced = volume of object (metal bob)In this case, volume of liquid displaced = volume of water rise in cylinder

Volume of water rise in cylinder = volume of metal bob4

n r2h = - n r3

n x r 2 x 1 = 36n

n r2 = 367T

144

r = V36 = 6 cmr z = 36

Example 12Consider a balloon of mass 0.030kg being inflated with a gas of de Y 9What will be the volume of the balloon when it just begins to " * “ * d, 'i y1.29 Ar^m-3? (g = 10ms-2) WAEC 2008SolutionLet V be the volume of the balloon

massDensity = —:-----volume

mass = density x volume

Mass of air in balloon = V xdensity of gas

= 0.54K

Mass of balloon = 0.030kgMass of air displaced = volume of balloon x density of air

=1.29VFor the balloon to begin to rise, the mass of the air displaced must be equal to the mass of the balloon and the air inside it

1.29V = 0.54V + 0.030 1.29V-0.54V = 0.030

0.75V = 0.0300.030

V = 0.75- 0.04m3

EXERCISES 4.

1. A block of material of volume 20cm"3 and density of 2.5gcm"3 is suspended from aspring balance with half of the volume of the block immersed in water. What is the reading of the spring balance? WAEC 1989 Ans: 0.4N or 40g2. A solid weighs 0.04N in air and 0.024N when fully immersed in a liquid of density 800 kgm~3. What is the volume of the solid? (g=10m/s2)

WAEC 1992 Ans: 2 x 1 a 6m33. A block of material of volume 2 xlO m and density 2.5 x 10"3kg/m3 is suspended from a spring balance with half volume of the block immersed in water. What is the reading of the spring balance? (Density of water = 1000kg/m3, g=l 0m/s2)

WAEC 1993 Ans: 0.40N4. A body weighing 10N in air is partially immersed in water. It displaces water of mass 0.3kg. What is the upthrust on the body? (g=10m/s2) WAEC 1995 Ans 3N5. An object weighs 2.7N in air and 1.2N when completely immersed in water.Calculate its relative density. WAEC 1997 Ans• 1 806. The apparent weight of a body wholly immersed in water is 32N and its weight in air is 96N. Calculate the volume of the body (Density of water = 1000kg/m3, g=10m/s2)

WAEC 2000 Ans: 6.4 x

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7- A piece of brass of mass 20.Og is hung on a spring balance from a rigid support and completely immersed in kerosene of density 8.0 x 102kgm \ Determine the reading on the spring balance (g=10m/s2, density of brass = 8.0 x 103 k g m ~3)

WAEC 2000Ans: 0.2N8. An object of weight ION immersed in a liquid displaces a quantity of the liquid. If the liquid displaced weighs 6N, determine the upthrust on the object.

WAEC 2001 Ans: 6N9. The apparent weight of a body fully immersed in water is 32N and its weight inair is 96N. Calculate the volume of the body. (Density of water = 1000kg/m3, g=10m/s2). WAEC2001 Ans: 6.4 x UFm10. A piece of metal of relative density 5.0 weighs 60N in air. Calculate its weight"'hen fully immersed in water. WAEC 2003 Ans: 48N11. A solid weighs 45N and 15N respectively in air and water. Determine the relativedensity of the solid. WAEC 2002 Ans: 1.512. A uniform cylindrical hydrometer of mass 20g and cross scctiona '1 0.54cmfloat upright in a liquid. If 25cm of its length is submerged, calculate the R e density of the liquid (Density of water = lg/cm3) W. lEC 2000 Ans: 1.4813. An object weighs 60.0N in air, 48.2N in a certain liquid X, and 44.9N in water.Calculate the relative density of X. WAEC 2004 Ans: 0.78214. Calculate the change in volume when 90g of ice is completely melted, (dens* y ofwater = lg/cm3, density of ice = 0.9 g/cm3) WAEC 2004 Ans: /■ i315. A block of volume 3 x 10"5m3 and density 2.5 x 103kgm"3 is suspended 1» ..n a spring balance with 21 of its volume immersed in a liquid of density 900kgt.i’3. Determine the reading of the spring balance. (g=10m/s2) WAEC 2006 Ans: 0.57N16. A piece of metal of density 3.9 x 103kgm"3 weighs 10N in air. Calculate theapparent weight of the metal when completely immersed in a liquid of dens; . l x 102kgm'3 (g=10m/s2). NECO 2005 Ans: 8.95N17. A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the mass of the body. NECO 2004 Ans: 0.02kg18. An object weighing 8N in air is partially immersed in water. It displaces water of mass 0.3Kg. What is the upthrust on the body? (g=10m/s2) NECO 2003 Ans: 3N

19. A body of weight 20N displaces 12N of the liquid in which it is immersed. Whatis the upthrust of the liquid on the body? NECO 2002 Ans: 12N20. A piece of stone weighs 180N in air and 150N in a beaker containing 160N of methylated spirit. What is the upthrust exerted on the stone? NECO 2000 Ans: 3ON21. The density of 400cm3 of palm oil was 0.9g/cm3 before frying. If the density ofthe oil was 0.6gcm'3 after frying, assuming no loss of oil due to spilling, what was its new volume? A. 360cm3 B. 600cm3 C. 240cm3 D. 800cm3 E. 450cm3

JAMB 1979 Ans: 600cm322. An object of mass m and volume V is totally immersed in a liquid of density p.What will be the tension of a spring holding the object? A. mg B. (m + pV)gC. (m - pV)g D. pVg E. m g / (m - pV) JAMB 1980 Ans: C23. The relative densities of zinc, brass, copper, gold and silver are respectively 7.1, 8.5, 8.9, 19.3 and 10.5. A metal ornament which weighs 0.425kg and can displace50 x l O 'V of water is made up of which of the above listed metals? A. Zinc B. brass C. copper D. gold E. silver JAMB 1980 Ans: Brass (8.5)24. A cube of side 10cm and mass 0.5kg floats in a liquid with only V5 of its heightabove the liquid surface. What is the relative density of the liquid? A. 0.125 B. 0.250 C. 0.625 D. 2.500 JAMB 1987 Ans: 0.62525. A copper cube weighs 0.25N in air, 0.17N when completely immersed in paraffinoil and 0.15N when completely immersed in water. What is the ratio of upthrust in oil to upthrust in water? A. 4:5 B. 3:5 C. 13:10 D. 7:10

JAMB 2002 Ans: 4:5

146

, , ro /I V lr v W i '3 W il l have the same massWhat volume of alcohol with a density of 8 .4 x 10 p . 3)2kEm J? A. 1.4m3 B. 3.6m C. 4.9m

JAMB 1984 A ns: 3.6m26.as 4.2m3 of petrol whose density is 7.2 x lO^kgm

27.5 ° " a ,es,E,uhe8of3,ad,„s 1.0cm is loaded .0 8.8g. If,. isthe depth to which it would sink. (g=10m/s~, density of w a t e r ^ ^ ^ 8 ^ Ans' 2 8cm

? (Density of B. 0.45m3 j

B. 5.2cm C. 25.5cm D. 28.0cm \ a i28. The change in volume when 450kg of ice is completely me teice = 900kg/m3, density of water = 1000kg/m3) A. 0.05mC. 4.50m’ D. 0.50m3 JAM B 200429. A body whose mass is 2kg and has a volume of 500cm just ^ encompletely immersed in a liquid. Calculate the density of the liquid. A. 4. x 9™B. 4.0 x 103kgm~3 C. 1.0 x I O ^ itT 3 D. 1.0 X 106^ m

JAMB 1988 Ans: 4.0 X 103kgm30. If a plastic sphere floats in water (density = 1000kgm J) with 0.5 of its \olume submerged and floats in oil with 0.4 of its volume submerged, what is the density of t e oil? A. 800 kgm~3 B. 1200 kgm "3 C. 1250 kgm "3 D. 2000kgm

JAMB 1992 Ans. 1250 kgm 331. An object of mass 400g and density of 600kg/m3 is suspended with a spring sothat half of it is immersed in a paraffin of density 900kg/m3. What is the tension in the spring? A. 1.ON B. 3.0N C. 4.0N D. 5.0N JAMB 1993 Ans: IN32. If it takes 5.0hrs to drain a container of 540.0m3 of water, what is the flow rate ofwater from the container in kg/s. (Density of water = lOOOkgm3). A. 32.5 B. 3 1.5C. 30.8 D. 30.0 JAMB 1994 Ans 30kg/s33. If a solid X floats in liquid P of relative density 2.0 and in liquid Q ol relative density 1.5, it can be inferred that the

A. Weight of P displaced is greater than that of QB. Weight of P displaced is less than that of QC. Volume of P displaced is greater than that of QD. Volume of P displaced is less than that of Q

JAMB 1994 Ans: D34. A cube of sides 0.1m hangs freely from a spring. What is the upthrust on the cubewhen totally immersed in water? (Density of water is lOOOkgm'3, g=T0m/s2). A. 1000N B. 700N C. 110N D. 10N JAMB 1997 Ans: 10N35. A solid of weight 0.60N is totally immersed in oil and water respectively. If the upthrust in oil is 0.210N and the relative density of oil is 0.875, find the upthrust in water.A. 0.600N B. 0.360N C. 0.240N D. 0.180N JAMB 1998 Ans: 0.2436. A solid weighs 10.0N in air, 6.0N when fully immersed in water and 7.0N when fully immersed in certain liquid X. Calculate the relative density of the liquid A 5 /3B. 4 /3 C. 3 /4 D. 7 /10 JAMB 1999 Ans: 0.75 or 3 /437. Two liquids Li and L2 are contained in a U-tube. The height and the density of Li are 8cm and 103kgm'3 respectively. If the density of l 2 is 800kg/m3, what is its height measured from the same level? A. 8cm B. 16cm C. 12cm D. 10cm

JAMB 2005 Ans: lUan.38. A 3m volume of liquid W of density 200kgm' is mixed with another liquid 1 of volume 7m3 and density 150kgm. What is the density of the mixture? A . 2 6 5 k n ’n 3B. 165 kgm ~3 C. 100 k g m '3 D. 350 kgm "3 JAMB 2005 Ans: I fJ i ■ '39. If the relative density of gold is 19.2, what is the volume of 2.4k« • | >(Density of water = 103kgm'3). A. 1.92 x 10_4m 3 B. 4.61 x 10 ,C. 8.00 x 10~3m 3 D. 1.25 X 10~4m3 JAMB 2006 Ans: 1.25 x 10 ■ .„ 340. A density bottle has a mass of 50.0g when empty. When it is filled with llu usmass is 1 OO.Og. Calculate the mass of the bottle when it is filled with kerosene vedensity of kerosene = 0.8) NECO 2007 Ans41. A liquid of volume 2.00m3 and density 1.00 x 103kgm"3 is mixed w ith - 0m ofanother liquid of density 8.00 x 103kgm‘3. Calculate the density of the mixture V .,,me there is no chemical reaction] WAEC 2007 Ans: 5.20x J 3

147

42. A piece of iron weighs 250N in air and 200N in a liquid of density lOOOkgm .The volume of the iron is A. 2 .0 x l0 3m3 B. 5.0x10 m ( .4 .5 x 1 0 mD. 2.5x10 V [g=I0m/s2| JAMB 2007 Ans: 5.0 X10 3m343. An empty density bottle weighs 2N. If it weighs 5N when filled with water and 4N when filled with olive oil, the relative density of olive oil isA. 1/3 B. 2/3 C. 1/5 D. 2/5 JAMB 200818 Ans: 2 /344. A piece of stone weighs 0.8N in air, 0.60N when completely immersed in water and 0.66N when completely immersed in a liquid X. Calculate the relative density of X.

NECO 200810 Ans ■ 0.7045. A plastic sphere floats in water with 50% of its volume submerged. If it floats in glycerine with 40 % of its volume submerged, the density of the glycerine isA. MOOkgnf3 B. 1250 kgm'3 C. 500 kgm'3 D. 1000 kgm 3

(Density of water = 1000 kgm'3) JAMB 200917 Ans: 800 kgm 346. A body of volume 0.046m3 is immersed in a liquid of density 980kgm 3 with % of its volume submerged. Calculate the upthrust on the body, (g = 10ms'2)

WAEC 20094 Ans: 338.10N

148

5CIRCULAR AND SIMPLE HARMONIC MOTIONS

CIRCULAR MOTION circle or circular path. AnCircular motion is defined as the motion of an object round a - a circular pathobject is said to undergo uniform circular motion when it 1 &at constant speed. Circular motion has three basic characteristics.

1. constant speed2. changing or variable velocity3. centripetal acceleration. • . •

Centripetal acceleration is experienced by any object undergoing circu ar moalways directed towards the centre of the circular path.

Centripetal acceleration, a =

where V = speed or velocityr = radius of circular path

Centripetal force is defined as the force that keeps an object moving along a circular path at constant speed.Remember, force = mass(m) x acceleration(a).Therefore, centripetal force = mass x centripetal acceleration

mV2That is, Centripetal force, F = ------r

Example 1An object of weight 150N moves with a speed of 4.5m/s in a circular path of radius 3m. Calculate its centripetal acceleration and the magnitude of the centripetal force. [Take g as 10m/s2]SolutionWeight of object, W = 150N; Speed or velocity, V = 4.5ms-1; radius, r = 3m

F = W = mg mass of object, m W 150N7 ‘ - ï F = 1 5 k s

V2 4.52Centripetal acceleration, a = — = ——- = 6.75ms-2

Centripetal force, F = mV2 15 x 4.52= 101.25N

Example 2If an object of mass 4kg goes round a circle of radius 0.5m in 3 14?s what is the force towards the center? [re = 3.142] 6 °Solution

The force towards the centre is the centripetal force, F m V2r

149

The distance round a circle is the circumference of a circle, 2n

Time taken to go round circle = 3.142sdistance

Velocity of object = ------------- V —2tt

F =

time2 x 3.142

3.142 mV2 4 x 22

3.142

= 2ms-1

0.5= 32N

Example 3A stone of mass 500g tied to a rope 50cm long is whirled at an angular velocity of 12.0 rads'1. Calculate the centripetal force.SolutionMass, m = 500g = 0.5kg; radius, r = 50cm = 0.5mAngular velocity (NOT linear velocity), to = 12 rads'1

Linear velocity (V) is related to angular velocity i.e. V = cormV2

Substitute V = cor into centripetal force, F =

F -

rm(cur)2 = mo)2r

r r

F = mu>2r = 0.5 x 12.02 x 0.5 = 36N

Example 4A body moves along a circular path with uniform angular speed of 0.6 rads1 and at a constant speed of 3.0ms1. Calculate the acceleration of the body towards the centre of the circle. WAEC 1993SolutionAngular speed or velocity, co = 0.6rads'!; Linear speed, V = 3.0ms_1

V2Acceleration towards the centre of the circle is same as centripetal acceleration, a = —

From V = cor, the radius (r) of the circular path can be foundV 3

r = — = — = 5m cj 0.6

V2 32Centripetal acceleration, a = — = — = 1.8ms-2r 5

Example 5A force F is required to keep a 5kg mass moving round a cycle of radius 3.5km at a speed of 7m s-1. What is the speed, if the force is tripled? A. 4.0ms-1 B. 6.6ms-1C. 12.1ms-1 D. 21.0ms-1 JAMB 2008sSolutionMass , m = 5kg; speed, v = 7m s-1; radius, r = 3.5m

The force in action is centripetal forcem V 2 5 x 72 5 x 49

F = = 70 Nr 3.5 3.5" . . . i f the force is tripled...”, the centripetal force F becomes 3F or 3 x 70 = 2 10N

Substitute F = 2 10N; m = 5kg; r = 3.5m

150

F =

210 =

mVr

S x V? 3.5

V1 =210 x 3.5 = 14?

y = Vl47 = 12.1 m**

second, calculate the velocity ofExample 6If a wheel 1.2m in diameter rotates at one revolution per - ^ p y Sm s'the wheel. A. 3.6ms-1 B. 3.8ms 1 C. ■ rns JAMB 20086

Solutiondiameter 1.2m

radius = -----------= —-— = 0.6m

revolution per second... ') to radian per second asconvert angular velocity ( ‘‘...at one follows:1 revolution, 9 = 360° = 2n radians; time for 1 rev, t = Is

0 2 TTAngular velocity, or = - = — = 2 x 3.14 = 6.28 rads

Linear velocity, V - rco = 0.6 x 6.28 = 3.8ms

SIMPLE HARMONIC MOTIONA body is said to undergo simple harmonic motion if:

(i) its acceleration is always directed toward a fixed point.(ii) Its acceleration is directly proportional to its distance from the point.Amplitude of a simple harmonic motion is maximum displacement of a body from its central or equilibrium position.

Period of a simple harmonic motion is the time taken to complete one cycle, oscillation or vibration.

Period, Ttime taken t

number of oscillations ~ nFrequency of a simple harmonic motion is the number of complete oscillation per

second.

Frequency, / = number of oscillations time taken

Frequency is measured in cycle per second (s’1) or Hertz.

nt

The following equations can be applied in problems involving simple harmonicmotion.

d1. O) = —

t 2. V = oj r

3. eu = 2 n f 4._ 2ttT = —

CO

5. a = u>2r or a = a>2A 6. a — a r

151

Where

7. V = (Oyj A1 — X2O)

8 f =

9. 360° = 2n rad1

10. T = -

11. f = -

co = angular velocity (rads-1) t = time in seconds (s)0 = angle turned by body (rad) V = linear velocity (m/s)A, r = amplitude or radius (m) a = linear acceleration m/s'f — frequency (s’1) or Hertz (Hz) T = period (s)a = angular acceleration (rads 2) x = displacement of object from centre of motion

The following examples will illustrate how any, or a combination of these equations can be applied to solving problems.

Example 7An object of mass 0.40kg attached to the end of a string is whirled round in a horizontal circle of radius 2.0m with a constant speed of 8ms’1. Calculate the angular velocity of the object. WAEC1997SolutionRadius, r = 2.0m; linear speed, V = 8ms'1; mass, m = 0.40kg Linear velocity (V) is related to angular velocity (to) by, V = cor

— = 4 rads 2- l

Example 8The period of oscillation of a particle executing simple harmonic motion is 47t seconds. If the amplitude of oscillation is 3.0m, calculate the maximum speed of the particle.

WAEC 2000SolutionPeriod, T = 47t sec; amplitude, r = 3.0m; speed, V = ?

T2itto

2tr4tt

Substitute into V = co r, to obtainV = 0.5 x 3 = 1.5ms-1

Example 9A body moving with simple harmonic motion in a straight line has velocity, V and acceleration, a, when the instantaneous displacement, x in cm, from its maximum position is given by x = 2.5 sin0.47it, where t is in seconds. Determine the magnitude of the maximum; (i) velocity (ii) acceleration WAEC 2005SolutionThe equation x = Asincot represents the motion of an object undergoing simple harmonic motion and can be compared with the given equation as follows.

x = Asincot x = 2.5sin 0.4;tt

Therefore, the amplitude or radius, r = 2.5cm = 0.025m

152

Angular speed, to ** 0.4 n rads-1(0 velocity, V - tor - 0.4n x 0.025 “ 3.14 x 10 rns

(ii) acceleration, a - toJr ■ (0.4«): x 0.025 ■ 3.95 X 10

„-1 v z,

A body is rotating in a horizontal circle of radius 2.5m with n g W A E C 2 0 0 6 Calculate the magnitude of the radial acceleration of the body.Solution _2

acceleration, a - co2r - (5)J x 2.5 ” 62.5ms

Example 11A particle moves in a circular orbit of radius 0.02m. \ calculate its frequency in cycles per second.A. 2.0 B. 7.0 C. 8.8 D. 14.0SolutionRadius, r = 0.02m; linear speed, V = 0.88ms"1

V = wr angular speed, u)

to = 2nf, a frequency, f

If the speed of the particle is 0.88ms'

E. 17.6 J A M B 1 9 8 4

1 2 2 = 44 rads"1 0.02

= — = 7.00s"1 2tt

SIMPLE PENDULUMThe motion of a simple pendulum is a typical example of a simple harmonic motion. All equations given above can be applied.

The period of a simple pendulum is given as T 1

Where T = period of the pendulum in seconds (s) l = length of pendulum in meter (m) g = acceleration due to gravity in ms-2

The period Ta and length /, of a particular pendulum can be compared with the period T2 and length l 2 of the same or different pendulum in the same location. The length of a pendulum is its main distinguishing factor from other pendulums.

Square both sides of T = 2n to obtain

9

For the 1“ pendulum, equation (2) becomes

T2 _ 4ir2*iT l ~ 9 ...................

4tt2 T2Rearranging, — = — ..................

2

3

•4

153

4tt2is a fixed constant for all pendulum of whatever length in a particular location.

For the 2nd pendulum, equation (2) becomes

7» 2 _ **’ «2 2 ~ —

9 U4jt2 T2

Rearranging, -----= —

Equate equation (3) and equation (4) to obtain

HlxT? Ti

T2 iRearranging, = j-T2l 2

ht2

T2

Example 12What is the period and frequency of a simple pendulum that makes 40 oscillations in 35 seconds.Solution

Period, Ttime of oscillation

number of oscillationstn

35— = 0.875s 40

Frequency, f =

or_ 1 _ 1

f ~ T ~ 0.875

number of oscillations time of oscillation

= 1.14Hz

n 40■ = - = — = 1.14Hz t 35

Example 13A boy timed 30 oscillations of a certain pendulum thrice and obtained lmin 10s, lmin. 12s and 1 min. 7s respectively. The mean period of oscillation of the pendulum is A. 0.14s B. 0.43s C. 2.32s D. 6.97s JAMB 1991Solution

Convert all tim es into seconds and substitute into T = —n

1 min. 10s 70sTi = ------------- = — = 2.33s1 30 30

1 min. 12 s 72 sT? = ------------- = — = 2.4s2 30 30

1 min. 7s 67 sT3 = ---- —— == — = 2.23s3 30 30

154

. , 7\ + r 2 + 7*3 2.33 + 2.40 + 2.23 _ 6S*_ = 7 „ sMean period = ---------------= --------------j --------------- 3

. . . . tj + ta + ta 70 = — = 69.6sAlternatively, mean time, t - ------ ------- = ---------- 3

Mean period =mean time 69.6

number of oscillations 30= 2.32s

Example 14The bob of a simple pendulum takes 0.25s to swing from its equilibrium position to one....______, ^ F . . . 6 WAF.C 2000extreme end. Calculate its period.Solution

Fig 5.1

If it takes 0.25s for the bob to swing from O to A, it will then take 4 x 0.25s to undergo one oscillation, i.e. 1-2-3-4 or O-A-O-B-O.

Period, Tt 4 x 0.25s Is- = -------------= — = 1.00sn 1 1

Example 15A simple pendulum 0.64m long has a period of 1,2s. pendulum 0.36m long in the same location.SolutionSubstitute both instances into

T24rr21

9

Calculate the period of a similarNECO 2005

1S1 instance 2nd instance

1.2247Tz x 0.64

9T2 =

47T2 x 0.36 9

Make g the subject in each case and then equate both g’s

47T2 x 0.64 4tt2 x 0.369 = IJ2 9 - ^2

47T2 x 0.64 4tt2 x 0.36 T22 “ T2

155

Cross multiply and make T 2 subject

4tt2 x 0.64 X T 2 = 1 .2 2 x 4tt2 x 0.36

T 2

T 2

1.22 x A n 2 x 0.36 4tt2 x 0.64

1.22 x 0.36 064 0.81

T = V0i8l = 0.9s

The above method is long and more difficult, so you are advised to use the following quicker and easier method for this and other similar questions.Period of first pendulum, T| = 1,2s Length of first pendulum, /j= 0.64 Period of second pendulum, T2 = ?Length of second pendulum, l2 = 0.36

hT2

T2 = 1.2 x 0.360.64 = 1.2 x

36 V36 „ 6 7.2- = 1 .2 x 7 = = 1 . 2 x - = - = 0 .9 S

Example 16A simple pendulum with a period of 2.0s has its length doubled. Its new period is,A. 1.00s B. 1.41s C. 0.35s D. 2.83s E. 4.00s JA M B 1985 SolutionPeriod of first pendulum, 7\ = 2.0s Length of first pendulum, = l Period of second pendulum, T2 =?Length of second pendulum, l2 = 21 (.... length d o u b le d ...)

H _ 11 T2 _ Tl X *2T22 h " 2 lx

2 2 x 21 A x 21 81

/ l “ T

.-. r = V8 = 2.83s

Example 17A simple pendulum has a period of 4.50s when the length of the pendulum is shortened by 1.2m the period is 2.50s. Calculate the original length of the pendulum.SolutionPeriod of first pendulum, 7\ = 4.5sLength of first pendulum, l t = lPeriod of second (shortened) pendulum, T2 = 2.50sLength of second (shortened) pendulum, l2 = l - 1.2

T2 lSubstitute into —2 = j - to obtain

h 2

156

4.52 l2.502 _ / - 1-2 20.25 _ l 6.25 ~ l - 1-2

20.251 - 24.3 = 6.25/ 20.25/ - 6.25/ = 24.3

14/ = 24.3

Z =24.3

141.74m

ENERGY OF SIMPLE HARMONIC MOTIONThe energy of a simple harmonic motion is derived when the motion of a

spiral spring is considered. The following equations, in addition to t ose given ,can be applied to solving problems involving loaded spiral spring and energy o simp e harmonic motion.

[m 4nzm1 . T = 2n — or T ~ — Z—K2. F = Ke or mg = Ke

e , An2e3. T = 2n - or T2 = -----------

A T9

4. (j) = —yj™

1 i K 7 K5. f 2;i j mor 4n2m

6. W = ^K A 2 = ^ tticj2A2

Where T = period in seconds (s)m = mass of body in kilogram (kg)K = force constant of spring in Nm'1 e = extension of spring in meter (m) g = acceleration due to gravity (ms'2)F = force, load or weight in Newton (N) to = angular speed or velocity (m/s) f = frequency of motion in s'1 or Hz W = total work done by spring or energy stored in spring (J) A = amplitude of motion (m)

157

Example 18 , ,A body of mass 500g suspended from the end of a spiral spring which o eys °° e law, produced an extension of 10cm. If the mass is pulled down a distance o cm released, calculate;

(i) The force constant of the spring(ii) The frequency of oscillation(iii) The period of oscillation(iv) The angular speed of the body

SolutionMass, m = 500g = 0.5kg, extension, e = 10cm = 0. lm

(i) From Hooke’s law, F = Ke,F mg 0.5 x 10

force constant, K = — = ---- = — —— = 50Aime e 0.1

(ii) Frequency, f = —2 tt.

(iii) Period, T = 2nJ j f = 2n

H = ± — = — VlOO = — x 10 = 1.59s- m 2 tt„ 0 .5 2 tt 2 r r

— = 2nV0.01 = 2tt X 0.1 = 0.628s 50

or 7 = } = ^ = 0.628s

(iv) Angular speed, co =50 ,----- ,— = V100 = 10 rads 10.5

EXERCISES 5.1. A stone tied to a string is made to revolve in a horizontal circle of radius 4m withan angular speed of 2 radians per second. W ith what tangential velocity will the stone move off the circle if the strings cuts. WAEC 1989 Ans: 8m/s2. A body weighing 100N moves with a speed of 5ms'1 in a horizontal circular path of radius 5m. Calculate the magnitude of the centripetal force acting on the body.[ g=10ms'2] WAEC 1999 /Ins; 5ON

3. An object of mass 5kg moves round a circle of radius 6m. If the period of themotion is 7t s, calculate the force towards the centre. NECO 2004 Ans: 120N4. A particle of mass 10'2kg is fixed to the tip of a fan blade which rotates withangular velocity of 100 rads1. If the radius of the blade is 0.2m, the centripetal force is: A. 2N B. 20N C.200N D. 400N JAMB 1999 Ans: B. 20N5. A body moves along a circular path with uniform angular speed of 0.6 rads'1 andat a constant speed of 3.0ms1. Calculate the acceleration of the body towards the centre of the circle. NECO 2005 Ans: 1.8m/s:6. A body executing a simple harmonic motion has an angular velocity of 40 radians per second. If it has a maximum displacement of 6cm, calculate its linear speed.

NECO 20081' Ans: 2.40ms'17. The angular speed of an object describing a circle of radius 4m with a linearconstant speed of 10ms1 is? WAEC 1995 Ans: 2.5 Or ads'1

158

8- A body executing a simple harmonic motion has an angular velocity of 22radi per second. If it has a maximum displacement of 1 Ocm, what is its linear velocity. ;

NECO 2002 Ans: 2.2ms9. A particle in circular motion performs 30 oscillations in 6 seconds. It s angularvelocity is? A. lOrc rads'1 B. 57trads~' C. 6rads_1 D. 5rads

JAMB 2002 Ans: JOn rads10. The amplitude of a particle executing simple harmonic motion is 5cm while itsangular frequency is lOrads’1. Calculate the magnitude of the maximum acceleration o the particle. WAEC 2001 Ans: 5m/s11. A loaded spring performs simple harmonic motion with an amplitude of 5cm. Ifthe maximum acceleration of the load is 20cms"2, calculate the angular frequency o I e motion. WAEC 2004 Ans: 0.32s1 Hint: a = co2A a n d f = “A«12. A simple pendulum makes 50 oscillations in one minute. What is its period ofoscillation? WAEC 1990 Ans: 1.2s13. A boy timed 20 oscillations of a certain pendulum three times and obtained 44.3s, 45.5s and 45.7s respectively. Calculate the mean period of oscillation of the pendulum.

WAEC 1993 Ans: 2.26s14. A student found out from a simple pendulum experiment that 20 oscillations were completed in 38 seconds. What is the period of oscillation of the pendulum?

WAEC 1995 Ans: 1.9s15. Two simple pendula x and y makes 400 and 500 oscillations respectively in equal time. If the period of oscillation of x is 1.5 seconds, what is the period of oscillation of y?

WAEC 1996 Ans: 1.2s H int: T,n, = T:n216. A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period of the motion.

WAEC 1999 Ans: 0.083s17. The length of a simple pendulum is increased by a factor of four. By what factoris its period increased? NECO 2005 Ans: 218. The time ti, t2 and t3 for 20 complete oscillations of a simple pendulumexperiment are 32.0s, 34.6s and 35.5s respectively. Calculate the mean period of the pendulum. NECO 2004 Ans: 1.70s19. Calculate the length of a simple pendulum that oscillates with a frequency of0.4Hz [g=10ms‘2, 7r=3.142] NECO 2004 Ans: 1.44m20. A simple pendulum with a period of 2.0s has its length doubled. What is its newperiod? NECO 2003 Ans: 2.83s21. In a simple pendulum experiment, a boy observed that the times for 30 oscillationsare 70.0s, 72.0s and 67.0s respectively. Calculate the mean period of oscillation of the pendulum. NECO 2000 Ans: 2.32s22. If in a simple pendulum experiment the length of the inextensible string is increased by a factor of four, its period is increased by a factor ofA. 4 B. 7 2 C. Va D. 2 7i E. 2 JAMB 1979 Ans: 223. A simple pendulum, 0.6m long, has a period of 1.5s. What is the period of a similar pendulum 0.4m long in the same location?

A. 1.5J f s B. 1.5J ^ s C. 2.25s D. 1.00s E. 2.00s JAMB 1984 Ans: A24. The length of a displaced pendulum bob which passes its lowest point twice everysecond is A. 0.25m B. 0.45m C. 0.58m D. 1.00m [g=10ms'‘)

JAMB 1995 Ans: 0.25m25. A simple pendulum has a period of 17.0s. When the length is shortened by 1 5m its period is 8.5s. Calculate the origin of length of the pendulum.A. 1.5m B. 2.0m C. 3.0m D. 4.0m JAMB 2000 Ans: 2.0m26. An oscillating pendulum has a velocity of 2ms’1 at the equilibrium position O and velocity at same point P. Using the diagram below in fig 5.2, calculate the height h of P above O. [Take g= 10ms’2 ].

159

Fig 5.2

27.

O

WAEC 1990 Arts:0.2m Hint

Fig 5.3O

A simple pendulum of mass m moves along an arc of a circle radius R in a vertical plane as shown in the figure above. What is the work done by gravity in a downward swing through the angle 0 to O?A. m gR sin 6 B. m g R (l — cos 6) C .m gR D. m gR (l - sin 9) JAMB ¡989 Ans: B28. A mass m attached to a light spiral spring is caused to perform simple harmonicmotion of frequency, f = ^ ;J ^ , where K is the force constant of the spring. If m =0.30kg, K = 30Nm'' and the maximum displacement of the mass from the equilibrium position is 0.015m, calculate the maximum

(i) kinetic energy of the system;(ii) tension in the spring during the motion [g = 10ms'2,7t = 3.142]

WAEC2005 Ans: (i) 3.38x1 ff3J (ii) 3.45N29. When a mass is hung on a spring, the spring stretches 6cm. Determine its period of vibration if it is then pulled down a little and released. [Take k = 22/7 and g =10m/s2]

NECO 2003 Ans: 0.49s. Hint. T = 2nJ^30. A car o f mass 1500kg goes round a circular curve of radius 50m at a speed of 4 0m s1. The magnitude of the centripetal force on the car isA. 1.2x 103N B.' 4.8x 1 04N C.1.2x102N D. 4.8x1 03N JAMB 2007 Ans: 4 .8x10^31. A gramophone record takes 5s to reach its constant angular velocity of 47trads'‘ from rest. Find its constant angular acceleration.A. 0.8jt rads'2 B. 20.On rads'2 C. 0.471 rads'2 D. 1.3^ rads'2

JAMB 2007 Ans: 20.On rads '32. The frequency of vibration of an oscillator is 4Hz. If it makes 12 vibrations in tseconds, calculate the value of t. NECO 2007 * Ans: 3.00s33. A mass attached to a string is moving in a circular path. If the speed is doubled, the tension in the string will beA. doubled B. halved C. four time as greatD. one-fourth as much WAEC 2008'‘ Ans: C

160

34. What is the frequency of vibration if the balance wheel of a wrist watch makes 90revolutions in 25s? A. 0.01 Hz B. 0.04Hz JAm b 200810 Arts: 3.60HzC. 2.27Hz D. 3.60Hz JAMo ¿v o35. A block of mass 4.0kg causes a spiral spring to extend y ■unstrctchcd position. The block is removed and another bo y o ma®s • from the same spiral spring. If the spiral spring is then stretchc an re eas angular frequency of the subsequent motion? (g = 10ms ) _

WAEC 200814 Ans: lOVS rads36. The period of a simple pendulum of length 80.0cm was found to have doubledwhen the length was increased by X. Calculate X. WAEC 200 ns■ • 07137. A body executing simple harmonic motion has an angular speed o n ra ians.What is the period of oscillation? (7r = 3.14) NEC02009 ns. . s38. Calculate the magnitude of the centripetal force on a particle of mass^ ^5.0 x 10~6kg revolving round the earth with radial acceleration of 6.0 * 1 0 ms ■

NECO 2009s Ans:300N

161

6LINEAR MOMENTUM

NEWTON’S LAWS OF MOTION• its st&tc of* rest orNewton’s first law of motion states that, every body continue 1 force to act

of uniform motion in a straight line unless compelled by some ex otherwise. f omentum of a

Newton’s second law of motion states that, the rate of change ° ^ which thebody is proportional to the applied force and takes place in the irec 1 force acts. e body, a*1

Newton’s third law of motion states that whenever a force acts on ^on there is equal and opposite force acts on some other body. In other words, to eve an equal and opposite reaction.

FORCE, IMPULSE AND MOMENTUMMomentum of a body is defined as the product of its mass and its velocity. The

S.I unit is kgms'1 or Ns.Momentum = mass (m) x velocity (v)

From Newton’s 2nd law, we can derive equation for force and impulse.

a) Force. If an object of mass m, with initial velocity u, is acted upon by a force, F, it will attain a final velocity, v in time t.

Therefore, initial momentum of object = mu Final momentum of object = mv

Change in momentum = final momentum — initial momentum = mv — mu

mv — muRate of change in momentum = ----- ------

By Newton’s force,

That is.

2nd law, the rate of change of momentum is proportional to the applied

mv — mu m(v - u)= F “ — —

„ , change in velocity v - uRemember, acceleration, a - --------- ;------------ -- --------

time t

Therefore, F oc ma ••• F = ma

Force (N) = mass (kg) x acceleration (ms~2). kgms~2 is also a unit of force.

b) Impulse: Impulse is defined as the product of a force and the time during which it acts on a body. It is also equal to the change in momentum produced by the force From Newton’s 2nd law,

„ change in momentumForce = ----------------------------time

162

F =mv - mu Ft = mv — mu

Also, Force =mass x change in velocity

F =

time m( v — u) Ft - m (v — u)

Impulse = Ft = m v - mu = m (y - u)

Impulse = force x time = mass x acceleration x time

The unit for impulse is Ns or kgm s~r

Example 1A net force of magnitude 0.6N acts on a body of mass magnitude of the resulting acceleration.SolutionNet force, F = 0.6N; mass of body, m = 40g = 0.4kg

40g, initially at rest. Calculate the WAEC1999

Force (F) = mass (m) x acceleration (a)

F _ 0.6 m ~ 0.04

15 ms 2

Example 2A truck of mass 3200kg moving with a velocity of 10ms-1 increases its velocity to 20ms-1 in 5s. Calculate the magnitude of the force exerted on the engine. NECO 2008 SolutionInitial velocity, u — 10m s-1; final velocity, v = 20m s-1; time taken, t = 5 s Mass,m = 3200kg

From Newton’s 2nd law. F =m ( v — u)

t3 2 0 0 (2 0 - 10) 3200x 10

••• F = -------------------= --------------= 6400N

Example 3A body of mass 20kg is set in motion by two forces 3N and 4N acting at right angles to each other. Determine the magnitude of its acceleration. WAEC 1997SolutionThe two forces, 3N and 4N, must be resolved into a single net force as follows.

Fig 6.1

4N

Apply Pythagoras theorem, F2 = 32 + 42 = 9 + 16 = 25

mass, m = 20kg

F = ma ••• acceleration, a

F = V2S = SN

L - J Lm 20 0.25ms 2

163

Example 4A moving car of mass 800kg experiences a frictional force of 200N. If f t2ms , calculate the magnitude of the force applied to the car.SolutionAcceleration, a = 2ms~2\ mass, m = 800kg; frictional force = 200N.

Net force required to accelerate, F = ma = 800 x 2 = 1600N.Net force = Force applied — Frictional force

Force applied = Net force + frictional force = 1600 + 200N = 1800N

Example 5A body of mass 2kg moving vertically upwards has its velocity increased uniformly from 10ms to 40ms 1 in 4s. Neglecting air resistance, calculate the upward vertical force acting on the body. [g=10ms'2]A. 15N B. 20N C. 35N D. 45N JAMB 1997SolutionMass of body, m - 2kg; initial velocity, u = 10ms~rFinal velocity, v = 40ms'1; time, t = 4s

Substitute into m(v — u)

2(40- 10) 2 x 30 604 “ 4 “ T 151V

Example 6Two bodies have masses in the ratio 3:1. They experience forces which impart to them accelerations in the ratio 2:9 respectively. Find the ratio of the forces the masses

JAMB 1999experience. A. 1:4 B. 2:1 C. 2:3 D. 2:5Solution

1st body 2nd M y

mass of bodies, m = 3m lmacceleration of body, a = 2a 9aforce, F = ma 3m X 2a lm x 9adivide each by 3ma 6ma 9maratio of forces = 2 3

Ans: 2:3

Example 7A rope is being used to pull a mass of 10kg vertically upward. Determine the tension in the rope if, starting from rest, the mass acquires a velocity of 4ms'1 in 8s. [g=10m/s2]A. 105N B. 95N C. 50N D. 5N JAMB 2000SolutionInitial velocity, u = 0; final velocity, v = 47ns"1 g=10ms 2Mass, m= 10kg; time,1 ~The tension (T) in the rope is the difference between the weight (W) of the object and the upward pulling force (F) i.e. T — W — F

m(v - u) 10(4 - 0) 40Upward pulling force, F = -----------= ------------= — = 5N

164

weight, W = mg = 1 0 x 1 0 - 100N Therefore, Tension, T = W - F = 100-5 = 95N

Example 8 - , -iA ball of mass 5.0kg hits a smooth vertical wall normally with a speed ot ¿ms rebound with the same speed. Determine the impulse experienced by the ball.

and

W A R C 1 9 9 6 .

SolutionMass of ball, m = 5.0kg; initial velocity, u = 2 m s"1Final velocity, v = —2ms"1 {"...the direction changes on rebound... *)

Impulse = mv - mu = m(v - u) = 5[2 - (—2)] = 5(2 + 2) = 5 x 4 = 20kgms

Example 9A ball of mass 0.15kg is kicked against a rigid vertical wall with a horizontal velocity of 50ms'1. If it rebounded with a horizontal velocity of 30ms'1, calculate the impulse of the ball on the wall. A.3.0Ns B. 4.5Ns C. 7.5Ns D. 12.0Ns JAMB 1998SolutionMass, m = 0.15kg; initial velocity, u = 50ms"1;final velocity, v = —30ms"1 {on rebounding, direction changes)

Impulse = mass x change in velocity = m(v - u)= 0.15 x [50-(-30)] = (50 + 30) x 0.15= 0.15 x 80 = 12Ns

Example 10A force of 100N is used to kick a football of mass 0.8kg. Find the velocity with which the ball moves if it takes 0.8s to be kicked.A. 32ms'1 B. 50ms'1SolutionForce, F = 100N;Initial velocity, u = 0;

Substitute into

C. 64ms D. 100ms JAMB 2003

mass, m = 0.8kg; final velocity, v = ?

m{v - u)F = ------------

time, 1= 0.8s

100 =0.8(v - 0) 0.8v

08 — ~08~ v = 100ms”1

Example 11A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20ms'1. The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the (i) magnitude of the resistance.

(ii) distance moved by the bullet in the wood WAEC 1999SolutionMass, m = 120g = 0.12kg; initial velocity, u = 20ms'1Final velocity, v = 0m s"1 (“... brought to rest... "); time, t = 0.1 s

(i) The magnitude of the resistance is equivalent to the deceleration force the bullet experiences

From 1sl equation of motion, v = u + at

165

Deceleration,O - u) _ 0 - 20 _ _20 _ 200ms-2

3 “ l “ 0.1 0.1Resistance = deceleration force = F = ma = 0.12 x 200 = 24N (ii) From 3rd equation of motion, v2 = u2 + 2as,

v 2 — it2 O2 — 202 -4 0 0 _ ., distance. s = _ _ = _ _ _ 5 _ — _ l m

Example 12A 0.05kg bullet travelling at 500ms'1 horizontally strikes a thick vertical wall. It stops after penetrating through the wall a horizontal distance of 0.25m. What is the magnitude of the average force the wall exerts on the bullet?A. 25N B. 50N C. 250N D. 5000N E. 25000N JAMB 1985 SolutionMass, m = 0.05kg; initial velocity, u = 500ms’1Final velocity, v = 0; distance penetrated, s = 0.25mFrom 3rd equation of motion, v2 = u2 + 2as,

deceleration,v 2 _ u 2 02 _ 5002 -250000

a = 2s = 2 x0.25 0.5(The negative sign indicates deceleration)

= 5 x 10sms-2

The average force = decelerating force, F = ma

= 0.05 x 5 x 10s =25000N

CONSERVATION OF LINEAR MOMENTUMA combination of Newton’s 2nd and 3rd law results in the law of conservation of

linear momentum which states that, when two or more bodies act upon one another, their total momentum remains constant, provided no external forces are acting.

The following examples shows how the momentum, velocity and kinetic energy of colliding bodies are calculated.

Example 13An object of mass 5.0kg moving with a velocity of 12ms'1 due north hits a stationary body of mass 7.0kg. If they stick together after collision and move with velocity, v due north calculate the magnitude of v. NECO 2005Solution

m-L = 5kg

Vj = 12ms'

Fig 6.2

m2 = 7kg

v2 = 0

5kg 7kg

V----------►

momentum before collision = momentum before collision

mxVi + m2v2 = (mx + m2)v

5 x 12 + 7 x 0 = (5 + 7)v

60 + 0 = 12v

12v = 60

166

I'6 012

Sms 1

A balTp ofmass 0.25kg, losses one-third of its velocity when it makes a head on collision with an identical hall Q at rest. Alter the collision. Q moves off with a speed1 olI-m* :in the original direction of P. t'alcnlate the initial velocity of P. rSolution

m, - 0.25kg

tti = i>

itij 0.25kg m x = 0.25kg m 2 = 0.25kg

Fig 6.3

Let initial velocity of P be t> ms~x. After collision the velocity ot P.23 V

Momentum before collision = momentum after collision

mlu1 + m2u2 = m,i>j + m2v2

P and Q have the same mass, therefore, m, = m2 = m

mu, + mu2 = me, + mv2

m(u, -f u2) = m(p, + v2)

Dividing both sides by m to obtain

»! + u2 = Vx + t’2

Substituting. v + 0 = + 2

V ~ l v = 2

0 = 2 x 3 = 6ms"1

Example 15A tractor of mass 5.0 x 10'kg is used to tow a car of mass 2.5 x 10'kg. The tractor moved with a speed of 3.0ms'1 just before the towing rope becomes taut. Calculate the(i) speed of the tractor immediately the rope becomes taut.(ii) loss in K .li o f the system just after the car has started moving.(iii) impulse in the rope when it jerks the car into motion.SolutionMass of tractor, ni| 5.0 x 10 kg; initial velocity of tractor, ut = 3.0ms'1Mass of car, m» = 2.5 x 10'kg; initial velocity of car, u> = 0ms'1Let v be the common velocity of tractor and car.

(i) The speed of the tractor immediately the rope becomes taut is velocity of tractor and car when they both began to move.

Momentum before movement momentum after movement

m,u, + m 2u2 = (m, + m ,) v

5.0 x I03 x 3 + 2.5 x 10' x 0 ** (5.0 x 10' + 2.5 x Ift V

IVAEC2003

the common

167

1.5 x 104 = 7.5 x 10\-

1.5 x 104v = = 2 m s'

7500(ii) Loss in K.E = final K.E - Initial K.E

Initial K.E = K.E of the moving tractor = ^ rrijUj (from KE = - m v 2)

= ^ x 5.0 x 103 x 32 = 2.25 x 104 J

final K.E = K.E of the moving tractor and car = ^ (ntj + m 2) v 2

= i(5 .0 x 103 + 2.5 x 103) x 22

= ^ (7500) x 4 = 15000 = 1.5 x 104J

loss in K.E = 1.5 x 104- 2.25 x 104 = -7500J

= 7500JThe negative sign signifies loss in K.E

(iii) Impulse = change in momentum

= final momentum — initial momentum

— (m 2 + m 2)v — (m 1u1 + m 2u2)

= 7500 x 2 - 1.5 x 104 + 0

= 1.5 x 104 - 1.5 x 104

= ONs or 0 kgm5_1

Example 16An arrow o f mass 0.3kg is fired with a velocity of lOOm/s into a wooden block of mass 0.7kg. Calculate the final K.E after impact, given that the wooden block can freely move. SolutionMass o f arrow, m j = 0.3kg; velocity of arrow, v x = lOOm/s;Mass o f block, m 2 = 0.7kgLet v be the common velocity of arrow and block.

Momentum o f arrow - momentum of arrow and block m jiT = (m i + m 2)v

0.3 x 100 =(0.3 + 0.7)v 30 = lv V = 307715"1

K.E after impact = + m 2) v 2

= i(0.3 + 0.7) x 302 = ^ x 1x9002V 2

= 450J

EXPLOSION: RECOIL OF A GUN

When a gun is fired the following statement and equation applies.

168

Momentum of bullet momentum ol gunMass of bullet x bullet’s velocity mass of gun x recoil velocity

ntjU, = MVwhere rrtj, are muss and velocity of bullet respectively and M, V arc mass and recoil velocity of gun respectively.

Example 17A sub machine gun of mass 20kg (ires a bullet of mass lOOg due South with a velocity ot 250ms1. What is the recoil velocity of the gun?SolutionMass of bullet, mt = 100g = 0.1kg Velocity of bullet, vt = 250m/s Mass of gun, M = 20kg Recoil velocity of gun, V = ?

Momentum of bullet = momentum of gunmj t>j = MV

0.1 x 250 = 20 x V

V =0.1 x 250

20 1.25ms 1

The recoil velocity is always in the opposite direction of the bullet’s velocity (due south), therefore the submachine gun recoil velocity is 1.25m/s due North.

JET AND ROCKET PROPULSIONA jet engine or rocket bums and expels a stream of hot gas at extremely high

velocity. According to Newton’s 3rd law, a momentum, equal and opposite to the expelled gas is experienced by the rocket or aircraft. Thus,

Rate of momentum of expelled gas = Force (thrust) experienced by rocket or aircraft Mass of gas expelled or consumed per second x velocity = Force on rocket

/. M xV = F

Example 18Fuel was consumed at a steady rate of 5.0 x 10'2kg per second in a rocket engine and ejected as a gas with a speed of 4 x 103ms '. Determine the thrust on the rocket.

WAEC 2005SolutionThrust (force) = rale of change of momentum

F = mass of fuel per second x velocity F = 5.0 x 10-2 x 4 x 103 = 200N

Example 19A jet engine develops a thrust of 270Ns when the velocity of the exhaust gases relative to the engine is 300ms1. What is the mass of the material ejected per second?

169

S o lM ioT 6 B 900kB C ° 90k8 D ° ° * ‘SVelocity, v = 300ms-1; force, F-270N

Mass of gas ejected per second X velocity = thrust on engineM x V = F

. . F 270M = - = -----= 0.9 kaV 300 y

JAMB 1987

W EIGHT OF A BODY IN A LIFTThe following instances and their equations exist.

If a lift is at rest or moving up or down with uniform velocity, the weight of the body is the same as when it is at rest on the earth’s surface.

2.

3.

4.

5.

W — F — mg

If a lift is accelerating upward (ascending), the weight of the body is

F = mg + ma = m(g + a)

If a lift is accelerating downwards (descending) with acceleration a, less than g, the weight of the body is

F = mg — ma = m(g — a)

If a lift is moving downward with acceleration a greater than g, the weight of the body is

F = ma — mg

If a lift falls freely, when, a = g then the weight of the object is

F = mg — mg = ON

Example 20A spring balance which is suspended from the roof of a lift, carries a mass of 1kg at its free end. If the lift accelerates upward at 2.5ms'2, determine the reading on the spring balance [g=10m/s2]. WAEC2001SolutionMass, m = 1kg; g=10m/s ; acceleration of lift, a = 2.5ms'The reading on the spring balance, F = mg + ma - m(g + a)

F = 1(10 + 2.5) = 1 X 12.5 = 12.5N

Example 21 2A 1000kg elevator is descending vertically with an acceleration of 1.0ms' . If theacceleration due to gravity is 10.0ms'2, the tension in the suspending cable is A 10N B. 10.ON C. 9000.0N D. 11000.0N JAMB 1986Solution 2g=10ms'2; mass, m = 1000kg; a - l.Om/sFor a descending lift the force (tension), F = mg - ma = m (g - a) = 1000(10 - 1)

= 1000 x 9 = 9000N

170

Example 22A mass of 5kg is suspended from the ceiling of a lift with a light inextcnsible string. As the lift moves upward with an acceleration of 2ms'2, what’s the tension in the string.7 |g 10ms'2] NEC02004

SolutionMass, m 5kg; g=IOms'2; acceleration of lift, a = 2ms

For ascending lift, force (tension), F = mg + ma = tn(g + a) = 5(10 + 2) - 60N

Example 23An elevator of mass 4800kg is supported by a cable which can safely withstan a maximum tension of 60 000N. The maximum upward acceleration the elevator can have is A. 2.5ms'2 B. 5.0ms 2 C. 7.5ms'2 D. 10.0ms'2 [g=10ms'2] J A M B 1987SolutionForce or tension, F = 60 000N; mass, m = 4800kg; g=10m/s2;upward acceleration, a = ?

Substitute into equation of ascending lift F = mg + ma

60.000 = 4800 x 10 + 4800 x a

60.000 = 48,000 + 4800a

4800a = 60,000 48,000

4800a = 12000 12000

" “ 4800 2.5m.s 2

EXERCISES 6.

1. A net force of 15N acts upon a body of mass 3kg for 5s, calculate the change inthe speed of the body. IVAEC 2003 Ans 25m/s2. A stationary object of mass 4kg is set in motion by a net force of 50N. If the object attains a speed of 5ms'1 in time, calculate the value oft. WAEC 1996 Ans: OAOsec3. A constant force acts on a body of mass 50kg and changes its speed from 20ms'1 to 90ms'1 in 10 sec. Calculate the magnitude of the force applied. NECO 2005 Ans: 350N4. A body of mass 4kg is accelerated from rest by a steady force of 9N. What is itsspeed when it has travelled a distance of 8m? NECO 2002 Ans:6m/s mm: a = F/m; J = + 2ns5. A force of 16N applied to a 40kg block that is at rest on a smooth, horizontal surface. What is the velocity of the block at t = 5 seconds? A. 4m/s B. lOm/sC. 20m/s I). 50m/s E. 80m/s JAMB 1983 Ans: 20m/s6. A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when it has moved 50m from O is 500ms'1, What is the value of F? A. 0.4N B. 100.ON ( ’.250.ON D. 1000.0N

JAMB 1986 Ans. lOON mmF = r = J + 2as7. A force of 200N acts between two objects at a certain distance apart. The value of the force when the distance is halved isA. I00N B. 200N C. 800N D. 400N JAMB 2005 Ans: 100N8. An engine of a car of power 80KW moves on a rough road with a velocity of 32ms’1. The force required to bring it to rest is A. 2.50 x 106N B. 2.56 x 10ftNC. 2.50 x I0’N I). 2.80 x lO’N JAMB 2006 Ans: 2.50 x l& N Hint: P = Fv

171

9. A constant force acts on a body of mass 50kg and reduces its speed from 90msr‘ to20ms in 20s. Calculate the magnitude of the force. NECO 2006 Arts. 175N10. An object of mass 2kg moves with a uniform speed of 10ms for 5s along a straight path. Determine the magnitude of its acceleration. WAEC. 2007 Ans. 4ms11. An external force of magnitude 100N acts on a particle of mass 0.15kg for 0.03s.( alculate the change in the speed of the particle. WAEC 2007 Ans: 20ms

2. A car of mass 800kg moves from rest on a horizontal track and travels 60m in 20s with uniform acceleration. Assuming there were no frictional forces, calculate the accelerating force. WAEC 2007 Ans: 240.00N13. A body of mass 5kg initially at rest is acted upon by two mutually perpendicular forces 12N and 5N as shown below. If the particle moves in the direction OA, calculate the magnitude of the acceleration.A. 0.40ms'2 B. 1.40ms'2 C. 0.26ms'2 D. 2.60ms"2 E. 3.40ms'2

Fig 6.4

14. A ball of mass 0.1kg approaching a tennis player with a velocity of 10ms is hit back in the opposite direction with a velocity of 15ms'1. If the time of impact between the racket and the ball is 0.01s, calculate the magnitude of the force with which the ball is hit.

WAEC 1997 Ans: 250N15. A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2ms’1 and rebounds with the same speed. Determine the impulse experienced by the ball.

WAEC 1998 Ans: 20kgms~'16. A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2ms'1.Determine the magnitude of the resulting impulse. WAEC 2003 Ans: 20kgms‘17. A force acting on a body causes a change in the momentum of the body from ^kgm s '1 to 16kgms'' in 0.2s. Calculate the magnitude of the impulse.

WAEC 2006 Ans:4kgms‘18. What change in velocity would be produced on a body of mass 4kg, if a constantforce of 16N acts on it for 2s? WAEC 1992 Ans: 8.0ms'119. A force acts on a body for 0.5s changing its momentum from 16.0kgms'‘ to21 .Okgms' . Calculate the magnitude of the force. WAEC 2003 Ans: 10N20. A force acting on a body causes a change in the momentum of the body from 12kgms_1 to 1 Okgms'1 in 0.2s. Calculate the magnitude of the force.

WAEC 2004 Ans: 20N21. When taking a penalty kick, a footballer applies a force of 30.ON for a period of 0.05s. If the mass of the ball is 0.075kg, calculate the speed with which the ball moves off. A. 4.50ms'1 B. 11.25ms'1 C. 20.0ms'1 D. 45.00ms'1

JAMB 1988 Ans: 20.00 ms 122. A body of mass lOOg moving with a velocity of 10.0ms'1 collides with a wall. If after the collision, it moves with a velocity of 2.0ms' in the opposite direction, calculate the change in momentum.A.0.8NS B. 1.2Ns C. 12.0Ns D. 80.0Ns JAMB 1991 Ans: 1.2Ns23. A constant force of 5N acts for 5s on a mass of 5kg initially at rest. Calculate thefinal momentum of the mass. NECO 2003 Ans: 25kgms‘24. A body of mass Mi moving with a velocity, U collides with a stationary body of mass M2 and both move with a common velocity, V. If linear momentum is conserved, which expression correctly represents V?

M?U „ Mi U n MyU m2UB. --------— c. —-----— D. -rr— r r E.A.

Mi + M2MXU Mi - M2 Mi — M2 Mi +M 2 M i+ M 2

172

NECO 2002 Ans: D25. A ball of mass 0.5kg moving at lOm/s collides with another ball of equal mass at rest. If the two balls move off together after impact, calculate their common velocity.

WAEC 1988 Ans: 5 ms26. A body of mass 4.2kg moving with 10ms'1 due east, hits a stationary body of mass2 8kg If they stick together after collision and move with velocity v due east, calculate the value of v. WAEC 1996 Ans: 6ms27. A trolley of mass 4kg moving on a smooth horizontal platform with a speed of 1.0ms'1 collides perfectly with a stationary trolley of the same mass on the same platform. Calculate the total momentum of the two trolleys immediately after the collision.

WAEC 1999 Ans: 4. ONs or 4.0kgms28. A body of mass 5kg moving with a velocity of 10ms'1 collides with a stationarybody of mass 6kg. If the two bodies stick together and move in the same direction after the collision, calculate their common velocity. WAEC 2005 Ans: 4.55ms29. A ball of mass lOOg travelling with a velocity of 100ms’1 collides with anotherball of mass 400g moving at 50ms'1 in the same direction. If they stick together, what will be their common velocity? NECO 2000 Ans: 60ms30. An arrow of mass 0.1kg moving with a horizontal velocity of 15ms is shot into a wooden block of mass 0.4kg lying at rest on a smooth horizontal surface. Their common velocity after impact isA. 15.0 ms'1 B. 7.5ms'1 C. 3.8ms'1 D. 3.0ms'1 JAMB 1997 Ans: 3ms31. A lead bullet of mass 0.05kg is fired with a velocity of 200ms'1 into a lead blockof mass 0.95kg. Given that the lead block can move freely, the final K.E after impact is A. 50J B. 100J C. 150J D. 200J JAMB 1999 Ans: 50J32. A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20ms’1. If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the

(i) magnitude of the resistance(ii) distance moved by the bullet in the wood.

WAEC 2006 Ans: (i) 24N (ii) 1.0m33. A bullet fired at a wooden block of thickness 0.15m manages to penetrate the block. If the mass of the bullet is 0.0025kg and the average resisting force of the wood is7.5 x 103N, calculate the speed of the bullet just before it hits the wooden block.A. 450ms'1 B. 400ms'1 C. 300ms'1 D. 250ms'1 JAMB 1998 Ans:300ms 134. The driver in a motor car, of which the total mass is 800kg and which is travellingat 20ms'1, suddenly observes a stationary dog in his path 50m ahead. If the car brakes can exert a force of 2000N, what will most likely happen?

A. The car will be able to stop immediately the driver notices the dog.B. The car will stop 30m after hitting the dog.C. The car will stop 20m in front of the dog.D. The driver will quickly reverse the car.E. The car will stop 5m beyond the dog. JAMB 1981 Ans B

35. A machine gun with a mass of 5kg, fires a 50g bullet at a speed of 100ms'1. The recoil speed of the machine gun is A. 0.5m/sec B. 1,5m/sec C. lm/secD. 2m/sec E. 4m/sec JAMB ¡978 Ans: C36. A gun of mass 2.0kg fires a bullet of mass 1.6 x 10'2kg due East. If the bulletleaves the nozzle of the gun with a velocity of 150ms'1, what is the recoil velocity of the gun? A. 150ms'1 due West B. 1.2 x loAns'1 due West C. 1.2ms'1 due WestD. 1.2ms'1 due East E. 150ms'1 due East JAMB 1984 Ans- C37. A gun of mass 3kg fires a bullet of mass 20g with a velocity of 500ms'1Calculate the recoil velocity of the gun. NECO 2004 Ans: -3 33ms'138. A rocket bums 0.01kg of fuel each second and ejects it as a gas with a velocity of5,000m/s. What force does the gas exert on the rocket? A. 500.000N b . 500NC. 50N D. 5,000N E. 50,OOON JAMB 1982 Ans: 50N

39. A rocket bums fuel at the rate of lOkgs'1 and ejects it with a velocity of

173

5 X l03ms'1. The thrust exerted by the gas on the rocket is A. 2.5 x 10 N 13. 5.0 x ^ ^ C. 5 .0 x 102N D. 2 .0 x 103N JAMB 1993 A n s : 5.0 x IO N40. A girl whose mass is 55kg stands on a spring weighing machine insi c a i t.When the lift starts to ascend, its acceleration is 2ms'2. What will be the rca ing on tie machine? [Take g=10m/s2] WAEC 1990 Ans: 66kg41. A body of mass 2kg is suspended from the ceiling of a lift with a light inextensi i estring. If the lift moves upwards with acceleration of 2ms 2, calculate the magmtu c o the tension in the string. [g= 10ms'2] WAEC 1996 Ans. 242. A force of 21 .ON acts on a particle of mass 3.0kg in a medium whose resistance is2-ONkg'1. Calculate the magnitude of the acceleration of the particle. ,

NECO 2007 6 Ans: 5.00ms-43. A bullet of mass m is fired from a gun of mass M with a velocity v. The recoil velocity of the gun is expressed as

B _MV_ c D jnv_ E nw NEcO 2007 (Ans: E)M -m ' M+m ' M-m M . .

An object of mass 2kg moving with a velocity of 3m s'1 collides head-on wi another object of mass 1kg moving in the opposite direction with a velocity of 4ms the objects stick together after collision, calculate their common speed.

WAEC 2008 3 Ans:0.67ms45. A body of mass 4kg resting on a smooth horizontal plane is simultaneously acted upon by two perpendicular forces 6N and 8N. Calculate the acceleration of the motion.

, MvA. —

44.

A. 2.52m s' B. 3.0ms' C. 4.0ms" D. 4.5ms~'i JAMB 20087 Ans: 2.5ms~2

46. A body of mass 80kg moving with a velocity of 6m s-1 hits a stationary body ofmass 40kg. If the two bodies stick together after the collision, calculate the magnitude of the velocity with which they move. NECO 200815 Ans: 4m s-147. A body of mass 12 kg travelling at 4.2 ms'1 collides with a second body of mass 18kg at rest. Calculate their common velocity if the two bodies coalesce after collision.A. 1.5 ms'1 B. 1.4 m s1 C. 2.1 ms'1 D. 1.7 ms'1 JAMB 20097 Ans: D48. A body of mass 5kg with a velocity of 30ms'1 due East is suddenly hit by anotherbody and changes its velocity to 50ms'1 in the same direction. Calculate the magnitude of the impulse received. WAEC 200913 Ans: lOONs49. A ball of mass 0.1 kg moving with velocity of 20ms'1 is hit by a force which acts on it for 0.02s. If the ball moves off in the opposite direction with a velocity of 25ms'1,calculate the magnitude of the force. NECO 2009'° Ans: 25N50. A body of mass 4.0kg moving with a speed of 2.5ms'1 collides with a stationarybody o f mass 0.2kg. If the two bodies stick together after collision, calculate the magnitude o f the velocity with which they move. NECO 2009s3 Ans • 2 38msJ

174

7MACHINES

A machine is any device or tool by means of which a force or effort(E) applied at one point can be used to overcome a force or load(L) at some other point. The following terms and formulae are common to all types of machines.

, w , , Load Output forceA. Mechanical Advantage = Force Ratio = ——— = —---- -7-------5 Effort Input force

That is, M.A. = F.R = - E

n „ distance moved by effort eB. Velocity Ratio = —------------------- ---------= -

distance moved by load l

distance between effort and fulcrumdistance between load and fulcrum

1 1 distance moved by load _ lVelocity Ratio V. R distance moved by effort e

„ „„ work outputC. Efficiency = --------------- x 100%work input

The relationship between mechanical advantage (M A ), velocity ratio (VJl) and efficiency (£) is derived as follows.

work outputEfficiency = ------;---------x 100%

work input

Generally, work *= force (load or effort) X distance

Hence,

But

load x distance moved by loadefficiency = ——-------- ------------------------------- - x 100%

effort x distance moved by effortL x l

x 100E x eL l

= — x - E e

LM.A.= -

Eand

1V7r

Efficiency = M. A x —— x 100% V. R

M.AEfficiency = —— x 100%

V. K

mechanical advantageEfficiency = -------- :——----- ---------x 100%velocity ratio

£ = — x 100%V.R

175

Where £ = efficiency in percentage (%)M.A. - mechanical advantage V.R = velocity ratioL = load in Newton (N) or kilograms (kg)^ - effort in Newton or kilograms (kg) e ~ distance moved by effort in meter (m) l = distance moved by load in meter (m)F.R = force ratio

Example 1A machine has an efficiency of 80%. If the machine is required to overcome a load of 60N with a force of 40N, calculate its velocity ratio. NECO 2003nSolutionEfficiency £=80% ; Load, L = 60N; Effort, E = 40N Velocity ratio, V.R. = ?

m v, ■ i aj load L 60N 1 cMechanical Advantage, M. A = — — = - = : = 1.5effort E 40N

M A 1.5Velocity Ratio, V. R. = —— x 100% = — x 100 = 1.875

£ 80

Example 2The efficiency of a machine is 80%. Determine the work done by a person using this machine to raise a load of 200kg through a vertical distance of 3.0m (Take g = 10ms')

WAEC'1988“SolutionEfficiency, £ = 80%; Load, L = 200kg = (200 x 10)N = 2000N;Distance moved by load, / = 3m

Efficiency =work done by machine

---------------------------------- -------------x 100%Work done on machine or by person

L x /£ = ----- — ------------------ x 100%

Work done by person

L x / x 100 2000 x 3 x 100 600000Work done by person = ------------------ = ----------™---------= — —— = 7500 J

Example 3The velocity ratio of a machine is 5 and its efficiency is 75%. needed to lift a load of 150N with the machine?A. 50N B. 40N C. 30N D. 20NSolutionVelocity ratio V.R =5; efficiency, £ = 75%; Load. L =

£ =M.A------x 100%V.R

M.A.

M.A.

£ x V.R 100

75 x 5 100

What effort would be

JAMB 200010

15ON; effort. E = ?

176

= 3.75

A l s o , M . A = — E

r 1 - 1 ^ - = 40/VE f f o r t , E - — A - 375

Example 4A load of mass 120kg is whose efficiency is 100%.

aised vertically " '" “ÌLbth e^ ach teby a machine

WAEC 200216[g = 10ms'2]Solution . * = 30s; efficiency, r. - 100%.Mass, m = 120kg; height (distance) s - 2m, im

Force x distance F ~x s _ m g x s P0Wer = ------- T i ^ ------- = ~ t

Emclency,£ = P°Wer° UtPUft ^ x 100%Power input (P. 1)P.0

£ = ------ x 100%P.I

If £ = 100% then the above equation becomesP.0

100 = — X 100

100 _ P.0 Too ~~pT

P.01 “ ~pT

Rearranging, P.I = P.0

Simply put, if a machine has 100% efficiency then the power input output.

m g x s 120 x 10 x 2Therefore, power output by machine = --------- = --------—-------=

is equal to power

80W

TYPES OF MACHINES

These include the levers, the inclined plane, the screw or screw jack, the wheel and axle, and gear wheels.

1. The Pulley or Block and Tackle System.

In a block and tackle or pulley system, the velocity ratio (V.R) is equal to the number ol pulleys in the system.That is V.R = number of ropes supporting the pulley

= number of pulleys in the system.

Example SThe diagram below represents a block and tackle o ,IS just able to lift a load of weight W. If he ^ ^ ,value of W emciency of the machine is 40%, find the

177

I VA EC 1991Fig 7.1

SolutionEffort, E = 50N; Load. L = W; Efficiency, £ = 40%

From diagram, velocity ratio. V'.R = number of pulleys = 6

10

M.A£ = v ^ x lo o %

LAlso, M.A = —

M.A. =£ x V.R

10040 x 6

100= 2.4

L = W = M.A x E = 2.4 x 50 = 120/V

Example 6The velocity ratio and efficiency of a system of pulleys are 6 and SO1’© respective!) lo'v much effort is required to lift a load of mass 120kg with this system [g=10ms *]

II.IEC 1996SolutionVelocity ratio = V.R = 6; :. L = mg = 120 x 10 =

M.A£ = _ x i o °o/o

Efficiency, £ = S0°o; Mass.m = ’21200N

M.A.£ x V.R

10080 x 6

100= 4.8

Also,L

M.A — — EL 1200

E ffo r t.E = — = - j j p = 250N

Example 7A block and tackle system is used to lift a load of 20N through a vertical height of 10m. If the efficiency of the system is 40%, how much work is done against friction .1

JAMB 19$9*SolutionLoad, L = 20N; Distance moved by load, / = 10m Efficiency, £ = 40%; Work output = L x / = 20x10 = 200N

work output load x distance moved by loadEfficiency = ------r r ----- - x 100% = ----------------- —----- :---- 1-------x 100%work input

Work input =

Work input =

work input load x distance moved bv load

efficiency 20 x 10 x 100 20000

x 100%

40 40= SOON

Work output = work input - frictional forces:. Work done against friction = Work input - work output

= 500 - 200 = 300N

178

Example 8

i l )

The figure 7.2 represents a frictionlcss pulley system in which a weight W is in equilibrium with

a weight of’40N. Find the value of W.

A. 13.3N B. 20.ON

D. 80.0N

C. 40.0NJAMB 199'/

Fig 7.2Solution || v R = M AVelocity ratio, V.R = number of pulleys = 2; for a friction ess puTherefore, M.A = 2. Effort, E = W; load , L = 40N.

LM A = 1

e = W = J L = ^ = 2 Q I VM.A 2

2. The Inclined PlaneThe inclined plane is used to lift heavy load into vans or trucks. Example arc, a

slope, a hill, a ramp e.t.c.Y

Velocity ratio of an inclined plane is given by:distance moved by effort length of inclined plane l distance moved by load height of inclined plane ~ h

opposite hFrom figure 7.3, sin0 = --------------= -hypotenuse l

1 lRearranging, — r = t° sin 9 h

l 1Therefore, V. R = - = ——:h sin0

For a Frictionless inclined plane, V. R = M. A = ——-sin 0

Furthermore, Work output = load x distance moved by load

= load x height of inclined plane

= L x h

Work input = Effort x distance moved by effort

179

Efficiency of inclined plane,

= effort x length of inclined plane = E x l

work outputE = ----------- — x 100%work input

L x h L h ^nnn,£ = ----- - x 100% = — x — X 100%

E x l E l

Example 9 _ :An inclined plane of angle 15° is used to raise a load of 4500N through a height of 2m. If the plane is 75% efficient, calculate ,(i) velocity ratio of the plane (ii) work done on the load WAEC 1999sSolutionAngle of inclination 0 = 15°; load, L = 4500N; height of plane, h = 2m;Efficiency, £ = 75%(i) Velocity ratio, V. R = — = —-— = 3.86J sin e sin 15 --------

(ii) Work done on the load is the work input

Efficiency =

work outputWork input = ——----------Efficiency

Lx h= — x l ° 0 =

work output------------— x 100%work input

load x height of planex 100% = --------- ----------- -— -efficiency

X 100%

4500 x 2 x 100 75~~

= 120007

Example 10A plane inclined at 30° to the horizontal has an efficiency of 50%. Calculate the force parallel to the plane required to push a load of 120N uniformly up the plane.

... . i:i>. . . . . .WAEC200717Solution ......... .Angle of inclination 0 = 30°; ••• V. R = —— =° cm f)s i n S s i n 3 0

= 2 ;load, L=120N; Efficiency;E = 50%

/Is C - • 'The force parallel to the plane required to push the load is the effort E.

' M. AFirst, we obtain M. A from £ = -r— - x 100%, ... V. KLC‘ 0 ; x :j ,'v. .i

• = E-X”V: R -50 x 2- 100' M A = ----- = ------- 4 = -----= 1

100 100 100

Also, M.A = — EE =

120M.A

= 120 N

.ninrrvjllii.

Example 11 — . = ' / ’/ i - ; ; ^ = »rjahtUd iv|VA man pulls up a box of mass 70kg using an inclined plane of effective length'5rrf6rtto aplatform 2.5m high at uniform, speed. . Jf.the friction^ forcq,between the box and the plane is 100N, draw the diagram of all the forces acting on tne box. when In motion and calculate the , x s-, .x

(i) minimum effort applied in pulling up the box.(ii) Mechanical advantage of the plane

180

(Hi)(iv)(V)(vi)(vii) given

Velocity ratio of the plane Efficiency of the plane Energy lost in the system Work output of the manTotal power developed by the man . nLatfonn is

that the time taken to raise the box on P 5<K<Ä VSolution

Mass of box, m =70kg; frictional force, F — 100N; g=10ms , Height of plane, h = 2.5m; length of plane, / = 5m.The load or weight of the block acts downwards and has two components.

a) mg sin 9, which is parallel to the planeb) mg cos 6, which is perpendicular to the plane.

h 2.5sin0 = j = — = 0.5

6 = sin-1 0.5 = 30°

(i) For the object to be pulled up the plane, it must overcome the opposing frictional force, F and the parallel component of the load, mg sin 8. Therefore, the minimum effort, E = mg sin 8 + F.

E = 70 x 10 x sin30 + 100 = 700 x 0.5 + 100 = 350 + 100= 450N

(ii) Mechanical Advantage or Force ratio = -------effort

M.A = * = ^ E E

70 x 10 700450 — 450

l 5Oii) Velocity ratio, V. R = - = —— = 2 n 2.5

Alternatively, V. R = 1sind

1 1sin 30 0.5 ^

(iv)M. A

Efficiency = tt- - x 100% =V. i\1.56 x 100 156

= 78%

(v) Energy lost in the system = work input - work output= Effort x distance moved by e ffo rt- lo a d x distance moved by load= ( E x Q — ( L x h ) J= (450 x 5) — (700 x 2.5)

= 1.56

181

= 2250- 1750 = 500J

(vi) Work output of the man = force (load) x height= mg x h= 70 x 10 x 2.5 = 1750J

, ... _ , . work output 1750(vu) Total power developed = ----------------= —— = 35W

time taken 50

3. The Screw

Examples of screw machines are bolts and nut, car screw jack, Engineer’s vice e.tc. All screws have thread. The distance between successive thread on a screw is called its pitch. For one complete revolution, the screw travels a distance that is equal to its pitch.For a screw, work done = work ouput

Effort x distance moved by effort = load x distance moved by load

Effort x circumference of handle = load X pitch E x 27tr = L x p

.. . . 2mHence, the velocity ratio, V. R = ----P

For a perfect or frictionless screw, M.A = V.R

2ttrHence, M.A = V.R = ----

PFor a screw jack, r is the length of the handle or Tommy bar of the screw jack.

Example 12A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000kg through a height of 20cm. The length of the Tommy bar of the jack is 70cm. If the efficiency of the jack is 80%, calculate

(i) the velocity ratio of the jack(ii) mechanical advantage of the jack(iii) effort required in raising the body(iv) work done by the effort in raising the body [=10ms'2; 7t = 22/7] WAEC 1993El

SolutionPitch, p = 4.4mm = 0.0044m; mass, m=8000kg; g= 10m/s2

.-. Load, L= mg = 8000 x 10 = 80000N; distance load moved, l = 20cm = 0.2m; length o f Tommy bar, r = 70cm = 0.70m; efficiency, £ = 80%

2m- 2 x 22/ ? x 0.70 4.4(i) The velodty ratio, V.R = — = ------= iooo

M.A(ii) From £ = — x 100% we obtain

V. K£ x V. R 8 0 x 1 0 0 0

Mechanical advantage, M. A = = ----- — -----= 800

182

OH) M A = loadeffort

LÊ

Effort =80000 U

= ÌOON

(iv)

M. A 800

Work done by effort = work input

Efficiency = w o rk o u tp u i X 1 0 0 %work input

... . . work outputWork input = ------------ — x 100%

efficiencyload x distance moved by load

efficiencyload x distance moved by load

efficiency

:l r

x 100%

x 100%

L x l x 100 80000 x 0.2 x 100£ 80

1600000= ----—----= 20000/ = 20A780

Example 13Two spanners X and Y of lengths 15cm and 20cm respectively are used in turn to* give ascrew of pitch 2mm one complete rotation. If R* and Ryare the respective velocity ratiosof the spanners, what is the ratio Rx: Ry? WAEC19967Solution

X Y „ ;Length of spanner, r = 15cm : 20cmPitch, p = 2mm 2mmConvert mm to cm (x 10) 20cm : 20cm ’ /'Velocity ratio, V.R = Rx Rv

Substitute into V. R 2 nr 2 n r ; rP P

2n x 15 2n x 2020 20

Divide both sides by 2n 2n x 3 4 2n

Multiply both sides by 4 34 1

3 x 4T " : 1

3 : 4Therefore, ratio Rx : Ry is 3 : 4

183

Example 14A screw jack with a Tommy bar of length 12cm is used to raise a car jhrough a \ertical height of 25cm by turning the Tommy bar through 50 revolutions. Calculate the approximate velocity ratio of the jack, [a = 3.14] WAEC 2001SolutionLength of Tommy bar, r = 12cm; number of revolution, n = 50:Height raised or distance load moved, h = 25cm;

velocity ratio,

, " h 25cmTherefore, pitch, p — — = = 0.5cm

V.R =

n 502trr 2 x 3.14 x 12cm 75.36cm

p 0.5cm 0.5cm= 150.72 = 151

4. The wheel and Axle

The wheel and axle is a simple machine used to raise water or sand from very deep hole or pit. For a wheel of radius R and axle of radius, r; it follows that:

Work input (work done by effort) = work output (work done by load)Effort x distance moved by effort = load x distance moved by load Effort x circumference of wheel = load x circumference of axle

E x 2nR = L x 2tit

, distance moved by effortVelocity ratio = —------------------—— :— —

distance moved by load

V.R =2ttR2nr

Rr

For a frictionless wheel and axle. M. A = V. R = —r

work outputEfficiency of wheel and axle = ------—----- — x 100%work input

L x 2ttrE x 2ttR

x 100%

L r£ = — x — x 100

E R

Example 15A wheel and axle is used to raise a load of 500N by the application of an effort of 250N. If the radii of the wheel and the axle are 0.4cm and 0.1cm respectively, the efficiency of the machine is? A. 60% B. 50% C. 40% D. 20% JASiB 20026SolutionLoad, L = 500N; E = 250N; Radius of wheel, R = 0.4cm; Radius of axle, r = 0.1cm

Efficiency,L r

E = - x - x 100%E R

500 0.1 5000£ = 25iX M X100% = l o r =50%

184

Example 16 . . anj axle are 40cm and HemIn a wheel and axle mechanism, the diameter ot t k " .-«■ rt ¡s required to lift « toadrespectively. Given that the machine is 80° o efficient, ^ JAMB I MBof 100N? A. 20N Solution

B. 25N C. 50N

Radius = radiuS of wheel, R = 7 = 20cm

Radius of axle, r = efficiency, £ = 80%2

Load. L ~ 100N,

Effort, E - ?

L T£ = t: X - x 100%t A

Effort, E =L x r x 100

£ x f i

E =100 x 4 x 100 _ 40000 _

80 x 20 1600

5. Gear WheelsGear wheels are used in cars and bikes. In gear wheels, the effort an

applied to the shafts o f the gears.Therefore:

number of teeth in driven wheel Velocity ratio = number 0f tecth in driving w heel

Example 17A 20-toothed gear wheel drives a 60 toothed one. If the angular speed ot the smaller wheel is 120 rev s’1, the angular speed of the larger wheel is?A. 3 rev s’1 B. 40 rev s' C. 360 rev s 1 D. 2400 rev s 1 JAM B / VcStfSolutionNumber of teeth of driving wheel speed of driving wheelNumber of teeth of driven wheel speed of driven wheel

20 ________ 120_______60 speed of driven wheel

„ r j . , L , 6 0 x 120 7200Speed of driven or larger wheel = ----- —------= — = 360 re v s -1

EXERCISES 7.

1. A machine has an efficiency of 60%. If the machine is required to overcome a load o f 30N with a force o f 20N, calculate its mechanical advantage WAEC ¡995')Ans 152. A simple machine with an efficiency o f 75% lifts a load o f 5000N when force of 500N is applied to it. Calculate the velocity ratio o f the machine.

WAEC 2001 13 An.s: ¡3.33

185

3. A machine has a velocity ratio of 6 and an efficiency of 75%. Calculate the effortneeded to raise a load of 90N. NECO 2002 Arts: 20N4. A machine whose efficiency is 60% has a velocity ratio of 5. If a force of 500N is applied to lift a load P, what is the magnitude of P?A. 750N B. 4166N C. 500N D. 1500N JAMB 2004" Arts: I SOON5. A machine has a velocity ratio of 4. If it requires 800N to overcome a load of 1600N, what is the efficiency of the machine?A. 2% B. 40% C. 60% D. 50% JAMB 200729 Ans:50%6. A block and tackle system has 6 pulleys. If the efficiency of the machine is 60%,determine its mechanical advantage. WAEC 200217 Arts: 3.67. A block and tackle system has six pulleys. A force of 50N applied to it lifts a load of weight W. If the efficiency o f the system is 40%, calculate W.

WAEC20062' Ans:120N8. A block system of five pulleys has an efficiency of 70%. Calculate the effort thatwill be required to raise a load of42.00N. NECO 200015 Ans: 12N

9. The figure 7.5 represent a block and tackle pulley system on which an effort of W newtons supports a load of 120N. IF the efficiency of the machine is 40%, then the value of W isA. 28.0N B. 48.ON C. 50.0N D. 288.0N JAMB 198612 Ans: 50.ON

10. The diagram in Fig 7.6 is a block-and-tackle pulley system in which an effort of 80N is used to lift a load of 240N. The efficiency o f the machine is?A. 60% B. 50% C. 40%D. 33% JAMB 20014 Ans: 50%

11. A plane inclined at an angle of 30° to the horizontal has an efficiency of 50%. The force parallel to the plane required to push a load of 120N uniformly up the plane is?

WAEC 1990s Ans: 120N12. A body of mass 7.5kg is to be pulled up along a plane which is inclined at 30° tothe horizontal. If the efficiency of the plane is 75%, what is the minimum force required to pull the body up the plane? [g=10ms‘2]. WAEC 19954 Ans: 50N13. What is the velocity ratio of a plane which is inclined at an angle of 30° to thehorizontal? NECO 200411 12 13 14 15 Ans: 214. A plane inclined at an angle of 30° to the horizontal has an efficiency of 60%. The force parallel to the plane required to push a load of 120N uniformly up the plane isA. 60NB. 100N C. 120N D. 200N E. 240N JAMB 198420 Ans: 100N15. An inclined plane which makes an angle of 30° with the horizontal has a velocityratio of? A. 2 B. 1 C. 0.866 D. 0.50 JAMB 1990'2 Ans: 2

186

f ,A.i The inclination of the r lane inclined at angle 0 has a velocity ratio cos q = / |0

plane to the horizontal is given by A. tan 0 = 1 10 B. cot ^ 10lb. A pk

JAMB 1993 Arts: sin 0 = MOu . sin 0 - / 10 ' • o the horizontal. If the17. A 20kg mass is to be pulled up a slope inclined 3 - , ,sefficiency of the plane is 75%. the force required to pull the loa upA. 13.3N B.73.5N C. 133.3N D. 533.2N [£ j ^ 991 '° Ans: 133.3N

. a „ion? of effective length 5m18. A man pulls up a box of mass 70kg using an inclined p ,onto a platform 2.5m high at a uniform speed. If the frictional force e %the plane is 1000N; , . .. - mntinn

(i) draw a diagram to illustrate all the forces acting on the ox(see Example 11)

(ii) Calculate theI. minimum effort applied in pulling up the box.II. Velocity ratio of the plane if it is inclined at j»0 to the honzontaIII. Force ratio of the plane

WAEC 2007e" Arts: I. 1350N II. 2 III. 0.5-19. A screw jack, 25% efficient and having a screw pitch 0.4cm is used to raise a load through a certain height. If in the process, the handle turns through a circle o f radius 40cm, calculate

(i) velocity ratio o f the machine(ii) mechanical advantage of the machine(iii) effort required to raise a load of 1000N with the machine.

[Take n = 3.14] WAEC 2002r" Arts: (i) 628 (ii) I57(iii) 6.37N20. Calculate the velocity ratio of a screw jack of pitch 0.3cm if the length o f theTommy bar is 21cm. A. Vuort B. 14a C. 70a D. 140a

JAMB I994,: Arts: 740a21. A wheel and axle of radii SOOmrn and 200mm respectively is used to raise a body of weight SOON by the application of 250N. Calculate the efficiency o f the machine.

WAEC 2005' Arts: 80%22. A wheel and axle have radii 80cm and 10cm respectively. If the efficiency o f the machine is 0.85. an applied force of 1200N to the wheel will raise a load o fA. 8.0N B. 6.8N C. S160.0N D. 9600.0N JAMB 1991s Arts: 816 0 S23. In an ideal wheel and axle system, R stands for the radius o f the wheel and r is theradius of the axle. The mechanical advantage is A. R/r B. (r/R)’ C . r RD. (R/rf JAMB 2005:? Arts. R/r24. Which of the following levers has the greatest mechanical advantage.

, 'L ZX I h

Z5 i E<— 1 cm — ------ 3cm ------*■

' k A k

n ______________ ~ ~ * t Ei L

Fig 7.7

JAMB 1995 Arts: C

187

25. A particle of weight 120N is placed on a plane inclined at an angle 30 to t e horizontal. If the plane has an efficiency o f 60%, what is the force required to push the weight uniformly up the plane?A. 175N B. 50N C. 2 ION D. 100N JAM B 2007 Ans: WON26. Calculate the velocity ratio o f a plane inclined at an angle o f 60% to thehorizontal. NECO 2007* Ans: 1.15

27. Which o f the following equation for the efficiency o f a machine is correct?» r■ ■ velocity ra tioA. E ffic iency = — — ——— — — x 100%velocity ra tio

m echanical advan tage

. load distanceC. E ffic iency = effort distance x 100%

B. E ffic ien cy = x 100%' ’ output

_ „ mechani cal advantage nno/D. E ffic iency = ------- ;--------- :------x 100%' ' ' velocity ratio

WAEC 20 0 8 '7 Ans: D

28. The efficiency o f the pulley systems shown in Fig 7.8 is 80%. Find the effort E required to lift a load of 1200N.A. 275N B .325N C. 375ND. 573N JAM B200S14 Ans: 375N

29. If a heavy barrel is rolled up a plane inclined at 30° to the horizontal, its velocity ratio will beA. 3.0 B. 3.1 C. 3.2 D. 2.0 JAMB 200912 Ans: D30. A machine o f efficiency 80% is used to lift a box. If the effort applied by the machine is twice the weight o f the box, calculate the velocity ratio o f the machine.

WAEC 200916 Ans: 0.6331. An effort P applied at one end o f a crowbar just overcomes the resistance W at the lid o f the tin. The mechanical advantage o f the crowbar is expressed asA .W + P B . P - W C. W / p D. P/ w E. WP

NECO 200914 Ans:C

188

8TEMPERATURE AND ITS MEASUREMENT

T E M P E R A T U R E SC A L E S f .Temperature is the degree of hotness or coldness of a body an is a rr'f^SU ,

average kinetic energy of the molecules in a body. Temperature is measure y types of thermometers in any of these scales.

1. Absolute or Kelvin scale.2. Celsius or centigrade scale.3. Fahrenheit scale.No matter the scale used, every thermometer has an upper fixed point (steam point), a

lower fixed point (ice point) and a fundamental interval as shown in figure 8.1.Kelvin Celsius Fahrenheit

Fig 8.1

Fundamental interval = steam point — ice point.E.g fundamental interval for Kelvin scale is 373 - 273 = 100K

Conversion Between Temperature Scales:(i) To convert from Fahrenheit (°F) to Celsius (°C), use this formula.

5(°F — 32) 9

Example 1What is the equivalent o f a temperature of 120°F in degree Celsius?

°F = 120°F5 ( 1 2 0 - 3 2 ) 5 x 88

°C = -----------------= - 9 = 48.89°C

(ii) To convert from degree centigrade (°C) to degree Fahrenheit (°F), use this formula.

°F = ^ + 32 o r °F = 1.8°C + 32

Exam ple 2The freezing point o f alcohol is — 115°C. What is this temperature in degree Fahrenheit?

°C = -1 1 5 °C °F = 1.8 x ( - 1 1 5 ) + 32

189

= -2 0 7 + 32 = —175°F

(iii) To convert Celsius (°C) to Kelvin and Kelvin (K) to Celsius, use these formulas,

K = °C + 273 and °C = K - 273

Example 3The temperature of a body increases from 30°C to 70°C, what is the temperature change of the body in Kelvin?

Temperature change = 70°C — 30°C = 40°C

°C = 40°C temperature in Kelvin, K = 40 + 273 = 313K

T H E R M O M E T R IC C A LC U L A T IO N S

Liquid -in-Glass ThermometersEven if a liquid-in-glass thermometer is ungraduated, faulty or it’s graduate' in

millimeter instead of a particular scale (°C,°F, K), it can still be used to me *re temperature if its reading is compared (“corresponded”) with the reading of a sca.'ed functioning thermometer. The following examples will illustrate how.

Example 4A liquid-in-glass thermometer is graduated in millimeters. The upper and low sed points of the thermometer are 230mm and 80mm respectively. What is the temperature in degree centigrade when the thermometer reads 95mm.SolutionBecause the answer is required in the Celsius or Centigrade scale, the thermometer is “corresponded” or compared with the Celsius scale as follows.Let T be the recorded temperature.

A 230mm 100°C Gi

c 95 mm

i

T E

1T 80 mm o 0 n fB D F H

Fig 8.2

Take proportions as follows from fig 8.2CD _ EF AB — GH

C - D E - F A - B ~ G - H

Substituting letters (or figures we have,95mm - 80mm T - °C230mm - 80mm 100°C — 0°C

190

15 mm _ 1150mm 100"C

Cross multiplying, we have

150mm x T = 15mm X 10015mm x 100°C

T = 150mm= 10°C

Example 5 . f50m mA temperature scale has an upper fixed point of 260mm and a lower fixe potn o What will be the reading on this scale when a thermometer reads 125 F.Solution h tIn this case, the temperature scale is compared with or ‘corresponded to the a ren scale.

260m w O ._______________ 2.12°F

125°F

50 mm 32°F

Fig 8.3Taking proportions in similar manner with example 4, we have

x — 50mm 125°F - 32°F260mm — 50mm _ 212°F - .3 2 ° £ ,

x - 50mm 93°F210mm = 180°F

180(x — 5 0 ) = 210 x 93

1 8 0 x - 9000 = 19530 ■9

1 8 0 x = 19530 + 9000

180x = 28530o •• 28530

x = = 158.5mmloUs

Example 6The ice point and steam point o f a mercury-in-glass thermometer are marked X and 180mm respectively. During an experiment, the mercury meniscus in the thermometer reads 45mm. If the corresponding reading on a Kelvin scale is 283K, calculate the value of X. SolutionThe thermometer scale is corresponded with the Kelvin scale.

Fig 8.4

191

Taking proportions, 45mm - * 283K - 273K180m m - X 373K - 273

4 5 - * 10180 - * “ Too

Cross multiplying, 100(45 - * ) = 10(180 - X)4500 - 100* = 1800 - 10*

4500 - 1800 = 100* - 10*

2700 = 90*

2700 A — -------------90

* = 30 mm

Example 7The ice and steam points of a certain thermometer are - 20° and 100° respectively. Calculate the Celsius temperature corresponding to 70° on the thermometer. WAEC 2002 SolutionThe thermometer scale is corresponded with the Celsius scale.

Fig 8.5

Taking proportions;70° - (-2 0 ° ) t - 0°C

1 0 0 ° - (-2 0 ° ) ~ 100°C - 0°C

70 + 20 t100 + 20 ~ Too

90 _ tTIo" ~ Too

120t = 90 x 100

t9000120

t = 75°C

Example 8 o nThe ice and steam points o f a faulty mercury-in-glass thermometer are 0.3 C and 99.2 Crespectively. When used to measure the temperature of a medium, it recorded 47.5°C. What is the correct temperature of the medium?SolutionThe fixed point and recorded reading of the faulty or inaccurately calibrated thermometer is corresponded with a correct or accurately calibrated Celsius thermometer as follows.Let 0 be the correct temperature when the faulty thermometer reads 47.5°C.

192

99.2°C n i p o ° c

45 5°C

Fig 8.6

Taking proportions;

().3°C J 0°C

47.5°C - 0.3°C 0 - 0°C^99.2 - 0.3°C

47.2

100°C - 0°C 6

1ÔÔ98.90 x 98.9 = 47.2 x 100

47209 = 98.9

= 47.7°C

Fxamplc 9The fundamental interval of a Celsius thermometer is 250mm. What is the temperature when the mercury level is 45mm above the ice point.SolutionRemember that the fundamental interval of a thermometer is the difference between the steam point and the ice point. The thermometer is corresponded with the Celsius scale.

Steam point

250mm

Ice point

45 mm

r :

100°C

0

o°c

Fig 8.7

'faking proportions;45mm 0 - 0°C

250mm 100°C -0°C45 _ 0

250 ~ 100°C 45mm x 100°C

250m m18°C

Resistance Thermometer

There is a proportionally direct relation between the resistance of metallic conductor (e.g platinum) and its temperature. Because there is a corresponding relationship between resistance and temperature, we can apply the same procedure used in liquid-in-glass thermometer. In addition we could use the following equation to find temperature 0 when the resistance is /?.

193

x 100°C9 = R - ReR100 — R o

Where R100 and R0 are the resistance at the upper fixed point (100°C) and the lower fixed point (0°C) respectively.

Example 10A resistance thermometer has a resistance of 10i2 at 0°C and 700 at 100°C. If it records 55i2 in a certain medium, calculate the temperature of the medium.SolutionThe resistance thermometer can be corresponded with the Celsius scale as follows.

7 0 f i n ------------------- n .io o °c

55n

Fig 8.8 io n

Taking proportions;

Cross multiplying.

6

U 0°C

5 5 n - 1 0 f i 0- O°C7o n - i o n i o o ° c - o ° c

4 5 _ B60 — Too

0 x 60 = 45 x 100

4500^ ~ ~60~ = 7 5 ° C

Alternatively, we could use the equation for resistance thermometer.

B =■ R - R o Ruin ~ R<

x 100°C

R — 550, Ro — 10n, Rioo — 70Ì1

55 - 100 = 7 F ^ X1OO°C

45= — x 100 = 75°C

60

Example 11A platinum resistance thermometer records a resistance of 4 il at 32°F and 10Q at 212°F. If resistance changes uniformly with temperature, what is the resistance of the thermometer when the temperature is 75 F.SolutionThe resistance thermometer is corresponded with the Fahrenheit scale as follows

ion n ________ n.2.i2°F

Fig 8.9 4 n

75°F

L i32°F

194

Taking proportions;

Cross multiplying,

R -A H 75°F - 32°F 10/2 - 4/2 “ 2 12°F — 32°F

R - 4 436 ~ 180

180(R - 4) = 6 X 43

180R - 720 = 258180R = 258 + 720 = 978

„ = ^ = 5.431! 180

example i l3 5D*°Wer *1XCC* P°'nl anc uPPer fixed point of a resistance thermometer are 0.5Q and thi* m»4S')CCtlV<!^ ' records 2.512 in a certain environment, what is the temperature of UK medium in Kelvin?SolutiontemneratiiStanCe t*lermome,er >s corresponded with the Kelvin scale because the temperature is required in Kelvin.

Fig 8.10

Taking proportions;

Cross multiplying,

2.5n - 0.512 6 - 273K3.5/2 - 0.512 " 373K - 273K

2 0 - 2 7 33 ~ 100

3(0 - 273) = 2 x 100

30 - 819 = 200

30 = 200 + 819 = 1019 1019

0 = - y - = 339.67 K

Example 13What is the temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20°C and 80°C respectively. A. 3.0 X 102 Km-1 B. 3.0 x 103/fm _1 C. 5.0 x 103Km_1 D. 3.0 x 104Km_1 JAMB 199926 Solution

Difference in temperatureTemperature gradient = — ——------------ -------------

Difference in length

8 0 - 2 00.02

60002 = 3 x 103oCm" or 3 x 103/fm _1

195

Gas Thermometers

There is a linear or directly proportional relation between the temperature of a constant volume gas thermometer and its pressure. As temperature increases, the pressure of the gas increases. Also, as temperature decreases, the pressure of the gas decreases.

Calculations involving gas thermometers arc done in similar ways like those of liquid-in-glass and resistance thermometers.

Example 14A constant volume gas thermometer records a pressure of 320mmHg at the ice point and 400mmHg at the steam point. What would the temperature be in degree Celsius when the gas thermometer records a pressure of 360mmHg?SolutionThe fixed points pressures (320mmHg and 400mmHg) and the recorded pressure (360mmHg) on the gas thermometer is “corresponded” with the Celsius scale as shown below.Let 9 be the recorded temperature.

100°C

e

o°c

360mmHg — 320mmHg 6 — °C Taking proportions; 40ommHg - 320mmHg = 100°C - 0°C

4 0 _ e80 “ Too

Cross multiplying, 80 x 0 = 40 x 100 = 400400

Therefore, 9 — - 50°C

Fig 8.11

400mmHg

360mmHg

320mmHg

Example 15A constant volume gas thermometer is used to measure the room temperature during an experiment. If the gas pressure at 273K is 220mmHg and 290mmHg at 373K, what is the room temperature if the thermometer records a pressure of 234mmHg?SolutionThe lower and upper fixed points pressures (220 and 290mmHg) and the recorded pressure (234mmHg) of the gas thermometer is “corresponded” with the Kelvin scale.

Fig 8.12

290mmHg -

234mmHg - •

220mmHg

373K

0

273K

196

_ . . 234m mHg - 220mmHgTak.ng proportions; 290mmHg _ 220m^ iTg

1470

6 - 273K 373K — 273K

0 - 273 100

Cross multiplying, 70(0 — 273) = 14 x 100

700 - 19110 = 1400

700 = 1400 + 19110 = 20510

0 =20510

70= 293K

EXERCISES 8.1. A thermometer has its stem marked in millimeter instead of degree Celsius. Thelower fixed point is 30mm and the upper fixed point is 180mm. Calculate the temperature in degree Celsius when the thermometer reads 45mm. WAEC1989 Ans: 10 C2. A mercury-in-glass thermometer reads -20°C at the ice point and 100 at the steam point. Calculate the Celsius temperature corresponding to 70° on the thermometer.

WAEC 1995 Ans: 75°C3. A platinum-resistance thermometer has a resistance of 50 at 0°C and 9 0 at 100 C.Assuming that resistance changes uniformly with temperature, calculate the resistance of the thermometer when the temperature is 45°C. WAEC 1996 Ans: 6.8Q4. An object is heated from 30°C to 57°C. What is the increase in its temperature onthe Kelvin scale? WAEC 1998 Ans: 300K5. The lower and upper fixed points of a mercury-in-glass thermometer are markedY and 180mm respectively. On a particular day the mercury meniscus in the thermometer rises to 45mm. If the corresponding reading on a Celsius scale is 10°C, calculate the value of Y. WAEC 1998 Ans: 30mm6. The ice point of an ungraduated mercury-in-glass thermometer is X, while itssteam point is 90°. This thermometer reads 60° when the true temperature is 40°C. Calculate the value of X. WAEC 1999 Ans: 40°7. A resistance thermometer has a resistance of 200 at 0°C and 85Q at 100°C. If its resistance is 520 in a medium, calculate the corresponding temperature.

WAEC 2001 Ans: 49.2°C8. Which of the following value on the absolute scale of temperature is the ice point?A. OK B. 32K C. 100K D. 273K WAEC 2004 Ans: 273K9. A platinum resistance thermometer has a resistance of 4 0 at 0°C and 100 at 100°C. Assuming the resistance changes uniformly with temperature, calculate the resistance o f the thermometer when the temperature is 45°C. WAEC 2003 Ans: 6. 7010. The lower and upper fixed points of a thermometer are 30mm and 180mmrespectively. Calculate the temperature in degree Celsius when the thermometer reads 45mm. WAEC 2005 Ans: 10°C11. A platinum-resistance thermometer has a resistance of 10O at 0°C and 180 at 100°C. Assuming that resistance changes uniformly with temperature, what is the resistance o f the thermometer when the temperature is 45°C. NECO 2002 Ans: 13.6°C12. What is the equivalent o f a temperature of 20°C in degree Fahrenheit? A. 36°FB. 68°F C. 11.1°F D.43.1°F E.25°F JAMB 1978 Ans: 68°F13. The ice and steam points of a mercury-in-glass thermometer of centigrade scalerind o f uniform bore correspond respectively to 3cm and 19cm lengths of the mercury thread. When the length is 12cm, what wjjl the temperature be? A. 32(,C B. 48(,CC. 56°C D. 65°C E. 75°C JAMB 1981 Ans: 56°C

197

14. The distance between the fixed points of a centigrade thermometer is 20cm. What is the temperature when the mercury level is 45cm above the lower mark?A. 22.5°C B. 29.0°C C. 90.0°C D. 100.0°C JAMB 1987 Arts: 22.5 C15. The resistance of a platinum wire at the ice and steam points are 0.7512 and 1.0512 respectively. Determine the temperature at which the resistance of the wire is 0.9012.

JAMB 1987 Ans: 22.5°C16. The resistance of a platinum wire at the ice and steam points are 0.7512 and 1.0512 respectively. Determine the temperature at which the resistance of the wire is 0.912A. 43.0°C B. 50.0°C C. 69.9°C D. 87.0°C JAMB 1990 Ans: 5(fC17. A thermometer with an arbitrary scale, S, of equal division registers -30°S at the ice point and +90°S at the steam point. Calculate the Celsius temperature corresponding to 60°S. A. 25.0°C B. 50.0°C C. 66.7°C D. 75.0°C JAMB 1991 Ans: 75°C18. A platinum resistance thermometer wire has a resistance of 512 at 0°C and 5.512 at 100 C. Calculate the temperature of the wire when the resistance is 5.212.A. 80 C B. 60°C C. 40°C D. 10°C JAMB 1992 Ans: 4 (fC19. The melting point of naphthalene is 78°C. What is this temperature in Kelvin?A. 100K B. 351K C. 378K D. 444K JAMB 1994 Ans: 351K20. A temperature scale has a lower fixed point of 40mm and an upper fixed point of 200mm. What is the reading on this scale when a thermometer reads 60 C?A. 33.3mm B. 36.0mm C. 96.0mm D. 136.0mm JAMB 1995 Ans: 136mm21. A platinum resistance thermometer records 3.012 at 0°C and 8.012 at lOC^C. If it records 6.012 in a certain environment, what is the temperature of the medium?A. 80°C B. 60°C C. 50°C D. 30°C JAMB 1998 Ans: 6(fC22. A platinum-resistance thermometer has a resistance of 412 at 0°C and 1212 at 100 C. Assuming that the resistance changes uniformly with temperature, calculate the resistance o f the thermometer when the temperature is 45°C. NECO 2003 Ans:7.6Q23. The electrical resistances of the element in a platinum resistance the: inometer at100°C, 0°C and room temperature are 75.000, 63.000 and 64.99212 respectively. Use these data to determine the room temperature. WAEC 2004 Ans: 16.6°C24. A temperature scale has a lower fixed point of 40mm and an upper fixed point of 200mm. What is the reading on this scale when a thermometer reads 60°C.

JAMB 1995 Ans: 136.0mm25. A platinum-resistance thermometer has a resistance of 412 at 0°C and 1212 at 100°C. Assuming that the resistance changes uniformly with temperature, calculate the resistance o f the thermometer when the temperature is 45°C. NECO 2003 Ans: 7.61226. The ice and steam points on a mercury-in-glass thermometer are found to be90.0mm apart. What temperature is recorded in degree Celsius when the length of the mercury thread is 33.6mm above the ice point mark. WAEC 1994 Ans: 37.33°C27. The height o f the mercury thread in a mercury-in-glass thermometer when inmelting ice and then in steam are 3cm and 18cm respectively. What would be the height o f the thread at a temperature of 60°C? A. 7.5cm B. 9cm C. 10.8cmD. 12cm E. 12.6cm M M B 1979 Ans: 12cm28 The lower and upper fixed points marked on a mercury-in-glass thermometer are 2 10mm apart. The end of the mercury column in the tube is 49mm above the lower fixed noint in a room What is the temperature of the room in degree Celsius? A. 55.3°CB 23 3°C C. 49.0°C D. 16.1°C E. 76.7°C JAMB 1984 Ans: 23.3°C29 The pressure on the gas of constant gas thermometer at the ice point is 325mm ofmercurv and at the steam point 875mm of mercury. Find the temperature when the pressure o f the gas is 490mm of mercury. A. 30K B. 243KP 303K JAMB 1989 Ans:303K30 The length o f mercury thread when it is at 0°C, 100°C and at an unknown temoerature 0 is 25mm, 225mm and 175mm respectively. What is the value of 0?A 8?.0»C B.80.0°C C.75.0°C D. 70.0°C JAMB 1997 Ans: 75°C.

198

M TU. 4 , the steam and ice points arc SOOmniHg and, '• 1 ,le readings on the pressure scale at the steam an > •300mmHg respectively. De.emiine the corresponding temperature m degree ( cls.us" hen it reads 450mmHc. NF.CO 2007" (Ans: 3(f)32. Hie ice and steam points on a mcrcury-in-glnss thermometer are cm an . cm respectively. Calculate the temperature in degree Celsius, when the nicicury meniscus isat the 14cm mark. WAEC 2007 (Arts: 2(fC)33. The lower and upper fixed points of a thermometer arc X0 and X100 respectively.A temperature t on the Celsius scale corresponding to Xt on the scale is given by A. l 0°(-vioo— -Y0) g Ay —.Vp .V(-.v0 p ioo(x0->rf)

' lOO(Xioo-Xo) ■ Af,oo-^n ' (*J00-*< )- 1/1/)

(Xioo_Ah)

ÑECO 200817 Ans. EIOO(Xjoo-X o) XI0.)-X0

E 100(Xt-Xo)

34. ^'^A^faulty thermometer registers 102.5’Ca, 100‘C. error, what will it register at 55°C?

199

9MEASUREMENT OF HEAT ENERGY

SPECIFIC HEAT CAPACITY AND HEAT CAPACITY

When heat is added to or removed from a body, the temperature change experienced depends on the mass of the body and its specific heat capacity as illustrated by this equation:

Q = m c(e2 - Bt) or Q = mc\0

Where Q = quantity of heat energy in Joule( J) m = mass of substance in kilograms (kg)02 = final temperature in degree Celsius(°C) or Kelvin (K)0i = initial temperature in degree Celsius(°C) or Kelvin (K) c = specific heat capacity in )kg~1K~1 or//cg~10C-1The heat capacity (thermal capacity), C, of a substance is defined as the heat

required to raise the temperature of the substance by 1°C or IK. The S.I. unit is/°C_1 or JK'\

The specific heat capacity, c of a substance is defined as the heat required to raise the temperature of a unit mass of the substance by IK or 1°C. The S.I. unit is Jkg~1K~1 or Jkg-^C-'

Heat capacity (C) is related to specific heat capacity (c) as follows:

Heat capacity = mass x specific heat capacity C = me

Quantity of heat (Q) is also related to heat capacity as follows:

Quantity of heat = heat capacity x temperature change

Q = C(62 - 0X) or Q = CA9

Example 1What is the quantity of heat that will be given out if a bar of brass of mass 350g is cooled from 95°C to 20°C. [specific heat capacity of brass = 380 Jkg~1K~1]SolutionMass, m = 350g = 0.35kg; c = 3S0Jkg~1K~1; Tempt change, A0 = 95 - 20 = 75°C

Heat energy given out = mass of substance x specific heat capacity x temperature change

Q = mcA9

Q = 0.35 x 380 x 75 = 9975 J

Example 2How much heat energy will be needed to change the temperature of 275g of paraffin oil by 75K [specific heat capacity of paraffin oil = 2130.1kg'1 K'1]SolutionMass, m = 275g = 0.275kg; temperature change, A0 = 75KSpecific heat capacity, c = 2130 Jkg~1K~l\

Substituting into Q = mcA8, we have

200

Q = 0.275 x 2130 x 75 = 43931.25 J

Example 3 0Calculate the final temperature of 2kg of alcohol at 25 C when 20160 Joules of energy is added to it. [specific heat capacity of alcohol = 2520J kg ' 1 K~1]SolutionQuantity of heat, Q = 20160J; mass of alcohol, m = 2kg; final tempt., 02 = ?initial temperature, 0i = 25°C; specific heat capacity, c = 2520 J k g ^ K '1;

Substituting into Q = mc(02 - &i), we have

20160— 2 X 252O(02 - 2 5 )

2O16O = 5O4O(02 - 2 5 )

20160 = 504002 ~ 126000

20160 + 126000 = STO4O02

146160 = 5O4O02 146160

02 = - - - - - - = 29°C

heat

Example 4500g of water is heated so that its temperature rises from 30°C to 72°C in 7 minutes. Calculate the heat supplied per minute. (Specific heat capacity = 4200 Jkg~1K~1')

WAEC1994SolutionMass of water, m = 500g = 0.5kg; initial temperature, 0i = 30°C;final temperature, 02 = 72°C; specific heat capacity, c = 4200 Jkg~1K~1;

Substituting into Q = mc(02 - #i)= 0.5 x 4200 x (72 - 30)

= 0.5 x 4200 x 42 = 88200 J88200J is the total heat supplied in the 7 minutes, therefore heat supplied per minute is

12600)

Example 5The temperature of a piece of metal of mass 9g is raised from 10°C to 110°C when it absorbs 108J of heat energy. Determine the specific heat capacity of the metal in Jk g -'IC 1 WAEC 1999SolutionMass of metal, m = 9g = 0.009kg; initial temperature, 61 = 10°C;final temperature, 02 = 110°C; heat absorbed, Q = 108J.

Q = mc(02 - 01) Specific heat capacity, c = Qm(02 - 0i)

1080.009(110- 10)

1080.009 x 100 120 J k g - 'K - 1

201

Method o f MixturesWhen a hotter substance and a colder substance (solid/solid, solid/liquid or

liquid/liquid) come in contact with each other, the hotter substance loses heat to the colder substance. The transfer of heat from the hotter to the colder substance will continue until the two substances attain the same temperature.

Therefore, we can state the following:

Heat given out by hotter substance = Heat gained by colder substance That is, Qh = Qc

Let 6h be the temperature of the hotter substance

0C be the temperature of the colder substance

9 be the final or equilibrium temperature of both substances or mixture.The temperature change of the hotter substance is:

A0h = 9h — 9

The temperature change of the colder substance is A9C = 9 - 9C

Heat given out by hotter substance, QH = mcA9H

Qh = mc{9„ - 0)Heat gained by colder substance, Qc = mcA9c

Qc = mc(0 - 9C)If heat given out = Heat gained

Then QH = Qc

mcA9H = mcAOc

mc(9H — 0) = mc(0 — 9C)

The masses and specific heat capacities of the hotter and colder substances could be different or the same.

Example 6Calculate the final or equilibrium temperature of the mixture of 250g of water at 65°C is

to 150g of colder water at 5°C. Neglect heat abMfcai by the surrounding, [specific heat capacity of water =42000Jkg~xK~x].SolutionMass o f hot water, m = 250g = 0.25kg, mass of cold water, m = 150g = 0.15kgTemperature o f hot water, 0h = 65°C. Temperature of cold water, 0c = 5°CSHC of water, c = 4200J k g ^ K '1, SHC of water, c = 4200]kg 1K 1

Let 8 be the final or equilibrium temperature.Heat given out by hot water = Heat gained by cold water

Qh = Qc

mc(0„ - 0 ) = mc{9 - 9C)

02.5 x 4200 x (65 - 0) = 0.15 x 4200 x (0 - 5)

Divide both sides by 4200 to obtain

202

0.25(65 - 0) = 0.15(0 - 5)

16.25 — 0.250 = 0.150 — 0/75

16.25 + 0.75 = 0.150 + 0.250

17.0 = 0.40 . -

:>y

Example 7 i . , Tf ,Hot water is added to four times its mass of water at 25°C and thoroughly stn"rc final temperature of the mixture is 40°C, calculate the initial temperature of t e ot \\ a er. Solution . .fjIUL>- -¡'j ;c,v'!Mass of hot water = m mass of cold water, = 4mTemperature of hot water, 0H Temperature o f col tf water, 9C— 25°CSHC of hot water = c SHC of cold water = cFinal temperature of mixture, 0 = 40°C.

tsL.i 1o yi:ic.’i- y « l

Heat given out by hot water = Heat gained by cold water - ;

Qh — Qc T* '•. .( 1J V ■ . . . '

mc(0H - 0) = mc(0 -—dOc)

m c(9H - 40) = 4mc(40 - 25)

Divide both sides by me to obtain

0W —4 0 —4 x 1 5 .

9H - 40 = 60

0*. = 60 + 40 = 100°C

H ea t:

Example 8 ' KA tap supplies water at 15°C while another supplies water at 90 ’C. ■ If a man wishes to bathe with water at 30°C, calculate the ratio of the mass of cold water to the mass o f hot water.Solution ,t -AatsifujiMass o f hot water = m HTemperature o f hot water, 9H — 90dC; - ' Teiiiperature o f col® Water, Qfc'fc 159C ■ " > SHC of water — c ¿ISHC of watfer = vc ' ' ’"¡-k w :Final temperature o f mixture of cold/hot water, 0 = 30°C. aoouuir

• ,>0 i!> «'¿8® - aOiS - m ?JCgiven 'out by hot Wafer = tteat gained by ¿oldwafe# *’ ie.*i o auiisnaroT:.!

0 '1' - - ' ley/ b +;< (ifw L - - i • ’ w. •Qh = Qc ' o 3 iu '

ra a \ w faadO oJ■ rnh c{ffH ir Q) p m cc(3; ,^c) < rrjvr.^teaH

Substituting and dividing both sides by’c we have

m „ ( 9 0 '- 30) = mc 0 (T -

6 0m V = I5mftc

Because we are asked to find ratio of cold water to hot water, we make m c the subject of equation.

1203

60m

Rearrange as follows

rnc _ 60 4mH 15 T

Ratio oi mass of cold water to mass of hot water is 4:1

On the other hand, if you were asked to calculate the ratio of mass of hot water to mass of cold water, you would proceed as follows:

From 60m H = 15mc, make m H subject of the equation

Rearrange;

m H =

m c

15 m c 60 15 _

= 6Ö“14

Therefore ratio of mass of hot water to mass of cold wateT is 1:4

Example 9When 100g of liquid Li at 78°C was mixed with Xg of liquid L2 at 50°C, the final temperature was 66°C. Given that the specific heat capacity of L2 is half that of L|, find X. A. 50g B. 100g C. 150g D. 200g JAMB 1994SolutionMass of Llt m H = 100g = 0.1kg mass of L2, m c = XTemperature of Lx, 8H = 78°C Temperature of L2, 8C = 50°CSHC of Lx = c SHC of l 2 = ¿cFinal temperature of mixture of Li and L2 is 0 = 66°C.

Heat given out by liquid, L| (QH) = Heat gained by liquid L2 (Qc)

Qh ~ Qcm Hc(.GH - 0) = m cc{6 - 0C)

0.1 x c x (78 — 66) = X x —c x (66 — 50)

10.1 x c x 12 = X x - c x 16

1.2c = X x 8c

Divide through by c1.2 = X x 8

1.2X = — = 0.15kg or 150g

8

Example 10When two objects P and Q are supplied with the same quantity of heat, the

temperature change in P is observed to be twice that in Q. If the masses of P and Q are the same, calculate the ratio o f the specific heat capacities of P to Q. WAEC 1994. Solution

P : Q

Same quantity of h e a t Q '• Q

204

Temperature change, A0 20 0

P and Q have same mass m m

QSpecific heat capacity, c = — —

mA9

Substituting into c for each Q2m6

Q_7710

multiply both sides by 2

2 \md)

12

1

: l ( ±

: 1

: 2

Ratio of specific heat capacities of P to Q is 1 : 2

Example 11Two metals A and B lose the same quantity of heat when their temperatures drop

from 20°C to 15°C. If the specific heat capacity of A is thrice that of B, calculate the ratioof the mass of A to that of B. NECO 2005Solution

A : B

Same quantity of heat lost, Q Q = Q

Same temperature change, A0 = 20 -- 15 = 5°C :. 5 : 5

Specific heat capacity, c 3c : C

Qv cA0

substituting, we have Q Q3c x 5 c x 5

Q Q15c 5c

1 (Q] •lS V c/ 5 \ c )

1 115 5

multiplying both sides by 15 15 Q : 15 ©i : 3

Ratio of mass of A to that of B is 1:3

Example 12heated to 100°C andA piece of copper of mass 0.1 kg is then transferred to a

lagged copper can of mass 0.05kg containing 0.2kg of water at 10°C. Calculate the final

205

steady temperature of the mixture. | Specific heat capacity of water : 4 .2 x l0 1yk.9 1 ’Specific heat capacity of copper ■= 4.0x 102 J kg 11( |. NECO 2005Solution

1 leul lost by copper I lent guined by calorimeter + water

l et the final steady temperature be 0

Heat lost by copper, Qc = me AQ = 0.1 x 4.0 x 102 x (1 0 0 - 0)

Heat gained by calorimeter, Qcaiorimeter = "icAO = 0.05 x 4.0 x 1O2(0 - 10)

Heat gained by water, Qwater = mcAO = 0.2 x 4.2 x 1O3(0 - 10)

" Qc Qcaiorimeter "b Qwater

0.1 x 4.0 x 102 x (100 - 0) = 0.05 X 4.0 x 1O2(0 - 10) + 0.2 X 4.2 x 1O3(0 - 10)

40(100 - 0) = 20(0 - 10) + 840(0 - 10)

4000 - 400 = 200 - 200 + 8400 - 8400

4000 + 8400 + 200 = 200 + 8400 + 400

12600 = 9000

12600e = ~ 9 0 0 - = 1VC

Example 13A metal o f mass 0.5kg is heated to 100°C and then transferred to a well-lagged

calorimeter of heat capacity 80)K~1 containing water of heat capacity 420JK'1 at 15°C. If the final steady temperature of the mixture is 25°C, find the specific heat capacity of the metal. A. 92 Jkg~ 1K~l B. 133 Jkg~ 1K ~1 C. 286Jkg~ 1K ~1D. 877 J k g - 'K - 1 JAMB 2000SolutionHeat energy gained or lost = heat capacity x temperature change

Q = CA0

Heat lost by metal = Heat gained by calorimeter + water

Heat lost by metal, Q = CAB — C x (100 - 25)

Heat gained by calorimeter, Qc = CAB = 80 x (25 - 15)

Heat gained by water, Qw = CA0 = 420 x (25 — 15)

Q — Qc + Qw

C x 75 = (80 x 10) + (420 x 10)

75C — 800 + 4200

75C = 5000

5000C =

75= 6 6 .6 7 //T 1

I leal capacity (C) = mass (m) x specific heat capacity (c)

Mass of metal, m = 0.5kgC

■■■ Specific heat capacity of metal, c - —66.67] K~l

0.5 kgI 3 3 3 3 jk g - 1K ~1

206

Electrical Method rh.'inter 7 of Rook I will heA quick revision of electrical energy am I™'“ b electrical method.

very helpful and relevant to calculation of specific * Substance or substances (liquid.When an electrical heater is used to heat P • substanCe(s) is equal to the

solid/liquid or solid), the heat energy (Q) gained by t electrical energy (H) supplied by the heater. That is

Heat gained by substance - heat supplied by electrical I

Q = H

mcA9 = IVt Or m c(02 ~ # i) _ ty*-'

mcA9 = Pt Or m c(02 — 9X) = Pt

mcA9 = l 2Rt Or m c(02 — 9X) = l 2Pt

V2tmcA9 = — -

AOr

V2tm c(9 2 - 9 X) - R

Where m = mass of substance in kilograms (kg)c = specific heat capacity in Jkg~ 1K ~1 or Jkg °C

A9 = change in temperature in K or °C 9Z= final temperature after heating in K. or °C 9X= initial temperature before heating in K. or °CI = current in Amperes (A)V = potential difference or voltage in volts (V) t = duration of current flow in seconds (s)P = IV = electrical power (power rating) in watt(W)II = electrical heat energy in Joules (J)R = resistance of heating coil in ohm (i2)

Any of the above equations can be used to solve problems but this will depend on the data given in the question.

Example 14How long will it take to heat 3kg of water from 28°C to 88°C in an electric heater taking a current of 6A from an e.m.f. source of220V?[specific heat capacity of water =4200 Jkg~ 1K~1]. WAEC 1990SolutionMass of water, m = 3kg; initial temperature, 0| = 28°CFinal temperature, 02 = 88°C; current, I = 6AVoltage, V = 220V; c = 4200 J k g - 'K ' 1

Heat energy supplied by heater = heat gained by water

IVt = m c(92 - 0 ,)

6 x 220 x t = 3 x 4200(88 - 28)

1320 x t = 12600 x 60

756000

Example 15How long does it take a 750W heater operating at full rating to raise the temperature of 1 kg of water from 40"C to 70°C? [Take the specific heat capacity o f water as 4200Jkg 'K 1 and neglect heat losses], WAEC 199}

207

SolutionMass of water, m = 1 kg; initial temperature, 0, = 40°C; final temperature, 02 = 70°C; power rating of heater, P = 750W; specific heat capacity, c = 4200 J k g - 'K ' 1

Heat supplied by heater = heat gained by wdter

Pt = m e (62 - 0j)

750 x t = 1 X 4 2 0 0 (7 0 -4 0 )

750 x t = 4200 x 30

126000

Example 16A 500W heater is used to heat 0.6kg of water from 25°C to 100°C in fi second. If another 1000W heater is used to heat 0.2kg of water from 10°C to 100°C in t2 seconds, find tj/t2 A- 50 B. 5 C. 5 /4 D. 1 /5 JAMB 1998Solution

From Pt = m e (02 — 0%) we make t subject of the equation.

mc{02 — 0X)P

0 . 6 x c x ( 1 0 0 - 2 5 ) 0.6c x 751st instance, = ----------- 7-77----------- -- — 777— = 0.09c

500 5000.2 x c x (100 - 10) 0.2c x 90

2nd instance, t 2 = -----------7_________ = —____ — = 0.018c10000.09c 0.09 90

Therefore, - = = — = - = 5

1000

Example 17An immersion heater rated 400W, 220V is used to heat a liquid of mass 0.5kg. If the temperature o f the liquid increases uniformly at the rate of 2.5°C per second, calculate the specific heat capacity of the liquid (Assume no heat is lost). WAEC 2005SolutionPower rating of heater, P = 400W; temperature change, A0 = 2.5°C;Mass o f liquid, m = 0.5kg; time, t = 1 sec. (.... rate of2.5°Cper second)-.Voltage, V = 220V.

Heat energy supplied by heater = ■ heat gained by liquid

Pt - mcAO

400 x 1 = 0.5 x c x 2.5

400 , ,c = o l T I s = 320l k9 K

Example 18An electric current o f 3 A flowing through an electric heating element of resistance

20Q embedded in l,000g of an oil raises the temperature of the oil by 10°C in 10 sec. What is the specific heat capacity of the liquid? A. 1.8 J /g B. 0.6/ / gC. 0.18//g°C D. 1.8 J/g°C E .0.06J/g°C JAMB 1983

208

SolutionC'unvnt. I = 3 A; resistance. R -0Q,temperature change, AO = 10 (

mass of oil, m - 1000g, time, t = 1 Os

Meat supplied by heater = heat gained by oil

l 2Rt = mcAO

l 2Rt 32 x 20 x 10 Specific heat capacity, c = —~tr = m nn * 10~

Example 19 . ,Calculate the power rating of an immersion heater used for 10 minutes ^ ,temperature of 10kg of water bv 15K [specific heat capacity of water = 4200Jkg |.If'AEC /WV SolutionTime, t = 10mm = 10 x 60sec = 600s; mass of water, m = 10kg; _ xTemperature change, A0 = 15K.; specific heat capacity of water, c = 42OOJkg K

mcAO — Pt

Therefore,mcAO

P = -------t

10 x 4200 x 15 600

630000 600 _

= 10501V

Example 20A heating coil of resistance 2012 connected to a 220V source is used to boil a certain quantity of water in a container of heat capacity 100JK ' for 2 minutes. If the initial temperature ot the water is 40°C, calculate the mass of the water in the container.| specific heat capacity of water = 4.2 x 10* ]kg~' K~1-, assume boiling point o f water = 100'C]. WAEC2002SolutionNote that that generated by the heater will be absorbed by the water and the container, therefore:

Heat generated by heater = heat absorbed by water + heat absorbed by container

Resistance of heater, R 2012;heat capacity of container, C = 100 J Kfinal tempt.. (V r 100°C (boiling point of water).Specific heat capacity of water 4200/ kg' 1 K~l

voltage, V - 220V;time, t 2min 2 x 60s = 120s;initial tempt., 0| - 4()°C;I et m be the mass of water

Substitute intoV2t~ j f = lmC(02 ~ B^Later l y »co n ta in er

(Rcnicmbei quantity of heat (Q) = heat capacity(C) x temperature change(AO))

2202— ■- x 120 = m x 1200 (100 - 40) + 100(100 - 40)

290400 = m x 4200 x 60 + 100 x 60

290400 = tu x 252000 + 6000

240400 - 6000 = 252000m

2H4400 = 252000m

209

_ 284400~ 252000 = i U k 3

Example 21A 1-2KW immersion heater is used to supply energy for 75s. The energy supplied is used to heat SOOg of paraffin in calorimeter of heat capacity 42.5 / / f “ 1 from 20°C to 66°C. Calculate the specific heat capacity of the paraffin. ÑECO 200821SolutionPower rating of heater, P = 1.2KW = 1200W; time taken, t = 75smass of paraffin, m = 800g = 0.8kg heat capacity of calorimeter, C = 42.5 JK _1initial tempt., 9X = 20°C final tempt., 92 = 66°Cspecific heat capacity of the paraffin, c =?

Pt = [mc(02 - 9x)]paraffin + [C(f?2 P l)lc a lo r im e te r

1200 x 75 = 0.8 x c x (66 - 20) + 42.5(66 - 20)

90000 = 0.8 x c x 46 + 42.5 x 46

90000 = 36.8c + 1955

9 0 0 0 0 - 1955 = 36.8c

88045 = 36.8c

88045 , ,c = — = 2392.53 = 2393Jkg-'K ~136.8

TEMPERATURE DIFFERENCE IN A WATERFALLWhen a mass of water falls from the top of a waterfall to the bottom, the potential

energy it possesses at the top is transformed to kinetic energy and internal molecular energy. These energy transformation eventually result in a slight increase in temperature at the bottom of the waterfall.

Example 22A waterfall is 1260m high. Calculate the change in temperature of a quantity of water that falls from the top to bottom of the water fall. (Neglect heat losses to surrounding, take gas 10m/s2 and specific heat capacity as 42W Jkg~ 1K~1) WAEC 1998SolutionHeight of waterfall, h = 1260m; g = 10m/s2; specific heat capacity, c = 4200 Jkg~ 1K ~1

The following assumption can be made. All the potential energy (PE) the water possesses at the top is converted to heat energy at the bottom. Therefore,

Potential energy = heat energy

m gh = mcA9

substituting we have, m X 10 x 1260 = m X 4200 X A0

10 x 1260 = 4200 x A9divide through by m,10 x 1260

A° = ~~42Ô Ô~ = 3°C

210

Example 23 „cnrw 'kr-1 falls through a vertical distanceA piece of substance of specific heat capacity 450Jkg K Viittinu the aroundOf 20m from Calculate the rise in temperature of the substance on hmmg the |roundwhen all energies are converted into heat. A. 2 /9 C B. / iooqD .9 /2 °C JSolution _ A zn ib n - 'K - 1Vertical distance, h = 20m; g=10m /s2; specific heat capacity, c J 9

Potential energy = heat energy

mgh = mcAO

substituting we have, m x l 0 x 2 0 = m x 4 5 0 x A #

divide through by m, 10 X 20 = 450 X Ad

1 0 x 2 0 _ 200 _ 4 = 0_44oC450 450 9

LATENT HEATLatent heat is defined as the heat absorbed or given out by a substance during a change of state at constant temperature.

Latent heat of fusionSpecific latent heat of fusion (l) of a substance is the quantity of heat (Q) required to change a unit mass (m ) of the substance at its melting point to a liquid without a change in temperature.

That is, Q = ml or l = — where Q = quantity of heat in joules (J)

m = mass of the substance in kg l = specific latent heat of fusion in Jkg’1

However, when the heat supplied (Q) is from an electric heater, the following equation apply:

From Q = ml we have

IVt = ml

Pt = ml

I2Rt = ml

V2t— = ml

See page 112 for the definition of the term on the left hand side of the above equations.

Example 24Calculate the quantity of heat required to change 200g of ice to water at 0°C. [specific latent heat of fusion of ice = 336/^-1]. NECO 2006gSolutionMass of ice, m = 200g; specific latent heat of fusion, Z = 336/ g -1

Quantity of heat, Q = ml = 200 x 336 = 67200J.

211

Example 25A 90W immersion heat is used to supply energy for 5min. The energy supplied is used to completely melt l60g of a solid at its melting point. Calculate the specific latent heat of fusion of the solid. WAEC 1991SolutionPower rating of heat, P = 90 W; time, t = 5mm = 5 x 60 = 300s;mass of solid, m = 160g = 0.16kg.

Pt = ml I = = 90 X 300 = 16875O7/C0'1 or 168.75/0-1m 0.16

\

Example 26What amount of current would pass through a 10ft coil if it takes 21s for the coil to just melt a lump of ice of mass lOg at 0°C if there are no heat losses? [latent heat o f fusion of ice = 336 J g -1]. WAEC 1993SolutionResistance of heating coil, R = 10ft; time, t = 21 s;mass of ice, m = 1 Og; ' Latent heat of fusion, 1 = 336J g -1

From I 2Rt = ml we obtain I2 = —Rt

1336 x 10 J 10 x 21

3360= V l6 = 4A

Example 27An electric heater is used to melt a block of ice, mass 1,5kg. If the heater is powered by a 12V battery and a current of 20A flows through the coil, calculate the rime taken to melt the block of ice at 0°C. [Specific latent heat of fusion of ice = 336 x 103 Jg~1]A. 76.0 min B. 35.0 min C. 21.0 min D. 2.9 min

JAMB 1989SolutionMass o f ice, m = 1,5kg; potential difference, V = 12 V; current, I = 20A

From IV t — m l we derive,

m l 336 x 103 x 1.5 504000Time taken, t = — = ----- ----- = — = 2100s = 35m insIV 20 x 12 240

Latent Heat of VaporizationSpecific latent heat of vaporization (L) of a substance is the quantity of heat (Q)

required to change a unit mass (m) of the substance from liquid at its boiling point to vapor without a corresponding change in temperature.

QThat is, Q = mL or L = —

m

where Q = quantity o f heat in Joules(J) m = mass of substance in kg L = specific latent heat of vapourization in Jkg'1

If the heat (Q) supplied from an electric heater, the following equations apply.

212

From Q = mL behave

IVt - m l

Pt - m l

l 2Rt = mL

V2tR

= m ithe left hand side of the above

See page 112 for the definition of the term onequations. •_. Question.The equation used for a question will depend on the data given

AXq u Z % of steam at 100”C condenses to water at the same t e m p e r e by r e ly in g M x 104J of energy. Calculate the mass of the condensed steam [speci 1 2006vapourization of water = 2.3 x 10 OJkg"1]SolutionQuantity of heat released by steam, Q = 6.9 X 10 J Latent heat of vapourization, L = 2.3 x 10bJkg 1

From Q = mL we derive

Mass of condensed steam,Q 6.9 x l O 4

m — — = ----------- 7 — 0.03 kgL 2.3 x 106

Example 29106J of heat is required to boil off completely 2kg of a certain liquid. Neglecting heat lost to the surrounding, what is the latent heat of vapourization of the liquid?A. 5 .0 x 1 0 sJkg~ l B. 2 . 0 x 1 0 5Jk g ~1

SolutionQuantity of heat supplied, Q = 106J;

From Q = mL we obtain

Latent heat of vapourization,

C. 5.0 x lO 6/ / ^ “1 D. 2.0 x 1 0 6yfc£-1 JAMB 2005

mass of substance, m = 2kg

, Q 106i = m = — = 5 x 1 0 5 1 * * - 1

Example 30A heating coil rated at 1000W is used to boil off 0.5kg of boiling water. What is the time taken to boil off the water? [Specific latent heat of vapourization of water = o 7 v1 f \ 6 I t ___ — 1 3 A 1 - I C w r'l -i -I r- "7 Z . . J XlO6; ^ " 1] A. 1.15 x 109s D. 1.15 x 103s SolutionPower rating, P = 1000W; mass of water, m = 0.5kg;

B. 1.15 x 107s C. 1.15 x 105s

Pt — mL ■■ Time taken to boil off water, t = mL

JAMB 1997

L = 2.3 x 10b] k g ~ x]

0.5 x 2.3 x 106ÎÔÔÔ

- 1150sec or 19.17min

213

Example 31All the heat generated by a current of 2A passing through a 6Q resistor for 25s is used to evaporate 5g of a liquid at its boiling point. What is the specific latent heat of the liquid?

WA EC 1989SolutionCurrent, 1 = 2 A; mass of liquid, m = 5g = 0.005kgresistance, R = 6fi; time, t = 25s;

Rt 22 x 6 x 25l~Rt = mL ■■■ specific latent heat of vapourization, L = ------= — ——7—

m 0.005600

0.005= 1.2 x 10sJkg 1 o r 1207#'

The problems treated so far involve only one of these for each question; (a) specific heat capacity (b) specific latent heat of vapourization (c) specific latent heat of fusion. However, there are many other instances that a combination of two or three of the specific heats could be used in solving problems as illustrated by the following examples.

Example 32What is the total amount of heat energy required to convert 0.5kg of ice at -15°C into steam at 100°C? [Specific heat capacity of ice = 2100 Jkg~1K~1, specific heat capacity of water = 4200)k g ~ 1K~1, specific latent heat of ice = 336000)k g ~ l , specific latent heat of steam = 2.26 x 10bJk g * 1]SolutionQuantity of heat required to convert ice at -15°C to ice at 0°C

Q1 = mcAd = 0.5 x 2100 x 15 = 1.575 x 104y

Quantity of heat required to convert ice at 0°C to water at 0°C

Q2 = m l = 0.5 x 336000 = 1.68 x 105/

Quantity of heat required to convert water at 0°C to water at 100°C

Q3 = mcAO = 0.5 x 4200 x 100 = 2.10 x 10sy

Quantity of heat required to convert water at 100°C to steam at 100°C

<?4 = mL = 0.5 x 2.26 x 106 = 1.13 x 106/

Total quantity of heat, Q = Q3 + Q2 + <2 3 + (?4

Q = 1.575 x 104 + 1.68 x 10s + 2.10 x 10s + 1.13 x 106

= 1.52 x 1067

Example 330. 5kg of water at 10°C is completely converted to ice at 0°C by extracting 188000J ofheat from it. If the specific heat capacity of water is 4200 J k g '^ C ' 1, calculate the specific latent heat o f fusion of ice. WAEC 1988SolutionMass o f water, m = 0.5kg; heat extracted, Q = 188000JConversion of water at 10°C to ice at 0°C involves two different processes;

1. Conversion of water at 10°C to water at 0°C. Because there is a change in temperature, the specific heat capacity of water is used.

mcAd = 0.5 x 4200 x 10 = 2100/

214

2 n , r\°r npranse there is NO change ofConversion of water at 0 C to ice at 0 C. Because

temperature, the specific latent heat of ice is used.

Therefore total heat extracted toform ice is equal to heat extracted during processes 1 and2.

Q = me Ad + ml 188000 = 21000 + 0.5/

188000 - 21000 = 0.5/167000 = 0.5/

/ = 16J °° = 3340007/C5-1

cAduipic j <i __ rator if the refrigeratorA cup containing lOOg of pure water at 20°C is placed in a re ge for t^e water toextracts heat at the rate of 840J per minute, calculate the time „„nac,tv of waterfreeze. [Neglect heat capacity of the material of the cup] [speci ic = 4.2 Jkg~1K~1] [specific latent heat of fusion = 336/5 ]Solution !Mass of substance, m = lOOg; c = 4.2Jg K ; o/ = 336J g '1 ; temperature change, A0 = 20 C

Total energy required for freezing, Q — me Ad + ml

Q = 100 x 4.2 x 20 + 336 x 100 = 8400 + 33600 = 42000/

If 840J is extracted in 1 minutes

42000J will be extracted in x minutes.42000

840 = 50min

Example 35A copper calorimeter of mass 30g contains 50g of oil at 20°C. Some dried ice at 0°C is added to the oil. If the final steady temperature of the calorimeter and its contents is 0°C, calculate the mass of ice added. [Specific heat capacity of copper = 400 ]k g ~ 1K~1] [Specific heat capacity of oil = 2400 Jkg~1K~1] [Specific latent heat of fusion o f ice =336000 J k g '1] NECO 2002SolutionFrom the question we deduce the following:

Mass of calorimeter, mc = 30g = 0.03kg

Mass o f oil, m0 = 50g = 0.05kg

Temperature change for calorimeter & oil, A0 = 20 - 0 = 20°C

Specific heat capacity of calorimeter, cc = 400Jkg~1K~1

Specific heat capacity of oil, c0 = 2400]kg~ xK~x

Latent heat of fusion of ice, / = 336000Jkg~x

Let m be the mass of ice added

Heat gained by ice = heat lost by calorimeter + heat lost by oil

(m /)ice = (tncA0)caiorjmeter 4" (^cA 0)ojj

m l = m cccA0 + m ocoA0

215

m x 336000 = 0.03 x 400 x 20 + 0.05 x 2400 x 20

336000m = 240 + 2400

336000m = 2640 2640

m = 336000 = 0 00786kg ° r 7 '86g

Example 36How many grams of water at ] 7°C must be added to 42g of ice at 0°C to melt the ice completely? [Specific latent heat of fusion of ice = 3.4 x 105 J k g '1; specific heat capacity of water = 4200/fcig~1] A. 200g B. 300g C. 320g D. 400g

JAMB 1998Solution

Heat absorbed by ice = heat lost by water m,Z = mcA6

Mass of ice, m i = 42g = 0.042kg

Temperature change of water, A0 = 1 7 - 0 = 17°C

Latent heat of fusion of ice, l = 3.4 x 10s/k g -1

Specific heat capacity, c = 4200Jkg~ 1K~1

Let m be the mass of water

Substituting into m,/ = mcA6 we have,

0.042 x 3.4 x 10s = m x 4200 x 17

0.042 x 3.4 x 10s 14280m = 4200 X 17 = = 0 2 k a 0T 200«

Example 37A piece of copper of mass 300g at a temperature of 950°C is quickly transferred to a vessel of negligible thermal capacity containing 250g of water at 25°C. If the final steady temperature of the mixture is 100°C, calculate the mass of the water that will boil away. [Specific heat capacity of copper = 4.0 X 102 Jkg~1K~1] [Specific heat capacity ofwater = 4.2 x 102 Jkg~1K~1] [Specific latent heat of vapourization of steam = 2.26 x 106 J k g - 1] WAEC1990Solution

Heat lost = heat gained

Heat given out by copper = heat absorbed by changing water at 25°C to water at 100°C +

heat absorbed by changing water at 100°C to steam at 100°C

m cccA6c = m wcwA6w + mL

mass of copper, m c = 300g — 0.3kg

specific heat capacity of copper, cc = 4.0 x 102 )k g ~ 1K~l

temperature change of copper, A0C = 9 5 0 -1 0 0 = 850°C

mass of water, m w = 250g = 0.25kg

Specific heat capacity of water, cw = 4.2 x 103 Jkg 1K 1

Temperature change of water, A0W = 100 - 25 = 75°C

Latent heat o f vapourization of steam, L = 2.26 x 106 J k g -1

216

Let m be the mass of water that will boil awaySubstituting into mcccA9c = TnwcwA9v) + mL

0.3 X 4.0 X 102 X 850 = 0.25 X 4.2 X 10* X 75 + m X 2.26

102000 = 78750+ 2.26 x 106 X m

_ 23250 _ = o.0103kg = 10.3gm 2.26 x 106

Example 38 j 4 ^ Qf waterWhat is the difference in the amount of heat given out by 4kg o steam steam =when both are cooled from 100°C to 80°C? [Specificaten y A v r 1996 2,260,000 Jkg~xK~1] [Specific capacity of water = 4200 Jkg ]Solution

Heat given out by steam = mi + mcA9

= 4 x 2,2^0,000 + 4 X 4200 (100 - 80)

= 9040000 + 336000 = 9376000J

Heat given out by water = mcA9

= 4 x 4200 x 20 = 336000J

Difference in heat given out = 9376000 — 336000

= 9040000J

RELATIVE HUMIDITY

Relative humidity =mass of water vapour in a given volume of air(m)

mass of water vapour required to saturate same volume of air at the same temperature (M)

saturation vapour pressure of water at the dew point

x 100%

saturation vapour pressure(S. V. P)of water at the original air temperature

x 100%

vapour pressure (partial pressure) of the air (V. P)S V. P at the temperature of the air * 10° O/,°

m V.PRelative humidity = — x 100% = ^ — - x 100%

Example 39The mass of water vapour in a certain volume of air is 0.20g at 30°C. What is the relative humidity of the air if the mass of water vapour required to saturate it at the same temperature is 0.35g.SolutionMass of water vapour, m = 0.20g;mass of water vapour required for saturation, M = 0.35g

217

Relative humidity = - x 100% M0.20Ô35

x 100% = 57.14%

Example 40The table below shows the saturation vapour pressure against in a particular city. If on a certain day, the vapour pressure in this city at 30°C is 22.0mmHg, what is the relative humidity?

r c 0 10 20 30 40 50 60

S.V.P(mmHg) 4.6 9.2 17.5 31.8 55.1 92.5 149.0

Solution

Vapour pressure of air at 30°C, V.P = 22.0mmHg

Saturation vapour pressure of air at 30°C, S.V.P = 31.8mmHg V P 22 0

•• Relative humidity = —— x 100% = — - x 100% = 69.2%S.V.P 31.8

Example 41On a certain day, the dew point is 10°C when the temperature of the air is 32°C. Calculate the relative humidity of the air. (S.V.P. of water at 32°C and 10°C are 16.2mm and 9.8mm of mercury respectively). NECO 200824 SolutionS.V.P. at dew point (10°C) = 9.8mmHg S.V.P. at air temperature (32°C) = 16.2mmHg

Relative humidity —S. V. P at dew point

S. V. P at air temperature9.8

x 100% = x 100% = 60.5% 16.2

EXERCISES 9.1. A tap supplies water at 26°C while another supplies at 82°C. If a man wishes tobath with water at 40°C, what is the ratio of the mass of hot water to that of cold water required? WAEC 1991 Ans: 1:32. A tap supplies water at 30°C while another supplies water at 86°C. If a manwishes to bath with water at 44°C, calculate the ratio of the mass of hot water to that of cold water required? WAEC 1994 Ans: 1:33. Two thermos flasks of volume Vx and Vy are filled with liquid water at an initial temperature of 0°C. After some time the temperatures were found to be 0X, 0y respectively. Given Vx/Vy = 2 and 0x/0 y = What is the ratio of heat flow into the flasks. A. 1 B. j C. 4 D. 1 E. 2 JAMB1985 Ans: 1:1 or 14. A tap supplies water at 25°C while another supplies water at 75°C. If a manwishes to bath with water at 40°C, what is ratio of the mass of hot water to that of cold water required? A. 1:3 B. 15:8 C. 7:3 D. 3:1

JAMB 1988 Ans: 7:35. Hot water at a temperature of t is added to twice that amount of water at a temperature of 30°C. If the resulting temperature of the mixture is 50°C, calculate t.

WAEC 1995 Ans: 9(fC

218

, . nc\°r If they are properly mixed400g o f cold water is added to 200g of water a temrierature o f the cold

and the temperature of the mixture is 30°C, calculate the mi 1 2000 Ans: l ( fCwater. [Neglect the heat absorbed by the container] hire Qf 80°C have7- Two liquids, P a, a ,cmpf ran,re of 20°C and 0 Ircq - ' ™ »scs ° fspecific heat capacities o f 1 .OJkg'1 °Cl and 1.5Jkg C respectiv y. A 44°cP and Q are mixed in a lagged calorimeter, what is the eqtiilibnurn \ cm . 5A°CB.50°C C.56-C D. 60°C E. 70‘C J A M B ¡ T ' ^8. Hot water is added to three times its mass of water at temperature is 20°C. What is the initial temperature of the hot waterA. 100°C B 80°C C 50°C D 40°C ^ M B 1987 Ans^ 5 ffC

9. 200g of water at 90°C is mixed with 100°C of water at 30°C. What i s t h e m a ternperamre? A. 50°C . B. 60°C C. 70°C D. 80°C JAMB1993 A n ..10. What mass of water at 100°C should be added to 15g of water at 40 to ma temperature of the mixture 50°C. [Neglect heat losses to the surrounding]

L NECO 2006 Ans: 3g11. Water of mass 120g at 50°C is added to 200g of water at 10°C and the mixture iswell stirred. Calculate the temperature of the mixture. [Neglect heat surrounding] Ans: 25°C WAEC 200212. How much heat is given out when a piece of iron of mass 50g and specific heatcapacity 460Jkg'1K'1 cools from 85°C to 25°C? WAEC 1990 Ans: 1.38 X 1(TJ13. A piece of copper of mass 30g loses 60J of heat energy. If the specific heat capacity of copper is 400 Jkg 'K"1, calculate the change in temperature of the copper.

WAEC 1997 Ans: 5K14. How much heat is emitted when a body of mass 200g cools from 37°C to 31 C?[Specific heat capacity of the body = 0.4 Jg 'K"1] WAEC 1998 Ans: 480J15. A piece of metal of mass 50g is cooled from 80°C to 20°C. Calculate the amount of heat lost. [Specific heat capacity of the material of metal = 450 J k g 'K 1)

WAEC 2000 Ans: 1350J16. A body of mass, m has a specific heat capacity, s and a heat capacity, C. If the temperature of the body changes by 0°C, which of the following equations is correct?A. ms0 = me'1 B. ms = c C. ms = s0 D. ms = c0 WAEC 2005 Ans: B17. A copper ball of heat capacity 400JK'1 is heated from 20°C to 100°C. Calculatethe quantity of heat absorbed. NECO 2005 Ans: 3.2 x 1 ( f j or 32KJ18. A body of mass 40g loses 80J of heat energy. If the specific heat capacity of the body is 400Jkg'lK '1, calculate the change in temperature of the body.

NECO 2004 Ans: 5. OK19. How much heat is emitted when a body of mass 50g cools from 80°C to 20°C?[Specific heat capacity of the body = 460 Jkg''K ']. NECO 2002 Ans: 1380J20. 22000J of heat is required to raise the temperature of 1.5kg of paraffin from 20°Cto 30°C. Calculate the specific heat capacity of paraffin. A. 1466.7 Jkg'1 °C'1B. 2933 Jkg ' °C ‘ C. 4400 Jkg'1 °C'1 D. 5866 Jkg'1 °C''

JAMB 1987 Ans: 1466.7 Jkg'1 “C '121. How much heat is absorbed when a block of copper of mass 0.05kg and specificheat capacity 390Jkg''K'' is heated from 20°C to 70°C? A. 3 98 x 10_1 /B.9.75X 102/ C. 3.98 x 103/ D. 9.75 x 103/ JAMB 1992 A ns-975J22. When two objects P and Q are supplied with the same quantity of heat the temperature change in P is observed to be twice that in Q. If the masses of P and Q are the same, calculate the ratio of the specific heat capacities of Q to P. WAEC 1995 Ans -2 123. When two objects P and Q are supplied with the same quantity of heat the temperature change in P is observed to be twice that in Q. The mass of P is half that of Q. What is the ratio of the specific heat capacities of Q to P? WAEC 1986 Ans 1124. A mass of liquid at 30°C is mixed with a mass of the same liquid at 70°C and thetemperature o f the mixture is 45°C. Find the ratio of the mass of the cold liquid to the mass of the other liquid, A. 3:5 B. 5:3 C. 3:7 D. 7:3

JAMB 1997 Ans: 5:3

219

25. A piece of copper ball of mass 20g al 200nC is placed in a copper calorimeter olmass COg containing 50g of water at 30°C, ignoring heat losses, calculate the final steady temperature of the mixture. [Specific heat capacity of water = 4.2Jg K , Specific heat capacity of copper = 0.4Jg‘! K.'1 ] WAEC 1994 Arts: 35.6'C26. An iron rod of mass 2kg and at a temperature of 280<’C is dropped into somequantity of water initially at a temperature of 30°C. If the temperature of the mixture is 70°C, calculate the mass of the water. [Neglect heat losses to the surrounding]. [Specific heat capacity of iron = 460Jkg 'K 1, specific heat capacity of water = 4200 Specific heat capacity of water = 4.2 Jk g 'K '1 ] IVAEC 1995 Ans: 1.15kg27. A piece o f copper block of mass 24g at 230°C is placed in a copper calorimeter of mass 60g containing 54g of water at 31°C. Assuming heat losses are negligible, calculate the final steady temperature of the mixture. [Specific heat capacity o f water = 4200 Jkg 'K 1. Specific heat capacity of copper = 400 Jkg 'K '1] WAEC2001 Ans: 38.34TC28. A calorimeter o f thermal capacity 80J contains 20g of water at 25°C. Water at100°C is added so that the final temperature of the set up is 50°C. What is the amount of water added? [Heat capacity of water = 4.18 Jg'1 °C''] A. 20g B. 25gC. 45g D. 50g E. lOOg JAMB 1980 Ans: 2Og29. 250g of lead at 170°C is dropped into lOOg of water at 0°C. If the final temperature is 12°C, what is the specific heat capacity of lead? [Specific heat capacity of water 4200 Jkg'1 °C''] A. 39.5 Jkg'1 °C ' B. 50.4 Jkg'1 0C '‘ C. 127.6 Jkg'1 ° C ‘D. 154.6 Jkg'' °C'' E. 173.4 Jkg'1 °C'' JAMB 1982 Ans: 127.6 J k g 1'30. Calculate the time taken to heat 2kg of water from 50°C to 100°C in an elc. ic kettle taking 5A, from a 210V supply. [Specific heat capacity of water = 4200 Jkg 'K ’1]

WAEC 1994 Ans: 400s or 6.67niins31. An immersion heater is rated 120W. How long does it take the heater to raise the temperature of 1.2kg of water by 15°C. [Assume heat lost to the surrou d is negligible. Specific heat capacity of water = 4200 Jk g 'K '1]

WAEC 2000 Ans: 10.5 minutes or 630 seconds32. How long will it take to heat 3kg of water from 28°C to 88°C in an electric kettle taking 6A from a 220V supply? [Specific heat capacity of water = 4180 Jkg 'K '1]

WAEC 2006 Ans: 570s or 9.5 min.33. How long does it take a 750W heater to raise the temperature of 1kg of water from20°C to 50°C. [Specific heat capacity of water = 4200 J k g 'K 1] A. 84s B. 112sC. 168s D. 280s JAMB 1991 Ans: 168s or 2.8 min34. An electric kettle with negligible heat capacity is rated at 2000W. If 2.0kg of water is put in it, how long will it take the temperature of the water to rise from 20°C to 100°C? [Specific heat capacity of water = 4200 Jkg 'K 1] A. 420s B. 336sC. 168s D. 84s JAMB 1995 Ans: 336s or 5.6 min35. An immersion heater rated 2.0A, 240V is used to boil water from temperature 52°C to 100°C. If the mass of water is 2.5kg, determine the time taken to boil the water. [Specific heat capacity of water = 4 2 x 103 Jkg 'K '1] A. 1.05 x 104s B. 1.05x 105sC. 1.05 x 103s D. 1.05 x 102s JAMB 2006 Ans: 1050secor 17.5 min36. A metal o f mass 1.5kg was heated from 27°C to 47°C in 4 minutes by a boilingring of 75 W rating. Calculate the specific heat capacity of the metal. [Neglect heat loses to the surrounding] WAEC 1992 Ans: 6.0 x /O3 J k g ° C ’37. A 400W immersion heat is used to heat a liquid of mass 0.5kg. If the temperatureof the liquid increases by 2.5°C in one second, calculate the specific heat capacity of the liquid. [Neglect heat losses to the surrounding] WAEC 1996 Ans: 320 Jkg 1IC138. Heat is supplied uniformly at the rate of 100W to 1.0 x 10':kg of a liquid for 20sec. If the temperature of the liquid rises by 5°C, then what is the specific heat capacity of the liquid? A. 2.0 x 102Jkg 'K ' B. 2.0 x 102Jkg ' C. 4.0 x 104Jkg 'K '1D. 4.0 x 104Jkg ' E. 8.4 x 103Jkg 'K ' JAMB 1984 Ans: 4.0 x l& Jkg 'lC '39. A 2000W electric heater is used to heat a metal object o f mass 5kg initially at 10°C. If a temperature rise o f 30°C is obtained after lOmin, what is the heat capacity of the material'7 A. 6.0 x lO4/ 0 1 B. 4.0 x lO4/ 0 1 C. 1.2 x 104/ 0C_1

220

D. 8.0 x 103y°C-1 ' f -mss 5ku. II in to mm40. A 50W, electric heater is used to heat a metal bloc o i c f the metal?a temperature rise of 12°C is achieved, what is the specific iea c ¿ n o J k t f ' K 1

JAMB 2003 Ans: 4.0 x 104J " C If in 10 minutes

i temperature A. 500 Jkg 'K '1 B. 130J k g 'K ' C. 390 Jk g 'K

JAMB 2004D. 400 Jkg

Ans: 500Jkg 1C

Calculate the difference ,nNeglect heat losses. [g=10m/s41. A waterfall is 420m high,

between the top and bottom of the waterfall heat capacity of water = 4.2 x 103 Jkg 'K ’1)42. A body of specific heat capacity 450Jkg 'Ka vertical height of 20m. Assuming conservation of energy, _ 0 44°C

striking the ground level [g=10m/s] WAEC m s -. . ■_

Ite-I

specificWAEC 1991 A n s . 'H fC

falls to the ground from rest throughcalculate the change in

temperature of the body on striKing ine grounu ic»u lb ------- - ■ f thc water43. Water falls through a height of 50m. Determine the tempera tv of water isat the bottom of the fall. [Neglect energy losses, specific heat capacl l i c f r4200Jkg‘lK '1, g=10m/si j. A « - «44. A waterfall is 1260m high. Calculate the difference in temperature ,between the top and the bottom of the water fall. [Neglect heat 0S;,^S „ ® 77sCspecific heat capacity of water = 4.20 x 103 Jkg’lK’1] NECO 2006 Ans. ot45. How much heat is required to convert 20g of ice at 0 C to water at e sa temperature? [Specific latent heat of ice = 336Jg '] WAEC 1995 (Ans. 6. x )46. The latent heat of fusion of ice is 80 cal g'1. How much heat is require to c angelOg of ice at 0°C into water at the same temperature? A. 80 cal B. 10 caC. 8 cal D. 800 cal E. 400 cal JAMB 1978 Ans: 800 cal47. The melting point of a solid is given as 80°C. If 105J of heat energy is required atthis temperature to melt lOg of the solid, what is the specific latent heat o f fusion o f the solid? A. 1.00 x 103Jkg_ 1 B. 1.25 x 105Jkg_1 C. 1.00 X 107Jkg_1D. 8.00 x 108Jkg-1 JAMB 1993 Ans: 1.00 X lÓ’j k g '48. A 90W immersion heater is used to supply energy for 5 minutes. The energy supplied is used to completely melt 180g of a solid at its melting point. Neglecting energy losses to the surrounding, calculate the specific latent heat of fusion of the solid.

WAEC 1998 Ans: 150J g '49. A 10Í2 coil takes 21s to melt lOg of ice at 0°C. Assuming no heat losses, determine the current in the coil. [Specific latent heat of fusion of ice = 336Jg‘']

WAEC 1999 Ans: 4A50. Heat is supplied to a test tube containing lOOg of ice at its melting point. The ice melts completely in 1 min. What is the power rating of the source of heat? [Latent heat of fusion of ice = 336Jg '] A. 336W B. 450W C. 560W D. 600W

JAMB 1994 Ans: 560WCalculate the quantity of heat released when lOOg of steam at 100°C condenses to [Take the specific latent heat of vapourization of water as 2.3 x 106Jkg'lK"1]

WAEC 1997 Ans: 2.3 X l& J How much heat is required to convert 50g of water at 100°C to steam at the same

51.water.

52.

53.

temperature? [Specific latent heat of vapourization of water = 2260Jg ']NECO 2003 Ans: 1.13 X IQrJ

All the heat generated in a 5Q resistor by 2A flowing for 30 seconds is used to evaporate 5g of a liquid at its boiling point. What is the correct value of the specific latent heat of the liquid? A. 120 J B. 60 Jg-1 q 1 2 0 |p _1 *D. 1500 J E. 1500 Jg“1 8JAMB 1979 Ans: 120Jg‘54. A heater marked 50W will evaporate 0.005kg of boiling water in 50 secondsWhat is the specific latent heat of vapourization of water in J/kg? a 5 0 x 106B. l .O x lO 6 C. 2.5 x 106 D. 2.5 x 1 0 s E. 5.0 x 10s

JAMB 1982 Ans: 5 X 10sJ k g '55. All the heat generated in a 5Q resistor by 2A flowing for 30s is used to evapnn, 5g of a liquid at its boiling point. What is the specific latent heat o f vapourization >i i

221

liquid? A. 120 J B. 60 Jg_l C. 120JC"1 D. 1500 J E. 1500 Jg ,JAMB 1 984 Ans: l20Jfi

56. Steam at 100°C, is passed into a container of negligible heat capacity, containing 20g of ice and lOOg of water at 0°C, until the ice is completely melted. Determine the total mass ol water in the container. [Specific latent heat of steam = 2.3 x U) Jg • specific latent heat of ice = 3.4 x 10‘Jg*1, specific heat capacity of water = 4.2 Jg K )

IVA EC 2005 Alls: 122.5g ^57. Calculate the heat required to convert 20g of ice at 0°C to water at 16 C . [Specific latent heat of fusion of ice = 336Jg specific heat capacity of water = 4.2 Jg"K ] WAEC 1994 Arts: 8064J

58. Calculate the heat energy required to change 0.1 kg of ice at 0HC to water boiling at 100°C. [Specific heat capacity of water = 4200 Jkg 'K specific latent heat o f fusionof ice = 336,000Jkg' WAEC 1999 Ans: 75.600J59. The specific heat of a substance in the solid state is Ci; its specific heat in the liquid state is C: and its latent heat of fusion is L. If a mass M of the substance is changed from the solid state at temperature T, to the liquid state, also at temperature T, the amount of heat required will be:A. M(Cj +C2) x T b . ML C. L D. M(Cx + C2) x T + MLE. ML [^y 1] x T NECO 2000 Ans: B60. Calculate the amount of heat required to convert 2kg of ice at —2°C to w'at at 0 C. [Specific heat capacity of ice = 2090Jkg'' °C specific latent heat o f fusion o. ce = 333Jkg' ] A. 666 J B. 8360 J C. 666000 J D. 674360

JAMB 1987 Ans: 674360Jlkg of copper is transferred quickly from boiling water to a block o f ice.

Calculate the mass of ice melted, neglecting heat loss. [Specific heat capacity of courier = 400Jkg lK '1 and latent heat of fusion of ice = 333 x lO’jkg '1] A. 60g . j /gC. 120g D. 133g JAMB 1990 Ans: 120g or 0.12kg62. What is the difference in the amount of heat given out by 4kg of steam and 4kg of water when both are cooled from 100°C to 80°C. [Specific latent heat of steam =

61

2,260,000jkg'\ specific heat capacity of w'ater is 4200Jkg‘ K 1].B. 2,260,000 J C. 9,040,000 J D. 9,380,000 J JAMB 1991 Ans: 9.040.000J63. Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss. [[Specific heat capacity of copper = 400 J k g 'K 1, latent heat of fusion of ice = 3.3 x 105Jkg '] A. 2-kg

A .4200 J

B64.

^ k g C. ^ k g D. ^-kg JAMB 1999 Ans: % kg or 0.24kgCalculate the heat energy required to vapourize 50g of water initially at 80°C if

the specific heat capacity of water is 4.2Jg K‘ water = 2260Jg']

[Specific latent heat of vapourization of WAEC 1989 Ans: 117200J

65. An electric heater immersed in some water raises the temperature of the waterfrom 40°C to 100°C in 6 minutes. After another 25 minutes, it is noticed that half the water has boiled away. Neglecting heat losses to surrounding, calculate the specific latent heat o f vapourization of water. WAEC 2000 Ans: 2.1 x 106J k g l66. Calculate the energy required to vapourize 50g of water initially at 80°C.[Specific heat capacity of water = 4.200Jg 'K' , specific latent heat o f vapourization of water = 2260Jg‘ ] WAEC 2002 Ans: 117.200J67. The mass of water vapour in a given volume of air is 0.05g at 20°C, while themass o f water vapour required to saturate it at the same temperature is 0.15g. Calculate the relative humidity of the air. WAEC 2000 Ans: 33.33%68. The table below shows the saturation vapour pressure against temperature in a certain town. If the vapour pressure in this town at 20°C is lOmmHg, what is the relative humidity?

T°C 0 5 10 15 20 40 60S.V.P(mmHg) 4.58 6.51 8.94 12.67 17.50 55.10 149.00

57.0%

222

0.4Jg 'K*’**• A body of ninss 200$: mid specilli' beni capacity’ ‘ ■ iscil by the body. BA IX

1 cools rioni WC 111 2007 Ans -/AYA/

is i|iiu'kly Iriunirmelie 1V . Calculate the quantity of heat ivlense.. ... , kly umis,r.rc.70. A piece ol copper ol muss JOOg nl u lempcmliui ol . | ( ,|)0 fjntllinto a vessel ol negligible (hcni.nl enpneily continuing ?5<>g ol " i 'l‘ 1 11 m |)()||steady temperature of the mixture is l()0"( , ciilcnlute the muss o wnaway. [Specific heat capacity of copper 4.0x10 1kg'K I ,, ......[Specific heat capacity of water 4 .2 \lo ’.lkg 'K '|. |Specil1c Intent boat « I o f steam 2.26xlOfl Jkg 11 H'AEC 2007 Ans: O.OIO+876kx < >71. 2kg of water is heated with a heating coil which draws 3.5/ lor two minutes. What is the increase in temperuturo ol the water!

.>00V minus

A. 10°C B. 15"C r . 25‘V O. .!()"( '[Specific heat capacity of water 4200 Jkg 'K 11 JAMB -00 os. I ( / ‘Cc I • • • v i i v i t i u i | i i u i t ^ v»l *V<1IVI f - o ' i . i n j i , • w I |

72. 1 he temperatures of water from tap A and tap H arc 25 ( and 75 11 sl11 J lvl ^If a mixture of water at 40°C is required, calculate the ratio of the mass of water from lap A to that from tap B. NECO 2007 Ans: 7:373. A 420W electric heater is used to heat water of mass 50kg liom 25 ( to its boilingpoint. How long, in hours, does the electric heater work? |Spceifie heat capacity o water = 4.2x10* Jkg'lC‘l] NECO 2007 Ans: l0.4Ur74. A well lagged copper calorimeter of mass 120g contains 70g ol water and lOgol ice both at O'C. Dry steam at 1 ()()”( ' is passed in until the temperature ol the mixture is 40 t . C alculatc the mass of the steam condensed.[Specific latent heat o f fusion of ice 3.2 x It)’ Jg 11 [Specific latent heat of vaporization of steam 2.2 x It)1 Jg 11 [Specific heat capacity of copper 4.0 x 10 1 Jg ‘K 11[Specific heat capacity of water 4.2 Jg 'K 11 N E('() 200 f ' Am: 7.5(>n75. Water ol mass 1.5kg is heated from 30°C to 80”C using an electric kettle which israted 5A. 230V. Calculate the time taken to reucli the final temperature. (Specific heat capacity of water = 4200 Jg 'K ') WAE( ' 2008‘‘ Ans: 4.57 min76. A block of aluminium is heated electrically by a 25W heater. If the temperature risesby 10°C in 5 minutes, the heat capacity of the uluminium is A. 850 JK 1 H. 750 JK 1C. 650 JK-1 D. 500 JK"1 ./AMU 2008” Arts 750 ¡K 177. If the partial pressure of water vapour at 27°C is 18mn.l Ig anil the saturated vapour pressure o f the atmosphere at the same temperature is 24 mmllg, the relative humidity at this temperature is A. 25% B. 33% C. 75% D. 82%

JAMIi 2008u Ans: 75%78. An electric heater rated 220V, 1000W is immersed into a bucket full o f water. Calculate the mass o f water if the temperature changes from 30"C to 1 ()0°C and the current flows for 300 seconds.A. 4.28 kg B. 42.86kg C. 1,02kg I). 7 .14kg

(specific heat capacity o f water = 4200JkglK r) JAMB 2009**A ns: ( '79. A heating coil rated 1000W is used to boil off completely 2kg o f boiling water.The time required to boil off the water is A. 1.15 x 104sB. 1.15 x 103s C. 4.6 x 104s D. 1.15 x 104s (Specific latent heat ofvapourization of water = 2.3 x 106Jkg_1) JAMB 20092' Ans: I)80. It takes 4 minutes to boil a quantity of water using un electrical heating coil. I low long will it take to boil the same quantity o f water using the same heuting coil if the current is doubled? [Neglect any external heat losses]

WA/'X 200(J‘° Ans: 2 mins.81. An immersion heater is rated 1500W. How long does it luke to raise thetemperature o f 1kg o f water by 30°C? (Assume heut lost to the surrounding is negligible specific heat capacity of water = 4200Jkg ' K '1) NECO JOB1? 1 Ans: 84s82. Calculate the heat energy lost when lOg o f boiling water changes to ice ai 0"< [Specific latent heat o f ice = 336Jg ', specific heat capacity o f water 4.2Jg 1K ' |

, NECO 200V2'Ans: 7500./

223

10GAS LAWS

Gases arc described by three basic properties; volume, temperature and pressure. I he pressure o a gas is the force the gas exerts perpendicularly per unit area. That is

Pressure = -^orcc = IT|assx acceleration due to gravity Area A ^a

- density xvolumexg / mass 'tArea [ Dcns,ty volume J

_ density X Area X height X g (Volume = Area X height)

Pressure = density x height x gP = phg is known as the pressure formula.

Pressure is measured in N/m21.013 x 10 N/m2 = 1 atmosphere = 1.013 x l0 5Pa = 760mmHg

GAS PRESSURE MEASUREMENTThe pressure of a gas is measured using a manometer which could be a water or

mercury manometer as shown below.Atmospheric Pressure (P)

The gas pressure (p) at X = pressure in the liquid at Z

= Atmospheric pressure + pressure due to water column YZ

Gas pressure, p = P + h

Fig 10.2

224

Atmospheric pressure (P) acting at N - pressure in the liquid at f= pressure of gas(p) + pressure due to mercury column LM

P = p + h

Therefore, pressure of gas, p = P - h

Example 1

The water manometer shown above is measuring the pressure of the gas supply. If the specific gravity of mercury is 13.6 and the atmospheric pressure is 70cm of mercury, what is the total pressure of the gas supply in cm of water.A. 67cm B. 73cm C. 949cm D. 952cm E.955 cm JAMB 1982Solution

Specific gravity is the same as relative density.

density of substance (Hg)Relativedensity = ---------------------------------

density of water

Therefore, density of mercury = Relative density of Hg x density of water

= 13.6 x 1000

= 13600kgm’3

The equivalent of the atmospheric pressure (70cmHg) in cm of water is calculated by- equating pressure formular for water and mercury as follows.(phg)water ~ (phg)mercurydensity of water x height of water x g = density of Hg x height of Hg x g

1000 x h x 10 = 13600 x 70 x 10

, 13600x70x10 952000n —------------------- = ------------— 95 "'em1000x10 1000

The atmospheric pressure, P = 952cm of water

Total gas pressure, = Atmospheric pressure (P) + pressure due to 3cm of water

952 + 3

955cm of water

225

Example 2 m theIf the atmospheric pressure is 760mmHg, what is the pressure ot the gas sup diagram above ?

Gas supply

Fig 10.4

Solution

Gas supply

h = difference in levels o f mercury in the two arms of the manometer (1 5 -4 .5 = 10.5cm)10.5cm = 105mm

Gas pressure = Atmospheric pressure + pressure due to 105mm of mercury P = p + h

= 760+ 105 = 865mmHg

BOYLE’S LAWHoyle’s law states that the volume (V) of a fixed mass o f gas is inversely proportional to its pressure (P), provided the temperature remains constant.

1 — V = — or PV = Kte V aP

or

226

P,V = p,v,

Initial gas volume, V , - 30cm

Example 3 lOcm3 What willA fixed mass of gas at a pressure of 650mmHg occup.es a volunK he the volume of the gas at 760mmHg if the temperature remains SolutionInitial gas pressure, P| = 650nimHg Final gas pressure, P2 = 760mmHg

p , v ,

From P|V | = P2V2, final gas volume, V 2 — —* 2

650mmHg x 30cm _ 25 .66cm 3760mmHg

Example 4 r f . rA gas at a volume V„ in a container at pressure P„ is compressed lo one 1 1 volume. What will he its pressure if it maintains its original temperature T.A. K/ B. jP 0 C. P0 D. 5P0 JAMB 1999SolutionInitial gas pressure, P| = P0 Initial gas volume, V| = V0 Final gas volume, V2 = ]/$V0

FromPiVi = P2V2, final gas pressure, p -2 V,

P V 1 5P - ° 0 - p V - - V =P V X---- = 5P2 1/ V 0 0 s 0 0 0 V

Application of Boyle’s Law to Gas Trapped in a Cylindrical Tube by a Column of Mercury

When a tube of uniform cross-section closed at one end contains a fixed mass of gas or air which is sealed by a column of mercury of length Wcm, the pressure exerted on the gas depends on the plane of inclination of the tube, as shown below.

(a) Vertically, with closed end at the bottom.Atmospheric pressure, P

^ --------- Capillary tube

Mercury thread

Gas

Fig 10.6The pressure, p, of the gas is due to the sum of pressures exerted by Acm column of mercury and the atmospheric pressure, P.

Therefore, p = P + X (cm Hg)

(b) Inverted, with open end underneathThe gas pressure, p is due to the difference o f the atmospheric pressure (P) and the pressure exerted by the X cm column of mercury.Therefore, p = P - X (cm Hg)

227

i

V; cm

.Y cn \ f

t

Atmospheric pressure, I’Fig 10.7

(c) Horizontally

Atmospheric pressure, P

Fig 10.8

The gas pressure, p is due ONLY to the atmospheric pressure P, because when the tube is in the horizontal position, the column o f mercury exert very negligible pressure the gas.

Therefore, p = P (cm 1 Ig)

Generally, the gas volume = length o f column of gas(v) x cross sectional area (A) o f the tube.Therefore, the gas volumes for figures a, b and c are v;A, y.A and VjA cm' respectively. However, because the cross sectional area of the tube is uniform and the same in each case, the volume o f the gas is taken as the length of the column of gas. That is, volume V is proportional to length y.

So, the volume of gas in Fig 10.6 isyicnv

The volume of gas in Fig 10.7 is v-cnv'

The volume of gas in Fig 10.8 is y.,cnr

Boyle’s law (PiY'i = P2V2) can be applied in deriving a relationship between Jig 10.6 and f ig 10.7 as follows

P

Fig 10.9(al Vertical (hi Inverted

From Fig. 10.9a, Initial or vertical pressure, P, - P + A

228

Initial or vertical volume, V i = v<A

From Fig. 10.9b, Final or inverted pressure. P2 = P - A

Final or inverted volume, V2 = y^A

Substituting into Boyle’s Law, P,V, = P:V2 we have

(P + X)y,A = ( P - A > ;A

divide tlirough by A to obtain

(P + X)y, = (P-A 0y3

Boyle's law can also be applied in obtaining a relationship between Fig. 10.6 and Fig.10.8

P

From Fig 10.10a, Initial or vertical pressure, P, = P + X

Initial or vertical volume, V, = y ,A

From Fig 10.10b, Final or horizontal pressure, P, = P

Final or horizontal volume, V, =_yjA

Applying Boyle’s Law, P,V, = P3V, we obtain

(P + X)J>;A= ? y , \

Dividing both sides by A, we have

(P + X )y, = P^jSimilarly, Boyle's law can be applied for Fig. 10.7 and Fig. 10.8 as shown below.

Fig 10.11 b. Horizontal

From Fig 10.11a, Initial or inverted pressure, P: = P - X

Initial or inverted volume, V, = y :A

From Fig 10.1 lb, Final or horizontal pressure, P3 = P

Final or horizontal volume, V., = y,A

229

Applying Boyle’s Law, P»V2 = P.|V3 we obtain

(P - X) y 2 A = P y, A

Dividing both sides by A, we have

(P - X)y2 = Pj>,

Please note that the length of gas column or column of mercury could be in units other than centimeter (cm).Finally, Boyle’s law can be applied to derive a general equation connecting Fig 10.6, 10.7 and 10.8.

P

ii

X cm ,k

..... iy ,cm

f w

Fig 10.12 a Vertical

1

_y2cm p — +■m )

X cm tn “1 —H -----------^X cm y 3cm ;

P. b. Inverted c. Horizontal

From Boyle’s Law, P,V, = P2V2 = P3V3 we obtain for the three diagram above,

(P + X)y,A . = ( P - ^ ) y ,A = P y jA

Divide through by A to obtain

(P + A )y , = (P -A )y : = P y 3

Example 5A uniform capillary tube, closed at one end contained dry air trapped by a thread of mercury 8.5 x l(Tm long. When the tube was held horizontally, the length of the air column was 5.0 x 10'2m, when it was held vertically with the closed end downwards, thelength was 4.5 x 1 0 'm. Determine the density of mercury = 1.36 x 104kgm'3]. Solution

Dry air

CL ^— P

jV /b.05m 0.085ma. Horizontal

Fig 10.13

of the atmospheric pressure. [g=10m/s'2, WAEC 2004

At horizontal position, gas pressure is due only to atmospheric pressure, P

From Fig 10.13 a. above, horizontal gas pressure, P, = P

Horizontal gas volume, V, = 0.05A

Where A is cross sectional area.

230

At vertical position, gas pressure is due to atmospheric press P

mercury column.From Fig 10.13 b. above, vertical gas pressure, P: = P + 0.085

vertical gas volume, V, = 0.045A

Substituting into Boyle’s law. P,V, = P:V; we obtain

P x 0.05A = (P + 0.085) x 0.045A

Divide both sides by A

P x 0.05 = (P + 0.085) x 0.045

0.05P = 0.045P + 0.003825

0.05P - 0.045P = 0.003825

0.005P = 0.003825

0.003825P = -0.005

= 0.765mHg

The atmospheric pressure (0.765mHg) so obtained is in length of mercury and has to be converted to N/m* or Pa using the pressure formula, P = pgh

Given values: density of mercury, p = 1.36 x 104kgm'’; g = lOm/s" Calculated value: height of mercury, h = 0.765mHg

Atmospheric pressure, P = pgh

= 1.36 x 104 x 10 x 0.765

= 104040Nm"2 or 1.04 x 105Pa

Example 6A thread of mercury of length 12cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the closed end at the bottom, the length of the trapped air is 25cm. Calculate the length o f the air column when the tube is held(a) horizontally(b) vertically with the closed end of the top.

(Atmospheric pressure = 76cm of mercury)Solution

P = 76cmHg Air

(m)Fig 10.14 (') (ii)

Let P be atmospheric pressure

Let volume be proportional to length.(a) From Fig 10.14 (i) above, vertical air pressure, P, = 76 + 12 = 88 cntHu

231

Vertical air volume, V, = 25cm

From Fig 10.14 (ii) above, Horizontal air pressure, P2 = 76 cmHg

Horizontal air volume, V2 = ?Applying Boyle’s Law, P, V, = P,V2 and substituting above values, we have

88 x 25 = 76 x V2

88x25••• v 2 =

76= 28.95cm

(b) Considering Fig 10.14 (i) and (iii) above, we have

Vertical air pressure, P, = 88cmHg Vertical air volume, V, = 25cm

Inverted air pressure, P3 = 76 - 12 = 64cm Inverted air volume, V3 = ?

Applying Boyle’s Law, P,V, = P3V, we obtain

88 x 25 = 64 x V3

V 388x25

6434.38cm

Example 7An air column 10cm in length is trapped into the sealed end of a capillary tube by a 15cm column o f mercury with the tube held vertically as shown below. On inverting the tube, the air column becomes 15cm long. What is the atmospheric pressure during the experiment? A. 90cm B. 76cm C. 75cm D. 60cm E. 50cm JAMB 1979

Fig 10.15

Solution

Hg — p

Air

15cm

10cm

Fig 10.16

.....ÂSSSSiSISÉ 1

..... i

u 1

15cm

10cm

(i) Vertical (ii) Inverted

Let P be the atmospheric pressureFrom (i) Vertical air pressure, P | = (P + 15)cm

232

Vertical air volume, V| = 10cm

Assuming volume is proportional to length.

From (ii) above, inverted air pressure, P2 = (P - 15)cm

Inverted air volume, V2 = 15cm

Applying Boyle’s Law, P,V, = P2V2 we obtain

(P + 15) x 10 = ( P - 15) x 15

10P+ 150 = 15P-225

15P - 10P = 225 + 150

5P = 375

P = = 75cmHg

Application of Boyle’s Law to Air Bubbles in a Water BodyAn air bubble at the bottom of a body of water (swimming pool, dam, drum of

water or sea) is under the influence of two pressures.(i) Atmospheric pressure, P(ii) Pressure due to height of water, H

Atmospheric pressure, P

atmospheric pressure ONLY.The equivalent of atmospheric pressure (760mmHg = 76cmHg = 0.76mHg) in height of water (H) is obtained by equating the pressure formular for water and mercury.

( p h g ) water (pbg)m ci-cury

i.e, density o f mercury x height of mercury x g = density of water x height of water x g

Let height o f water, h - H

13600 x 0.76 x 10 = 1000 x H x 10

13600x0.76x101000x10

10.34m

One atmosphere o f pressure therefore supports a height of about 10m of water.

The pressure, P, o f the air bubble at the bottom isP , = P + H o rP ,= 10 + H

The pressure, P2 of the air bubble just below the surface is P2 = P

Boyle’s law can be applied to air bubble rising from the bottom to top of a water bi as follows:

233

From P,V, = P2V2 we obtain

(P + H)V, = PVj

(10 + H)V, = PV,

Where P = atmospheric pressure in height of water (1 Om) H = depth or height of water body V, = volume of air bubble at the bottom of water V2 = volume of air bubble just below water surface

Example 8An air bubble of volume 6cm3 rises from the bottom to the top of a swimming pool which is 15m deep. What will be the volume of the air bubble just below the pool’s surface, if the atmospheric pressure is equivalent to 10m of water?Solution

P= lOmH.O

Surface

Swimming pool

Bottom

At the bottom: P, = P + H = 10+15 = 25mH20

V, = 6cm3

At the top: P2 = P = lOmffjO

V2 = ?

Applying Boyle’s Law, P,V, = P2V2 we have

25mH20 x 6cm3 = 10mH2O x V2

, 25 m H ,0 'x 6 cm i 150 }V, - --------- ------------- = ------= 15 cm

2 10 m H 20 10

Example 9The volume o f air bubble changed from 3cm3 to 12cm3 when it rise from the bottom to the top of a hydroelectric power dam. Calculate the depth of water in the dam if the atmospheric pressure is equivalent to 10m of water.Solution

P =10 mH20

Fig 10.19

At the bottom: Pi = P + H = 1 0 + H

234

V| = 3cm3

At the top: P2 = P = 1 0mH2OV2 = 12cm3

Applying Boyle’s Law, P|Vi - P2V2 we have

(10 + H) x 3 = 10 x 12

30 + 3H = 120

3H = 120-30

3H = 90

H = — = 30m 3

CHARLES LAWCharles’s law states that the volume of a given mass of gas at a constant pressure is directly proportional to its absolute temperature.

VThat is, V a T or — = constant

TV V

Therefore, —- = —-T T

Where V, and T, are initial volume and temperature, while V2 and T2 are final volume and

temperature of the gas, respectively.

Always convert temperature in °C to K when using Charles’s law by adding 273.

Example 10The volume of a given mass of gas is 40cm3 at 27°C. What is its volume at 90°C if its pressure remains constant? NECO 2003SolutionInitial gas volume, V1 = 40cm3Initial gas temperature, T] = 27°C = (27 + 273) = 300KFinal gas temperature, T2 = 90°C = (90 + 273) = 363K

V, V2 , , „ V.T, 40x363 „„ 3From —L = — , final gas volume, V ,= —1 -------------= 48.40cm

T, T2 T, 300

Example 11Dry oxygen is trapped by a pellet of mercury in a uniform capillary tube which is sealed at one end. The length of the column of oxygen at 27°C is 50cm. If the pressure of the oxygen is constant, at what temperature will the length be 60cm. WAEC 1 998SolutionThe volume of the gas (oxygen) is assumed to be proportional to the length of the column of gas, i.e. V a /Initial gas volume, V, = 50cmInitial gas temperature, T, = 27°C = (27 + 273) = 300KFinal gas volume, V2 = 60cm

235

Fromv, V2 „ , ^ T ,V 2 3 0 0 x 60— = ----, final uas temperature, I = --------= ---------- —T, T2 . V, 50

360K or 87°C

Example 12A fixed mass of gas of volume 600cm3 at a temperature of 27°C is cooled at constant pressure to a temperature of 0°C. What is the change in volume? WAEC 1991SolutionInitial gas volume, V, = 600cm3Initial gas temperature, T, = 27°C = (27 + 273) = 300KFinal gas temperature, T2 = 0°C = (0 + 273) = 273K

From final gas volume, V2 = v ,t 2

T,

600x273300

= 546cm3

Change in volume = volume at 27°C - volume at 0°C

i.e. AV = V, - V2 = 600 - 546 = 54cm3

Example 13A gas occupies a certain volume at 27°C. At what temperature will its volume be doubled assuming that its pressure remains constant? NECO 2000SolutionInitial gas temperature, T, = 27°C = (27 + 273) = 300K Initial gas volume, V, = V Final gas volume, V2 = 2V ( ....... its volume be doubled ........)

Substitute above values into to obtain

V _ 2 V 300 “ T2

300 x2V =30Qx2 = 600K or 327°C 2 V

Example 14At what value o f x, y and z is the equation, PxVyTz = constant, a statement of (i) Charles’s law (ii) Boyle’s law (iii) Pressure law (iv) General gas law SolutionThe following laws o f indices can be applied to a problem of this kind, a. ’/X = x '1 b. x = x' c. x° = 1

V(i) Charles’s law; — = constant

X = X_ - v ’T '1T T'

yP xV yT z becomes P°V 'T 1 or l x V T = —

Therefore, for x = 0, y = 1 and z = -1, PxVyT2 = constant is Charles’s law.(Any term P, V or T which is not present in the equation is assigned a superscript o f zero, hence P°).

(ii) Boyle’s law: PV = constant

PV = p ' v 1

... p * v yr becom esP'V 'T” or P 'V 'x l = PV

236

P(tii) Pressure law: —= constantT

P P 1— = —- = P 'T 1 T T 1

pp*V yT z becomes p 'V °T 1 or P 'x l x T ’1 = -

Therefore, for x = l, y=0 and z= -1, PxVyTz = constant is Pressure law.

PV(tv) General gas law: ---- = constant

Therefore, for x = 1, y = 1 and z = 0, P'V/ I / = constant is Boyle s law,

PV _ P 'V 1 T T 1

P 'V 'T '1

. P xV>Xz becomes P V‘T 1 = ----T

Therefore for x = 1, y = 1 and z = -1, p 'V yTz = constant is general gas lav/.

PRESSURE LAWPressure law states that the pressure of a fixed mass of gas at constant volume is proportional to the absolute temperature of the gas.

PThat is, P a T or — = constant

T

P PTherefore, — = —2- T, T,

Where P, and T, are initial gas pressure and temperature, while P and T2 are final gas pressure and temperature, respectively.Always convert temperature in degree Celsius (°C) to Kelvin (K) when applying pressure law.

Exam ple 15Before starting a journey, the tyre pressure of a car was 3 x 105Nm'2 at 27°C. At the end of the journey, the pressure rose to 4 x 105Nm'2. Calculate the temperature o f the tyre- after the journey assuming the volume is constant.A. 400°C B. 300°C C. 273°C D. 127°C JAMB 1997SolutionInitial gas pressure, P, = 3 x lO^Nm "Initial gas temperature, T, = 27°C = (27 + 273) = 300K.Final gas pressure, P: = 4 x 103Nm

P P, P,TFrom , — = — final temperature, T , = — L T, T, - P,

4 x l0 » x 3 0 03x10

or (400 - 273)°C = 127°C'

Exam ple 16A gas at pressure P Nm'2 and temperature 27°C is heated to 77°C at constant volume.

237

Wliat is thè new pressure? A. 0.85PNm ' B 0 86PNm'2 C. 1.16PNm‘‘D. 1.18PNm E. 2.851’N m 2 JAMB 1978Solutionlnitial gas pressure, P| = p NnTInitial gas temperature, T | = 27"C (27 + 273) 300KFinal gas temperature, T 2 = 77°C = (77 + 273) = 350K

p [)From - r = ~ f . final pressure, 1 \ = I t - i i = P x 3 5 ° = i 161’ N m 2

*i T, 300

Example 17A elosed inexpansible vessel contains air saturated with water vapour at 77 °C. The total piessure in the vessel is 1007 mmHg. Calculate the nevv pressure in the vessel il he temperature is reduced to 27 °C . (The S.V.P. o f water at 77 °C and 27 °C respectively are 314mmHg and 27mmHg. lreat the air in the vessel as an ideal gas). WAEC2008 SolutionBased on the law ol partial pressures, the pressure of air alone in the vessel is equal to the difference between the total pressure in the vessel and the S.V.P. o f water at the same temperature.

. initial air pressure, P ,= 1007 - 314 = 693mmHg Final air pressure, P: = ?Initial temperature, T, = 77 °C = 77 + 273 = 350K Final temperature, T2 = 2 7 "C = 27 +273 = 350K.

Because the vessel is inexpansible, its volume does not change, therefore pressure , can be applied.

Substitute into ! l = I ii ; t2

to obtain

693 _ P2 350 300

1\693X300 =594j|)m/

350The pressure in the vessel at 27 "C is equal to the sum ofthe pressure ol airand the S.V.P. o f water at 27 "C.Vessel pressure at 27 "C = air pressure + S.V.P. o f water at 27 UC.

= 594 mmHg t- 27 mmHg 621 mmHg

GENERAL (JAS LAW

A combination o f Boyle’s, ( harles’s and Pressure laws results in the general eas law for

ideal gas.PV

That is. constant

Therefore, I’.V,T,

238

Example 18Ehe volume and pressure of a given mass of gas at 27°C are 76cm3 and 80cm of mercury respectively. Calculate its volume at s.t.p. WAEC 2000solution ^ (-73K) and 760mmHg (1.013 x 105 Nm :) arc called standard temperature and

pressure, usually abbreviated, s.t.p.Finn^ volume, V] = 76cni3 Initial gas pressure, P, = 80cmHgFinnl §aS V° Ume’ V- = ? Initial gas temperature, T, = 27°C = (27 + 273) = 300K

gas temperature, T; = 273K Final gas pressure, P2 = 760mmHg = 76cmHg

Substituting above values into the ideal gas equationPIV! _ P,V,T “ we have,

1 A :80cmHg x 76cm3 76cmHgxV,

3°0K 273K

y _ 80cmHg x 76cm3 x273K _ ^ gcmr 2 ~ 76cmHg x 300K

Example 19A given mass of an ideal gas occupies a volume V at a temperature T and under a pressure P. If the pressure is increased to 2P and the temperature reduced to 'AT, then what is the percentage change in the volume of the gas?A 0% B. 25% C. 75% D. 300% E. 400%SolutionInitial gas volume, V, = V Initial gas pressure, P, = P Final gas pressure, P2 = 2P

JAMB 1984

Initial gas temperature, T, = TFinal gas temperature, T; = 'AT = 0.5TFinal gas volume, V2 = ?

Substituting into P,v, P,V,T, T,

P xV 2PxV,

we obtain

0.5T

v _ PxVxQ ,5T _ Vx0.5 0.5V2PxT 2

Percentage change in volume

2

V! ~ V2 V,

x 1 0 0

V -0.25 VV

0.75VV

x 1 0 0

xlOO =0.75x100 = 75%

Example 20The pressure of a given mass of a gas changes from 300Nm': to 120Nm'2 while the temperature drops from 127°C to -73°C. What is the ratio of the final volume to the initial volume? A. 2:5 B. 5:4 C. 5:2 D. 4:5 JAMB 2001SolutionInitial gas volume, P, = 300Nm " Initial gas volume, V, = V,Initial gas temperature, T, = 127°C = (127 + 273) = 400KFinal gas pressure, P; = 120Nm' Final gas volume, V2 = V:

239

Final gas temperature, T, = -73°C = (-73 + 273) = 200KP V. p vSubstitute into ——L _ f_2v 2T. T,

300 xV, = 120V,400 200

Rearranging, we have 300x200 _ V,400x120 v (

60000 _ V,48000 '

60 _48 “ V,

Divide both numerator and denominator by 12

_y_2_

4 v .

The ratio of final volume (V2) to the initial volume (V,), V2: V, = 5:4

W O R K DONE BY AN EXPANDING GAS ATC O N ST A N T PRESSURE

Work done = Force x distance (length) ie. W = Fx/ ............................... (])

~ F o r c e _ FP r e s s u r e = ---------- i .e . P = — .......................................... (2)

A r e a A

. Force = Pressure x Area i.e. F = PA ................................... (3)

Substitute equation (3) into equation (1)

If F = PA

Then W = F x / becomes

W = PA / .....................................................................................(4)

Volume = Area x length i.e. V = A / .......................................................(5)

Substitute equation (5) into equation (4)

If V = A /

Then W = PA / becomes

W = PV ......................................................................................(6)

When a gas expands the volume (V) changes.

Therefore work done by expanding gas = Pressure x Change in volume.

Equation (6) becomes, W = PAV or W = P(V; -V ,)

Where W = work done by expanding gas in Joules (J)AV = (V: - V,) = change in volume V2 = final volume after expansion V, = initial volume before expansion

240

Example 21 ^The volume o f a fixed mass of gas increased from 0.04m to 0.32m while the p remains constant at 1.5 x 10'Nm *. I low much work was done by the gas. Solution 3Initial volume, V| = 0.04m3 Final volume, V2 = 0.32mConstant pressure, P = 1,5 x 105Nm "

Work done = constant pressure x change in volume

W = P(V2- V , )

W = 1.5 x 105 (0.32 -0 .0 4 )

W = 1.5 x 105 x 0.28 = 4.2 x 104Joule

Example 22The work done by a gas when it expand to eight times its initial volume o f 0.5 x is 35.5J. At what constant pressure did the expansion occur?SolutionWork done, W = 35.5J Initial volume, Vi = 0.5 x 104m‘Final volume, V2 = 8 x (0.5 x 104 ) = 4 x lO^m3 From W = P(V2- V |)

Constant pressure, P = ----------V2 -V ,35.5

0.00035

35.54x1 O'4 -0 .5x10 '" '

1,014x10s N m 1

1 O^nr

CUBIC AND PRESSURE EXPANSIVITY OF A GASCubic Expansivity o f a gas at constant pressure is the fraction of its volume at 0°C

by which the volume o f a given mass of gas expands per Kelvin change in temperature.

. . change in volume from 0°CCubic expansivity, y= — -------- ——— --------:-------------------

volumeatO Cxchangetn temperature

i.e. Y = Vq- V 0V0xG

Where y = cubic or volume expansivity of gas measured in 0°C"' or K '1 V0 = volume o f gas at 0°C V0 = volume o f gas at 0°C 0 = change in temperature of gas

Because the length o f a gas column in a tube of uniform cross-sectional area can be tak> 11 as being proportional to its volume (i.e. V a I), cubic expansivity can also be written ;.

Y =W o

/„X0

Where /e = length o f gas column at 0"C l 0 = length o f gas column at 0°C 0 = change in temperature of gas

241

Example 23The volume of air in a capillary tube is 56cm3 at 0°C and 75cm3 at 100°C. Calculate the cubical expansivity of the air at constant pressure.SolutionVolume of air at 0°C, V0 = 56cm3 Volume of air at 100°C, Vo = 75cm3Change in temperature, 0 = 100°C

Cubic expansivity, y = ^2—Y°. - 2 } - . l 9._Vox0 56x100 5600

Y= 3.39 x 10'3K '! or 3.39 x lO '^C '1Example 24The cubic expansivity of a fixed mass of gas trapped in a capillary tube of uniform cross- sectional area is 3.68 x 10 3K 1. If the gas column at 0°C is 13.69 x 10 2m long, calculate its length at I00°C.SolutionCubic expansivity, y = 3.68 x 10"3K '! Length of gas column at 0°C, lB = 13.69x 10 2m Length of gas column at 100°C, /9 = ? Change in temperature, 0 = 100°C

Substitute into y = —----— to obtain/ox0

3 .6 8 x 1 0 -^ /o - 13-69f ° " 2 13.69xl0~2 xlOO

/e - 13.69 x 10'2 = 3.68 x 10‘3 x 13.69 x 10'2 x 100 /e - 13.69 x 10'2 = 0.0503792

/, = 0.0503792 + 13.69 x 10'2 /„ = 0.1873m or 18.73 x 10'2m

Example 25The cubic expansivity of a certain gas at constant pressure is If a given mass of

273the gas is held at constant pressure and its volume at 0°C if 273m3, determine the volume of the gas at 273°C. WAEC 2008SolutionCubic expansivity, y = —L K'1 Volume of gas at 0°C, V0=273m 3

Volume of gas at 273°C, Ve = ? Change in temperature, 0 = 273°C = (273 + 273)K= 546K

The change in temperature 0 given in centigrade (°C) had to be converted to Kelvin (K) because the cubic expansivity is given in K‘ .

V — VSubstitute into y = —-— -f- to obtain

V x0

1 VB- 273273 273x546

Rearranging,273x546

273 ~ °-273

546 = V0—273

/. Volumeof gas at 273°C, V0 =546 + 273 = 819m

242

Pressure hxpansivity of a gas ai constant volume is the fraction o f its pressure at 0 ( l>y " Inch the pressure ol a given mass of gas increases per Kelvin change in temperature.

Pressure expansivity, p =change in pressure fromO'C

pressurcat ()"Cx change in temperature

p — pPressure expansivity, P = —-----—

Po* 0

Where P0 = gas pressure at 0°C Ph = gas pressure at 0HC 0 = change in temperature of gas P = pressure expansivity of gas

Example 26

What is the pressure expansivity of a gas whose pressure changes from 83mml!g at 0 to 105mmHg at 70°C?Solution

Gas pressure at 0°C, P0 = 83.6mmHg Gas pressure at 70°C, Pe = 105mmHg Change in temperature, 0 = 70°C

D . . _ P0- P , 105- 83.6 21.4 , , _Pressure expansivity, P = —-----2- = ------------- = -------= 3.66x 10 (P„x0 83.6x70 5852

243

EXERCISE 10

1.

When one arm of a U-tube manometer is connected to a gas supply, the levels o f rr 'rcury in the two arms of the U-tube are as shown in the diagram above. If the atmo > ieric pressure is 76.0cmHg, what is the gas pressure?A. 62.6cmHg B. 72.5cmHg C. 79.5cmHg D. 85.9cmHg JAMB 1988 Ans:85.9cmHg2.

In the diagram above, if the atmospheric pressure is 760mm, what is the pressure in the chamber G?A 660mm B. 690mm C. 830mm D. 860mm JAMB 1994 A ns: 830mm3. A thread of mercury of length 15cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length o f the trapped air column is 20cm. Calculate the length of the air column when the tube is held

(i) horizontally(ii) vertically with the open end underneath.

(Atmospheric pressure = 76cm of mercury) WAEC 1988 Ans: (i) 23.95cm (ii) 29.84cm4.

Capillary tubei i

r 15cmMercury thread .....1 i .....30cm

f ........Dry air

Fig 10.22

244

The set-up illustrated above shows a capillary tube of uniform cross sectional area in two different arrangements. Using the data in the diagrams, calculate the pressure of t e atmosphere. WAEC 1991 A n s : 75cm o f mercury5. A thread of mercury of length 16cm is used to trap some air in a capillary tube of uniform cross-sectional area and closed at one end. When the tube is held vertically, with the closed end at the bottom, the length of the trapped air column is 30cm. Calculate the length of the air column when the tube is held,

(i) horizontally(ii) vertically with the open end underneath. [Atmospheric pressure - 76cm of

mercury] NECO 2006 Ans: (i) 36.32cmHg (ii) 46cmHg

Fig 10.23

In the figure above, a thread of mercury 76cm long is used to trap some dry air in a capillary tube of uniform cross-section closed at one end. Calculate the ratio o f the pressure P2/P\ [Atmospheric pressure is 76cm of mercury] NECO 2004

Ans: P: _ 76+76 = 1 52 _P, 76 76

7.

In the J-tube above, Y and X are on the same horizontal level and 30cm3 of air is trapped above Y when the atmospheric pressure is 75cmHg. Calculate the volume of air trapped above Y when 15cmHg is now poured into the limb above X.A. 15cm3 B. 25cm3 C. 35cm3 D. 45cm3 JAMB 1999 Ans: 25cm38. An air bubble o f volume 2cm’ is formed 20m under water. What will b e itsvolume when it rises to just below the surface of the w'ater if the atmospheric pressure is equivalent to a height o f 10m o f water? WAEC 1995 Ans: 6cm39. An air bubble of volume 4cm3 is formed 20m under water. What will be itsvolume when it rises to just below the surface of the water if the atmospheric pressure is equivalent to a height o f 10m of water? NECO 2002 (Ans: 12cm3)10. An air bubble rises from the bottom to the top of a water dam w hich is 40m deep The volume o f the bubble just below' the surface is 2.5cm3. Find its volume at t h e b o u m

o f the dam, if atmospheric pressure is equivalent to 10m of water.A. 10.0cm3 B. 2.0cm3 C. 1.6cm3 D. 0.625cm3 E. 0.5cm3 JAMB 1979 Ans: 0.5 f11. The pressure o f a fixed mass of gas is 2 . 0 x 1 0 N m at a k n o w n t e m p e r a i

Assuming that the temperature remains constant, what will be the pressure of t h e ga.->

volume is halved? WAEC 1996 (Ans: 4.0 x h: V/;;

245

12. A fixed mass of gas occupies a volume of 20cm3 at a pressure of 700mml(g.Assuming that the temperature remains constant, what will be the volume of the gas at 750cml Ig? WAEC 1997 (Ans: 18. 7cm )13. When the pressure of a fixed mass of gas is doubled at constant temperature, the volume of the gas is;A. increased four times B. doubled C. unchanged D. halved WAEC 20<)3Ans: D14. Dry hydrogen is trapped by pellet of mercury in a uniform capillary tube closed atone end. If the length of the column of hydrogen at 27"C is I .Om, at what temperature will the length be 1,20m? WAEC 1990 Ans: 360K or 8 / 'C15. A gas which obeys Charles’s law exactly has a volume of 283cm3 at 10°C\ Whatis its volume at 30°C? WAEC ¡992 Ans: 303cm}16. A gas has a volume of 546cm3 at 0°C. What is the volume o f the gas at 100"C ifits pressure remains constant? WAEC 1993 Ans: 746cm317. A gas occupies a certain volume at 27°C. At what temperature will its volume be three times the original volume assuming that its pressure remains constant?

WAEC 1994 Ans: 900K or 627°C18. A balloon containing 546cm3 of air is heated from 0°C to 10°C. If the pressure iskept constant, what will be its volume at 10°C? WAEC 1995 Ans: 566cm319.

The diagram above shows the variation of volume V of a gas with temperature 0 in a Charles’s law experiment. What is the value of the temperature at the point X?

WAEC 2000 Ans: -273°C20. The volume of a given mass of gas is 273cm’ at 0°C. What is the volume at273°C if its pressure remains constant? NECO 2006 Ans: 546cm321. A gas occupies a volume of 300cm' at a temperature of 27°C. What is its volume at 54°C when the pressure is constant?A. 150cm' B. 273cm' C. 327cm' D. 600cm' J.AMB 1986 Ans: 327cm322. A column o f air 10.0cm long is trapped in a tube at 27°C. What is the length of the column at 100°C?A. 12.4cm B. 13.7cm C. 18.5cm D. 37.0cm J.-IMB 1989 Ans: 12.43cm23. The equation P 'V yTz = constant, is Charle’s law when

A. x = l , y = -1 ,2 = 1 B. x = 0 ,y = 1 ,z = -1B. x = 1, y = 0. z = -1 D. x = 0 ,y = l , z = 1 J.iMB 1995 Ans: B

24. A gas has a volume of 100cm ' at 27°C. If it’s heated to temperature T until a finalvolume of 120cm ' is attained, calculate T.A. 33°C B. 60°C C. 87°C D. 114°C JAMB 1998 Ans: 360k or S ^ C25. A given mass o f gas has a pressure ot SONm ‘ at a temperature o f 47°C. If thetemperature is reduced to 27°C with the volume remaining constant, what is the new pressure? WAEC 19SS Ans: 75.Vm '26. The pressure of air in a tyre is 22.5N’m ' at 27°C. If the air in the tyre heats up to47°C. calculate the new pressure o f the air. assuming that no air leaks out and that the change in volume o f the air can be neglected. WAEC 1994 Ans: 2 4 \m '27. A given mass of an ideal gas has a pressure of 500N m ' at -13°C. If its volumeremains constant, calculate its pressure at 247°C. If AEC 2001 AnslOOO.Xm'

246

2». I I,c lyre p r e n r e o f . car wa.s f,.„,»l » I* I * ® " ' '» ' h' * ‘lempenrrure o f 27',C. Hard driving in lire afternoon raised lire lempera 57"C. The tyre pressure had by afternoon; n «ir-mll»A. increased by 15.0cmHg B. decreased by 15.0cmI Ig C. mcrt***f m u2 Ans AD. decreased by 22.50cml Ig E. remained constant. • f ■29. A motor tyre is inflated to pressure of 2.0 x 10 Nm when t e temperais 27°C. What will be the pressure in it at 87°C assuming that the volume of the yrdocs not change? A. 2.6x10’Nm ’ B. 2 .4 x l0 ’ Nm 1 C. 2.2x l 0'N m

JAMB 1994 Ans: 2.4 x I(fN m30. A given mass of gas at a temperature of 30"C is trapped in a tube of volume V. C alculatc the temperature of the gas when (he volume is reduced to two-thir o its original value by applying a pressure twice the original value.

WAEC 1992 Ans:404K or 131 (31. 500cm'1 of a gas is collected at 0°C and at a pressure of 72.0cm o f mercury. Whatis the volume of the gas at the same temperature and at a pressure o f 76.0cm o f mercury. A. 76 x 500cm' g 72x 500cm’ Q 72x76cm’ |) 76cm’ £. 72cm’

72 70, 500 72x500 76x50072x500WA PC 1()()4 Ans: B

32. The volume of an ideal gas at a pressure of 77cmHg and temperature 60 C is240cnv. If the temperature and pressure arc increased to 98°C and 81cmHg respectively, calculated the new volume of the gas. WAEC 2001 Ans: 254.2cm333. The volume of a given mass of an ideal gas at 327K and 9.52 x 104Pa is 40cm '. Calculate the volume of the gas at 273K and 1.034 x 105Pa. WAEC 2003 Ans: 30.75cm334. The temperature of 900cm1 of an ideal gas at a pressure of 114cmIIg is 27°C.Calculate its volume at 76cmHg and 0°C. WAEC 2004 Ans: 1228.5cm335. A quantity o f sungas at -133HC occupies '/3m1 under a pressure 1.4 x 105Nm"7 If the gas occupies 31m1 at 37°C, what will its pressure be?A. 1.26xl03Nm 1 B. 10‘Nm 1 C. 1.35xl0JNm ' D. 2.2x10‘ N m 1 E. lffNm2

JAMB 1980 Ans: 3.3 x ltfN m . no correct option36. If the pressure on 100cm1 of an ideal gas is doubled while its Kelvin temperature is halved, what then will become the new volume of the gas?A. 25cm1 B. 50cm1 C. 100cm1 D. 200cm3 E. 400cm3 JA M B 1983 A ns: 2 5 cm 337. A quantity o f gas occupies a certain volume when the temperature is -73°C and the pressure is 1.5 atmospheres. If the pressure is increased to 4.5 atmospheres and the volume is halved at the same time, what will be the new temperature of the gas‘>A. 573°C B. 327°C C. 300°C D. 110°C E. 27°C JAMB 1985 Ans: 300K or 27°C38. A mass o f gas at 7°C and 70cm of mercury has a volume of 1200cm3. Determine its volume at 27°C and pressure of 75cm of mercury.A. 1200cm’ B. 1378cm’ C. 4320cm1 D. 4629cm1 JAMB 1989 Ans: 1200cm339. A gas with initial volume 2 x 10'6m3 is allowed to expand to six times its initial volume at constant pressure of 2 x lO^Nni'2 what’s the work done?A. 2.0J B. 4.0J C. 12.0J D. 1.2J JAMB 2001 Ans: 2.0J40. A small circular membrane is 10cm below the surface of a pool o f mercury when the barometric height is 76cm o f mercury. If the density of mercury is 13600kgm‘3, what is the pressure on the membrane in Nm'2? [g=10m/s2] WAEC 1992 Ans: 1.17 x 10iNm':41. As a result o f air at the top of a barometer the height of the mercury column is 73.5cm when it should be 75.0cm; the volume of the space above the mercury is 8.0cm3. Calculate the correct barometric height when the barometer reads 74.0cm and the volume o f the space above the mercury is 6.0cm3.A.72.0cm B.74.5cm C.75.lent D.76.0cm JAMB 1987Ans:76cm42. The pressure o f 3 moles o f an ideal gas at a temperature of 27°C ha\ ing a volumeof 10"nr is? [R=8.3J molK '1] A .2.49xlO ^N m 2 B .7 .4 7 x l0 6Nm C.2.49 x 10s N m " D. 7.47 x 10s N m '2

JAMB 2002 Ans:7.47 x 106NnC Hint, use - nR

247

43. A uniform capillary tube of negligible expansion scaled at one end, contains air trapped by a pellet of mercury. The trapped air column is 13.7cm long at 0°C and 18.7cm long at 100°C. Calculate the cubical expansivity of the air at constant pressure.

WAEC 1993 Arts: 3.65 x l ^ K '44. The volume of a given mass of gas is 40cm3 at 27°C. What is its volume at 90°Cif its pressure remain constant? NECO 2003 Ans: 48.4cm345.

JAMB 2006 Ans: 86cmHg46. 1 he pressure of a fixed mass of an ideal gas at 27°C is 3Pa. The gas is heated at a constant volume until its pressure is 5Pa. Determine the new temperature of the gas.

WAEC200727 (Ans: 227°C)47.

Atmospheric pressure = 76cmHg

Using the above diagram, calculate the pressure of the gas supply. NECO 2007~l Ans: 79cmHg

48. A sealed flask contains 600cm3 of air at 27°C and is heated at 35°C at constant pressure. The new volume isA. 508cm3 B. 516cm3 C. 608cm3 D. 616cm3 JAMB 2008 Ans: 616cm349. The pressure o f two moles o f an ideal gas at a temperature of 27°C and volume 10‘2 m3 isA. 4.99 x 105N m '2 B. 9.80 x 103Nm-2 C. 4.98 x 103Nm~2D. 9.80 x 105Nm-2 [R = 8.313 J m o l 1fC1 ] JAMB 200930 Ans: A50. A thread o f mercury of length 20cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air column is 15cm. Calculate the length of the air column when the tube is held

(i) horizontally (ii) vertically with the open end underneath.[Atmospheric pressure = 76cmHg] WAEC 2009EL Ans: (i) 18.95cm (ii) 25.71 cm

248

51.The equation PxVyTz = Constant is Charles’ ifA. x = 0, y = 1 and z = —1B. x = 0, y = —1 and z = 1C. x = 1, y = 1 and z = 0D. x = 1 y = 1 and z = — 1E. x = 1, y = 1 and z = 1 NECO 2009'7 Ans: A

?n°rhe ^ressure ot a certam mass ot gas at constant volume increases from 250mn to 400mmHg at 50°C. Calculate the pressure expansivity of the gas.

NECO 2009E'2 Ans: 0.02°C_1

249

PRESSURE IN FLUIDSPressure is defined as the force acting perpendicularly per unit area.

Pressure = Force FThat is, P = —

Area AAlso,

Pressure = WeightThat is, P = ™

Area A

Note: F - W - mg. Where m is mass (kg) and g acceleration due to gravity (m/s2). Pressure is measured in Nm'2. 105Nm'2 = 105Pa = lbar

Example 1A rectangular water tank of weight 4.5 x 103N measures 2.0m by 1.5m by 1.2m. Calculate the minimum pressure it can exert when resting on a horizontal surface.

N E C 0 2 <6.SolutionMinimum pressure is obtained when the tank rest on the greatest area.From, 2.0m by 1.5m by 1.2m, the greatest area, A = 2 x 1.5= 3.0m2 Weight, W = F = 4.5 x 103N

Pressure = — = = 1 .5x l03A/>»~2A 3.0

Example 2A rectangular block of dimensions 2 .0m xl.0m x0 .5m weighs 200N. Calculate the maximum pressure exerted by the block on a horizontal floor. WAEC 2008SolutionMaximum pressure is obtained when the block rest on the least area.From, “ 2 .0m xl.0m x0.5m ” the least area, A = 1,0m x 0.5m = 0.5m2 Weight o f block W = F = 200N

F 200NPressure exerted P = — = ------- = 400Nm'2A 0.5

Example 3A rectangular tank contains water to a depth of 2m. If the base is 4m x 3m calculate the force on the base. (Density o f water = 103kgm'3, g=10ms'2) A. 2 .4 x l0 4N B. 2 .4 x 104N C. 2 .0 x l0 4N D. 1 .7 x l0 5N

JAM B 1986.SolutionThe force (F) on the base o f the tank will depend on the weight (W) of the water.

Volume o f tank = 4 x 3 x 2 = 24m3

Density o f water = lOOOkgm 3

Density = mass_ Mass o f water, m = Density x Volumevolume

m = 1000 x 24 = 24000kg

Weight o f water, W = F = mg = 24000 x 10 = 2.4 x 105N.

250

FLUID PRESSUREPressure in fluids (liquids or gases) depends on the depth (height) and density

A

P =

P =

P =

P =

Where

h

mgA

px V x g A

p x A x h x gA

phgP = pressure (Nm ‘), m = mass (kg),A = area (m2)

= height or depth(m)

F = force (N)g = acceleration due to gravity(m/s ) p = density (kgm '3)

Example 3A reservoir is filled with a liquid of density 2000kgm’3. Calculate the depth at which t e pressure in the liquid will be equal to 9100Nm"2 (g=10m/s2) WAEC 2002SolutionDensity of liquid, p = 2000kgm’3; Pressure, P = 9100Nm’2; g = 1 0 m /s2

P = /?hg, /. depth, h = - ^ - = — — = 0.455mp g 2000x10

Example 4A 5m x 4m x 3m vessel of negligible weight is filled with a liquid of density 2500kgm‘3. If the vessel is placed on a fiat surface, what is the maximum pressure it can exert? SolutionThe maximum pressure is exerted when the vessel rests on its least area.

From 5m x 4m x 3m, the least area, A = 4 x 3 = 12m2.Volume of vessel, V = l x b x h = 5 x 4 x 3 = 60m3 Density of liquid p = 2500kgm '3

From p = m/v, mass, m = p x V = 2500 x 60 = 150000kg W = F = m g= 1.5 x 105kg x 10 = 1.5 x 106N

P_EA

1.5xl06N12

= 12500 = 1.2 x l0 5N/m2

Alternatively, we could apply the pressure formular, P = phg because pressure in a liquid depends on depth or height. I f the vessel rests on its least area 4 x 3 , then the height(h) becom es 5m.

P = phg = 2500 x 5 x 10 - 1.25 x 105N/m2

ATMOSPHERIC PRESSUREAt sea level the atmospheric pressure is 760mmHg or 1.013 x 10sN/m2. Pressure reduces as you go higher into the atmosphere (above sea level). Pressure increases as you o lower (below sea level). The pressure formular can be applied in both cases.

251

Example 5Calculate the length of the liquid in a barometer tube that would support an atmospheric pressure of 3.06 x 10'Nm2 if the density of the liquid is 1.36 x 104kgm 3 (g= 10m/s2). SolutionDensity of liquid, p = 1.36 x 104kgm‘3 Pressure, P = 3.06 x 105Nm'2; (g=10m/s2).

From pressure formula, P = phg, Lenght of liquid, h = — = 3 06x10— = 2.25mH p g 1.36x10 xlO

Example 6A pilot records the atmospheric pressure outside his plane as 63cm of Hg while a ground observer records a reading of 75cm of Hg with his barometer. Assuming that the density of air is constant, calculate the height of the plane above the ground. (Take the relative densities of air and mercury as 0.00136 and 13.6 respectively). WAEC 1998SolutionPressure change for air = pressure change for mercury

p x H x g = p x h x g

or p x H = p x h

ie. plir x H = pHg x h

Where ps,r and pHg are densities or relative densities of air and mercury respectively.

0.00136 x H = 13.6 x (75 - 63) = 13.6 x 12

Height above ground, H = 13.6x12 _ j2000cm = 1200m0.00136

Example 7The atmospheric pressure due to water is 1.3 x lO^Nm"2.. What is the total pressure at the bottom of an ocean 10m deep? (Density of water = lOOOkgm 3, g=10m/s2)A. 1.3x107Nrn1 B. 1.4xl06Mn~2 C. 1.4x1 Nrn2 D. l.OxUfNm'2 JAMB1998 SolutionAtmospheric pressure, P = 1.3 x lO ^m 2.Depth of ocean, h = 10mDensity of water, p = lOOOkgm'3; g = 10m/s2Total pressure = P + phg

= 1.3 x 106+ 1000 x 10 x 10 = 1.4 x lO ^m '2

PASCAL PRINCIPLEPascal principle states that when a fluid completely fills a vessel and a pressure is applied to it at any part of the surface, that pressure is transmitted equally throughout the whole of the enclosed fluid and the walls of the containing vessel.Pascal principle is applied in a hydraulic press as follows.

P r e s s u r e , P =F,A ,

o r f 2 =a 2f ,

A ,

252

Where Ft = effort or force applied on smaller piston A, = cross sectional area of small piston F2 = load or force applied on larger piston A2 = cross-sectional area of larger piston

The velocity ratio (V.R.) of an hydraulic press is

X A = X 2A, or — = —X 2 A,

Where X| = distance moved by the small piston X2 = distance moved by the larger piston

Example 8A hydraulic press has a large circular piston of radius 0.8m and a circular plunger o f radius 0.2m. A force of 500N is exerted by the plunger. Find the force exerted on the piston. A. 8000N B. 4000N C. 2000N D. 3 IN JAM B 1995Solutionr, = 0.2m; r2= 0.8m; F, = 500N

Area of large piston, A2 = n r \ = ;rx O.82 = 0.6471 Area of small piston, A, = nr^ = n x 0.22 = 0.4jt Force exerted on small piston, F, = 500N

F FSubstitute into — = —- to obtain

A, A2500 _ F2

0.04;z 0.64^

Force on larger piston, F , = --------------- = 8000N2 0.047t

HARE’S APPARATUSThe densities o f two liquids can be compared by means o f Hare’s apparatus a Applying the pressure formular on the figure below, the following equ: obtained.

hiPi g = hip:g

253

•l| Pi = ll:p2

— = — — relative density of p, (liquid X) if p2 (liquid Y) is the density 01 water. p2 h,

Example 9In an experiment using Hare’s apparatus, the height of water column was 10.5cm and that of glycerine is 8.1cm. (a) Calculate the density of the glycerine if the density of water is lOOOkgm'1. (b) If the density of was not given in (a) above, show how you would calculate the relative density of glycerine.Solution(a) Height of glycerine column, h, = 8.1cm

Density of glycerine column, p, = ?Height of water column, h. = 10.5cm Density of water column, p2 = lOOOkgm3

Substitute into — =— ---- ..P 2 h.

(b)

Pi _ 10.5 1000 ~ 8.1

Pi =-1000x10,5

H= 1296.3 = 1300kgm-

Height of water column, h2 = 10.5 Height of glycerine column, hi = 8.1

Ik = relative density o f glycerine, if h2 is the height o f water column.b|

h, 10.5 , ,Relative density, — = —— ~ 1 3

h, 8.1

254

EXERCISE 11.. ~r -i 7 v l(V:N on a groove of1. The stylus of a phonograph record exert a force ol /

radius lO'm. compute the pressure exerted by the stylus on the groove.^ ^A. 2.45x10s Am'2 B. 3.45x10*Am'2 C. 4.90x10s Am D. 2.42 x 10 Nm . ,

JAMB 2003 Ans: 2.45 ^2. A man exerts a pressure of 2.8 x 103 Nm" on the ground and has 4 x 10 mfeet in contact with the ground. What is the weight of the man.’ . . _A. 112N B.140N C. 102 D. 70N JAMB 2004 A ns:H 2N3. The horizontal door of a submarine at a depth ol 500m has an area 0 ■ *Calculate the force exerted by the sea water on the door at this depth. [Relative ensity o sea water = 1.03] [Atmospheric pressure = 1.0 x 105Nm2] [Density of pure water lOOOkgm3] [g=10m/s2]. WAEC2002 Ans: 2.06 x I(iN4. A diver is 5.2m below the surface of water density lOOOkgm ■ If the atmospheric pressure is 1.02 x 105Pa, calculate the pressure on the diver (g=10m/s ).

WAEC 2004 Ans: 1.54 x ¡O'Pa5. Two divers G and H are at depths 20m and 40m respectively below the water surface in a lake. The pressure on G is Pi while the pressure on H is Pr- If tl*c atmospheric pressure is equivalent to 10m of water, what is the value of P2/P1?

WAEC 1985 A ns:'//6. A liquid of mass 1.0 x 103kg fills a rectangular tank of length 2.5m and width 2.0m. If the tank is 4m high what is the pressure at the middle of the tank? [g=10m/s ]A. 1.0xl04Am~2 B. 2.0xl0, Am'2 C. 1.5xltf Am'2 D. 1.0x103Am'2

JAMB 1997Ans: 1 x l& N m :7. A weightless vessel of dimensions 4m x 3m x 2m is filled with a liquid of density lOOOkgm3 and sealed. What is the maximum pressure this container can exert on a flat surface? [g=10m/s2] A. 9x10*Am'2 B. 4 x l0 4Am'2 C. 3x l0 4Am'2 D. 2 x l0 4Am'2

JAMB 1993 Ans: 4 x l& N m 28. Normal atmospheric pressure at seal level is lO'^Nm'2 and the acceleration due togravity is approximately 10m/s. If the atmosphere has a uniform density of lkgm°, what is its height? Hint use p = phg A. 100m B. 1000m C. 10,000m D. 100,000mE. 1000,000m JAMB 1984 Ans: 10,000m or 1 tfm9. A faulty barometer reads 72.6cmHg when the atmospheric pressure is 75.0cmHg. Calculate the atmospheric pressure when this barometer reads 72.0cmHg.

WAEC 2004 Ans: 74.38cmHg10. Calculate the length of the liquid in a barometer tube that would support anatmospheric pressure of 204000Nm ", if the density of the liquid is 5200kgm'3 [g=10m/s2] NEC0 2004 Ans: 3.92m11. What is the length of the liquid in a barometer tube that would support anatmospheric pressure of 102000Nm'2 if the density of the liquid is 2600kgm'3 [g= 10m/s2] A. 0.75m B. 0.76m C. 3.92m D. 39.23m JAMB 1990 Ans: 3.92m12. A pilot records the atmospheric pressure outside his plane as 63cm of Hg while a ground observer records a reading of 75cm of Hg for the atmospheric pressure on the ground. Assuming that the density of the atmosphere is constant, calculate die height of the plane above the ground. (Relative density of Hg = 13.6 and that of air = 0.00013)A. 1200m B. 6300m C.7500m D. 13.800m JAMB 1991 Arts. 120,000cm = 1200m13. The hydrostatic blood pressure difference between the head and feet o f a boy standing straight is 1.65 x lO^Nm'. Find the height of the boy. [Density of blood = 1 1 \ 103kgm'3, g= 10m/s2] Him: P = phg A. 0.6m B. 0.5m C. 2.0m D.1.5m

JAM B 2002 Ans: 1.5m14. The areas of the effort and load pistons of a hydraulic press are 0.5m2 and 5m:

respectively. If a force Fi of 100N is applied on the effort piston, what is the force F-- on the load. A.10N B.100N C.500N D.1000N E.5000N JAMB 1985 Ans.lOOON

255

15. In a hydraulic press, a force of40N is applied on the effort piston of area 0.4m If the force exerted on the load is 400N, what is the area of the large piston?A. 8m2 B. 4m2 C. 2m2 D. lm 2 JAMB 2004 Arts: 4m

16.

The figure above shows the heights of two liquids X and Y when some air is sucked out of the apparatus through the pump P. The diameter of the tube in X is twice that o f the tube in Y. What is the relative density of liquid X with respect to liquid Y.A. 1/3 B. 2/3 C. 3 D. 6 JAMB 1992 Ans. %17. A reservoir 500m deep is filled with a fluid of density 850kgm'3. If the atmospheric pressure is 1.05 x 105Nm'2, the pressure at the bottom of the reservoir isA 4 .2 8 x l0 m n '2 B. 4.72 x O'Tvim'2 C. 4.25x lO 'W 2 D. 4.36x lO^m "2

[g= 10ms’2] JAMB 2007 Ans: 4.36x lOpNm218. In the Hare’s apparatus, water rises to a height of 26.5cm in one limb. If a liquid rises to a height of 20.4cm in the other limb, what is the relative density of the liquid?A. 0.8 B. 1.1 C. 1.2 D. 1.3 JAMB 2008 Ans: 1.319.

The diagram above illustrates a Hare’s apparatus. represent densities and h,, h:heights o f column o f liquids. Which of the following equations is correct?

1 . , = ^ B . h , = ^i i

h. = ^ - D. h. =h J .

WAEC2008 Ans: B19. In a hydraulic press, a force of 40N is applied to the smaller piston of area 10cm2. If the area of the larger piston is 200cm2, calculate the force obtained.

WAEC 20093 Ans: 800N

256

12WAVES

GEN ARAL CHARACTERISTICS OF W AVESA wave is a disturbance which travels through a medium and tranifeTS energv fro point to another, without any permanent displacement of the medium.A typical wave is defined by the following tenns:

a. Amplitude!A). This is the maximum displacement of the particles of a w av efrom its mean or equilibrium position. It's measured in meter.

b. W’avelength ('/.). This is the distance between two successive troughs or two successive crests. It is also the distance covered by a wave when it completes one cycle, vibration or oscillation. It is measured in meter. A wavelength can be subdivided into half ('/2), three-quarters ("''It) and one-quarter ('74).

c. Period (T). This is the time taken by the particle of a wave to perform one complete cycle or oscillation. It is also the time required by a wave to travel one wavelength. It is measured in seconds and can be stated thus:

time taken _ time takennumberof cycles number of wavelengths

d. Frequency (f) It is the number of cycles or oscillations which the wave completes in one second. Frequency is measured in Hertz(FLz) or s '1

_ numberof cycles _ number of wavelengths time taken time taken

Frequency and period are related as follows:

e. Wave Speed or Velocity (V). —period T

or V = wavelength X frequency = /.f

257

. . Xie V = - or V = Xf

Example 1

SolutionWave velocity, V = 300ms"1; frequency, f =?For a particular wave, the number (n) of wavelength and their displacement (d) are related by

Hi = £kd, d2

where ri|, di = one wavelength and its corresponding distanceIl2,d2 = any number of wavelength and its corresponding distance.

The diagram shows 2Zi or 2.5 wavelengths with a corresponding length of 3m. Therefore, ri2 = 2.5X. and d2 = 3m. For one wavelength, il| = lA. and d) = ?

Substitute into — = — to obtaind, d 2 U = 2.5/1 d, 3m

cross multiplying, d,x2.5X = 3m xU

, 3 m x R 3md, = --------- = — = 1.2m1 2.5X 2.5

V 300From V = A.f, frequency, f = — = —— = 250HzX 1.2

Example 2

258

The diagram above shows two waveforms X and Y. If the frequency o \s - ,2is the frequency of Y? A.12Hz B. 19Hz C. 60Hz D.75HzSolution

o - number of wavelenghtWave frequency, f ------------------------------time taken

X has one wavelength and frequency of 30Hz. Therefore time taken by X is

_ number of wavelenght _ 1 frequency 30

From the diagram, wave Y with 2.5 wavelengths also takes V30 sec, same as X.

Therefore, frequency of Y = number of waveleng_hts = T5time taken Xo

= 2.5x30 = 75HzAlternatively, we could derive an equation relating the frequency (fx) and number of wavelengths (nx) of wave X with the frequency (fy) and number of wavelength (ny) of wave Y, as follows:

I k = ^f f* y

. . . 1 2.5substituting, ------- = —30Hz f y

frequency of Y, f y=30H zx2.5 = 75Hz

Example 3

The diagram above shows a waveform in which energy is transferred from X to Y in a time of 2.5 x 10'3s. What is the frequency of the wave? NECO 2002uSolutionNumber of wavelengths = 2Vi or 2.5 Time taken =2.5 x 10'3s

number of wavelenghts 2.5Frequency =time taken 2 .5x l0“3s = l.OxlO 'Hz

Example 4A wave travels a distance of 20cm in 3s. The distance between successive crests of the wave is 4cm. What is the frequency of the wave? NECO 200034SolutionWavelength, X = 4cm; wave distance = 20cm; time taken = 3s.

, wavedistance 20cm ,wavespeed = ----------------- - = —-— = b-b /cm

timetaken 3s

From V =fX , frequency, f = — = ^ = 1.67HzX 4cm

259

Example 5 g ,1 tie wavelength of ultraviolet radiation is 400nm. If the speed of light in air is 3x10 ms , then the frequency o f the ultraviolet radiation is A. 1.3 X lO'^IIz B. 7.5 X 10 Hz C. 1.2 X 10 'Hz D. 7.5 X 10I4Hz , mSolutionWavelength, X = 400nm = 400 X 10'Vi; speed, V = 3 X 1 0 W

From V = (X, frequency, f - X - ^ x t0 = 7 5 X1QhHzA 400x10’

Example 6

D/cm

Fig. 12.51 he diagram above represents the displacement D versus time, t, graph of a progressive wave. Deduce the frequency of the wave. WAEC 200318SolutionFrom the graph, number of wavelength = 2.5À.; time taken by the 2.5 wavelengths =0.25s._ number of wavelenghts 2.5 ,„ rIFrequency = ------------------------ 2— = -------= 10Hz

time taken 0.25

Example 7A radio station broadcast at a frequency of 400KHz. If the speed of the wave is 3 x 108ms_I, calculate the wavelength of the radio wave. NECO 2005'5SolutionFrequency, f= 400KHz = 400000Hz; Speed, V = 3 X 108m s'1

V 3 x l0 8From V - Xi. wavelength X = — = ---------- = 7.5 x 102 m

f 400000

Example 8The diagram below illustrates a variation of the displacement y of a wave particle with time t. If the velocity o f wave is 250ms"1, calculate the distance between two successive particles which are in phase. WAEC 199640

y/m iv

0

Fig. 12.6 SolutionThe distance between two successive particles in phase is equal to the wavelength X. Wave speed, V = 250m/s.

260

yniA

Fie. 12.7

The distance OA is equal to one wavelength. The time taken to travel OA is 0.0*.s.

Frequency^ jjSP'bn-of wavcIengNs = _ 50Hztime taken 0.02

V 250From V = A.f, wavelength X = — = = 5,0m

f 50

Example 9 ,A periodic pulse travels a distance of 20.0m in 1.00s. If its frequency is 2.0 x 10 Hz. calculate the wavelength. WAEC 2000~Solutionn i , distance 20.0m „Wave speed V = ----------- = --------- = 20ms

time 1,00sV 20From V = iA , wavelength. A= — = ---------- - = 0.01 = 1.0 x 10 ' m/ 2.0X103

Example 10A wave has a frequency of 2Hz and a wave length of 30cm. The velocity o f the wave is. A. 60.0ms'1 B. 6.0ms’1 C. 1.5ms’1 D. 0.6ms’1 JAMB 198721SolutionFrequency, f = 2Hz; wavelength, X = 30cm = 0.30m Velocity, v = fX = 0.30 X 2 = 0.60ms’1

Example 11A boat at anchor is rocked by waves whose crest are 100m apart and whose velocity is 25ms’1. At what interval does the wave crest reach the boat?A. 2500.00s B. 75.00s C. 4.00s D. 0.25s JAMB 19871%SolutionWavelength, X = 100m; velocity, v = 25ms"1. Interval at which boat reaches crest is period, T = ?

X . , „ X 100 ,From V =—, period, T = — = — - - 4sT v 25

Example 12The figure below represents a displacement - time graph of a travelling wave moving with a speed of 2ms’ .

0.2

0 I

0

-0.1

- n?

Fig. 12.8

261

Determine, from the graph, the (0 amplitude (ii) period (iii) wavelength o f the wave.SolutionWave speed, V = 2m s'1

(il Amplitude. A = 2.0m as shown in the diagram below.(u) Period, T = 0.1 s i.e. the time for one wavelength. M - N.

(m) From V = wavelertgth, X = \ T = 2 X 0.1 = 0.2ms"1

E xam ple 13Infrared rays ol frequency 1.0 X 10* Hz have a wave length o f 3.0 x 10 m in vacuum.The wavelength ot X-ravs o f frequency 5.0 x 1016 Hz in vacuum will beA. 6.0 X 10;: m B. 6.0 X 10'10 m C. 6.0 X 10' mD. 5.0 X 1010 mE. 3.0 X 10 m S E C O 2008nSolutionFor infrared rays: f = 1.0 X 10IJ Hz: X = 3.0 X 10 * m

v = Xf= 3.0 X 10 5 x 1.0 X 10!- = 3.0 X 10s ms 1

For the X-ray: f = 5.0 x 1016 Hz; v = 3 x 10s m s'1

, , 3 »* 3 x 1 0 sv = Xf a = — -------------— = 6 .0x10 mf 5 .0x 10

None o f the options given is correct.

STATIONARY WAVESA stationary or standing wave is a wave that is formed when two progressive

waves o f equal frequency and amplitude travelling in opposite direction are combined. Stationary waves consist o f nodes and antinodes as shown in fig 12.10.

A node is a point on a stationary wave where there is no movement o f the medium, while an antinode is a point on a stationary wave where there is maximum displacement o f the medium.

Below' in Fig 12.10 is a diagram o f a stationary wave.

For a stationary wave the following applies.

Distance between a node and the next antinode, N - A = 4

Distance between two successive nodes (N - N) or antinode (A - A) = >

Distance between three successive nodes (N - N - N) or antinode (A - A - A) = X

262

Fig. 12.10

/ n

\ x +■ x r2 4 2

Example 13 ,The wavelength of a stationary wave is 36.0cm. What is the distance between a no ® the next anti-node? NECO 2002SolutionWavelength X = 36.0cmDistance between a node and next antinode = 'AX

= % x36 = 9cm.

Example 14A wave of frequency 10Hz forms a stationary wave pattern in a medium where the velocity is 20cms_l. The distance between adjacent node isA. 1.0cm B. 1.5cm C. 2.0cm D. 5.0cm JAMB 1986SolutionFrequency, f = 10Hz; wave velocity, v = 20cms''

From v = A.f, wavelength x = — = — = 2mf 10

A 2Distance between adjacent node = _ = _ = icni2 2

Example 15Radio waves emitted from an antenna are picked up by a radar after reflection from an aircraft in 4 x 10‘3s. How far is the aircraft from the antenna? (v=3xl03m s1).A. 6.0Xl02km B. 1.2Xl03km C. 3.0xl03km D. 6.0xl05km JAMB 1993™ SolutionVelocity, v = 3 X l 0 W ; time = 'A (4X10°) = 2 X 10'3s

(The time is divided into two because the wave was reflected).

Distance = Velocity X time = 3x10s X 2x10'3 = 6x105m or 6X l02km.

MATHEMATICAL REPRESENTATION OF

PROGRESSIVE WAVE MOTION

A progressive wave is any wave, transverse or longitudinal, that spreads out from a vibrating source and moves through a medium transferring energy as it travels. A graph o f displacement against time for a progressive waveform is as shown in fig 12.11 Mathematically, a wave can be represented by the following equation.

Y = Asm S ì ?

263

Fig. 12.11X

Where Y =X = A =x =

(27U-) _

X

vertical displacement of oscillating wave particle.wavelengthamplitudehorizontal coordinate, from the origin of the oscillating wave particle, phase difference of wave particle at distance x from the origin.

A wave can travel in either the positive r-dircction or negative .r-direction. After a particular time t, the displacement of the wave particle can be represented by the following equations: Positively travelling wave and negatively travelling wave,respectively.

2 71Y4.vc = Asm — (,v - vt)

A.or Y.vc . . 2n . .Asm — (vt - x )

X

2. Y+vc = A s in 2 ^ ^ -- f /

3. Y+Ve = Asin2^f — - —1 2 T

, v . . ( 2 n x4. Y ,ve = Asirw — ---- cot

or A sin2^.. xf t -----

À

or Y.ve = Asin2m — - — T À

or Y.vc = Asin cot -2nx

5. Y+ve= Asin* Y.vc = Asinrq t

Where to = 2n( = angular speed or velocity, f = frequency of wave.T = period of wave, t = time.v = wave velocity.

Students are advised to memorize just equations 1 and 2 as the others arc basically the same equation expressed in different forms. Just remember that: T= /f , V= X(, to = 2nf

Example 16A progressive wave has a wavelength of 50cm. Calculate the phase difference between two points at a distance of 20cm apart. IVALt 199/SolutionPhase difference o f a wave between two points = i z î l l

XWavelength, X = 50cm; distance from origin, .v = 20cm.

(27i.y) _ 20cm . 2 4Substitute into ------ = 2 ^ x ---------= 2 / t x — = —n

X 50cm 5 5

264

Example 17 , , fThe distance between two points P and Q, along a wave is 0.05m. I f t ie waveeny j the wave is 0.10m, determine the phase an die between P and 0 in radians.SolutionWavelength, A. = 0.1m; distance between point P and Q, x = 0.05m.

Phase angle = (2k.y)X

2n x 0.05m _ 0.1 n _ 0.1m " 0.1

Example 18 2A wave has an amplitude equal to 4.0m, angular speed '/37t rads'1 and phase angle /3 rad. The displacement y of the wave is given as? WAEC 20SolutionAmplitude A = 4.0m; angular speed co = 1/3ji rads"1; Phase angle 2n rad

The appropriate equation is y =

Substitute given values into the two equations.

or

Y+ve

Y.ve

l;rO . K t - , \ = — = 4siny ( 2 - 0

2 n \ . . n ,I = 4sin—(f -t- 2)

Both answers are correct depending on the options given. If it is an essay question and the direction of the wave is not specified, students are advised to state both answers in the same way as solution to a quadratic equation, i.e.

4 sin — ( 2 - / ) or — (t + 2)3, 3

Example 19A travelling wave moving from left to right has an amplitude of 0.15m, a frequency of 550Hz and a wavelength of 0.01m. The equation describing the wave is A. y = 0.15 sin 200^(jt - 5.5t) B. y = 0.15 sin 7r(0.0Ijc — 5.5t)C. y = 0.15 sin 5.57t(;c - 200t) D. y = 0.15sin 7t(550x - 0.0It)SolutionFrequency, f = 550Hz; amplitude, A = 0.15m; wavelength, X = 0.01m

:. Wave speed, v = Xf = 0.01 X 550 = 5.5ms 1The direction of the wave is stated, “moving from left to right". Because it is a

Inpositively travelling wave, we substitute into y - Asin— ( x - v t )

Y = 0.15 sin----- (x -5 .5 t)0.01

Y = 0.15sin200;r(jc - 5.5t)

Example 20The equation of transverse wave travelling along a string is given by y = 0.3sin (0.5.r - 50t) where y and x are in cm and t is in seconds, find the maximum displacement of the particles front the equilibrium position.A. 50.0cm B. 2.5cm C. 0.5cm D. 0.3cm JAMB 199428

265

Solution , amplitudcThe maximum displacement of the particles from equilibrium position is ic ^ ^ (A) of the wave. Comparing the given equation y = 0.3sin(0.5x - 50t) w iti o

- m It is obvious that the amplitude is 0.3cm.general equation y= Asinf ~n xl *

Example 21The equation of a wave travelling along the positive x-direction is given by y=0.25 x 10 sin(500t - 0 .025x). Determine the angular frequency of the wave motion.

A. 0.25X10 rads B. 0.25xl0 ''rads'' C. 5.00X 102rads'' D. 2.50X 102rads 1„ JAMB 199923SolutionThe given equation y = 0.25X10'3sin(500t - 0.025x) can be compared with y = Asin^ft*

Therefore, cot = 500t (Remember, co = 2*027tft = 500t

5001 In t

500111 -----= 2.50xl02rad 3

n

Example 22A wave is represented by the equation y = 2sin(0.5.x - 200t), where all distances aremeasured in centimeters and time in seconds. For this wave, calculate its(i) frequency (ii) wavelength (iii) speed. NECO 2000cl3Solution

Y = 2sin 7i(0.5.v - 200t)The above equation can be compared with any of the general equation (1 ,2 or 3 in page 168) as long as they are arrange in such a way that it can be compared with the given equation. The three instances are shown below.

2 71I. The first equation y = Asin —— ( v - vt) can be rearranged by taking 2/x into the

A.bracket to obtain y = Asini — - — I___aU x Jcomparable to y = 2 sin 7t (0.5x-200t).....B

from equation B, 0.5x2x—r in equation A

cross multiplying 0.5x X A = 2x

wavelength A = 2x0.5x

— = 4m 0.5

similarly, from equation B, 200/2v/ .— in equation A

cross multiplying and substituting X = 4m we obtain

2vt = 200t x 4

speed, v = :20 0 tx 4 _ 200x4 _ 800

2t _ 2 2= 400ms"

266

From v = Af, frequency, f = — = —— = 1001 Iz

Therefore, (i) 100Hz (ii)4m (iii) 400ms’1

II. The second equation y = Asin2;z/ — — ft (can be rearranged by taking 2 into theU J

bracket to obtain y = Asin^f — - 2f/ I_____ c which can be compared toV J

y = 2sin7i (0.5* -2 0 0 t)______ o

2xFrom equation C, — = 0.5* in equation DA

Cross multiplying, 0.5.* x A = 2x

Wavelength, A = - ^ - = — = 4m 0.5* 0.5

Similarly, from equation C, 2ft = 200t in equation D.

cross multiplying, frequency, f = - 00t = — = 100Hz2t 2

Speed, v = Af = 4 x 100 = 400ms"'

Therefore, (i) 100Hz (ii) 4m (iii) 400ms"1

III. The third equation y = Asin2.J * OU TJcan be rearranged by taking 2 into the

bracket

to obtain y = Asin 2t A T

y = 2sin^(0.5x - 200t)

Therefore: — = 0.5*;A

Also: 200t = —T

, compared with

wavelenght/l = 4m

Cross multiplying 200t x T = 2t

T =2t 2 1

200t _ 200 _ 100= 0.01s

Frequency, f = — = —— = 100Hz T 0.01

Speed, v = fA -1 0 0 x 4 = 400ms"1

(i) Frequency = 100Hz (ii) Wavelength = 4m (iii) Speed = 400ms"1

267

EXERCISE 12

l.

I lie diagram above shows a waveform in which energy is transferred from A to B in a mne of 2 .5x10 's. Calculate the frequency of the wave. WAEC 1992** Ans: 1.Ox IQ*Hz

I he diagram above represents a transverse electromagnetic wave travelling with speed 3.0x I O W . What is the frequency of the wave? A. 3 .0 x l0 7Hz B.9.0x 107HzC.l .Ox 109Hz D.3.0x 109Hz JAMB 19992* Ans:3.0x 109Hz3.

The diagram above illustrates the profile of a progressive wave in which energy is transferred from P to Q in 3.0x10'’s. Calculate the frequency of the wave.

WAEC 1997*' Ans: l.Ox 10*Hz4. In a ripple tank experiment, a vibrating plate is used to generate ripples in the water. If the distance between two successive troughs is 3.5cm and the wave travels a distance o f 31.5cm in 1.5s, calculate the frequency of the vibrator.

WAEC 199322 Ans: 6.0Hz5. A radio wave has a wavelength of 150m. If the velocity of radio waves in free space is 3 x l0 8ms’\ calculate the frequency of the radio wave.

WAEC 200225 Ans: 2 .0 x 1 0hHz6 A wave o f wavelength 0.30m travels 900m in 3.0s, calculate its frequency.

WAEC 200319 Ans: 1000Hz7. The distance between two successive crests of a water wave travelling at 3.6ms'1is 0.45m, calculate the frequency o f the wave. WAEC 20051'1* Ans: 8Hz8. The distance between the successive crests of a wave travelling at 20ms'1 is 25cm.Calculate the frequency o f the wave. WAEC 2006 Ans. 80.0Hz

9.

Fig. 12.15The wave motion shown above has a frequency of A. 0.1 Hz B. 0.2Hz C. 5Hz D. 10Hz E. 100Hz JAMB 198122 Ans: 5Hz

268

10.

I he frequency of the wave in the diagram above is .A. 0.25Hz B. 1.57Hz C. 0.17Hz D. 1.05Hz JAMB 2006 Arts: 0.25Hz11. I he period of a wave is 0.02 second. Calculate its wavelength if its SP^C ,s330ms '. WAEC 1995 Ans: 6.6m12. Calculate the wavelength of light travelling with a speed of 3x10 ms and afrequency of 6.25 x 10 I4I Iz. NECO 200423 Ans. 4 .8x1 (T m13. Radio waves have a velocity of 3x |0 ’lms'1. A radio station sends out a broadcast on a frequency o f 800KHz. The wavelength of the broadcast isA. 375.0m B. 267.0m C. 240.0m D. 37.5m E. 26.7m JAMB 1984 Ans: 375.0m14. Ihe wavelength of signals from a radio transmitter is 1500m and the frequency is 200KH/.. What is the wavelength for a transmitter operating at lOOOKHz.A . 7500m B. 300m C. 75m D. 15m JAMB 19952H Ans: 300m15. Sixty complete waves pass a particular points in 4s. If the distance between three successive troughs o f the waves is 15m, calculate the speed of the wave.

WAEC 19952* Ans: 112.5ms16. The distance between two successive crests of a water wave is 1 m. If a particle onthe surface of the water makes two complete vertical oscillations in one second, calculate the speed of the wave. WAEC 199637 Ans: 2 .0m s117. Ihe distance between two successive crests of a water wave is 0.5m. If a particleon the surface of the water wave makes two complete vertical oscillations in Is, calculate the speed of the wave. NECO 2003n Ans: 1.0ms'118. What is the frequency of a radio wave of wavelength 150m if the velocity of radiowaves in free space is 3x 10*ms '? WAEC I99724 Ans: 2 x J(/JHz19. Ihe distance between two points in phase on a progressive wave is 5cm. If thespeed o f the wave is 0.20ms'1, calculate its period. WAEC 200527 Ans: 0.25s20. If the wavelength of a wave travelling with a velocity of 360ms 1 is 60cm, the period of the wave isA. 6s B. 3.6s C. 0.17s D. 0.61s E. 3s JAMB 1979,} Ans. 1.67x 10T}s21. A wave travelling with a speed of 360ms'1 has a wavelength of 60cm. The periodof the wave in second is? NECO 200725 Ans: 1.7x l ( f 322. Ihe distance between two successive crests of a wave is 0.664m. If the wavetravels 43.2m in 1.2 seconds, calculate the (a) speed (b) frequency o f the wave.

NECO 2007t6 Ans: (a) 3 6 m s1 (b) 54.22Hz23. In the wave equation y = Eosin(200t - 7tx) E„ representsA. Amplitude B. frequency C. period D. wavelength. WAEC 200226 Ans: A24. Ihe equation o f a wave is y = 0.005sin[x(0.5x - 200t)] where x and y are in meters and t is in seconds. What is the velocity o f the wave?A. 4000m s'1 B. 400 m s1 C. 250ms 1 D. 40m s'1 JAMB 199220 Ans:250ms'1

Hint: y= Asin(2n*/X —cot).25. A progressive wave equation is represented by y = Asin(1507tt - nx/4). If the phase difference o f a progressive wave is 45", the value o fx in the equation isA. 2cm B. lem C. 4cm D. 3cm JAMB 2006s Ans: lem26. I he equation, y = 5sin(3jr 4t), where y is in millimeters, x is in meters and t is inseconds represents a wave motion. Determine the (i) frequency (ii) period and (iii) speed o f the wave. W AEC20041'3 Ans: (i) 0.6Hz (ii) 1.56s (iii) 1,34ms'1

269

27. The relationship between L and the wavelength X of the stationary wave isA- 2L = X B. L = k C. L = x/3 D. 4L = 3X Ans: D28. Determine the distance between consecutive antinodes XX if the wavelength is 60cm. A. 15cm B. 30cm C. 60cm D. 120cm Ans: B29. The equation y = Asin(fit- kx) represents a plane wave traveling in a medium along the x - direction, y being the displacement at the point x at time t.i. Given that x is the meters and t is in seconds, state the units of k and w.

ii. What physical quantity does — represent?k

iii. State whether the wave is traveling in the positive or negative direction xdirection. WAEC 2008Ans: (i) k(m ’), w (rad s '1, s’1 or Hz)

('i) "A = Speed of wave(iii) positive direction

30. Given the progressive wave equation y = 5 sin(200^ /- 0.4x), calculate the wavelength A. 12.4m B. 15.7m C. 12.5m D. 18.6m JAMB 2008 Ans: 15.70m31. The diagram below represents a stationary wave pattern.

Fig 12.18The distance XY represents

A. One and a half times the wavelength.B. The wavelengthC. Three times the wavelengthD. Twice the wavelengthE. Two and a halftimes the wavelength NECO 200826 Ans: E

32. A progressive wave is represented by y ~ 10 sin (lOOnt - Two layers of the wave separated by 153cm have a phase difference ofA. 270° B. 45° C. 90° D. 180° JAMB 200927 Ans: D33. If a light wave has a wavelength of 500nm in air, what is the frequency of the wave?A. 3.0 x 1014Hz B. 6.0 x 1014Hz C. 6.0 x 1012HzD. 2.5 x 1014Hz (c — 3 x 108m s_1) JAMB 2009*° Ans: B34. A boat is rocked by waves of speed 30ms'1 whose successive crests are 10m apart. Calculate the rate at which the boat receives the waves. WAEC 200() :* Ans: 3 s '35. A plane progressive wave is represented by the equationy = 0.2 sin(2007r - 0.5x) where x and y are measured in cm and / in s. What is the frequency of wave0 NECO 2 0 0 Ans: 100Hz

270

13REFLECTION OF LIGHT WAVES

Formation of images by a pinhole camera, plane and spheiical mirror P the facts that light travels in a straight line and can be reflected v P ane

THE PINHOLE CAM ERA: M A G NIFICATIO NThe magnification (M) of a pinhole camera is given as:

^ _ image height _ image distance from pinhole object height object distance from pinhole

lenght of camera object distance from pinhole

Any, or a combination of the above formulas can be used when solving problems involving a pinhole camera.

Example 1 . ,A boy 1.5m tall stands 2.0m in front of a pinhole camera of length 20.0cm. acu a e• height of h.s image. NECO 2006SolutionObject height = 1,5m; object distance from pinhole = 2.0m Camera length = 20cm = 0.2m; height of image = ?

Substitute into

.'. Image height =

image height _ camera lengthobject height object distance from pinhole

image height _ 0.2m1.5m 2.0m

1.5x0.2 A lr ----------= 0.15m or 15cm

R EFLEC TIO N O F LIG H T AT PLANE SURFACES

When a ray of light is incident upon a plane mirror AB shown above, the following are true.

1. LN is the incident ray MN is the reflected ray and ON is the normal.2. The angle of incidence (i) is equal to the angle of reflection (r). That is i = r

271

3. The angle (9) between the incident ray and the plane mirror is called the glancing angle, and is always equal to the angle between the reflected ray and the plane mirror, MNB.

4. The angle o f deviation d = 180 - (i + r). Because i = r, the angle o f deviation could be, d = 180 - 2i o r d = 1 8 0 - 2 r .

Example 2I

The diagram above shows a ray of light DC incident on plane mirror at K. Calculate the angle o f deviation o f the ray after reflection. W AEC 1994Solution ^

Therefore, angle o f deviation, d = 180 - 2i = 1 8 0 -2 x 5 5= 180 - 110 = 70°

Example 3A ray of light strikes a plane mirror at a glancing angle of 55°. Calculate the angle between the incident and reflected rays. NECO 200oSolution

Angle of incidence i = 90 - 55 = 35°

Angle of reflection r = 35°

Angle between incident and reflected ray is, i + r.

i + r = 35 + 35 = 70°

272

Example 4

Fig. 13.5

mi

Solution

Fig. 13.6C

Any time an incident ray is reflected on a plane mirror, a normal perpendicular (at 90°) to the mirror can be drawn.

EC is the normal between incident ray BC and reflected ray CD.EB is the normal between incident ray AB and reflected ray BC.Angle ABE = 90° - 30° = 60°Angle EBC =60° (Angle of incidence = Angle of reflection)Angle CBF =90° - 60° = 30°Angle BCF =180“ -(90° + 30°) = 180° -120° = 60° (sum of angles in a triangle is 180°) Angle BCE =90°-60° =30°Angle ECD =30° (Angle of incidence = Angle of reflection).Angle DCG =90° - 30° = 60°Therefore, angle of reflection at mirror m2 = ECD =30°

Example 5P

Fig. 13.7N

273

Two mirrors of the same length are arranged as shown in the diagram above. A ray of light NO strikes the system at O and emerges along PQ. The emergent ray has been deviated through

‘ A. 220° B. 200° C. 210° D. 180° E. 230° JAMB 198236

The incident ray ON that emerges along PQ has been deviated twice; through angle dj

and d2 as shown above. Therefore, the emergent ray will be deviated through di + d? — D.

Remember, angle of deviation, d = 180 - 2i or 180 - 2r.

The angles o f incidences and angles of reflection are found in ways similar to

those in example 4.

Angle o f incidence (i) or ray NO is 50°. Therefore, d| = 180 - 2i = 180 - 2 x 50 = 180 -

100 = 80°

Angle o f incidence o f emergent ray PQ is 40°. Therefore, d2 = 180 - 2i = 180 - 2 x 40 =

1 8 0 -8 0 = 100°

Angle of deviation of the emergent ray D = d| + d2 = 80 + 100 = 180°.

Exam ple 6An object is placed 10cm in front of a plane mirror. If it is moved 8cm farther away from the mirror, determine the distance of the final image from the mirror. WAEC 20073' SolutionWhen an object is placed in front of a plane mirror, the size of the image formed is the same size as the object. Also, the distance (u) of the object from the mirror is always equal to the distance (v) of the image from the mirror.That is u = v.Object distance from mirror, u = 10 + 8 = 18cm. Therefore, image distance v = 18cm.

Example 7A man stands 4m in front o f a plane mirror. If the mirror is moved lm towards the man, the distance between him and his new image is A. 3m B. 5m C. 6m D. 10m

JAMB 199920SolutionWhen the mirror is moved lm towards the man, the object distance becomes 4-l=3m .The image distance from the mirror is also 3m. Therefore, distance between the man and his image is 3 + 3 = 6m.

274

IMAGE FORMED BY INCLINED MIRRORSWhen two mirrors are inclined at an angle (0) to each other, the number (n)

image formed is given by360 ,

n = ------- 10

Exam ple 8 . r •j<yo facingA candle is place between two mirrors which are inclined at an ang e o each other. How many images will be observed in the two mirrors?Solution

Angle of inclination, 0 = 72°

Substitute into n = _ ]0

T/TANumber o f images, n - J _ _ i

72n = 5 - 1 = 4

Example 9At what angle 0 will 2 plane mirrors be inclined so that 11 images will be formed w en an object is plaoed in front o f the two mirrors?Solution

Substitute intoNumber of image n = 11

360 ,n = ------- 1036011 = — - 10

360 0

360 0

11 + 1 =

12 = « = “ 2 = 30 12

o

ROTATION OF PLANE MIRRORSWhen a ray o f light is incident on a plane mirror and the mirror is rotated through

an angle 0 while keeping the incident ray constant, the following are true.

is i + 0 or INN2 .

275

1. The angle between the initial normal (NN|) and the final normal (NN2) is equal to the angle of rotation 0.

2. The angle of reflection before rotation is r or R1NN 1. The angle o f reflection after rotation is i + 0 or R.2NN2 -

3. Angle between the incident ray and the final reflected ray is (i + 0) + (i + 0) = 2(i + 0).4. The reflected ray is rotated through twice the angle of rotation, 20, or R 1NR2 as shown

in the diagram above.5. The angle between the normal after rotation and the initial reflected ray is i-0 if angle

of rotation is less than initial angle of incidence or 0 - i if angle of rotation is greater than initial angle of incidence.

Example 10A ray of light is incident on a plane mirror at an angle of 20° This mirror is rotated through twice this angle. In this new position, the angle between the incident ray and the reflected ray is? WAEC19992'Solution.Angle of incidence i = 20°; angle of rotation 0 = 2 x 1 = 2 x 20 = 40° (...rotated through twice— ). Angle between incident ray and new reflected ray = 2 (i + 0) = 2 (20 + 40) = 2 x 60 = 120°

Example 11An incident ray is reflected normally by a plane mirror onto a screen where it forms a bright spot. The mirror and screen are parallel and lm apart. If the mirror is rotated through 5°, calculate the displacement of the spot. WAEC 200123SolutionLet x be the displacement of the spot. Angle of rotation 0 = 5°. Angle through which reflected ray rotates = 20 = 2 x 5 = 10°.

From triangle ABC, Tan 10° = y

x = Tan 10° x 1 =0.176m or 17.6cm

Example 12A ray o f light is incident normally on a plane mirror. If the incident ray is kept fixed while the mirror is rotated through an angle of 30°, determine the initial and final angles of reflection respectively. WAEC 200531SolutionWhen a ray of light is incident normally (at 90°) on a plane mirror, the angle of incidence and the angle of reflection are zero. That is, i = r = 0. If the mirror is rotated through 30°, the final angle of reflection 20 becomes 2 x 30 = 60 .

Therefore, the initial and final angles of reflection respectively are 0° and 60°.

276

Example 13 i nf 15" with “ *",xcd *’nc AY w^ ‘c*1In the figure below, an incident ray XY mnkes an »nK « as indicated by the arrowis normal to the surface of a plunc mirror. If the mirror isthrough 40°, what angle will the reflected ray make wi t JAMB 1982A. 25° B. 40° C. 15° *>• 55

Solution , - = 1 5 °XY is the incident ray. Angle of rotation 0 = 40°; angle of incidence t • The angle that the new reflected ray makes with AY (normal before ro ton i + 0 + 0 or i + 20 = 15 + 2 x 40 = 95°.

SPHERICAL MIRRORS: CONCAVE AND CONVEXThe following formulars, expressions and rules are used in solving concave and convex mirror problems.I. Focal length = Vi radius of curvature. That is f = r/2II. Relationship between the object distance u, image distance v and focal length f is

given by mirror equation:

u v f u + vMagnification M - ‘ma®e height _ *mage distance from lens

object height object distance from lens

That is, M = — u

• A convex mirror has negative focal length.• A concave mirror has positive focal length.• Image distance in a convex mirror is negative.• Image distance in a concave mirror is positive except when the object distance

is less than the focal length (see example 17).• Distances o f real objects and images are positive.• Distances of virtual objects and images arc negative.

Example 14The image o f an object is located 6cm behind a convex mirror. If ils magnification is 0 o

WAECcalculate ;the focal length of the mirror.SolutionImage distance, v = -6cm; magnification M = 0.6;

Object distance u is obtained by substituting into M = —u

focal length f - ?

0.6 =— 6cm

u— 6

u = — = -10 cm 0.6

277

Substitute above value into f = uv to obtainu + v

Focal length f =- 1 0 x - 6 60 60

- 1 0 + ( - 6 ) - 1 0 - 6 -1 6

f = -3.75cmThe negative sign shows it’s a convex mirror.

Example 15An object is placed in front o f a concave mirror whose radius o f curvature is 12cm. If the magnification o f the image produced is 1.5, how far is the object from the mirror?

NEC0200528SolutionMagnification M = 1.5; radius of curvature r = 12cm; focal length f = , 2 / 2 = 6cm;Object distance u = ?

Substitute into M = — to obtain 1.5 = 7 U.u

Therefore image distance, v = 1,5u

Substitute into f = ■ UV to obtain u + v

u x l.5 u _ 1.5u2 _ 1.5u u+1.5u 2.5u 2.5

2.5

1.5u = 6 x 2.5 = 15

.*. u = — = 10cm1.5

Example 16A concave mirror o f radius of curvature 40cm forms a real image twice as large as the object. The object distance is A. 30cm B. 40cm C. 60cm D. 10cm JAMB 200228 SolutionRadius o f curvature, r = 40cm f = r/2 = 40¡2 - 20cm.Object distance, u = ? Magnification, M = 2, because image is twice (2x) the object. w image size (v) ~ v .M = ------ ------------ = 4 - — .. image dis tan ce, v = 2u

object size (u) u

uvsubstitute into f = -------- to obtain,U + V

ux2u 2u220 = ------ = ----

u+2u 3u

32u = 20 x 3 = 60

60 anu = — = 30cm2

278

Example 17 of focal length 15cm. What is theAn object is placed 10cm in front of a concave mirror . g 6cm and realposition and nature of the image formed? A. 30cm an JAMB 198812C. 6cm and virtual D. 30cm and real.Solntlon _ Aictonr* v = ?Object distance, u = 10cm; focal length, f= 15cm, imag

Substitute into f =

15 =

uvu + v

lOxv

to obtain

10+v

Cross multiplying, 15(10 + v) = 10 x v

150+ 15v= lOv

150 = lOv- 15v = -5v

-150 no v = ------ = -3 0cm.

If the image distance of a concave mirror

Example 18Which of the following expressions gives the linear magnification produced by a concave mirror of radius of curvature r, if u and v are the object and image distances respectively?

A. - - 1 B. — - l C. - - 1 D. — -1 JAMB 199326r r r r

SolutionRadius of curvature = r Focal length, f = '/¡r; Linear magnification = v/u

Substitute into — = — + — f u v

i - = I + I/ 2 u v

1 = 1 + 1 r u v

multiply both sides by v

2v _ v r u y

2vr

make v/u subject of above equation to obtain linear magnification, v 2v

Linear magnification, — ------- 1u r

Example 19The driving mirror of a car has a radius of curvature of lm. A vehicle 4m from the mirror. Find the image distance behind the mirrorA. */7 B. % C. 9/2 D. 4/7 SolutionDriving mirrors are usually convex mirrors. Radius of curvature, r = 1 m

behind the car is

JAMB 2001'4

i=Zi

279

1. The angle between the initial normal (NN|) and the final normal (NN2) *s equal tangle of rotation 0. . ft

2. The angle of reflection before rotation is r or R|NNi. The angle of re ec ion a rotation is i + 0 or R2NN2.

3. Angle between the incident ray and the final reflected ray is (i + 0) + (i + 9) — 2(i + )■4. The reflected ray is rotated through twice the angle of rotation, 20, or R1NR2 as shown

in the diagram above.5. The angle between the normal after rotation and the initial reflected ray is i-0 if angle

of rotation is less than initial angle of incidence or 0 - i if angle of rotation is greater than initial angle of incidence.

Example 10A ray of light is incident on a plane mirror at an angle of 20° This mirror is rotated through twice this angle. In this new position, the angle between the incident ray and the reflected ray is? WAEC 199921SolutionAngle of incidence i = 20°; angle of rotation 0 = 2 x I = 2 x 20 = 40° (. ..rotated through twice....). Angle between incident ray and new reflected ray = 2 (i + 0) = 2 (20 + 40) = 2 x 60 = 120°

Example 11An incident ray is reflected normally by a plane mirror onto a screen where it ft--ns a bright spot. The mirror and screen are parallel and lm apart. If the mirror is rotated through 5°, calculate the displacement of the spot. WAEC 200123SolutionLet x be the displacement of the spot. Angle of rotation 0 = 5°. Angle through which reflected ray rotates = 20 = 2 x 5 = 10°.

x = Tan 10° x 1 = 0.176m or 17.6cm

Example 12A ray o f light is incident normally on a plane mirror. If the incident ray is kept fixed while the mirror is rotated through an angle of 30°, determine the initial and final angles o f reflection respectively. WAEC 20053'SolutionWhen a ray o f light is incident normally (at 90°) on a plane mirror, the angle of incidence and the angle o f reflection are zero. That is, i = r = 0. If the mirror is rotated through 30°, the final angle of reflection 20 becomes 2 x 30 = 60°.

Therefore, the initial and final angles o f reflection respectively are 0° and 60°.

276

Example 13 . f , 5*> wi,|, a fixed line AY whichIn the figure below, an incident ray XY makes an angle 1 mdicated by the arrowis normal to the surface of a plane mirror. If the mirror is tun c through 40°, what angle will the reflected ray make with A Y •A. 25° B. 40° C. 15° D. 55° E. 95 JAM »

Solution .. . . , ■ _ 15»XY is the incident ray. Angle of rotation 0 = 40 ; angle of mci UKC •The angle that the new reflected ray makes with AY (normal ie ore to a i + 9 + 0 or i + 20 = 15 + 2 x 40 = 95°.

SPHERICAL MIRRORS: CONCAVE AND CONVEXThe following formulars, expressions and ndcs are used in solving concave and convex mirror problems.I. Focal length = I/2 radius of curvature. T h a t i s f = r/2II. Relationship between the object distance u, image distance v and focal length 1 is

given by mirror equation;uv

III.

I 1 = 1U V ~ f

f = -u + V

Magnification M - 'ma8e height _ image distance from lens object height object distance from lens

That is, M = — u

• A convex mirror has negative focal length.• A concave mirror has positive focal length.• Image distance in a convex mirror is negative.• Image distance in a concave mirror is positive except when the object distance

is less than the focal length (see example 17).• Distances o f real objects and images are positive.• Distances o f virtual objects and images are negative.

Example 14The image o f an object is located 6cm behind a convex mirror. II its magnification is 0 <> calculate the focal length o f the mirror. 11 It ( /<)<)<SolutionImage distance, v = -6cm; magnification M = 0.6; focal length f '>

Object distance u is obtained by substituting into M = —u

- 6 c m - 60.6 = -------- .. u = ---- = — 10 cm

u 0.6

277

Substitute above value into (IV

11+ vlo oliiain

l ocal length Il O x 6

10 l ( 6)60 60

10 6 16

I '- 3.75cmThe negative sign shows it's a convex mirror.

Example 15An object is placed in front o f a concave mirror whose radius o f curvature is 12cm. If the magnification of the image produced is 1.5, how far is the object from the mirror?

NECO 20052HSolutionMagnification M = 1.5; radius of curvature r 12cm; focal length I = 1 ¡ i - 6cm;Object distance u = ?

Substitute into M = - to obtain 1.5 = v/u. u

Therefore image distance, v = 1,5u

Substitute into f = to obtainu + v

( _ u x 1,5u _ 1,5u2 _ 1.5u u + 1.5u 2.5u 2.5

1.5u = 6 x 2.5 = 15

u = —— = I Ocm1.5

Example 16A concave mirror o f radius of curvature 40cm forms a real image twice as large as the object. The object distance is A. 30cm B. 40cm C. 60cm D. 10cm JAM B 20022H SolutionRadius o f curvature, r = 40cm f = V2 = ¡2 = 20cm.Object distance, u = ? Magnification, M = 2, because image is twice (2x) the object.

M _ image size (v) = 2 = - /. imageclis tan ce, v = 2uobject size (uj u

uvsubstitute into f —-------- to obtain,u 4- v

20 =u x 2 u

u + 2u

2u^

3u

2u = 20 x 3 = 60

60 i nu = — = 30cm 2

278

Example 17An object is placed 10cm in front of a concave mirror of focal length 15cm. What is theposition and nature o f the image formed? A. 30cm and virtual B. 6cm and real

- - - - • JAM B 198a

i, f= 15cm; image distance, v

Substitute into f = io obtain

C. 6cm and virtual D. 30cm and real SolutionObject distance, u = 10cm; focal length

uv

■ = 9

15 =

U + V

lOx v10 + v

Cross multiplying, 15(10 H- v) = 10 x v

150 + 15v= lOv

150 = lO v - 15v = -5v

-1 5 0 7Av = -------= -30cm .5

If the image distance of a concave mirror is negative, it means that the image is virtual.

Example 18Which of the following expressions gives the linear magnification produced by a concave mirror of radius of curvature r, if u and v are the object and image distances respectively?

A. - - 1 B. - - ] C. - - 1 D. — -1 JAM B 199326r r r r

SolutionRadius of curvature = r Focal length, f = ‘/2 r ;

Substitute into 1 _ i 1— + —f u V

1 1 1' — — + —

X u V

2 i 1— = —+ -r u V

Linear magnification = 7 U

multiply both sides by v

2v r u

_ v v

2v v ,— = - + 1 r u

make 7U subject o f above equation to obtain linear magnification v 2vLinear magnification, — = - — 1 u r

Example 19The driving mirror of a car has a radius of curvature of lm. A vehicle behm 1 t], .4m from the mirror. Find the image distance behind the mirror.A. 8/7 B. % c . 7 2 d . 4/7 JAMB '0011'SolutionDriving mirrors are usually convex mirrors. Radius of curvature, r = lm ■ f = i

14REFRACTION OF LIGHT WAVESL A W S O F R E F R A C T IO N A N D R E F R A C T IV E IN D E X

Refraction is the bending or change in direction of a light ray as a result of change in the speed of light when it passes from one medium to another.

Fig. 14.1

ATI = incidence ray; i = angle of incidence BC = refracted ray; r = angle of refraction CD = emergent ray; e = angle of emergence d = angle of deviation of incidence ray = i —r

The laws of refraction are as follows:1. . The incident and refracted rays are on opposite sides of the normal at the point

of incidence and all three are in the same plane.2. The ratio o f the sine of the angle of incidence (i) to the sine of the angle of

refraction (r) is a constant for a given pair of media. This is also known as Snell’s law and is stated thus.

— = nsin r

f | = is known as the refractive index of the second medium with respect to the first medium. Air and glass are used as typical examples of a pair o f media. It could be different. General equations for refractive index are as follows:

(a)

(b)

(c)

Refractive index of glass, af lg

Refractive index o f air, ef | a =

sin i in air sin r in glass

sin i in glass sin r in air

_ 1 n 1Therefore, af]g = —— or *11,- —r -

gl la a*^ _ speed o f light in air . g

a B speed o f light in glassna « V,

^ _ wavelength of light in air . ^ fl = — i e w avelengthof light in glass B

284

While the speed and wavelength of light change Itoni one frequency remainsjincJianned.

medium to the other, (he

. nf 10° Find the angle of at an angle ot JU ■

JAMB I9 )3 29

Example 1A beam of light is incident front air to water refraction if the refractive index o f water is /}.A. 15° B. 18° C. 22° D. 24"SolutionAngle of incidence i = 30o: refractive index D = 4/y, angle of refraction

Substitute into f | = Sm< to obtain sinr

■ = 9

sin 30 sinr

cross multiplying, sin r x 4 = 3 x sin 30° 3x0.5 1.5sin r = -------- = — = 0.375

4 4Angle of refraction r =sin''0.375 = 22

Example 2

The figure above shows the path of light passing through a glass block. Calculate the refractive index of the glass. NECO 200027SolutionAngle of incidence i = 90 - 30 = 60°; Angle of refraction r = 90 - 60 = 30°

Refractive index n =sinisinr

sin 60°sin 30°

0.866

0.51.73

Exam ple 3

Normal

Fig 14.3

285

A ray of light is incident on a glass block as shown in the diagram above. ^and the refracted rays are perpendicular to each other, what is the refractivein e ^30 glass relative to air? "SolutionAngle NOB = 56° (angle of incidence AON = angle of reflection BON)Angle of refraction COD = 180 - (90 + 56) = 34°

i = 56° and r = 34°

Refractive index of glass, f | = ——-g sinr

sin 56° sin 34°

0.8290.559

1.48

Exam ple 4

Fig. 14.5

The diagram above shows an incident ray AO inclined at an angle o f 50° to the interface CB. The refractive ray OB is found to lie along the surface. What is the refractive index o f the medium X with respect to air?

sin 50° „ sin 40° „ sin 90° ^ sin 40° „ sin 90°sin 40°

B.sin 50°

C. D.sin 50°

Solutionsin 90°

E.sin 40°

WAEC 199032

i L

Fig. 14.6

A normal LM is drawn through O as shown in the diagram above.Angle o f incidence i = angle AOM = 90° -50° = 40°Angle o f refraction r = angle BOM = 90°If you were asked to find the refractive index of air, your answer will be:

286

PI _ sin of angle of incidence in mcJuim X _sin of angle of refraction in air sin‘X)

However, you are asked to find the refractive index ot medium X on!cs theair. So. based on the principle of reversibility of light, the refracteincident ray and the incident ray becomes the refracted ray.

sin of angle of incidence in air _ sin 90Retractive index of medium X. .1 I. = ------------------- -— -----: ' vsin of angle of refraction in A sin40

Note that . n =. a

Example 5A ray o f light is incident at an angle of 30l at an air-glass interface. ^

(i) Draw a ray diagram to show the deviation of the ray in the g-ass.(ii) Determine the angle of deviation, [refractive index of glass - 1 • J

IVA EC 20 0 ?SolutionAngle o f incidence i = 30°; = 1.5(t)

sin 30° sinr

0.5sm r= — = 0.333

1.5

r = sin~'0.333 = 19.5° Angle o f deviation d = i - r = 3 0 -19 .5= 105°

x \a v o'f hght is inclined at angle o f 30° on one top surface of a parallel-sided glass of refractive index 1 5 The ray finally emerges from the low surface. What is the angular deviation o f the emerged ray? A 60° B. 39“ C. 28” D O “ JAM B m *

When aligh t ray passes through a rectangular glass block, the incident and the emergent rays are parallel to each other. There is no deviation or change in the direction o f the emergent ray when compared with the incident ray. Therefore, the angular deviation of the emergent ray with respect to the incident ray is zero. (.Inv. fr )

Exam ple 7The absolute refractive indices of glass and water are ' : and J , respectiv ol\ \ " \

refractive index at the interface w hen a ray trav els from water to class isA. Vi B .% C . \ D. l7/,2 t . 2 ' M w ii i ov '

287

SolutionThe absolute refractive index of a medium is the value of fl when the first medium is a vacuum. Therefore, for glass ¡,0^ =3/2; for water aHw = 4/3.

Refractive index o f glass, aD , = _sPee< in air8 speed of light in glass

3 _ 3 x l0 82 Vt>

Speed o f light in glass, V = j j< 3 x l0 8 = 3 x l0 8x2 8 3 3

Refractive index o f water, af | w = _SPCC(* ° f 1*8^ *n a*rspeed o f light in water

4 _ 3x10s3 ~ Vw ~

Speed o f light in water, V = - X3xlQ w 4

Refractive index ol ray from water to glass, wfl = ?P.eed o f » 6 ^ in water = V„8 speed o f light in glass Vg

_ 3x3x10s 3 x l0 8 x24 3

3 x 3 x 108 3 3x3= ------------ x ------- r---- = -------4 3x10 x 2 4 x 2

Exam ple 8The velocities of light in air and glass are 3.0 x 108ms‘‘ and 1.8 x 108m s'! respectively. Calculate the sine o f the angle of incidence that will produce an angle of refraction of 30° for a ray o f light incident on glass. WAEC 1988:sSolutionVelocity o f light in air, V3 = 3.0 x 108ms 1 Velocity of light in glass, Vg = 1.8 x 10sms 1 Angle of refraction r = 30°Sine of angle of incidence, sin i = ?

pi _ sin i _ velocity of light in air a 8 sinr velocity oflight in glass

sini 3 x l0 8 ” sin30° ” 1 .8x l08

sini =sin30° x3 .0x 108

1.8 xlO80.5x3

1.8

Example 9The .'■peed of light in vacuum is 3 x 10 ms . If the refractive index of a transparent liquid is 4/'„ then the speed oflight in the liquid is A. 0.44 x 10sm s'1 B. 2.25 x 1 0 'W C. 3.0 x l O W 1 D. 4.0 x l O W E. 4.33 x 108ms"' JAMB 198346 SolutionSpeed oflight in vacuum Va = 3.0 x 108ms Speed oflight in the liquid Vl =?

288

Refractive index fl = 4/3 V

Substitute into fl = — to obtainV.

3x10

3x10s x3 _ 9 -?<;xl08ms‘Speed of light in the liquid VL= ------

Example 10 . , r rPfractive index fl. If theLight of velocity 3.0 x 108ms is incident on a material ot retrvelocity of light is reduced to 2.4 x lO ^ s '1 i n t h e material, what is j a u r 1Q0735A. 2.33 B. 2.25 C. 1.33 D. 1.25 JAM BSolutionVelocity of light at incidence Va = 3.0 x 108ms''Velocity of light after refraction Vg = 2.4 x 108ms''

D , . . n V, 3.0xl08 ,Refractive index (| = — = ---------r = I -25Vg 2.4 x10s

Example 11A glass plate 0.9cm thick has a refractive index of 1.50. How long docs it take for a pulse of light to pass through the plate? A. 3.0 x 10'los B. 3.0 x 10 "s C. 4.5 x 10 sD. 4.5 x 10'us [c=3.0 x 1 0 W ] JAMB 200720SolutionGlass thickness or length = 0.9cm; fl = 1.5; Va = 3.0 x 108ms‘'

n

1.5 =

velocity of light in air (Va) velocity of light in glass (Vg)

3.0x10sV

Vg =3.0x10s

1.52x10 ms

The speed of light in glass is V = 2 d 108ms' Length (distance) of glass is d = 0.9cm = 0.009m

distance distance(d)----------, time = ----------time velocity (v)

From velocity

Time taken to pass through glass, t = — =v 2x10s

Example 12Surface waves travelling in deep water at 15ms'1 are incident at a shallow water boundary. If the angles of incidence and refraction are 45° and 30° respectively calculate the speed of the waves in shallow water. ’SolutionWave speed in deep water VD = 15m/s Angle of refraction r = 30°-Angle of incidence i = 45° Wave speed in shallow water Vs = ?

, sin i VnSubstitute into -----= —sin r Vs

289

sin 45° _ 15 sin 30° ~ Vs

Wave speed in shallow water Vssin 30 ° x 15

sin 45°10.6ms

Example 13 , fA light wave travels from air into a medium of refractive index 1.54. If the wave e n g to the light in air is 5.9 x 10'7m, calculate its wavelength in the medium. WAEC 1 SolutionD f- , wavelenghtof ligth in airRefractive index = --------------- ------- -------------

wavelenght of light in meduim5 . 9 x l 0 “ 71.54 =

wavelength of light in meduim5 9x1 O'7wavelenght of light in meduim = —----- — 3 83 x 1 O'7 m

1.54

Example 14An electromagnetic wave of frequency 5.0 x 1014Hz is incident on the surface of water of refractive index 4/3. Taking the speed of wave in air as 3.8 x 108ms ', calculate the wavelength of the wave in water. WAEC 199839SolutionFrequency f = 5.0 x 10l4Hz; refractive index fl = 4/3 ; Wave speed V = 3.8 x 108ms_l

From V = Xf, , . . V 3.8x10swavelength in air, X = — = --------- —f 5.0x104

6.0xl0~7m

, nwavelength of wave in air (Xa )

wavelength of wave in water(Àw )4 _ 6.0X10"73 ~ X„, 6 - O x l O ' 7 K —

w 43x6.0xl0~7

4= 4.50 x 10“7 m

REAL AND APPARENT DEPTHRefractive index (fl) is related to real and apparent depth by the following

equation;_ Real (actual) depth

Apparent depth

- « RThat is, (1 = —A

Lateral displacement or apparent (upward) displacement of an object placed under a glass block or in a pool of liquid is given by;Apparent displacement = Real depth - Apparent depth

That is, a = R - A

For example, the apparent upward displacement of a point object placed under a rectangular glass prism of thickness d and absolute refractive index 0 is derived as

290

follows:Real depth = d; refractive index = 0

Refractive index = -

Apparent depth

Real depth Apparent depth

Real depthRefractive index f|

Apparent displacement = Real depth - Apparent depth

Q d -d d ( f l- l)n n

Example 15 ,A transparent rectangular block 5.0cm thick is placed on a black dot. The ot w en viewed from above is seen 3.0cm from the top of the block. Calculate the re rac index of the material of the block. WAESolutionReal depth R = 5.0cm; Apparent depth A = 3.0cm

Refractive index, D = — = — = 1.67 A 3

Example 16

Calculate the refractive index of the material of the glass block shown in the diagram above if YZ = 4cm. WAEC 199125SolutionReal depth or height (XZ), R = 10cm; YZ = 4cm

Apparent depth, A = XY = XZ - YZ = 10 - 4 = 6cm

^ R 10cm ,Refractive index, |1 = — = -------= 1.67A 4cm

Example 17A rectangular glass prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench, if the refractive index of the material of the prism is 1 5 calculate the apparent displacement of the mark. WAEC 2()0(jE,'}SolutionReal depth (thickness) R = 12cm; Refractive index IT = 1.5

For a real depth R and refractive index IT, the apparent displacement, a is given by R (H -l) 12(1.5-1) 12x0.5 ,a = ----------- = — ---------- - ------------= 4cm

291

SolutionRefractive index fl = 3/2;Angle o f minimum deviation Dm = ?

smy2{Dm+60°)

Prism’s refracting angle A - 60°;

Substitute int o 0 = -Yi

to obtain

3 _ sin X (An + 60° )2 s in 6

1 5 _ sinX (D w+60°) sin ^ (£> „+ 60°) sin 30 0.5

cross multiplying 1.5x0.5 = sin }<(£>„, +60°)

0.75=sin/K(£)m+60°)

0 75— = yAD„+60") sin

s in '10.75 =D m 60° 2 2

48.59° = —=- + 30 2

a

—— = 48.59 - 30 = 18.59° 2

Dm =2x18.59° =37.2"

An alternative method for ‘Example 30’ is as follows:

A = 60°; fl = 7 2 = 1.5; At minimum deviation r = A/2 = 60¡2 — 30°

~ s in 1Substitute into | ] = ------- to obtains i n r

sin30

sin i = 1.5 x sin 30° = 1.5 x 0.5 = 0.75

i= sin ‘0.75 = 48.75° ~ 48.6°

Angle o f minimum deviation Dm = 2i - 2r

= 2(48.6)-2(30)

= 37.2°

or Dm = 2i - A

= 2(48.6) - 60

= 37.2°

Exam ple 31Calculate the refractive index of the material for the glass prism in the diagram below

A ,V 2 B .7 , C. D .7 , JAMB 2002s

296

Fig. 14.10

SolutionRefracting angle of prism A = 60°; Angle of deviation I) = 30°

Substitute into f) = sin y2(A + D) sin 'AA

sin ^ (6 0 + 30) _ sin^(90) _ sin 45" sin>^(60) sin 30 sin 30°

From Trigonometry, sin45 = ^J/y^; sin30"= X

sin 45 _ 1 1 _ I x 2 _ 2sin30 V ? 2 yfi 1 4 l

multiply both numerator and denominator by -J22 V2 _ l 4 l _ 2V2 _ rz

; v T T T ï ï ' 2

Exam ple 32The figure below shows a plot of angles of deviation through a glass prism when light is incident at 0 degrees on the pnsm. The incident angle that produces the minimum deviation is A. 25° B. 18° C. 35° D. 20° JAM B 200539

SolutionAs seen from the graph, fig 14.12, at minimum deviation o f 18°, the angle o f incidence is25°

Ans: A

297

REFRACTION THROUGH CONVEX AND CONCAVE LENSESIn contrast with convex mirror, a convex lens converges rays of light. Similarly, a

concave lens diverges rays o f light as opposed to concave mirror that converges rays of light.

The following formulas, expressions and rules are used in solving convex and concave lenses problems.1. Relationship between the object distance u, image distance, v and focal length, f is given by the lens equation.

u v f u + v

2 .

3.

4.

Magnification M - ima®e ^e'Sht _ image distance from lens object height object distance from lens

That is, M = — u

(i)(ii)(iii)(iv)

A convex (converging) lens has a positive focal length.A concave (diverging) lens has a negative focal length. Distances of real object and real images are positive. Distance o f virtual object and virtual images are negative.

The power of a lens,I

P = ------— ----------------focal length in meters

The unit o f lens power is the dioptre (D)

That is, P lf

Example 33An object is placed 36cm from a converging lens of focal length 24cm. If a real ima«e which is 4cm high is formed, calculate the height of the object. WAEC 1988'0SolutionObject distance, u = 36cm; focal length, f = 24cm; Image height = 4cm; object height = ?

Substitute into f = to get image distance, v u + v

36v

298

Cross multiplying, 24 (36 + v) 36v24 x 36 + 24v » 36v 24 x 36 = 36v - 24v

24x3612v

12= 72cm

substituting;

^ _ image height _ image distance object height object distance

4cm 72object height 36

. . . . . 4x36 144 _object height = --------= ----- = 2cm72 72

Example 34A lens of focal length 15cm forms an upright image four times the size o an o jec . Calculate the distance of the image from the lens. WAhSolutionf = 15cm; M = 7 U = 4 From 7 U = 4; u = v/4; image distance, v = ?

Substitute into - = - If _ v u

J _ _ J_ J _ _ J _ + 415 v v / v v

/ 4

15 “ J___5 15 _ v

14-4 _ 5 v v

v = 15x5 = 15cm

Example 35An object placed 50cm away from the focus of an emerging lens of focal length 15cm, produces a focused image on a screen. Calculate the distance between the object and the screen. WAEC I9 9 (f‘2SolutionFocal length, f = 15cm; object distance, u = 50 + 15 = 65cm

Fig. 14.13 50cm- A -

15cm

Substitute into f =

15 =

uv to obtain image distance.u + v

65v 65+ v

15(65 + v) = 65v 975 + 15v = 65v

975 = 65v - 15v 975= 50v

975v = — = 19.5cm

50

Object distance from screen = u + v = 65 +19.5 84.5cm

299

SolutionRefractive index PI = 3/2; Prism’s refracting angle A = 60°;Angle of minimum deviation Dm = ?

Substitute int o fl = sin + 60 ) to 0\,tain Vi

3 _ smy2(Pm +60°)2 sin6! g ^ s in X ^ + e O 0) smYi {Dm +60°)

sin 30 0.5cross multiplying 1.5x0.5 = siny2(Dn + 60°)

0.75=sin>$(Dm+60°)

0.75 ,— - = /2(Dm+60°)

sin“10.75 = - + —

48.59° = —— + 30° 2

A.2

+ 30°

—— = 48.59 - 30 = 18.59° 2

Dm = 2x18.59° =37.2°

An alternative method for ‘Example 30’ is as follows:

A = 60°; fl = 3/2 = 1.5; At minimum deviation r = A/2 = = 30°„ sin/ , .

Substitute into | 1 = ------ to obtamsinr

sin i = 1.5 x sin 30° = 1.5 x 0.5 = 0.75 i = sin '0.75 = 48.75° ~ 48.6°

Angle of minimum deviation Dm = 2i - 2r= 2(48.6)-2(30)

= 37.2°

or Dm =2i -A= 2(48.6) - 60

= 37.2°

Example 31Calculate the refractive index of the material for the

A. a/2 b . 3/2

glass prism in the diagram below.

D. 4/3 JAMB 2002”

-296

Fig. 14.10

SolutionRefracting angle of prism A = 60°; Angle of deviation D - 30°

Substitute into f | =sin %(A + D)

sin / 2A

sin X (60 + 30) sin (90) sin 45° sinj^ôO) sin 30 sin 30°

From Trigonometry, sin 45 = sin 30° = Yi

sin 45 _ 1 1 _ 1 2 _ 2sin 30 ~ V ? ' 2 _ V2 l ” V2

multiply both numerator and denominator by V i ;2 V2 2V2 _ 2V2 _ nr

V 2 XV2 = “ 2 V

Example 32The figure below shows a plot of angles of deviation through a glass prism when light is incident at 0 degrees on the prism. The incident angle that produces the minimum deviation is A. 25° B. 18° C. 35° D. 20° JAMB 200539

SolutionAs seen from the graph, fig 14.12, at minimum deviation of 18°, the angle of incidence is

Ans: A

REFRACTION THROUGH CONVEX AND CONCAVE LENSESIn contrast with convex mirror, a convex lens converges rays of light. Similarly, a

concave lens diverges rays of light as opposed to concave mirror that converges rays of light

The following formulas, expressions and rules are used in solving convex and concave lenses problems.1. Relationship between the object distance u, image distance, v and focal length, f isgiven by the lens equation.

U V f u-f-v

_ image height image distance from lens2. Magnification, M = ------ ------= --------- --------------------------object height object distance from lensv

That is, M = — u

(i) A convex (converging) lens has a positive focal length.(ii) A concave (diverging) lens has a negative focal length.(iii) Distances of real object and real images are positive.(iv) Distance of virtual object and virtual images are negative.

The power of a lens, P = —— —---- —----------- That is, P = —focal length in meters f

The unit o f lens power is the dioptre (D)

3.

4.

Example 33An object is placed 36cm from a converging lens of focal length 24cm. If a real image which is 4cm high is formed, calculate the height of the object. WAEC 198$°SolutionObject distance, u = 36cm; focal length, f = 24cm; Image height = 4cm; object height = ?

uv ..Substitute into f = ------ to get image distance, v

u + v

24 =36v

36 + v

298

Cross multiplying, 24 (36 + v) = 36v24 x 36 + 24v = 36v24 x 36= 36v - 24v = 12v

v =24x36

12= 72cm

image height image distanceobject height object distance

. . . 4cm 72substituting, ------------ —— = —object height 36

, . , 4x36 144 _object height = — — = — - = 2cm

72 72

Example 34A lens of focal length 15cm forms an upright image four times the Calculate the distance o f the image from the lens.Solutionf = 15cm; M = 7 U = 4 From 7 U = 4; u = 74; image distance.

size o f an object. WAEC 1993-°

Substitute intof V u

15 v v / v v / 4

1 _ 14-4 _ 515 v v

— = — /. v = 15x5 = 75cm15 v

Example 35An object placed 50cm away from the focus of an emerging lens o f focal length 15cm, produces a focused image on a screen. Calculate the distance between the object and the screen. WAEC 1 9 9 (f2SolutionFocal length, f = 15cm; object distance, u = 50 + 15 = 65cm

Substitute into f = to obtain image distance.

15 =

u + v65 v

65 + v15(65 + v) = 65v975 + 15v = 65v

975 = 65v - 15v975 = 50v

975 -v --------- 19.5cm

50

Object distance from screen = u + v — 65 + 19.5 = 84.5cm

299

Example 36A converging lens of focal length 5cm is used as a magnifying glass by a man whose near point is 35cm. Calculate the magnification given by the lens. WAEC 1998Solutionf — 5cm; the near point is also the image distance v = 35cm; object distance, u = ?

uvSubstitute into f =

5 =

u + v 35u

35+ u5(35 + u) = 35u175 + 5u = 35u

175 = 35u - 5u175= 30u

175 ru = -----= 5.83

30

Magnification, M = —= - = 6 u 5.83

Example 37A lantern gives an image of 3m square of a slide 7.62cm square on a screen. If the screen is 10m from the projection lens of the lantern, calculate the focal length of lens.

WAEC 199832SolutionImage size or height, v = 3m = 300cm; object size or height, u = 7.62cm Magnification, M = v/u

w 300M = ------= 39.377.62

Image distance, v = 10m = 1000cm

From v/u = 39.37, object distance, u =

25.40x1000Focal length, f =

v _ 1000 39.37 ~ 39.37 25400

= 25.40

u + v 25.40+1000 1025.40= 24.8cm

Example 38An object 2.5mm long is viewed through a converging lens of focal length 10.0cm held close to the eyes. A magnified image o f the object is formed 30.0cm from the lens. Calculate the (i) distance of the object from the lens

(ii) size of the image(iii) power o f the lens WAEC 2002Eli

Solutionf = 10cm; image distance, v = -30cm (the final image seen by the eye is virtual);object height = 2.5mm = 0.25cm; object distance from lens u = ?

. 1 1 1 (iV Substitute into — = — + — v ' f u v

j _ = j _ j _ .10 ~ u + -30 “ 10 u 30

1 _ 1 _1_ = ^ + u _ 30 10 " u 30 10

300

— = 1 + 3u

30 _ 4 u = — = 1.5cmu 4

(")v 30M = - = ----= 4u 7.5

^ image height image heightobject height 0.25cm

image height(size) = 4 x 0.25 = 1cm or 10mm

(iii) Lens power, D = Vf = '/o.i = 10 dioptreNote that the focal length, 10cm is converted to metres (0.1m)

Example 39 .An object 40cm from a converging lens of focal length 20cm moves with a velocity o 5cms'' towards the lens. Calculate the image position after 2s. NECO 2004SolutionObject distance, u = 40cm; after 2s at 5cm/s the object moves 10cm, therefore the new

object distance, u = 40 - 10 = 30cm; f = 20cm.

1 - 1 1f u v

J _ = J _ 1 20 " 30 v

J__ J ___ 1_v _ 20 30

1 _ 3 -2 v 60

— = — .'. v = 60cm• v 60

Image distance, v = 60cm.

Example 40By what factor will the size of an object placed 10cm from a convex lens be increased if the image is seen on a screen placed 25cm from the lens?A. 15.0 B. 2.5 C. 1.5 D. 0.4 JAM B 200316SolutionObject distance, u = 10cm; image distance, v = 25cm Magnification, M = 7 U = 25/ jo = 2.5

Example 41If u is the object distance and v the image distance, which o f the following expressions gives the linear magnification produced by a concave lens o f focal length f?

A. — + f B. - 1 - f C. 1 - 1 D. 1 + 1 JA M B 20047v f f 1

SolutionMagnification, M = 7 U

301

1 = 1 + 1 f U V

Multiply both sides by v

v v v I u v v _ v v U f v

Linear magnification (M = 7 U) = j -1

Example 42An object is placed in front o f a converging lens of focal length 20cm. The image is virtual and has a magnification of 2. What is the distance of the object from the lens?A. 5cm B. 10cm C. 30cm D. 40cm JAMB 199435SolutionFocal length, f = 20cm; image distance = — v (negative because it’s virtual);M = — /u = 2 v = — 2u; object distance, u = ?

Substitute into f = uv u - V

2 0 = u x ( - 2 u) = - 2 u l = - 2 + = 2 u

u + (-2u) u - 2u - u

20 = 2u u = 20/2 = 10cm

Example 43A concave lens o f focal length 20cm form an image A the size of the object. The object distance is A. 100cm B. °°/9cm C. 60cm D. 60/7cm E.None of these values.

JAMB 1979s2Solution

y

f = —20cm; M = 7 U = y (.... Image lA size of object.....)

V = ‘Au; object distance, u = ?

! = - + ! f u v

1 = - + - ! — = — + — = — (l + 2)- 2 0 u y2u

-_1 = 3 20 u

u u u

u = 20x3 = 60cm

Example 44The image o f a pin formed by a diverging lens of focal length 10cm is 5cm from the lens. Calculate the distance of the pin from the lens. WAEC 1992Solution _Focal length, f = - 10cm (diverging lens); image distance, v = - 5cm;Object distance, u = ?

uvSubstitute into f = — -— u + v

302

10 =— 5u u-5

Cross multiply - 10(u - 5) — 5u - lOu + 50 = - 5u

5 0 = lOu - 5u 50= 5u

u = -^= 10cm 5

S Z i l b n - d * . converging ,ens of foca, .eng* f - - lens. Calculate the magnification produced.SolutionFocal length, f = f;

Substitute into

image distance, v = 2f; magnification, M

I = I + If u v

f u 2f

1 = 1 - — u _ f 2f

7u = 9

1 2-1 _ 1 u _ 2f 2f1 1

Magnification,

u

u

2f

— = 1.5 2f

u = 2f

Example 46A thin converging lens has a power of 4.0 diopters. Determine its focal length. A. 0.25m B. 0.03m C. 5.00m D. 2.5m J A M B 2 0 0 5SolutionPower = Vf 4 —Vf; f — Vi — 0.25

APPLICATION OF LIGHT W AVES

The Human EyeThe near point is the nearest distance (about 25cm) from the eye at which an

object can be seen clearly by the eye.The far point is the farthest point (infinity) at which an eye can see an object

clearly. The least distance of distinct vision is the distance from the near point to the eye.Long sightedness is corrected by convex lens. Short sightedness is corrected by

concave lens. The lens formula is applied in problems involving the human eye as exemplified by the following:

303

Example 47A tar-sighted person cannot see objects that are less than 100cm away. If this person wants to read a book at 25cm, what type and focal length of lens docs he need.’ ^A. convex, 20cm B. concave, 20cm C. convex, 33cm D. concave, 33cm JAMB ¡9S9~'s SolutionObject distance, u = 25cm;

Focal length, f = u% = u + v

image distance, v = 100cm25x100 2500~ ——— = -------= 20cm25+100 125

A ns A: convex, 20cm

Example 48A patient with a sight detect has a least distance of distinct vision of 150cm. For him to be able to read a material placed at a distance of 25cm. what is the focal length of the glasses he should wear?A. 15.0cm B. 17.6cm C. 21.4cm D. 30.0cm JAMB 1997}t>SolutionImage distance, v = 150cm; object distance, u = 25cm; focal length, f = ?

Focal length, f = = = — = 21.43cmu + v 150+25 175

Example 49A man wears convex lens glasses o f focal length 30cm in order to correct his eye defect. Instead o f the optimum 25cm, his least distance of distinct vision is A. 14cm B. 28cm C. 75cm D. 150cm JAMB 199834SolutionFocal length. f = 30cm; image distance, v = ?; object distance, u = 25cm;

1 = 1 If u v

— = — I30 ~ 25 v

J_ _ J ____1_

v ~ 30 25

1 5—6 —1v “ 150 “ 150

v = 150cm

Example SOA long-sighted man cannot see clearly objects that are less than 120cm away. What is the type and focal length o f lens required to make him read a book at 30cm away. NECO 200f: SolutionImage distance, v = - 120cm (negative because image formed by the eye is virtual); Object distance, u = 30cm.

f u v

f 3 0 + - 120

• U - U - Lf 30 120

1 4 - 1 3 1f ” 120 120 40

304

f = 40cm. Because fis positive, the lens is a converging or con

M ICRO SCO PES AND TELESCO PES AND M A G N IFY IN G G L A

In addition to the lens formular (Vu + '/„ = %). 11,0 lo,loWing &XC USC P involving microscopes and telescopes. _ ^ ^1. Magnifying power (M) o f compound microscope is M - i * 2

Where M| = magnification by the objective lens

M2 = magnification by the eye piece

2. Magnifying power of the telescope is

Where f0 = focal length of objective lens

fe = focal length of eye piece lens3. Distance (d) between objective lens and eye piece of an astronomical telescope is

d = f0 + fe4. Distance (d) between objective lens and eye piece of a Galilean telescope is

d = f0-fe

Example 51Converging lenses of focal lengths 120cm and 10cm are used to construct an astronomical telescope. What is the distance between the lenses at its normal adjustment?

NECO 200730SolutionDistance between lenses d = f0 + fe = 120 + 10 = 130cm

Example 52A simple magnifying glass is used to view an object. At what distance from the lens must the object be placed so that an image 5 times the size of the object is produced 20cm from the lens? WAEC 199428SolutionImage distance, v = - 20cm (image formed by magnifying glass is virtual);

vmagnification, M = — = 5

u

Substitute v = -20cm into

- 20 = 5u

Example 53Ln a compound microscope, the image formed by the objective lens is at a distance of 3.0cm from the eye lens. If the final image is at 25.0cm from the eye lens, calculate thernriil I math nt th<* f * \ n * puc ’ ™focal length of the eye lens. Solution WAEC 20023

Object distance, u - 3.0cm; image distance, v = -25cm (image is virtual)- Focal length, f = ? h

305

3x25f = uvU + V 3+ (-25)

753 -2 5

— = 3.41cm22

EXERCISE 14.1. A ray of light is incident on a body X as shown in the diagram below. What is

the refractive index o f the body?

2. When a ray o f light is incident normally on an air-glass interface, what is its angle o f refraction? WAEC 199629 Ans: 0°

3.

In the figure above, a ray o f light in our strikes a glass plate at an angle of incidence o f 60°. The reflected ray is observed to be perpendicular to the refracted ray. What is the refractive index o f the glass?A. 1.73 B. 150 C. 0.87 D. 0.57 JAMB 198627 Ans: 1.734.

The refractive index o f the medium M in the diagram above isA. ^ B. 2V3 C. V3 D. JAMB 2004 Ans: V3

306

5.to

I lie speed of light in air is I x 10 ms wiitcr is Yu then which of the following — - s \

s ,n h e r e f r a c t i v e index o f light from a i r -u ,'s the correct value of the speed of light in

„ . * 4 , 1 I). 2.25 x 10 ms ‘water? A. 4 x I 0 W H. 2.25 x I0"ms < ■ h x /( (y : Ans: 2.25 x 1 0 W

! ; 4 " i ^ . M i,* ...............* . \ n ( a a i n

n dr 8f ' m - iS ' A '■ 'r ’ln n "" .!" ' i W"” ‘ 10 JAMB 19*/' An*: DI). 1.82 x 10 ms L. 6.00 x 10 ms . , ■ 0f light in vacuum is 3.0 x7. fhc refractive index of a liquid is 1.5. II the vid y ..logins’1, the velocity of light in the liquid is A. 1.5x10 msC. 3-Ox IO V s-1 D. 4.5 x I 0 W E . 9.0 x I0xm s '• J A \

^ __1 w hat is the speed in glass with a

i o WIM»> _ oJAMH 1983 Arts: 2.0 x l ( f ms

X. The speed of light in air is 3.0 x 10 msrefractive index of 1.50 A. 1.5 x 10 ms B. 3.0 x 10 ms C. 2.0 x 10 ms'

I). 6.0 x 10 ms9. The velocities of light in air and glass are

JAMB 199228 Ans: 2.0 x l ( f ms '3.0 x 10W and 2.0 x 10 ms'

respectively. If the angle of refraction is 30°, the sine of the angle of ' sA. 0.33 B.0.50 C. 0.67 D. 0.75 JAMB 1 9 9 ? 4 Ans. 0.3310. Radio waves travel in air at 3.0 x i o W . If the waves enter water o f refractive index 4/3, calculate the speed of radio waves in water. WAEC 2005" Ans: 2.25 x W ms11. Light of wavelength 5000 x lOAun travels in free space with a velocity of 3.0 xlogins'. What is its wavelength in glass of refractive index 1.5? 8A. 3333 x 10"cm B. 5000 x 10 scm C. 6666 x 10 8cm D. 7500 x 10' cm

JAMB 198924 Ans: 3333 x 10' cm12. A light wave of frequency 5 x 101411z moves through water which has a refractiveindex of Y3. Calculate the wavelength in water if the velocity of light in air is 3.0 x 108m s'. A. 4 . 5 x l 0 ' 7m B . 6 .0 x l 0 ' 7m C. 1 . 7 x l 0 6m D . 2 . 2 x l O fim

JAMB 199021 Ans: 4.5 x 10'7m13. Light of frequency 6.0 x 10l4Hz travelling in air is transmitted through glass of refractive index 1.5. Calculate the frequency of the light in the glass.A. 4.0 x 10I4Hz B. 6.0 x 10l4Hz C. 7.5 x 10l4Hz D. 9.0 x 10l4Hz

JAMB 199434 Ans:6.0 x 10,4Hz14. A point object placed in contact with one surface of a glass block of thickness 1,6cm and of refractive index 1.61 is viewed along the normal to the opposite surface. By how much does the point object appears to be displaced. WAEC 1 9 8 v i Ans: 0.60cm15. A microscope is focused on a mark on a table. When the mark is covered by a plate of glass 3.00cm thick, the microscope has to be raised 1.18cm for the mark to be once more in focus. Calculate the refractive index of the glass. WAEC 1995P2 Ans: 1.6416. A rectangular glass prism of thickness 6cnt and refractive index 1.5 is placed onthe page of a book. The prints on the book are viewed vertically downwards from above. Determine the apparent upward displacement of the prints. WAEC 2001p2 Ans: 2cm17. A rectangular glass prism of thickness d and absolute refractive index fl is placed on a point object, which is viewed vertically downward from above the prism. Which o f the following expressions correctly defines the apparent upward displacem ent o f the object?A. d/n B. d n c . d/n.2 D. d(n' 1)/n W AEC 200734 Ans: D18. The displacement d produced in a glass block of thickness t and refractive index fl when an object is viewed through it isA. t - fl B. t(l+X,) C. t(l- yn) D. i(yn - l) JA M B 1998}2 Ans: C19. A coin lies at the bottom of a lank containing water to a depth o f 130cm. If therefractive index of water is 1.3, calculate the apparent displacement of the coin when viewed vertically from above. wAFC ¡ 99f - Ans: 30cm20. The horizontal lloor of a water reservoir appears to be 1.0m deep when viewedvertically from above. If the refractive index o f water is 1.35, calculate the real depth of the reservoir. WAEC 2 ( m u Ans . ]M m

307

21. A coin is placed at the bottom of a cube of glass t cm thick. If the refractive in e* of the glass is p, how high does the coin appear to be raised to an observer oo ing perpendicularly into glass?

A. 1t - / r

B. t (A-D C. t , + l D. — E. pt JAMB 198432 Ans B

21. Two rays of light from a point below the surface of water are equally inclined to the vertical and are inclined to each other at 60° in water. What is the an^le between the rays when they emerge into air? [Take the refractive index of water to be /3]

JAMB 1985" Ans: 83.6°22. A coin placed below a rectangular glass block of thickness 9cm and refractiveindex 1.5 is viewed vertically above the block. The apparent displacement o f the coin is A. 8cm B. 6cm C. 5cm D. 3cm JAMB 20021 Ans: 3cm23. A trough 12.0cm deep is filled with water o f refractive index 4/3. By h&w much would a coin at the bottom of the trough appear to be displaced when viewed vertically from above the water surface?A. 3.0cm B. 6.0cm C. 9.0cm D. 16.0m JAMB 199127 Ans:3.0cm24. The refractive index of a medium relative to air is 1.8. Calculate the critical anglefor the medium to the nearest degree. WAEC 199611 Ans: 34°25. Calculate the critical angle of a medium of refractive index 1.65 when light passesfrom the medium to air. WAEC 2005P~ Ans: 3 7.3°26. Calculate the critical angle for light travelling from water to air. [Refractive udexof water =1.33] NECO 2003‘8 Ans: 48.75°27. If the refractive index of water is 1.33, what is the critical angle for a water - airboundary? NECO 200230 Ans: 48.8°

28. If the refractive index of a medium is -\/2 , what is the critical angle?A. 45° B. 50° 121 C. 56° 251 D. 75° E. 90° JAMB 198224 Ans: 45°29. If the refractive index of a medium in air is 2.0, what is the critical angle for thismedium? A. 30° B. 42° C. 45° D. 50° JAMB 199T 9 A ns:3(f30. What is the approximate critical angle for total internal reflection for diamond if the refractive index of diamond is 2.42?A. 21° B. 22° C. 23° D. 24° JAMB 1995" Ans:24°31. A ray o f light is incident at an angle of 30° on a glass prism of refractive index1.5. Calculate the angle through which the ray is minimally deviated in the prism. (The medium surrounding the prism is air). WAEC 199126 Ans: 21.1°32. Calculate the angle of minimum deviation for a ray which is refracted through an equiangular prism of refractive index 1.4.A. 29° B. 60° C. 99° D. 90° JAMB 20045 Ans:29°33. If the critical angle of a glass-air boundary is C and the refractive index o f the glass is fl, which o f the following relationship is correct?

A. n = — — B. f | = — C. sin90sinC = n D .sinC = - E . n = —sinC 90 f | sin45

WAEC 199427 Ans: D34. A ray o f light experience a minimum deviation when passing through anequilateral triangular glass prism. Calculate the angle of incidence of the ray. [Refractive index o f glass = 1.5] NECO 200429 Ans: 48. ( f35. In an experiment to measure the focal length of a converging lens, object distance, u and corresponding image distance, v were measured and ’/u plotted against '/v to obtain the type o f graph illustrated below. How would f be found from this graph?

308

File. 14.17

A. f ■ the slope of the graphB. f ■ the intercept on the */u axisC. f line intercept on the '/v axisD. * f - reciprocal o f the slopeli. f* reciprocal o f the intercept in either axis. _ - sljdc bc p |aCcd in u

At which of the following distances from the lens .

IVA EC 1988:v Ans.E

36. At which of the following distances from the lens ¡»nuu.uslide projector, if f is the focal length of the projection lens? , tunn j fa i . r n r.......... .. or C. Greater than I but less than L\A. Less than f B. Greater than 2f WAEC 19912* A ns-C1). Equal to f E. Equal to 2f ' r «rnm ,.37. What is the image distance of an object placed at a < inns38 Ans i fconverging lens o f focal length (738. A converging lens has n local length of 5cm. Determine its ,,, n

* N WAEC 2003 Arts:+20 D39. An object is placed a distance 10cm in front of a concave mirror of focal length 15cm. Determine the characteristic of the image formed. W A E C 2 0 0 4 A r t s . V i r t u a l , v . U l a n

40. A converging lens of focal length 5cm forms a virtual image which is 10cm fromthe lens. How far from the lens is the object? WAEC 1996 Ans. J..h ni41. In an experiment to measure the focal length of a converging lens, object distance, u and corresponding image distance, v were measured and /u plotted against /v to obtain

How would f bc found from this graph?A. f - the slope of the graphB. f = the intercept on the Vu axisC. f = the intercept on the '/v - axisD. f = reciprocal o f the slopeE. f = reciprocal o f the intercept in either axis. WAEC 199225 Ans E

42. A real image of an object formed by a converging lens o f focal length 15cm is three times the size of the object. What is the distance of the object from the lens?

, , , WAEC 19922H Ans: 20cm43. A converging lens of focal length 15cm is used to obtain a real image magnified l'/t times. Calculate the distance of the image from the lens. WAEC 1997“ Ans 37 5cm44. I he real image of an object, formed by a converging lens of focal length I 5cm. is three times the size of the object. Calculate the object distance. WAEC 1 9 9 9 Ans-20cm45. An object is placed 20cm from a lens. If an image is formed on a screen 260cm away from the lens, calculate the magnification of the image. WAEC 2001:s Ans -1346. A converging lens of focal length 15cm forms a virtual image at a point 10cm from the lens. C alculate the distance of the object from the lens. WAEC 2002™ Ans 6cm

309

47. A converging lens produces an image four times as large as an object placed 25cmfrom the lens. Calculate its focal length. WAEC 2003 Ans:20cm48. A converging lens forms a real image of a real object. If the magnification is 2 and the distance between the image and the object is 90.0cm, determine the

(i) focal length of the lens;(ii) object distance for which the image would be the same size as the object.

WAEC 2006e13 Arts: (i) 20cm (ii) 40cm49. An object placed on the principal axis o f a convex lens o f focal length 10cm, produces a real image o f double magnification. The image distance from the lens isA. 30cm B. 25cm C. 20cm D. 15cm E. 10cm JAMB 198146 Arts: 30cm

50. A lens o f focal length 12.0cm forms an upright image three times the size o f a real object. The distance between the object and the image is A. 8.0cm B. 16.0cmC. 24.0cm D. 32.0cm JAMB 198628 Ans:64cm [incorrect options]51. An object 3.0cm high is placed 60.0cm from a converging lens whose focal length is 20.0cm. Calculate the size of the image formed.A. 0.5cm B. 1.5cm C. 2.0cm D. 6.0cm JAMB 1987s4 (Atis: 1.5cm)52. To obtain a magnification o f 2.5, how far should an object be placed from the pole of thin converging lens of focal length 0.20m?A. 0.13m B. 0.25m C. 0.28m D. 0.50m JAMB 198830 (Arts: 0.28m)53. What must be the distance between an object and a converging lens o f focr »ngth 20cm to produce an erect image two times the object height. JAMB 1990‘7 Ans. 'em54. If the focal length o f a camera lens is 20cm, the distance from the film at which the lens must be set to produce a sharp image o f an object 100cm away isA. 17cm B. 20cm C. 25cm D. 100cm JAMB 199327 Ans: 25cm55. A projection lantern is used to give the image of a slide on a screen. If the image is 24 times as large as the slide and the screen is 72.0m from the projecting le... vhat is the position o f the slide from the lens?A. 4.0m B. 3.5m C. 3.0m D. 0.3m JAMB 200034 Ans:3.0m56. A certain long-sighted person cannot see clearly objects placed 75cm from his eye. This defect can be corrected by the use o f a

A. converging lens of focal length 37.5cmB. converging lens of focal length 75cmC. cylindrical lens o f focal length 37.5cmD. diverging lens of focal length 37.5cmE. diverging lens of focal length 75cm NECO 200433 Ans: B

57. A person can focus objects when they lie beyond 75cm from his eyes. The focal length o f the lens required to reduce his least distance o f distinct vision to 25cm isA. 75.00cm B. 18.75cm C. 25.00cm D. 37.50cm JAMB 20063 Ans: 18.75cm58. A simple microscope forms an image twice the size o f the object. I f the focal length o f the lens o f the microscope is 20cm, how far is the object from the lens?

WAEC 199034 Ans. 10m59. When an astronomical telescope is in normal adjustment, the focal length o f theobjective lens is 50cm and that o f the eye piece is 2.5cm. What’s the distance between the lenses? WAEC 199230 Ans: 52.5cm60. At what distance from a simple microscope must an object be placed so that an image 5 times the size of the object is produced 20cm from the lens? WAEC 199723Ans: 4.0cm

61. An astronomical telescope, having an objective o f focal length 100cm and eyepiece o f focal length 10cm, is used in normal adjustment. Calculate the separation of the lenses. WAEC 199831 Ans: 1.10m62. A simple microscope forms an image 10cm from an eye close to the lens. If the object is 6cm from the eye, calculate the focal length o f the lens. WAEC 200030 Ans: 15cm63. Four lenses are being considered for use as a microscope objective. Which o f the following focal lengths is most suitable?A .-5m m B.+5m m C .-5 cm D .+5cm JAMB I98737 Ans: B

i 310

and 80cm respectively from an64. A converging lens and a screen are placed 2 c formed on the screen. If theobject in a straight line so tat a sharp image o e o WAEC 2008 Ans: 9cmobject is 3cm high, calculate the height of the image ° j„ce a real image four times65. I f , con v« le»S o f focal length 12Cm ¡5 usedd^P'Od ?

the size of the object, how far from the lens must e j JAMB 2008 Ans: 15cmA. 10cm B. 15cm C. 20cm D. 25cm with a magnifying glass of focal66. An observer with normal eyes view an obj ^ 6length 5cm. The angular magnification is _A. -6 • " 20Qg[least distance of distinct vision D = 25 cm) J liaht travelling from glass to66. Calculate the critical angle in glass for the g , . 5 Q\

water.(Refractive index of water = 1.33, refractive in ex^ _ 8 ,^ 200829 Ans: 41.8°

67. If the refractive index of crown glass is 1.51, its critical angle isA. 48.6o B. 22.5o c 41 5o D. 45.0° JAMB 20093 Ans: C68. When light passes through two media x and y of refractive in ices respectively, the speed of light inA. x is same as in y B. x and y is same as in vacuumC. x is higher than in y D. y is higher than in x JAMB 200¥ Ans.69. A near-sighted student has a near point of 0.1m and a focal length o f 5.0cm. w at is the student’s far point?A. 0.200m B. 8.000m C. 0.125m D. 2.100m JAMB 20093 Ans: A70. A thin lens is placed 50cm from an illuminated object. The image produced has a linear magnification of /*. Calculate the power of the lens in dioptres.

WAEC 200930 Ans: 10.0D71. A ray of light is incident on one face of an equilateral glass prism.(i) Draw a ray diagram to show the path of the ray through the prism.(ii) Calculate the refractive index of the glass if the angle of minimum deviation is 41 °

WAEC 2009*13 Ans: 1.5472. An object placed 15cm from a converging lens forms a real image of >magnification 3. Calculate the focal length of the lens.

NEC0200929 Ans: 11.25cm

311

15SOUND WAVES

VELOCITY AND REFLECTION OF SOUND WAVESSound wave is a longitudinal wave and also a form of energy produced by

vibrating bodies.As common with waves, the velocity or speed of sound wave in any media is

given by:

V = Xf or V = - T

Where V = velocity of sound wave (ms’1)X = wavelength of sound wave (m) f = frequency of sound wave (Hz)T = period of sound wave (s)

Also, speed of sound (V) in air is directly proportional to the square root o f the absolute temperature, 9.

That is, V = V&A quick review of Chapter Twelve with emphasis on stationary wave will be helpful and relevant to this Chapter.

Example 1A source of sound produces waves in air of wavelength 1.65. If the speed of sound in air is 330ms'1, the period of vibration in second is? WAEC 198833SolutionWavelength A. = 1.65m; speed of sound V = 330ms'1; period T = ?

Substitite into V = — to obtain T

330 = ^ - ^ Periodof vibration T = -1 - = 0.005sT 330

Example 2In a sound wave in air, the adjacent rarefactions and compressions are separated by a distance o f 17cm. If the velocity of the sound wave is 340ms'1, determine the frequency A. 10Hz B.20H z C. 100Hz D. 5780Hz JAMB 198922SolutionDistance between successive rarefaction and compression is equivalent to /2.

x/2 = 17 or X = 2 x 17, = 34cm = 0.34m; Velocity, V = 340ms'1

From V = Xf, frequency, f = y = y y = = 1000Hz

Example 3A sound wave of velocity 350ms'1 is directed towards the surface of water. If the ratio of the wavelength o f sound in water to that in air is 425:100, calculate the velocity of the wave in water. WAEC 1996JSSolutionSound wave velocity in air V| = 350ms'1 Sound wave velocity in water V2 = ?Wavelength o f sound in air X| = 100 Wavelength of sound in water X2 = 425

The frequency of a wave does not change when it travels from one medium to another.

312

f rom V = AT, wc derive f = —A.

T herefore — = - =- on substitution we obtain,k, X2

350 _ V,100 “ 425

Sound wave velocity in water, V,_ 350x425 _ |4g7 5ms

100

Example 4The velocity o f sound wave at 27°C is 360m s'1. Its velocity at 127"( is A. 120>/3ms 1 B. 240 ms ' C. 240>/3ins"' L> 720%/J msSolutionTemperature (0) and speed o f sound in air are related by V <1 7 0

■ vjV: fO ,

0| = 27 + 273 = 300K; V, 3 6 0 m s1 0; = ] 27 + 273 400K: V2 ?

[Always remember to convert °C to Kelvin |

3 6 0 7 3 0 0

V, 7400

3 6 0 y j

V2 “ 7 4

v _ 3 6 0 x 7 4 2x360 _ 7207 73 73

Multiply both numeratorand denominator by 73

720 x 73 72071 72073

Substituting

V, =7 3 x 7 3 73x3 79

720T3Velocity o f sound wave at 127°, V,3

= 24073ms

.j a m b J (m -

Example 5The speed o f sound in air is 330m s'1. How far from the centre o f a storm is an observe who hears a thunder clap 2s after the lightning flash? [Neglect the time taken by lielr > travel to the observer] ^ I ’ t .SolutionSpeed o f sound V = 330ms time taken t = 2s; distance of sto rm 's centre

From, speed (V)distance ( v)

timet t)x - V x t = 330x2 - 660m

EchoesLike any other wave, sound waves can be reflected

after a sound wave has been reflected from a hard plane suit I he time (t) taken for an echo to be heard in any me

_ 2a V

31

Where x 7 dislance between the source of sound anil the reflecting sur ace (wall, cliff, sea bed, hill c.t.c.)

V = velocity of sound in the medium (air, water, glass, metal c.t.c.)

Example 6A bat emits 'a sound wave at a speed of 1650.0ms 1 and receives the echoes 0.1 s a|^ r' Calculate the distance of the bat from the reflector. WAEC 1)SolutionSpeed of sound wave V = 1650ms"1; echo time t = 0.15s; Distance from reflector, x

Substitute into x = — to have 2

1650x0.15x = -------------- = 123.75m

Example 7A fathometer is used to send a wave down to the sea bed. The reflected wave is received after 0.5 seconds. Calculate the depth of the sea. [Speed of sound in water = 1500ms ]

NECO 2007.36

SolutionEcho time t = 0.5;

Depth of sea, x =

speed of sound V = 1500ms"1V x t 1500x0.5------ = -------------= 375m

2 2

Example 8A man standing 510m away from a wall sounds a whistle. The echo from 1 wall reaches him 3s later. Calculate the velocity of sound in air. NECO ¿000"SolutionDistance from reflector (wall) x = 510m; echo time t = 3s; Sound velocity V = ?

V t

Substituting, V = = 340ms"1

Example 9How far from a cliff should a boy stand in order to hear the echo o f his clap 0.9s later? [Speed o f sound in air = 330ms"1]. WAEC 199236SolutionSpeed of sound V = 330ms"1; echo time t = 0.9s; Distance from reflector (cliff) x = ?

. . . Vt 330x0.9 1AOenSubstitute into x - — = ------------= 148.50m

2 2

Example 10A sound pulse sent vertically downwards into the earth is reflected from two different layers o f the earth such that echoes are heard after 1,2s and 1,4s. Assuming the speed of the pulse is 2000ms"1, calculate the distance between layers. WAEC l9 8 (/ :Solutiondistance o f first layer = x,; first layer echo time t, = 1,2s;distance o f second layer = x2; second layer echo time t3 = 1,4s.Velocity o f sound pulse V = 2000ms"1;

2 x \/From t = — , distance from reflector (layer) is, x = —

314

Therefore, distance between between layers = x2 - x, =Vt, Vt, V (t2 - 1,)

x 2 ~x.=-V (t2 - t , )

_ 2 0 00 (1 -4 -1 .2 ) _ , n n n x Q 2 2

= 200m

Example 11 . . . •A man hears his echo from a nearby hill 2s after he shouted. If the frequency o is v is 260Hz and the wavelength is 1,29m, how far is the hill? iq q]31A. 330.0m B. 335.4m C. 660.0m D. 670.8m SolutionEcho time t = 2s; frequency f = 260Hz; wavelength X = 1 29mVelocity of sound V = ^.f = 1.29 x 260 = 335.4m/s

Distance from reflector (hill) x = — = - 335.4ms'12 2

Example 12As a ship approaches a cliff, its siren is sounded and the echo is heard in the ship after 12 seconds. 2.1 minutes later the siren is sounded again and the echo is heard 8 seconds later. If the speed o f sound in air is 340ms"1, calculate the velocity at which the ship is approaching the cliff. WAEC 200JEI3Solution ,First echo time t] = 12s; second echo time t2 = 8s;Sound speed V = 340ms"1; time interval between echoes T = 2.1 min = 2.1 x 60 = 126s

V x tx = ------

2

c l - j- „ , si , V x t. 340x12 „„„„Ship distance at 1 echo x. = ------ 1 = ---------- = 2040m2 2

. . . ,. .nd , V x t , 340x8 . . . .Ship distance at 2 echo x , = -------= = ---------= 1360m2 2

Distance between l sl and 2nd echoX = x , - x 2 = 2 0 4 0 - 1360 = 680m

, , distance between echoes X 680mVelocity o f approach = -------------------------------------- = — = -------- -- 5.4ms

time interval between echoes T 126s-

Example 13A hunter 412.5m away from a cliff, moves a distance x towards the cliff and fires a gun He hears the echo from the cliff after 2.2 seconds. Calculate the value o f x. [Speed ofsound in air = 330ms"1] NECO 2006™SolutionInitial distance from cliff x, =412.5; final distance from cliff xi; echo time t = 2 V Sound speed V = 330ms"1

Vt 330x2.2 2 2 2

Distance moved toward cliff, x = x, - x 2 - 412.5 - 363 = 49.5m

315

Example 14A man stands one-third of the way between two walls and clap his hands. He hears two distinct echoes Is apart. If the speed of sound is 330ms'1 the distance between the walls is A. 33m B. 110m C. 220m D. 330m E 495m JAMB 198(f Solution

£

Fig. 15.1t

x is the distance between two walls.

One third of x = x/3 and „ V _ lx ,x - n - It,

From V = — t

echo time, t = 2xV

Let t| be the echo time for x = x/3

Let t2 be the echo time for x = Zt/i

_ . . 2xSubstitute t = — V

to obtain

2 x At, 3 .

330 t, =■2 x 2-y:

330Time interval between echo is t2 - ti = Is

2x2A 2* /3330 330 330

Cross multiplying

4x 2x 1x330 = — - —

3 3

330 =4 x - 2 x

330 = —3

2x = 3x330 = 990

990x = -----= 495m2

Example 15A ship travelling towards a cliff receives the echo of its whistle after 3.5 seconds. A short while later it receives the echo after 2.5 seconds. If the speed of sound in air under the prevailing condition is 250ms how much closer is the ship to cliff?A. 10m B. 125m C. 175m D. 350mSolutionFirst echo time t| = 3.5s; second echo time t2 - 2.5s,

Vt, _ 250xL 5 = 437i5m

E. 1000m JAMB 198310

Speed of sound V = 250ms'1

Distance at first echo x, = —y -

Vt, 250x2.5Distance at 2nd echo x2 = = 312.5m

2 2

Distance from cliffx = x , - x 2 = 4 3 7 .5 - 312.5 - 125m

316

Example 16 1 eA plane sound wave of frequency 85.5Hz and velocity 342ms is r wall. At what distance from the wall does the wave have an antmo

D. 3m

fleeted from a vertical

B. 4m C. lmJ A M B 2001'

A. 2m SolutionFrequency f = 85.5Hz;

V 342From V = A.f, wavelength X = — = ----- = 4m

B f 85.5

Velocity of sound V — 342ms

An antinode is % of a wavelength

Therefore distance of antinode to wall = — A. = — x4= lm4 4

VIBRATION OF STRINGSFundamental Frequency. Harmonics. Overtones

Fundamental frequency of a vibrating string is the lowest possible frequency th can be obtained from a plucked string when it vibrates in a single loop as shown e ow.

Fundamental frequency, f0 or first harmonic

Fig. 15.2

The length / of the string is equal to V2, distance between ( N - N )

That is, l = X/2 orX = 2 lV

For a sound wave, V = Xf or Frequency, f = — .A

Fundamental frquency, f 0= —

the two consecutive nodes

Substitute X = 2 l to obtain

Harmonics are frequencies which are whole numbers multiples of the fundamental frequency f0.Overtones are frequencies above the fundamental frequency which may or may not be whole numbers multiples o f the fundamental frequency. Whole numbers overtones are also called harmonics.

The second harmonics or first overtone of a string is obtained from a plucked string when the string vibrate in two loops as shown below.

317

The ilislmicc between three enn.secutivc nodes is A and is equal to /, the length of the siting. That is / A

Vfrom, V Xf, I substitute/ A. to obtain.

Vhirst ovcrtonc/sccond harmonic, f, - —

' /The thiixl harmonic or 2'"1 overtone of a string is obtained when the string is plucked and made to vibrate in three loop as shown below.

L---------------------------------------- ► \ ------------------------------------------H

r o r ; 31 , 2/from the diagram, / = — or A. = —2 3

From V = A(, frequency, f = —

Therefore, 2nd overtone or 3rd harmonic f = v + — = —2 3 /

Thus, for a vibrating string or transverse wave,

Fundamental frequency (1M harmonic) f0=

First overtone (2m harmonic)

Second overtone (3™ harmonic)

Third overtone (411 harmonic)

fi = 2f0 = ^

f2= 3f0 = y 2l

f j= 4 f 0 = 2v/

Example 17A vibrator o f frequency 60Hz is used in generating transverse stationary waves in a long thing wire. If the average distance between successive nodes on the wire is 45cm, find the speed of the transverse waves in the wire.A. 27m s'1 B. 54ms'1 A. 90ms'1 D. 108ms'1 JAM B 198919SolutionFrequency f = 60Hz; length, / = 45cm = 0.45m

Fig. 15.5

From the diagram, X/i = / or X - 2 1 - 2 x 0.45 - 0.90m Speed, V = AT = 0.9 x 60 = 54ms'1

318

Example 18 . . . IA string is stretched tightly between two points 50cm apart. It is plucked .it its euit the velocity o f the wave produced is 300m s'1. Calculate the number ol \ '"/ y vythe string in one second.Solution

Number o f vibration in one second is same as frequency.

Frequency,V = 300_ = 300 _ 10()I Iz 21 2 x 0 .5 /

Exam ple 19Find the frequencies o f the first three harmonics o f a piano string ol length 1.5m. II the velocity o f the waves on the string is 120m s1.A. 40Hz, 80Hz, 120Hz B. 80Hz, 160Hz. 240Hz C. 1801 Iz. 3601 Iz, 54011/D. 360Hz, 180Hz, 90Hz IVA E C 2001 *SolutionLength, l — 1.5m; velocity V = 120ms’1

Fundamental frequency o f string, f = — - 120 = — = 4011/° 21 2x1.5 3

l 51 harmonic fc = 40Hz 2nd harmonic 2f0 = 2 x 40 801 Iz

3rd harmonic 3f0 = 3 x 40 = 120Hz . Ans: 40Hz, 80Hz, 1201 Iz

Exam ple 20I f tension is maintained on a stretched string o f length 0.6m, such that its fundamental frequency o f 220Hz is excited, determine the velocity o f the transverse wave in the string. A. 66m s’1 B. 132ms'1 C. 264m s'1 D. 528ms'1 J A M B 2 0 0 2 ' *

SolutionLength i = 0.6m; frequency f = 2201 Iz; velocity o f wave V = ?

Substitute into f.=V21

V220 = --------2x0 .6

V = 2 2 0 x 2 x 0 .6 = 264ms 1

Example 21The lowest note emitted by a stretched string has a frequency o f 40H z.Ilow many overtones are there between 40Hz and 15011/.? A. 1 B. 2 C. 3 I). 4 JAMU O S 7 u'SolutionOvertones o f fundamental frequency, f, o f a plucked string arc 2f0, 3f„, 4f„ c.t.c.

So, f0 = 40Hz, 2f0 = 2 x 40 801 Iz, 3f0 = 3 x 40 = 1201 lz, 4f„ 4 x 40 1601 lz.

Therefore, there are 2 overtones between 401 Iz. and 15011/. 16011/ is not included

because it is greater than 1501 Iz.

319

Example 22 f - s6 Hz inaCalculate the wavelength of a note which is one octave lower than a note ot ¡op#17 medium in which the speed of sound is 352ms 1 WAECSolutionAn octave of a note is a note o f twice the frequency. 9«HzTherefore a note which is one octave lower than a note of 256Hz is '/¡(256Hz) or 12

Frequency f = 128Hz; speed o f sound V = 352ms1

From, V = Xf, wavelength, X = X = 111 = 2.75m

SONOMETER: FACTORS AFFECTING FREQUENCY OF VIBRATING STRING

The frequency of the notes obtained from a string instrument is studied with the help of a Sonometer.

The frequency (f) of a vibrating string of tension T, length t and mass per unit length M obeys the following laws:

1. f a - / (at constant T and M)

2. f aV F (at constant I and M)

3.f t t VM

(at constant / and T)

Combining the above three relationships, we obtain f a - — or f = — J —/V M / V M

The velocity V o f a wave propagated along a string depends on its tension T and its mass per unit length M and is given by;

V VAt fundamental frequency, f = — = —

X 21i p F

Therefore, fa = fundamental frequency = Is* harmonic = —J —

1 p Ffi = first overtone = 2nd harmonic = 2fe = y J —

3 p Ff2 = second overtone = 3rd harmonic = 3f0 = — J —

2 p Ff3 = third overtone = 4th harmonic = 4f° =J ^

Example 23A plucked string produces a note of 200Hz when its length is 1.50m. Determine thefrequency o f the note produced with a length of 0.75m. [Assume constant tension]

NECO 2006u

Solutionfi = 200Hz; /, = 1.50; f2=? /2= 0.75m

320

At constant tension, f a - 1

Substituting,200 _ 0.75 f 2 ~ 1.50

200x1.50 Frequency, f 2 = — — — i2 2 . = 400Hz

0.75

Example 24 , , ,A Sonometer wire under a tension of 10N produces a frequency of 250Hz w en P uc e ■ Keeping the length of the wire constant, the tension is adjusted to pro^uce frequency of 350Hz. Calculate the new tension.SolutionOriginal tension Tj = 10N; original frequency f| = 250Hz New frequency f2 = 350Hz; new tension T2 = ?

From f a Vt or f = Vt we derive,

f! VCSubstituting,

250 _ VTo350 _ Vt7

^ - = 350x VT0 = 1 4 x ^ = 443 ^ 250

^/t7 = 4.43 Square both sides to obtain

(T j= (4 .4 3 )2

T, = 19.6

Alternative method:

f a j f or f = KVT (K is constant)Substitute T = 10N and f=250H z into f = K-\/t to obtain

250 = KxVlO

K ™ = 79.1Vio

Now, substitute f= 359Hz and K = 79.1 to find the new tension.f = k Vt

350 = 79.1x Vt

V T = ^ - = 4.43 79.06

Square both sides

( J r j = (4.43)2 T = 19.6N

Exam ple 25When the tension in a Sonometer wire is doubled, the ratio o f the new frequency to the initial frequency is A. f a B. 'A C. -J2 D. 2 JAMB 198921

SolutionIn a Sonometer, tension and frequency are related by f = Vt Let f, be initial frequency and f: the new frequency

321

L. - jLh f , V '',

or L . . J H' 7 T

irtcnKion is doubled, T, becomes 2T,

ff (note that, Tj T, T)

V2

A Sonometer wire ol linear density O.OHkgm'1 subjected to a tension o f SOON is p u Calculus the speed ol a pulse which moves from one end o f the wire to the 2()()3i0

Solutionlinear density M 0 . 0 0 8 k g m t e n s i o n T = SOON (linear density is also known as mass per unit length)

Speed o f pulse V = , /— = J-5^- = Vl 0000 = 100ms 1 1 VM V 0.08

Exam ple 27Under constant tension and constant mass per unit length, the note produced by a plucked string is 50011/ when the length of the string is 0.90m. At what length is the frequency 15011/? A. 3m B. 4m C. 5m L). 6m JAMB 1997‘Solutionf, 50011/.; /, 0.90m; f2 15011z; l2 - ?

. . . I f, /.At constant M and I, I a - V /

/ * 2 'l

Substituting, ---- — i—150 0.90

. , 500x0.90 .Ixngth ol string I, = ---- — -----= 3m

Exam ple 28I wo strings o f tlie same length and under the same tension give notes ol frequencies in

A. 2:1 Solution

B. 1:2 C. 1:4

1 or constant (same) length and tension

Substitute f] 4 and f2 J to obtain.

Square both sides;

(r2 ( iM ,

R atioo fM ,:M j =1:10

1). 1:I6

f (Jt1

Vmor f, .

u

4 [ m 7

1 V M .

161

M,M.\,

or

JAMB 1988-

M,

I16

Mj.M2

322

The frequency o f a plucked string is 800Hz when the tension is 8N. ( a cu ate(i) Frequency when the tension is reduced to % o f its original 'a ue^ ^ i(j()5rn(ii) Tension required to produce a note o f frequency 600Hz.

SolutionFrequency f = 800Hz; tension T = 8NSubstitute into f = k / t to find constant K.

800= K x / 8

^ _ 800 _ 800 -Js 800/ 8/ 8 = V 8 XV s ' V64

8(i) Frequency f = ? tension T = % (8N) = 2N (.Jension is % o f its original. ..);

K = 100/8Substitute into f = k / F to obtain

f = K / f = 100>/8 x / 2 = lO O jS xi = 100/16

f =100x4 = 400Hz

(ii) Tension, T = ? Frequency f = 600Hz

Substitute into f = k / t to obtain

600 = 1 0 0 /8 x / t 600 _ 6

100/8 yfS

Square both sides

62 _ 36t

368

4.5N

VIBRATION OF AIR COLUMNS IN PIPESAir column can be made to vibrate in closed pipes, open pipes or in a resonance

tube.Vibration in Closed Pipes

When an air column vibrates in a closed pipe, a node is ALWAYS formed at the closed end while an antinode is always formed at the open end as shown below.

Fig. 15.7 a. Fundamental note, f0 b. First overtone, f. c. Second overtone, f;

From Fig. 15.7a, /= X/4 (distance between node and antinode). Therefore. X = 4 /

323

From V = At, frequency, f _VA

Fundamental frequency or Is' harmonic f = v/0 /4i

From Fig. 15.7b, /= or A =

From V = At, frequency, f = ^-

First overtone or 2nd harmonic, f, = V + — =3 4/

f =3V /1 /4/

From Fig. 15.7c, / = or T = 4

VFrom V = /if, frequency, f = —

Second overtone or 3,d harmonic f , = V + — = —2 5 4/

Thus for vibration in a closed pipe:

Fundamental frequency (1 harmonic), f 0= Y ^

First overtone (2nd harmonic),

Second overtone (3rd harmonic),

Third overtone (4th harmonic),

f'= 3 « o = %

Example 30A pipe closed at one end is 100cm long. If the air in the pipe is set into vibration and a fundamental note is produced, calculate the frequency of the note. [Velocity of sound in air = 340m s1] WAEC 2007k'3solutionLength /= 100cm = 1.0m; velocity V = 340ms1

Fora closed pipe, fundamental frequency f = V/^

f =340 340

4x1.0 4.0= 85.0Hz

Example 31 . . .A pipe closed at one end is 1.0m long. I he air in the pipe is set into vibration and a fundamental note of frequency 85Hz is produced. Calculate the velocity of sound in the air. [Neglect end correction). WAEC 1992s9

L engthof pipe l = 1.0m; fundamental frequency f0 = 85Hz

For a closed pipe f =Velocity o f sound in air,

V = f o x 4/ y = f0 x 4 / = 8 5 x 4 x l

= 340ms'1

324

Example 32 Gf length 33cm isThe wavelength of the first overtone of a note in a dosed pipe o <■ JAMB 2006A 22cm M. 17cm C. 44cm D-33cm Solution l ength /= 33cmIn first overtone of a closed pipe, / =

Wavelength, A = ^ = 4 / x 33 = 44cm

Example 33If the fundamental frequency of a closed pipe organ on a day when the spec 340ms is 170Hz, then the length of the pipe is A. 50cm 13.70cm C. 100cm D. 150cm E. 200cm SolutionFundamental frequency f0 = 170Hz; speed of sound V = 340ms 1

JAMB ¡OSS'

From f = — 41

lenghtof pipe, / =----4f

/ = _ 3 4 0 _ = 340=0 5m or 50cm 4x170 680

V ibration in Open PipesWhen a column of air vibrates in an open pipe, the stationary waves formed

always have antinodes at both ends as shown below.

a. Fundamental note, f0 b. First overtone, f.Fig. 15.8 Vibration of open pipes.

From Fig. 15.8a, I = / or A = 21

frequency,f = —From V = Af,

Fundamental frequency or l sl harmonic f = V/

c. Second overtone, f;

From Fig. 15.8b, / = X

First overtone/2nd harmonic f = —

From Fig. 15.8c, / = ^/^/ <)r ^ =

Second overtone or 3rd harmonic, —2 X 21

Thus for vibration in an open pipe: Fundamental frequency ( IM harmonic) f V

2/

325

First overtone (2nd harmonic)

Second overtone (3rd harmonic) 3V

fi =2 f o = ^

f, = 3 f =21

Third overtone (4lh harmonic) f , = 4 f = 2 V

P. . . ^ n ort an P*Pe oi length 40cm opened at both ends sounds its fundamentaln° - Cn corrccl'ons* calculate the frequency of the note. [Speed o f sound inair - 340ms J NRC() 2Q06PSolution

Length o f pipe t = 40em = 0.4m; speed of sound V = 340ms*'

l or an open pipe, fundamental frequency, f _ V /

. 340f„ = - ------ = 42511/.

2x0.4

The Resonance Tube

A resonance tube is used to find the velocity of sound in air as show n below,

c \ V

xu

(a)

c end of correction

Kio. 15.0 First and Second Positions of Resonance.

I or the first position o f resonance, /, + «* — —

1 or the second position o f resonance, /, + c -

/ x 3/i ASubtracting, /, + c - (/, + ( ') - — ~

4 4

2 A/, + C - / , - c = —

3/i

A

326

>1 = 2 ( / , - / , )

substitute the above into V = Xf to obtain

V = 2 f( l2 - / , )

Where V = velocity of sound in air. f = frequencyl2 = second resonance position /, = first resonance position

exam ple , I f *A tuning fork of frequency 600Hz is sounded over a closed resonance u • j-and second resonant positions are 0.130m and 0.413m respectively, ca cu a 7 0 0 0 s3 sound in air.SolutionFrequency f= 600Hz; first resonant position /, = 0.130m;Second resonant position l2 = 0.413m Speed of sound V = 2 /( / , - /, )

V = 2 x 600(0.413-0.130)= 1200 x 0.283 = 339.6ms'1

Example 36The shortest length of the air column in a resonance tube at resonance is 0.1 lm and the next resonant length is 0.36m. Calculate the frequency o f vibration, given that the speed o f sound in air is 340ms'1. NECO 2007SolutionFirst resonance position /, =0.1 lm; second resonance position /,= 0.36m;Speed o f sound V = 340ms'1

VFrom V = 2f(/2 - /, ), frequency, f = — — —

2(/2 ~ l\)340

2(0.36-0.11)340

2x0.253400.5

= 680Hz

Example 37In a resonance tube experiment, if the fundamental frequency of the vibrating air column is 280Hz, the frequency o f the third overtone isA. 70Hz B. 840Hz C. 1120Hz D. 1960Hz J A M B 1 9 9 S 28

SolutionFundamental frequency f0 = 280Hz

For a resonance tube, 1st, 2nd, 3rd overtone are fi = 3f0, f2 = 5f0, f3 = 7f0 e .t.c.

The third overtone f3 = 7f0 = 7 x 280Hz = 1960Hz

Example 38A tuning fork vibrating at a frequency o f 512Hz is held over the top o f a ja r filled with water and fitted with a tap at the bottom. If the ja r is 60cm tall and the speed o f sound is 350m s'1, determine the possible resonance position(s). [Neglect end correction]

1V A E C 2 0 0 3 e i s

SolutionFrequency f = 512Hz; speed o f sound V = 350m s'1

Height o f resonance ja r or tube = 60cm = 0.6m

327

From V = Af, wavelenght, A = — = = 0.68mf 512

p - .■ c A 0.68First resonance position, f = — = ----- = 0.17mI 4 4

Second resonance position, 3f0 = — = — = 0.51 m4 4

T, , , . 5 A 5x0.68 „Third resonance position, 5f0 = — = --------- = 0.85m4 4

Because the resonance jar is 0.60m, the possible resonance positions are 0.17m and 0.51m. 0.85m is NOT a possible resonance position because 0.85m is greater than t e length o f the jar.

Example 39If the first position of resonance in a resonance tube is 18.0cm from the open end, the distance from the open end to the next position of resonance will be? NECO 200ff SolutionFirst resonance position f0 = x/4 = 1 8cm X = 18 x 4 = 72cm

-,nd ... 3 A 3x722 resonance position, 3f = — = -------= 54cm4 4

Example 40In a resonance tube experiment, a tube of fixed length is closed at one end and several tuning forks of increasing frequency used to obtain resonance at the open end. If the turning fork with the lowest frequency which gave resonance had a frequency fi and the next tuning fork to give resonance had a frequency f2, find the ratio f2/f,.A. 8 B. 3 C. 2 D. Vi E . ‘/3 JAMB 1983'SolutionIn a closed tube, frequency at first resonance; f x=-A

nd 3VFrequency at 2 or next resonance, f, =—J2 41

_ f f2 3V V 3V 4 / ,Therefore, — = ----+ — = ---- x — = 3.f, 41 41 41 V

Example 41

Fig. 15.10A. 1.85ms'1 B. 225ms'1

In a resonance tube experiment, the

frequency of the tuning fork is 750Hz and

the resonating length of air column is 0.3m

as shown. The velocity of the wave is

C. 300ms'1 D. 53 x 10'5ms'‘ E. 2,500ms'1JAMB 198247

328

Solutionf 75011/; / 0.3mThe diagram shows second resonant position. At second resonant position, f = ^

Make V subject of equation to obtain velocity of sound wave

v T x 4/ 7.^0X40 X 0 3 = 900 _3 3 3

Example 42A siren bus a disc of 64 holes and makes 20 revolutions per second. Calculate the frequency of the sound from the siren. WAEC 1996Solution

l'or a disc siren, frequency = number of revolution x numberof holes time taken for revolution in seconds

20x641

= 1280Hz

Also, frequency for a toothed wheel is given by

numberof revolutionx numberof teeth in wheelFrequency, f = -time taken in seconds

Beat FrequencyBeat frequency is a phenomenon that occurs as a result of interference of sound waves. Beats are heard whenever two sources of sound of nearly, but not exactly, the same frequency emits sound waves. If two instruments have frequencies f x and f 2 such that f2 is greater than f,, then the number of beats per second or beat frequency, f h , is given by;

f b = f 2- i

Example 43A Sonometer wire has a frequency of 259 Hz. It is sounded alongside a tuning fork of frequency 256Hz. Calculate the beat frequency WAEC 2008SolutionGreater frequency, f2 = 259Hz; lesser frequency, f, = 256Hz

Beat frequency, f b - f 2 ~ f|

i . = 259 - 256 = 3Hz

329

EXERCISE 15.1. A note of frequency 2000Hz has a velocity of 400m s1. What is the wavelength ofthe note? WAEC 199433 Ans: 0.2m2. The velocity of sound in air will be doubled of its absolute temperature isA. doubled B. halved C. constant D. quadrupled JAMB 2005 Ans.D3. The speed of sound in air at sea level is 340ms' while that of light is >00,000kms . How far (to the nearest meter) from the centre of a thunderstorm is an

observer who hears a thunder 2s after a lighting flash?A. 170m B. 340m C. 600m D. 680m JAMB 198827 Ans: 680m4. A note of frequency 2000Hz has a velocity of 400ms'1. Calculate the wavelengthof the note. W A E C 200629 Ans: 0.2m5. A boy standing some distance from the feet of a tall cliff claps his hands and hears an echo 0.5s later. If the speed of sound is 340ms l, how far is he from the cliff?

WAEC 198835 Ans: 85m6. A man standing 300m away from a wall sounds a whistle. The echo from the wall reaches him 1,8s later. Calculate the velocity of sound in air. WAEC 199137 Ans: 333.3m/s7. A man stands in front of a tall wall and produces a sound. If he receives the echoof the sound two seconds later, calculate his distance from the wall. [The speed of sound in air is 330ms"1] WAEC 199324 Ans: 330m8. A sound note of frequency 250Hz and wavelength 1.3m is produced at a pointnear a hill. If the echo of the sound is received one second later at the point, how far away is the hill from the point? WAEC 19953' Ans: 162.50m9. A sound note is produced by a ringing bell and the echo of the note from a nearbywall is received 0.5s later. If the frequency of the note is 400Hz and its wavelength lm, calculate the distance between the bell and the wall. WAEC 1997~9 Ans: 100m10. A pulse of a sound is transmitted from a ship and the reflection from the sea bed isrecorded after 0.2s. Calculate the depth of the sea. [Take speed of sound in sea water = 1560ms'1] WAEC20023' Ans: 156m11. A girl stands 80m away from a tall cliff and blows a whistle. If the speed of sound in air is 330ms'1, how long would it take for her to hear the echo of the sound?

WAEC 200433 Ans: 0.485s12. A body stands in front of a tall wall and produces a sound. If he hears the echo 3s later, calculate his distance from the wall. (Speed of sound in air = 330ms'1)

NEC 0 200633 Ans:495m13. How far from a hill should a boy stand to hear the echo of his clap 1.6s later?[Speed o f sound in air is 340ms'1] NECO 200434 Ans: 272m14. A man stands in front of a high wall and produces a sound. If he receives the echoof the sound four seconds later, what is his distance from the walls. [The speed o f sound in air is 330ms'1] NECO 2002' Ans: 660m15. The echo of a sounder from the bottom of an ocean is heard 2.5 seconds later. If the speed o f sound in water is 1400ms ’, the depth of the ocean isA. 1,750m B. 3500m C. 1400m D. 2800m JAMB 19826 Ans: 1.750m16. In order to find the depth of the sea, a ship sends out a sound wave and receives an echo after one second. If the velocity of sound in water is 1500m/s, what is the depth of sea? A. 0.75km B. 1.50km C. 2.2km D. 3.00km E. 3.75km

JAMB 198527 Ans: 0.75km17. A man clapping his hands at regular intervals observes that the echo of a clap coincides with the next clap. If the reflecting cliff is 160m away and the speed of sound is 320ms'1, what is the frequency of the clapping.A 1Hz b . 2Hz C .4H z D. 8Hz JAMB 19S6' Ans: 1H:18. The sound from a source travelled to the bottom of the sea and the echo was heard 4s later. If the speed o f sound in sea water is 1500ms the depth ot the sea isA. 6000m B. 3000m C. 1500m D. 375m JAMB 1994' Ans. 3000m19. A boy stands between two vertical wall and fires a rifle.(i) Under what condition will he hear a single echo from both walls?

330

.. -r .l pcho is heard 6s later,(n) Calculate the distance between the walls if the s 8 A between the[Speed of sound in air = 330m s'). NEC 0 2007' Ans. (i)walls at a point equidistant from the them (ii) 1980m 20. A wire is stretched between two¿». A wire is stretched between two points, lm apart. ....... ■ u „.-iigenerated on plucking the wire is 200m s1, what is the mlnl™U™r r? ™ ^ A n s : ¡00Hz resonate with the wire? WAE A .21. A note A from a guitar produces a wave of amplitude mm an 1000Hz. Another note, B from a whistle produces a similar waveform o amp i and frequency 2200Hz. If the two notes are compared,A. A has a higher pitch than B. B. A is louder than BC. B has a greater speed than A D. B is louder than AE. A and B have the same quality22. A steel wire o f length 0.50m is stretched between 2fundamental frequency is 200Hz. The speed of the wave in the wire is ;A. 100ms'1 B. 120ms'1 C. 200ms1 D. 250ms'1 JA M B 1998 Ans: 200ms23. - A string is fastened tightly between two walls 24cm apart. The wavelength o f the second overtone is A. 24cm B. 16cm C. 12cm D. 8cm JAMB 2001 Ans: 16cm24. ; All of the following frequencies are overtones of 320Hz EXCEPT A. 96QHz B. 640Hz C. 520Hz D. 1280Hz E. 1600Hz

JAMB 198124 Ans: 520Hz25. O f two identical tuning forks with natural frequency 256Hz, one is loaded so that 4 beats per second are heard when they are sounded together. What is the frequency of the loaded tuning fork?A. 260Hz B. 252Hz C. 248Hz D. 264Hz E. 258Hz JAMB 198416 Ans: A26. The lowest note emitted by a stretched string has a frequency o f 40Hz. How many overtones are there between 40Hz and 180Hz?A. 4 B. 3 C. 2 D. 1 JAMB 199919 Ans: 327. A Sonometer wire of length 100cm under a tension of 10N has a frequency o f 250Hz. Keeping the length of the wire constant, the tension is adjusted to produce a new frequency of 350Hz. The new tension isA. 5. IN B. 7.1N C. 14.0N D. 19.6N JAMB 199033 Ans: 19.6N28. A string of length 1.0m vibrates in 10 loops. If the total mass of the string is 1.0 x 10'3kg and the tension in it is 10N, calculate the frequency of the vibration.

WAEC 199635 Ans: 50Hz29. The fundamental frequency of a plucked wire under a tension of 400N is 250Hz. When the frequency is changed to 500Hz at constant length, the tension isA. 160N B. 1600N C. 40ND. 400N JAMB 2007n Ans:40N30. A transverse wave is applied to a string whose mass per unit length is3 x 10'2 kgm'1. If the string is under a tension of 12N, the speed of propagation o f the wave is A. 40ms'1 B. 30ms'1 C. 20ms'1 D. 5ms'1 JAMB 200424Ans: 20ms'131. The note produced by a stretched string has a fundamental frequency o f 400Hz If the length of the string is doubled while the tension in the string is increased by a factor of 4, the frequency isA .200H z B.400Hz C. 800Hz D. 1600Hz JAMB 199432 Ans: 400Hz32. A piano wire 0.50m long has a total mass of 0.01kg and is stretched with a tensiono f 800N. Calculate the frequency of the wire when it sounds its fundamental note A.200HZ B. 100Hz C. 4Hz D. 2Hz JAMB 199432 Ans: 400Hz

33. A pipe closed at one end is lm long. The air in the pipe is set into vibration and afundamental note is produced. If the velocity of sound in air is 340ms'1, calculate the frequency of the note. WAEC 1989E2 Ans: 85Hz34. An open pipe closed at one end produces its first fundamental note. If the velocity o f sound in air is V and 1 the length of the pipe, the frequency of the note is

JAMB 2003'8 Ans: C

If the speed o f the wave

N E C 0200435 Ans: B fixed points and its

A . 2 L/

B. — 51

c .L41

D. — 21

331

-• An organ pipe closed at one end is 80em long. Determine the frequency ot thea ^ nATj11*3 n°*e assum*ng that the speed of sound in air is 340ms’1.36 A B_ 213Hz C. 318Hz D. 425Hz JAMB 199&: Ans: 106H: n f th ° 4^cm is closed at one end. Calculate the fundamental frequencyrisipof ^°U 4 wave generated in the pipe if the velocity of sound in air is 360ms’ . [Neglect end corrections]3?.5'5Hf . B - 148-5Hz C. 200.0Hz D. 550.0Hz JAMB I994JI Arts: 200H: in air is v ^ ’r0^ 11,at en(*s’ produces a fundamental note. If the velocity o f sound tu„ nr.ti.9 rxt i 1 6 en®^ P'Pe, which o f the following expresses the frequency o fthe note? [Neglect end corrections]A ‘ T B ’ ^ - C ^ D. v E. v WAEC199733 Arts: V/2L

2L ^ 4L sl38. A tuning fork of frequency 340Hz is vibrated just above a cylindrical tube of height 1,2m. If water is slowly poured into the tube, at what minimum height will resonance occur? [Speed of sound in air = 340ms'1 ]A. 0.95m B. 0.6m C. 0.50m D. 0.45m JAMB 2003n Ans:0.95m39. In a resonance tube experiment, the effective length of the air column for the first resonance is 20cm when set into vibration by a tuning fork of frequency 480Hz. Neglecting end effect, the velocity of sound in air isA. 96m s'1 B. 255ms'1 C. 340ms'1 D. 384ms'1 JAMB 199038 Ans: 3 8 4 '-s '40. The shortest length of the air column in a resonance tube at resonance is * 2mand the next resonant length is 0.37m. Calculate the frequency of vibration given tnai the speed o f sound in air is 340ms’1 WAEC 199834 A ns: 68UHz41. If the position of resonance in a resonance tube is 16.50cm from the open end ofthe tube, calculate the distance from the open end to the next position where resonance occurs. [Neglect end correction] WAEC 1999~9 Ans. 30cm42. If the distance from a point source of sound is doubled, by what factor does the intensity decreases? A. 4.00 B. 2.00 C. 0.50 D. 0.25 JAMB 2003'4 Ans: D43. Two tuning forks of frequencies 256Hz and 260Hz are sounded close to each other. What is the frequency of the beats produced?A. 2Hz B .4H z C. 8Hz D. 258Hz JAMB 199130 Ans: B44.

c16cm

A. 128Hz B. 256Hz C. 512Hz 45.

In a resonance tube experiment which is

illustrated in fig 15.11, the velocity of

sound in air is 327.68ms’1, the frequency

of the tuning fork used is therefore

D. 768Hz JAMB I97923 Ans: 5I2Hz

In figure 15.12, a resonance tube experiment is performed using one tuning fork. As the water level is lowered the first resonance is obtained when the length of the air column/ - /■ The second resonance is obtained when / equals

332

a . x/2 jg 3Ay C A, D __46. What is the frequency o f the sound made by a siren having a disc with 32 hoand making 25 revolutions per second? 23 0n n u -,A .80H z B. 600Hz C. 800Hz D. 1600Hz JAM B 1992 A n s :800Hz47. When the length o f a vibrating string is reduced by one-third, its equency becomesA. Three times its former value B. Twice its former valueC. One-third o f its former value D. One-sixth o f its former va ue

3X, JAM B 1986s5Ans: B

JAMB 2008 Ans: A rnm-.f= / /48. A sound wave is produced from a source and an echo is heard t seconds afterwards. If d is the distance o f the reflecting source from the source, V the speed, the wavelength and T the period of wave, then

A. d = — B. d = — C. d = — D. d = — W AEC2008A n s:d = ^ ~2 IT At 21 11

Hint: substitute V — ^ into 2d Vt

49. A Sonometer wire has a frequency of 259Hz and is under a tension o f I200N. If a meter of the wire has a mass of 0.03kg, calculate the length / o f the wire when it is vibrating in the fundamental mode. WAEC 2008E,S Ans: 0.386m Him: f =

50. A note of frequency 300Hz is produced when the length of a wire is 80cm and thetension is 40N. What is the frequency if the length of the wire is halved and the tension is doubled? NECO 2008s4 Ans: 848.4Hz Mntj= ^51. In a resonance tube closed at one end, the first two lengths o f the air column that vibrate in resonance with a tuning fork of frequency 316Hz are 20.6cm and 73.1cm respectively. Calculate the velocity of sound in air to 3 significant figures.

NECO 2008s5 Ans: 332m s'52. A stretched wire of length 24cm when plucked, produces a note of the samefrequency as that o f a fork of frequency 256Hz. If the wire is then adjusted to 16cm and the tension kept constant, calculate the frequency of the fork which will be in tune with the wire. NECO 2008EIS Ans: 384Hz53. An observer heard the sound of thunder 5.8s after the lightning flash was seen.How far was he from the source? [speed of sound in air = 330ms'1]

NECO 2009s4 Ans: 1914m54. The first two lengths of air column that vibrate in resonance with a tuning fork offrequency / are and l2. If the resonance tube is closed at one end, express the velocityof sound in terms of/ It and l2A. l f ( l 2 ~ k ) B. f ( l 2 - lt ) C. 2f ( l 2 - lx) D. 3f ( l 2 - / ,)E. 4 /(72 - l2) NECO 200935 Ans: C55. A sound wave from a ship is received 8.4s later. What is the debt o f the sea?[speed of sound in water = 1500ms'1] NECO 200934 Ans: 6.30km56. A man, standing between two cliffs, claps his hands and hears two echoes after3.6s and 4.2s respectively. If the speed of sound in air is 330ms'1, calculate the distance between the cliffs. NECO 2009E6 Ans: 1287m57. A pipe closed at one end has a length of 15cm. Calculate the frequency of the

(i) fundamental note,(ii) first overtone [speed of sound in air = 340ms'1]

NECO 2009EIS Ans: (i) 566.6Hz (ii) 1700Hz

333

BOOK THREE

1

GRAVITATIONAL FIELD

(JnvfoM ial tWU is ihc recon around a body liavmp mass in ,hcforce of the Kviy can be experienced, Gravitational bnvc • • .bodies is governed by the .Vrwftw’x o f unhyrsal ¡¡runMion, " ,K ' s n .tx\o particles o f nutter attracts one another with a tdive I . which is propoi u product o f their masses m t& m; and inversely proportional to the squaiv o t icu r. apart

” 0 'r

■ 0Fig. l . iGravitational force bepveen two masses is civen bv

F =Gm^m

r-Where m t and m 2 are the masses of the two particles; r is the distance between masses ms and m2: G is the gravitational constant = 6.67 x 10 " Nmmky

Example 1Two spheres o f masses 100kg and 90kg respectively have their centers separateli by a distance of 1 Om.Calculate the magnitude of the force o f attraction between them. [G = 6.70 x 10-n NmJJt5~:] ” WAEC 200237Solution

Fig. 1.2

•' rn,

Force of attraction F - 6-70 x 1 0 » x iqq x 901- = 6.03 X 10_7/V

Example 2

A force of 200.V acts between two objects at a certain distance apart. The v alue o f the force when the distance is halved is A. 100/V B. 200N C. BOON D. 4 0 0 N

I A M S 20053iSolution

Initial force. Fx = 200N; final force (when distance is halved) F2 =?; \ mU!X\ distmrx = n final distance r2 = r / 2 (— the distance is holwti__ )Gravitational force is inversely proportional to the square o f the distance

F « 4

V i 2 = F2r i

200 x r = f 2 x ( 7 2)2

200 x r 2 = F2 x — 2 4

334

4 x 200 x r 2Rearranging,/*2 ------------ --------- p = 800N

r 2 2

Example 3I he gravitational force between two objects o f masses 1024 and 1027 is 6.67N.

Calculate the distance between thcm(G = 6.67 x 10-11 Nm2kg-2) NECO 200850 Solution

F =

m , = 1024fc<7

c e:-r 2 =

G m , m 2

m 7 - 1027k.qe2 _ 6.67 x 10 11 x 1024 x 1027

~ 6l67 'r = V lO 40 = 1.0 x 102Om

10_ n x 1024 x 1027 = lO -n+24+27 = 1040

Example 4A planet has mass 77ix and is at a distance rx from the sun. A second planet has i ss m 2 — lOrrij and is at a distance of r2 - 2rj from the sun. Determine the ratio o f the gravitational forces experienced by the planets.A. 1 :5 B. 2 :5 C. 3:5 D. 4 :5 JAMB 19978SolutionFirst planet: mass m = m t ; distance r = rxSecond planet: mass m = 10?^; distance r2 = 2rx; mass of the sun = m s

GmsmSubstitu te for each planet into gravitational force equation F = ----r—

1st planet 2nd planet

Fi =G x m s x mj

F7 =G x m s x 10mx

G x m s x 10m j4 ^

F2 G x m s x 10m! G x ms x m j Ratio of gravitational force, — = -------- ^ " -------- p.-------

F2 G x m s x lOmj r 2Fx 4rx2 G x m s x m 1

F2 G x m s x 10mx x r 2 10 R e a rra n g in g ,- = - - “ — x 4 r 2 = T

F2 _ 10 _ 5= T _ 2

Therefore, F2: Fx = 5: 2 o r Fx: F2 = 2: 5 Ratio of gravitational force is 5: 2 o r 2: 5

Example 5The force of attraction between two point masses is 10 when the distance between them is 0.18m. If the distance is reduced to 0.06m, calculate the force.

335

B. 3.3 x 10~5/V C. 3.0 x 10 4A/I). g.O x 1(T4A/

JAMB 199BA. l ' l X 10~SN

SolutionIs' case: F1 = 10"4A/; rx = 0.1 8m 2nd case: F2 = ? ; r2 = 0.06m

Gravitational force (F) is inversely proportional to square of the dista ( ) Therefore, Ft r 2 = F2r2 Substitute to obtain 10 4 X 0.18 — F2 *

_ 10“4 x 0.182 10“4 x 182 _ 10~4 x 324 _ q x2 “ 0.062 “ 62 “ 36

GRAVITATIONAL POTENTIAL , ,Gravitational potential, V, is the potential due to the gravitational He o t c ca . is defined as the work done in taking a unit mass front infinity to the point (a pa icu ar distance form the center of the earth).

CMGravitational potential V = — —-

Where M = mass of the earthG = gravitational constant r = distance of the point from center of earth

The negative sign indicates that the potential decreases as the object is moved from infinity towards the earth.

Example 6What is the gravitational potential at a point on the surface of the earth if G = 6 • 6 X 10-11Nm2/cg-2 and the radius and mass of the earth are respectively 6 • 4 X 106m and 5 • 98 x 1024kg'?Solution

M = 5 - 98 x 1024kg ; r = 6 - 4 x l 0 6m; G = 6 • 6 x 10_11Nm2A:^“2

Gravitational potential V = —GMr

6 • 6 x 1 0 '11 x 5 • 98 x 1024 6.4 x 106

3 • 95 x 1014= T ~ 4X 10^ - 6 2 x 107, k g -

ACCELERATION DUE TO GRAVITY/GRAVITATIONAL FIELD INTENSITY

The acceleration due to gravity (g ) is also known as the force per unit mass (m ) o f an object and is responsible for the weight or force an object experiences on the earth

GmemFrom F = — (m e and re are mass and radius of the earth respectively)

F GmeForce per unit mass — = —rr-

m r 2Acceleration due to gravity is equal to force per unit mass.

Therefore,F

9 = — or mGme

9 = r 2 'e

Hence, F = m g

336

Example 7The magnitude of the gravitational attraction between the earth and a particle is40N. If the mass of the particle is 4kg, calculate the magnitude of the gravitational field intensity of the earth on the particle WAEC 2002SolutionForce of attraction on particle, F = 40N; mass of particle m = 4kg

F 40Gravitational field intensity, g = — = — = 10N kg~x

Example 8The mass and weight of a body on earth are 8kg and 80N respectively. Determine the mass and weight of the body respectively on a planet where the pull of gravity is - that onearth. WAEC 2005Solution

Mass m = 8kg-, w eigh t w = 80N

Weight = m ass x acceleration due to g ra v ity

w = m g

Therefore, g = w/ m = = 10m s-2

1 , 1— o f¿7 = — x 10 = 1 • 25ms 2o 8m = 8k g mass of objects do not change whatever their location.

From w = mg; w = 8 x 1 • 25 = 10N Ans: 8kg, 10N

Example 9The earth is four times the size of the moon and the acceleration due to gravity on the earth is 80 times that on the moon. The ratio of the mass of the moon to that of the earth is

A. 1 :320 B. 1:1280 C. 1:80 D. 1:4 JAMB 200417 SolutionAcceleration due to gravity on moon g m = 1

Acceleration due to gravity on earth, g e = 80

radius of the moon, rm = 1

radius of the earth, re = 4 Gm

Acceleration due to gravity, g =

GmeFor the earth, g e — — '

re

GmmFor the moon, g m = — — •

rm

Therefore, m e =g ert.

Therefore, m m = 9m^m

The ratio o f the mass of the moon (m m) to that of the earth (m P) is:mm _ gmr^ gere _ 9mXm Gm e G G G g ere2

_ 9m rm wc obtainm e g er 2

337

I = 1 : 128011 X 1:110 x 4 2 110x16 12110

Therefore, tnm : m ,

WEIGHT OF OBJECTS IN SPACE•r near the surface o f the earth, il experiences the full extent of thederation due to gravity of the earth. However, if the object (for

• . — <v™n »he surface or center o fWlicn an object is on orforce of gravity or acceleration ciue ro g iav»/ ^ surfaCeexample a satellite) is launched into outer space at a distance ro' ,Ur~ e n r t h s g r a v i t a t i o n a l

The farther the distancethe earth, the weight o f the object reduces bccause the earth’s gravitational force on the object diminishes in proportion to its distance from the earth.from the earth, the lesser the weight.

Example 10ff the weight o f a satellite on the earth’s surface is 2000 N, a. What would be its weight when it is

(i) at a distance equal to the radius o f the earth from the surface of the earth(ii) at a distance o f 1.28 x 1077n from the surface o f the earth?(bi) at a distance equal to four times the radius R o f the earth, from the center

o f the earthb Fo l• or eac case in (a) above, derive an expression for the acceleration due to gravity

9 s m space with respect to the acceleration due to gravity (g e) on earth.LR adius o f ea r th R = 6 • 4 x 106ml

SolutionLet We = weight o f satellite on earth = 2000N

w s = weight o f satellite in space M = Mass o f the earth m = Mass of the satellite

g e — Acceleration due to gravity on earth 9 s — Acceleration due to gravity in space R = Radius o f the earth = 6 - 4 x 1 0 6m

a(i)R -o

Sa te lliteE arth

Fig. 1.3Distance o f satellite from center o f earth = R + RThe weight o f the satellite in space Ws is equal to the gravitational force o f attraction between the earth and the satellite in space.

F = Ws =GMm GMmGMm

(R + R )2 (2 R )2 4 R 2

Rearranging,m GM

m g e

Gs u b s t itu tin g g e = — _

lVs = — — = [su b s titu te We = m g e]

We 200NTherefore, = ~ — “ — 500/V

This means that the weight o f an object in space at a distance R (radius o f earth) from the surface o f the earth or a distance 2R from the center of the earth is one-fourth its weight

on earth: Ws = ^ '/4

338

(ii) If the radius of the earth R = 6 • 4 x 10‘'m,

Then 1 • 28 X 1 0 7 m = 2 x 6 • 4 x 106 = 2R

K_ 2 R

Fig. 1.4

In similar manner to a (i) above;

Earth

Weight of satellite in space, =

m GM

GMm

-oSatellite

GMm GMm(R+2R2) (3 R )2 9 R2

Til UM rRearranging Ws = - - — [substitu tege = — J

Ws =m g e

[substitute Wc = m ge]

W* = We/ 9

Therefore, Ws = 2000/V/9 = 222.22NHence, the weight of object in space at a distance 2R from the surface of the earth or a

wedistance 3R from the center of the earth is one-ninth its weight on the earth: Ws = —(«0

SatelliteEarth,

Fig.1.54 R-

Ws =GMm GMm m GM m gc Wr( 4 R)2 = 1 6 R2 ~ T g '~R2 = ~ 7 tT = T b

We 2000^ = i t = ~ u T = 125N

Therefore, at a distance 4R from the center of the earth or distance 3R from the surface ofthe earth, the acceleration due to gravity or the weight of an object is one-sixteenth of that

. weon earth: W — —1 6 WIi. 1 ruin a (i) the equation Ws = can be rewritten remembering that generally,Hf - wg

MU* =

9s =

K4

mg,,4

Hr

4At i distance 2R from the center o f the earth, an object experiences ' / | (mie fourth) of the earth a force of gravity i e. acceleration due to gravity

From a (ii)W

W -

13 J

From a (iii)

m 9 s =mge 9e

9s g

wew = —-

‘ 16

™ 9s = ' 16_

16

ESCAPE VELOCITYEscape velocity is the minimum velocity with inflwrr. o f the earth’s gravitational field.

2 CM

which an object would just «cape from the

Escape velocity Ve = j 2 g R — J~~g~

Where g = acceleration due to gravity o f the astronomical body (ea ) R = Radius of the earth (astronomical body)

M = Mass of the earth (astronomical body)

Example 11 .What is the escape velocity of a satellite launched from the earth’s surface? (Take Q as 10ms-2 and die radius o f the earth as 6 • 4 x 10*m) N E C O 2 0 0 2Solutiong = 10ms- 1 ; R = 6 - 4 x 106m

Escape velocity Ve = ^ 2 g R = V2 x 10 x 6 * 4 x 106

V' = V l - 28 x 108 = 1 - 13 x 104m /s = 11.3fcm/sExample 12Complete the following i file, assuming Newton’s law of universal gravitation. [G = 6.67 x 10~11 N m 2k g ~ 2\

Table 1.1

Mass

I’lanct A Planet B Planet C

2.0 x 1027kg 7.4 x 1022kg

Diameter 1.84 x l()8m 1.28 X 107m

Weight of 10kg ball 98 N 16N

Escape velocity

Planet A

Radius r =

Mass of planet M = 2 • 0 x 1027kg

diameter 1 • 84 x 108m -----------= --------- --------- = 9 • 2 x 107m

340

6 - 6 x 1 0 -11 x 2 • 0 x 10 27 S = ~ (9 • 2 X 107) z --------- = 15 -7 6 m / s ,u

W eight o f 10fc^ ball, W = m /7 = 10 x 15 - 7 6 = 1 5 7 • 60N) p i '- V . i X ,\Y {; i S .V : ; x j i i

Escape velocity Ve = JlgR = V2 x 15 • 76 x 9 • 2 x 1 0 7 = 5 • 39 x 10 4 m / s

Substitute into g = yp to obtain acceleration due to gravity in planet A

/\tViPlanet BM ass o f ball m = 10 kg-, w eight o f ball, W = 98N

irhjia // i !•'.<!

W 98F rom W = mg, a cc e le ra tio n d u e to g rav ity in p la n e t B is g,= — = tt :

? ' ! v A iu 'O I : ¡M i t * O f? . m .W«.*bv aqso«'-! ji ----- — . ' . 1. = 9 . 8 ms'2 y —r - l

d ia m e te rRadius r = ■ = 1 • 28 X 1 0 7/ 2 = 6 • 4 x 10 6m

i a a o n a / a*i: ■ 'tnUfdt tail.•/ . ‘Av m •*, :j : x >, ins:-.!t£)r3cn>.-iifc:!/•;*:$ aril u:.’l nw iDa = >M mass of planet B, . -M = - y - I 4 r~ w /n d fiortoi/mc aril» oA - ‘Of x " acm to

- 9 -8 X (6 -4 xl lO V2! ■ - -n; U’AOr. hns v M8 «asru 1o r^do ow: » T O 0 X 10i4fcg. , . n . )ncj.':rto*j i',noiij;itv»ng

i V1 * ■ i I tmv/tydE ist^ 'vdbcity ^V e ‘ JTgR ¿¿'<J2rX 9 • 8 x 6 -4 x 106- = 1 -12 * iV riis"1’oorrefc'b a ( J- SsjAistp« i nr rlA ovn «■j'av/.'srf rtoilasitics to sw /t arti ted) >•<):.!•'OrertA " * Oi ?>'F -k CI .U.1A 1.'. 7' «¡!

-■oiijt or!? <.s:-r rrvf 0 s.S'L> 01 Lns y>'V) ■ 2 to -'jijtiq,- u.v lP lanet C ■/ 4 4 9 1 v Tg • ; A x 00M ass o f p lanet M =- 7-4 x IQ22 kg o :

i< •- Mod norJociof; Vj■•a0* o i x og •: n a" oi x rc • f.o

M ass o f ball m = ioifc^; w eight o f ball W = 16NK ' f ; t i ar.ht j.ii ;k . .jrj /P.in >j'A0! \<> m c !lo S./IO; 'j fi ■ j ' t l n d i i > .K

F ro m W = mg, g = - = | ^ = 1 .6 m s " 2 • x * ;> - ^ 'i ” !) ■ 77t 10 ■ •)!/ A' * ’ 01 .< \ ■ d v>

lO -»•*. 1 >*: ya'l-'te < tlHLi O.'i! fl ■ ffTloq : li. Ii!!)fl370q Icnoii».!l/hrg 'Jfil y'lilusls'l .0 F ro m 51 = —£■, r 2 = ^ - ' x 0 ■ d = dnsoodi lo aeern jn a01 x I » rfnu» arb

' 40 I 2< 81. • 0 i p J i i i v.. .v. '7, t i ' A - i n A ' * 0 1 x V.A A

■\ fi.nsJn’b a .i. sit s '.•■v i a. ¡, u! -ai!.. anSn-jioq Icnounti/cii3 urb ,fGM-- 6.67 x 1 0 - i l x 7;4 x 1 0 22 -./?f

1.6= y f3 ^0 8 x W ^ = i.76 x 106m

1 i/b i. ; iitvloq tutlO!tL»i4/B4l4( Jftt ' .X

Diameter, — 2r = 2 x 1 —76 x 10®•=> 3 - S x 10®m ■ K j ■ tr, i- • A)A , iHr'M

Esiape velocity,' V‘ = ^ 2 ^ = V2 x 1 • 6‘x f ' 76 x 10® = V 5 - 6 3 x 10® "1! ,K T j lo r r i s n o d t U iv i r r j *jdj so v i i iu m m t>ffr <t ta r lW ./ . ‘O f > T ,'t. • i ¿\ L lo d

* / m f 8 . P Q ‘ "v^'A'Ot x l . d . 0 i ' t \ 4 2373ms-1 .• >'• .A

1 /•rrA ^ ¿ 9 0 1

nc oJ 3'iofiq^omit orit otfu rn tit et oofotiux tlrtn-j orb no A OOF ^nni-'D.v -lipyirn A Of LfTable M is redrawn;wthaheicdlpulaicd v&lues fdr planet A.sB aiid G ias dhbwahnitalalei,« * }(, :rf•»:*••••/ *j ri; ’«'J t ;!u o /< tcriW no iiu ir/oT y l t 2TV/n«o to ' 4s;j o u ;u i^ *j< r tii

1.2 below. »r znA r <?0 ' 7)43k N ' j b i j j i J i f i / d t jc

ru » « l

Planet A Planet B Planet C

Mass 2.0 x 1027kg 6.0 x 102Akg 7.4 x 1022fc<7

Diameter 1.84 x 108m 1.28 x I0 7m 7.4 X 1022*5

Weight of 10 kg ball

157.60/V 98 N 16 N

Escape velocity 5.39 x 104m /s 1.12 x 104m /s 2373 m /s

TabU 1.2

EXERCISE 11. Given that the gravitational constant is 7 x 10~11Nm2kg 2, what is the force ofthe attraction between 106kg mass of lead hanging one meter away from a 103 kg mass of iron? WAEC 199036 Ans: 7 x 1 0 '21V2. Two object of masses 80kg and 50kg are separated by a distance 0 • 2m. If thegravitational constant is 6 -6 x 10~11Nm2kg~2, calculate the gravitational attraction between them. WAEC 199142 Ans: 6 • 6 x 10-61V3. Given that the gravitational constant is G, Newton’s universal law of gravitationstates that the force of attraction between two masses Afx and M2 separated by a distance r is A. Gr2M,M2 B. GrM1M1 C. D. WAEC 200442 Ans: C4. Two spheres of masses 5 - 0 kg and 10- 0kg are 0 ■ 3m apart calculate the forceof attraction between them. A. 4 - 00 x 1021V B. 3 • 57 x 10_2/VC.3 - 71 x 10-8/V D.3 - 50 x 10-1°1V [G = 6 - 67 x 10~11Nm2k g -2]

JAMB 20 0735 Ans: 3 • 71 X 10~8N5. Calculate the force on a mass of 10kg placed on die earth’s surface.(Radius of the earth = 6 - 4 x 106m, mass of the earth = 6 - 0 x 1024fe , )G = 6 • 7 x lO~11Nm2kg ~2 NECO 200639 Ans: 98N6. Calculate the gravitational potential at a point on the earth’s surface. (Radius o f the earth = 6 • 4 X 106m, mass of the earth = 6 - 0 x 1024kg,G = 6.7 x 10~11Nm2kg~2) NECO 2W 6E7 Ans: 6 - 28 X 107J k g ~1

7. What is the gravitational potential due to a molecule of mass m at a distance rfrom it? (G = gravitational constant) A . B . — C . ^ - D .^ ^ - E —

r r r ‘ r cr2WAEC 198937 Ans: B

8. What is the gravitational potential due to a point mass m at a distance from it?(G = gravitatonalconstant) A. G™ B. Gm C.— ™ D .^^ - £ ~Cmr r Crz f ‘ 2

WAEC 199334 Ans: D9. The force experienced by an object of mass 60 - 0kg in die moon’s gravitational field is 1 • 002 x 102N. What is the intensity of the gravitational field?A. 0.6ONkg' 1 B. 1.67N kg ' 1 C. 6 .1 2 x 1 0 2N kg~1 D. g .e im s -2

JAMB 19937 Ans: 1.67AGt - 110. A missile weighing 400N on the earth’s surface is short into the atmosphere to analtitude of 6 • 40 x 106m. taking the earth as a sphere of radius 6 • 4 x 106m and assuming inverse square law of universal gravitation. What would be the weight o f the missile at the altitude? WAEC 19977 Ans: 100/V

342

11. A hotly moves in a circular orbit o f radius 4 R round the earth. Express theacceleration of free fall due to gravity o f the body in terms of9■ 1^ = radius of the earth , g — acclerationof free fall due to gravity ]

A.— B .— C - D .- E — NECO 2 0 0 7 3U A n s : g / \G9 16 q 4 8

12. I he gravitational force of the moon is one-sixth that o f the earth. If a body weighs 8 • ON on the moon, calculate its weight on the earth. NECO 20 0 4 41 Ans: 48/V13. Calculate the escape velocity for a rocket fired from the earth’s surface at a point where the acceleration due to gravity is 10m s”2 and the radius o f the earth is 6 X 10 m WAEC 198938 A n s : 1• 1 x 104 m /s14. C alculate the escape velocity of a satellite launched from the earth s surface.(T a ke g as 10m s~ 2and th e ra d iu s o f the e a r th as 6.4 x 106m)

WAEC 199 333 Ans: 1.13 x 104m s " 115. What is the escape velocity of a body on the surface o f the earth o f radius R if the gravitational constant is G and the mass o f the earth is Af?(Neglet energy losses to the

surrounding) A. yflGR B. C .y flG R i D. ¡ 2 2 E.yJlGM\ R v R . Q mWAEC 199535 Ans: D. H in t Ve = J l g R ; g = -¿r

16. A rocket is launched from the surface o f the earth. If the radius o f the earth is6 • 4 X 106m and the acceleration of free fall due to gravity is 10 m s-1 ca1 'ate the escape velocity o f the rocket. WAEC 1996s Ans: 1 • 13 x 104 -117. A rocket o f mass m is fired from the earth surface such that it just escapes from the earth 's gravitational field. If R is the radius of the earth and g the acceleration o f free fall due to gravity, the escape velocity o f the rocket is expressed as

A. y f lg R B .^ R /2 g C .y j lg /R D . f R 2/2 g W AEC 199935 A "-: A18. Calculate the escape velocity o f a satellite from the earth's surface, lake g as 1 0 m s-2 and the radius o f the earth as 6 • 0 x 106m

NECO 2 0 0 039 Ans: 1 • 1 x 104m s -119. Calculate the escape velocity o f a satellite from the earth 's attraction. [5 = 1 0 m s-2 , radius of the earth = 6 • 4 x 106m ] NECO 2005+o Ans: 1.1 x 104m s -120. The escape velocity o f a rocket at location R meters, above the surface o f the earth o f mass M and radius R. is given by

A. V2GM R B . ^ ~ - C D

NECO 2 0 0 8 16 Ans: C21. The values o f x , y and 2 respectively in the expression Mx Ly T z for the universal gravitational constant G areA. 2 , - 3 , - 2 B . - 1 . 3 . - 2 C . - 1.2, -3 D. -2 ,-1 .3 JAM B 2 0 0 9 'Ans: B22. The gravitational field intensity at a location X, in space, is two-fifths o f its valueon the earth 's surface. If the weight o f an object at X is 4.SON, what is its weight on the earth? NECO 2009’” Ans: 12.00N

343

2ELECTRIC FIELD: STATIC ELECTRICITY

has lost o r ' u a T C,ectros,atlcs is 'he study of charges at rest. A charge is an atom that charges experiences ^ e*ectric 's a region or space where a system o f electric

El2!nL ° M B S LA Wforce of repulsion"^0611 tW° C*1ar es IS governed by Coulomb's law which states that the product o f the two ^ |,ractlon between two point charges is directly proportional to the between the two ch a rg e .^ 65 ^ mverse*y proportional to the square o f the distance

T h a t* F a 2 i | t or F = K2!f e = _ l _r2 4 n£0 r 2

----------" F i R2

------------------------* ' r--------------------- ►F*'g- 2 /l ------------------------------------------------------Where F - magnifcde o f electric force Newton .

<h .Rz- charges m coulombs (C)J ~ ri 1StflnCe between charges in meter (m)

~ 9 X lo9N m2C~2 or 9 x 1 OqmF~1

e° = o ? s “ x r & ' ■ a 8 s » •»

Exam ple 1

An electron is at a distance of 5.0 X 10- n m away from the center of a proton. Calculate the electrostatic force on it. [Electron charge = -1 .6 x 10_19(7, proton charge - + 1.6 x 10_19C,—-— = 9.0 x 109/V-vn2C_zl

47T£0 JSolution

Distance r = 5.0 x 10~n m; q1 - q2 = 1.6 x 10_19C, = 9.0 x 109N t : C -

1 R1R1Electrostatic force F = — - x — —4HCO 7"2

9 x 10° x (1 • 6 x 10“ 1 ) 2 2 • 30-' x H r 28( 5 x l 0 - “ V ~~ 2-

= 9.2 x 10"8,V

9.2 x 10 ~eN is a force of attraction

Example 2if the distance between two point charges is increased by a fact'- o f 4 the electrostatic force between them will be A.p of its former v fue.

value. C.— of its former value. I). 4 times its former value.16

Solution

The inverse square law F — ~ relates electrostatic force (F) and Jistare

* rrrgm . - H ; o f its former

I A' I B 2003 '"

3 4 4

•ri r 1 1 1 11 herefore, electrostatic force F = — = —7 = —r 2 42 16

/ — 4 ( distance - is increased by a fa cto r o f fo u r)

ELEC TR IC FIELD INTENSITYElectric field intensity (E) also known as the strength of an electric jfield at any point, is defined as the force (/•’) experienced by a unit positive test charge (<j) at that point.

Electric force = Electric charge x Electric field intensity,

F = qE

F a K~qElectric field intensity E — — or E = — — - or E = —5-

q 4tt s0r 2 r 2

The unit of electric field intensity is Newton per coulomb, NC-1

Example 3In a uniform electric field, the magnitude of the force on a charge of 0.2 C is 4N. Calculate the electric field intensity. WAEC 199937SolutionForce; F = 4N; charge q = 0.2C

F 4 NElectric field intensity E = — = - - - - - = 20NC 1 —q 0 -2 C

Example 4Calculate the magnitude of the electric field intensity in vacuum at a distance of 25cm from a charge of 5 x 10_3C. fTake —— as 9 x 10')N m 2C~2]L 4 7TE0 J

AfECO 200442SolutionDistance r - 25cm = 0.25m; charge q — 5 x 10~3CNm2C~2 — = 9 x 109iVm2C -247T£0

£7 _ 5 x 10~3 x 9 x 10947Tf0r 2 (0.25)2

7.2 x lO8^ " 1

Example 5A charge of 1.6 x 10_1°C is placed in a uniform electric field of intensity2.0 x 105/VC-1, what is the magnitude of the electric force exerted on the charge?

WAEC 199536 SolutionCharge q = 1.6 x 10"10C; £ = 2 .0 x lO ^ C " 1

Electric force F = qE = 1.6 X IO"10 X 2.0 x 10s = 3.2 x 10~SN

Example 6Two point charges of magnitude +20 x 10~8C and —5 x 10_8C are separated by a distance of 10cm in vacuum as shown in the diagram below.

P+ 20 x 10-8C • ----------------------------------- • ----------------------------- • - 5 x 10"8C

5cm 5cm<------------------------------------------------------------------ ►

Calculate

(i) The electric fieid intensity at a point P, midway between the charges.t ' ;

345

(ii) The force on a —4 x 10 8C charge placed at I • [\jBCO 2 0 0 4 i14

Wake — = 9 x 109/Vm2C"2lL 4 ne0 JSolution

+ 20- 5 x 10 " 8C

(i) Field at P due to q1

= 1. = 20 x 1 0 - x 9 x 10’ = ? 2 x jo s jv c - i1 47T£0r 2 0.052

Field at P due to q2

f = _ 5 x 10' 8 x 9 x 10 = i.8 x 105/VC"12 An£0r 2 0.052

Resultant field at P is E = Er + E2

E = 1.8 x 105 + 7.2 x 105 = 9 x 105/VC"1

The field is directed towards the —5 x 10"8 Charge

q = - 4 x 10~8C; E = 9 x 105/VC_1

Force P = qE = 4 x 10-8 x 9 x 10s = 3.6 x 10-2 /V

Because the charge (—4 x 10~8C) is negative, it’s directed toward the positive charge

(+ 2 0 X 1 0 -8C).

Exam ple 7A student is at a height 4?n above the ground during a thunderstorm. Given that the potential difference between the thunder cloud and the ground is 107F, the electric field created by the storm is A. 2.0 x 106NC~1 B. 2.5 x 10°/VC~lC. 1.0 x 107/VC-1 D. 4.0 x 107/VC'1 IAM B 2 0 0 1 26Solution

Electric field intensity. E = - Potential difference V = 107; distance r = 4m

E = — = 2.5 x lO 'W C -1 or 2.5 x lO ^Fm ” 1 4

Exam ple 8The potential difference between two parallel plates o f a capacitor placed 3.5cm apart is 5V. Calculate the magnitude of the electric field intensity between the t w o 'pl ues*

Ñ E C O 200a40SolutionDistance between plates, r = 3.5cm = 0.035m Potential difference, V — 5V

V svElectrtic fieid intensity, E = — --------------= 1 4 2 .8 6 p m "1

r 0.035m

Exam ple 9A charge 50pC has an electric field strength of 36 0 A C "1 at a certain point | he -lectu field strength due to another charge 120/rC kept at the same distance apart and m ih ■

346

JAMB 20074same medium isA. 1 8 N C ~ 1 B.144A/C”1 C.864A/C"1 D . I S O N C ' 1

SolutionCharge q = 50qC = 50 X 10_6C; Electric field strength due to SOqC E = 360N

Electric force of the medium F = qE = 50 x 10"6 x 360 = 0.018A/.

Electric field strength due to 120gC(120 X 1 0 _6C)Charge,0.018F

E = - = 120 x 10'= 150A/C'

ELECTR1C POTENTIAL AND POTENTIAL DIFFERENCEElectric potential (y) at a point is defined as the work done by an electric bringing a unit positive charge from infinity to that point.

-1 ~ n1 9 „ _ w lElectric potential V = ----- or V — ft4tt£0 r rExample 10Calculate the electric potential at a distance of 20.0cm from a point charge o placed in air of permittivity, 9.0 x 109/Vm2C_2j. WAEC 1995Solution

Distance r = 20.0cm = 0.2m; charge q = 0.015C ; = 9-0 x 109Nm2C

1 q a 0.0Electric potentail V = --------- = 9 x 109 x ——

471 , r 0.

1.35 x 108

0.0^5 2~

V = 0.2 = 6.75 x 108K

Electric potential difference between point A and point B (VAB) is defined as the work done in moving a unit positive charge from A to B.

q------------------------------------- ►

A •---------------------------------- * B ---------------- ► E■*------------------r---------------- ►

Fig. 2.2Work done = charge xp.d between A and B

W = q V -----------------(1)W = q(VA - VB)

In general, work done = force x distance

Therefore, work done = electric force between A and B x distance betwen A and B

W - F x r

Remember, F = qE (E = electric field intensity)

W = q x E x r -----------------(2)

i.e. work done = charge x field intensity x distance

Equating both work done in equation in 1 and 2 we obtain

q x E x r = qV

E xr = V

347

VE = — r

This equation sums up the relationship between electric field inten ty potential V. Unit of field intensity is also Vm~1

E, and electric

Example 11If two charged plates are maintained at a potential difference in taking a charge of 600pC across the field is A. 1.8/ B. 18.0/ C. 0.8/ D. 9.0/

of 3 KV, the work done

JAMB 200740

SolutionPotential difference, V = 3KV = 3000K; charge q = 60OpC - 600 x 10 C

Work done = charge x potential difference

W = qV = 600 x 10"6 x 3000 = 1.8/

Example 12A work of 30 joules is done in transferring 5 millicoulombs of charge from a point B to a point A in an electric field. The potential difference between B and A is. WAEC 199039 SolutionWork done W = 30/ ; charge q = 5 millicoulombs = 5 x 10-3C ;Work done = charge x p. d.

W 30W = q x V V - — = -— —— = 6000Kq 5 x 1 0 -3

Example 13A point charge of magnitude 2.5pC is moved through a distance of 0.04m against a uniform electric field of intensityl5Km-1. Calculate the work done on the charge.

NECO 200741SolutionCharge,q = 2.5pC = 2.5 x 10-6C; distancer = 0.04m; field intensity E = 15Vm-1

Work done VP = ^ x £ 'x r = 2.5x 10-6 x 15 x 0.04 = 1.5 x 10 ~6J

Example 14Two parallel plates at a distance of 8.0 x 10_3m apart are maintained at a potential difference of 600 volts with the negative plate earthed. What is the electric field strength?A. 4.8Km-1 B.75.0^m_1 C^SOO.OKm'1 D^SOOO.OKm"1 SolutionPotential difference, V = 600K; distance r = 8.0 x 10-3m

JAMB 199537

V 600Electric field intensity E = — = — — - , = 75000^m_1r 8.0 x 10-3

ELECTRON VOLTElectron volt is defined as the quantity of energy gained by an electron when it accelerates through a potential difference of 1 volt. The unit of electron volt is joules

Work done by electron = charge x potential difference

348

Kinetic energy (K. E) = q x V

^ m v 2 = q x V

Where in = mass of electron in kilogram (kg) v = velocity of electron In ms-1 V = potential difference q = magnitude of electron charge

Example 15An electron of charge 1.6 x 10-19C and mass 9.1 x 10~3lkg is accelerated between 2 metal plates with a velocity of 4 x 107ms-1, the potential difference between the p ate is A. 4.55 x 10XV B.9.10 x 101V C .4 .55x l02V D .4 .5 5 x l0 3V JAMB 2000 SolutionCharge q = 1.6 x 1(T19C; mass m = 9.1 x 10~3lkg; Velocity v = 4 x 107m s_l; p .d V =?

From,

Potential difference. V =1 22 mv 0.5

Q7.28 x 10~16 1.6 X 10~19

x 9.1 x 10~31 x (4 x 107)21.6 x 1 0 '19

= 4.55 x 103P

Example 16An electron of mass m and charge e enters a uniform electric field between two metal plates P and Q separated by a distance d. P is maintained at a potential V while Q is earthed. Determine an expression for the magnitude of the acceleration of the electron through the field A .^ - B.— C.— £>.— WAEC 200639^ md meV eV VmdSolutiona = acceleration; F = electric force; q = charge = e

E = electric field intensity; m = mass; r = distance = d;V = potential difference

F = ma F = qE E = -

From F = ma and substituting above equations, we obtain

a _ F ^ q x v/ r m m m

Substitute q with e and r with d as given in the question.

a * * 7 d V 1 = e x - x —

d meVmd

CAPACITORS AND CAPACITANCECapacitors are devices for storing electric charges. The capacitance (C) of a capacitor is defined as the ratio of charge (Q) stored by the capacitor to the potential difference (V) between the plates.

VCapacitance is measured in farads(F).1 microfarads = 1 fiF = 10_6F or 1 x 10 6F

349

1 nanofarads = 1 nF = 10_9F or 1 x 10"9F 1 picofarads = 1 pF = 10_,2F or 1 x 10~12F

Capacitance (C) of parallel plate capacitor is also given by,

Where A = area of overlap of plates (m2)d = distance between plates (m) _2£ = pennittivity of material between plates Fm 1 or C N tn

A parallel plate capacitor of capacitance 600pF has a potential d if ference of 2000V between its plate. Calculate the charge on either plate of the capacitor.SolutionCapacitance C = 600pF = 600 x 10"6F

Potential difference V = 2000F

From C = £ , Q = CV = 600 x 10"6 X 2000 = 1.2C

Examplel8A parallel plate capacitor of area 10cm2 in vacuum has a capacitance of 10~2pF. what is the distance between the plates? [c0 = 9 x 10-12Fm-1]A. 9 x 10-13m B. 9 x 1 0 '7m C. 9 x 10“3m D. 9 x 107m JAMB 199340 SolutionArea A = 10cm2 = 10 x 10_4m2 permittivity e0 = 9 x 10_12Fm_1

Capacitance C = 10~2pF = 10-2 x 10_6F

From C £oA~d ' distance, d = 9 x 10"12 x 10 x 10“4

IQ"2 x lO “6 9 x 10"7m

SERIES AND PARALLEL ARRANGEMENT OF CAPACITORSParallel ConnectionWhen capacitors are arranged in parallel as shown below, the following apply

__ ______

___________ 2z| ^2__________

, pis-----

Fig. 2.31. The equivalent or combined capacitance C, is given by: C = C1 + C2 + C32. Cv C2 and C3 arc all at the same potential difference V3. Total circuit charge, Q = Qx + Q2 + Qi where Qx = QK, Q2 = c2V, Q2 = c2V

350

Example 19What is the equivalent capacitance of a 3uF capacitor and a 6/iF capacitor connected in parallel? WAEC 200641SolutionEquivalent capacitance CT = C3 + C2 = 3fiF + 6fiF = 9pF

Example 20

_2j |io^f

I—20KFig. 2.4

The diagram above shows two capacitors P and Q of capacitances 5[iF and 10 fiF. Find the charges stored in P and Q respectively. A. 200¡xC and lOOpCB. IOOjuC and 200/uC C. 4^C and 2/xC D. 2fiC and AfiC J A M B 200030 SolutionFrom C = ~ , charge Q = CV

QP = CP V = 5u F x 20 = 5 x 10-6 x 20 = 100 x 10~6C = 100/iC

Qq = CQV = 10u F x 20 = 10 x 10-6 x 20 = 200 x 10-6C = 200/rC

Series ConnectionWhen capacitors are arranged in series as shown below, the following apply

— *Q

1— J h -Q Q Q

- T i l ------------e Q

Vi v2

________ 1 1 ___

V3

Fig. 2.51 1 1 1

1. The equivalent or combined capacitance (C) is,C Lx C2 C3

c cEquivalent capacitance for two capacitors C1 and C2 is, C = - 1 2

2. Capacitors Cx, C2 and C3 all have the same charge Q.

3. Total circuit voltage, V = V1 + V2 + V3 where V1 = V1=j r ,V2 =^r ,V3 - ^ rL1 L2 i-3

Example 21Three capacitors each of capacitances 18/rF are connected in series. Calculate the effective capacitance of the capacitors. WAEC 200236

351

Solution— 11— 11— i Hr

18 /if 18/ti 18^fFig 2.6

1 1 1 1 1 + l + l _ 3 C = Î8 + Î 8 + Ï 8 _ 18 l 8

- = — .-. C = ^ = 6/tF C 18 3

Example 22

2 nF

6/J.F

3 nF

Fig. 2.7The diagram above shows three capacitors Cv C2 and C3 of capacitances 2fiF, 6fiF and 3fiF respectively. The potential differences across C1, C2 and C3 respectively are A. 4 V .6 V and 2V B. 2V,6Vand4V C. 6V,4Vand2V D. 6V, 2V and 4V

JAMB 200125Solution

i _ 1 , 1 , 1 1 1 1 3 + 1 + 2C ~ C 1 + C2 + C3 _ 2 + 6 + 3 _ 6

1 6— = - = 1 equivalent capacitance C = 1L OCircuit charge Q = CV = l x 12 = 12C

From C — — ,

Q 12Vx = -jr = — = 6V; 1 2

Ans: 6V, 2V and 4V

Q 12 ? 12Kj = - = t = 41,

Example 23Two capacitance of 6fiF and 8fiF are connected in series. What additional capacitance must be connected in series with this combination to give a total of 3fiF2 A. 3nF B. !6fiF C. 24/iF D. 30fiF JAMB 198740SolutionLet CT —total capacitance = 3fiF, x — additional capacitance

1CT

1 1 ~ 6 + 8 4

1 1 13 6 8

352

8 - 4 - 3 1 24 ~ x

J _ _ 1 24 ~ x

x = 24 fiF

Combined Series And Parallel Connection

Example 24

Fig. 2.8

The resultant capacitance in the figure above is A. 15.0fiF B. 9.8fiF C. 1.3¡J.F D. 0.8nF jAm b 199543SolutionCapacitor 2fiF, 3(iF and 4¡J.F are in series with one another and also in series with a parallel combination of capacitors 2 fxF and 4fiF.

1 1 1 1 1C ~ 2 + 3 + (2 + 4) + 4 1 1 1 1 1 - = - + - + -T + T

1 6 + 4 + 2 + 3 15C ~ 12 “ 12

1C

15 12— C = — = 0.8 uF 12 15 M

Example 25What is the total capacitance in the circuit represented by the diagram below?

I 12 u?

H t2uF

H h2/tF

Fig. 2.9

SolutionIn parallel: 2pF + 2fiF + 2/xF = 6fiF. 6(iF m senes with 2\iF

6 x 2 12 _. r ---------- = — = 1.5liFL 6 + 2 8

ÑECO 200046

353

Example 26 ■ the diagram below is 1-50\iFThe effective capacitance between points X and

t C

ÑECO 20024Fig. 2.10What is the value of the capacitance C?SolutionIn parallel: C + C + C = 3C. Then 3C in series with 2fiF

n v -ir f,rTotal capacitance, 1.50 =

Cross multiply 1.50 =

2 + 3C

6 C2 + 3 C

6 C = 1 .5 (2+ 3C) 6C = 3 + 4.5C

2 + 3C

to obtain

SC - 4.5C = 3

1.5C = 3 C = - = 2 ^

Example 27

Fig. 2.11—I I-----100 /iF

, AQfiF

■CP-60 uF

NECO 2007E8Calculate the current in the circuit diagram above.SolutionCombined parallel capacitance = 40 + 60 = 100/tF 100\iF in series with 100fiF

, , „ 100 x 100 10000.-. Total combined capacitance C = ——---------= ----------= 50uF

F 100 + 100 200 M

From C = ^ , charge, Q = CV = 50 x 10-6 x 24 = 1.2 x 10-3C

Q 1.2 x 10~3A Current / = — = ----------------

Example 28

Fig. 2.12 4 u F

3 5 4

What is the resultant capacity circuit above?A. 1.5gF B. 18.0nF C. 6.0gF D. 6.8gF E. 8.83gF JAMB 1982Z6SolutionlfiF and 5pF are in parallel, therefore combined capacitance = 1 + 5 = 6fiF 6 gF is in series with 4 gF and 4 gF

1 1 1 1 4 + 6 + 6 _ 16C ~ 6 + 4 + 4 ~ 24 “ 24

1 _ 16 C ~ 24

24 3■■■c = Te = 2 = 1^ F

Example 29In the network shown below, determine the p. d across the 5gF capacitor.A- 3l/ B. 6V C. 12V D. 18 JAMB 1993

5gF 15gF

Fig. 2.13

41

Solution

5gF and 15gF are series 5 x 1 5 755 + 15 ~ 20 3.75

Charge following through series capacitors is:

Q = CV = 3.75 x 24 = 90pC

The same charge (90pC) flow through 5pF

potential difference across 5pF,Q _ 90pC C “ 5pF 18E

ENERGY STORED IN A CAPACITOREnergy stored in a capacitor is equivalent to the work done in charging the capacitor

1 1 2 1 q2Energy store in capacitor W = - q V = -CV = - • —Any of these equations can be used depending on the term giving in the question.

Example 30Calculate the energy stored in a 20qF capacitor if the potential difference between the plates is 40V. W A EC 2004«SolutionCapacitance C = 20qF = 20 X 10 6F ; potential difference V - 40V

energy W = ^ CV2 = ^ x 20 x 1 0 '6 x 402 = 1.6 x 10~ZJ

3 5 5

Example 31 . ,The energy stored in a capacitor of capacitance 5pF is 40/. What is the voltage app u across its terminal? NECO 20Solutioncapacitance C = 5pF = 5 x 10-6F energy, W = 40/

1 ,W = - C V 2

240 = 0.5 x 5 x 10“6 x V2

V240

0.5 X 5 x 10"6= 1.6 x 107

V = V l.6 x 107 = 4 x 103V

Example 32Use the diagram shown below to answer the following question.

Fig. 2.14

6uF 6uF 6uF

— IHHH-il—100F

1. What is the effective capacitance in the circuit?2. What is the total energy stored by the capacitors? Solution, x „ 1 1 1 1 (1). Effective capacitance —= — + — + —-

C Ci C2 C3

1 _ 1 1 1 C ~ 6 + 6 + 6

I 1 + 1 + 1 3 1C ~ 6~I IC ~ 2

(ii) C = 2uF potential difference,1 l

Total energy W = - C K Z = - x 2 x 10~6 x 1002 = 0.01 o r 1 x 10~2/

6 2

C = 2\iF

V = 100K

Example 33

Fig. 2.15

_P | | 2 uF

4fiF

VThe diagram above shows two capacitors P and Q of capacitances 2uF and 4uF respectively connected to a d.c source. The ratio of energy stored in P to Q is

356

B. 2: 1 C. 1 :4 D. 1 :2A 4 :1 S o lu tion

w = - c v 22 \VP = ^ C PV 2

VVp _ \ c PV 2 _ cp _ 2pF _ 1

WQ ~ \ c QV > ~ r Q ~ W ~ 2

JA M B 2 0 0 123

W q2

WP:W Q = 1 :2

E x am p le 34

Fig. 2.16

1 w o capacitors Cx and C2 are connected as shown in the diagram above. The capacitance ot C2 is tw ice Ci. W hen the key is opened, the energy stored up in Cx is W . I f the key is la.er closed and the system is allowed to attain electrical equilibrium, the total energy stored in *he system will be A. W B . ^ W C. W D. 2W E. 3W

W AEC 199143S olu tion

Ci

Fig 2.17 VC2 — 2 C\ ( .............. capacitance ofC 2 is twice C i ............... )

Vi hen the key is opened, capacitor 2C1 is cut off.

Energy in Cx is IV = - C XV 2

When the key is closed both capacitors are involved, . ova capacitance = 2Ct + Cx - 3CX

Energy stored at equilibrium = - x 3Cl V2

'arranging, we obtain, 3^- C'i V2)

Subsntv.tr ic IV = - C , k 2, energy stored at equilibrium = 3VV

Exant pic 35The plates o f a parallel plate capacitor arc 5mm apart and 2 m 2 in area Ihe plates are in vacuum. A po ten tia l difference of 1000k is applied across the capacitor calculate the

lit CapacitanceChoree on each plate

{■ 1 lcctnc intensity in the space between them(is ; i nerpv stored in the capacitor.

357

(permittivity of free space e0 — 8.85 x 10 12C2N 1m 2) NECO 2000'SolutionDistance apart, d = 5mm = 0.005m ; area, A = 2m 2;

potential difference, V = 1000V; e0 = 8.85 x 1 0 '12C2A/_1m _z

£oA 8.85 x IO-12 x 2 1.77 x lO 11(0 Capacitance, C = ---------------------------------------------

d 0.005 0.005(ii) Charge Q = CV = 3.54 x 10~9 x 1000 = 3.54 x 10~6C

= 3.54 x 10 9F

..... V 1000(in) Electric intensity E = — = ------ - = 2 x 10sFm -1

a 0.0051 1 -i(iv) Energy stored W = - CV2 = - x 3.54 x 10-9 x 10002 = 1.77 X 10 J

Example 36An isolated electrically charged sphere of radius, r , and charge , Q, is supported on an insulator in air of permittivity, e0. a Write down

(i) An expression for the electric field intensity on the surface of the sphere.(ii) An expression for the electric potential at the surface of the sphere .(iii) A relationship between the electric field intensity and the electric potential

at the surface of the sphere.b. The plates of a parallel plate capacitor, 5.0x 10_3m apart are maintained at a potential difference of 5.0 x 104K. Calculate the magnitude of the

(>) Electric field intensity between the plate(il) Force on the electron(111) Acceleration of the electron

[Electric charge = 1.60 x 10_19C] [Mass of electron = 9.1 x lO-31/^ ] WAEC 2005E14

Solution

a. (i) Electric field intensity, E = ---------4 n e 0r 2

Q(ii) Electric potential, V = -------47Tc0r

V(iii) E = -

b. Distance between plate, r = 5.0 x 10-3m; potential difference, V = 5.0 x lOO4^;

electron charge, q = 1.6 X 10-19C; mass of electron, m = 9.1 X 10~31kg

V 5.0 x 104(i) Electric field intensity E = — = - - x 1Q_3 = 1 x 107Km 1

(ii) Force on electric F = qE = 1.6 X 10 19 x 1 x 107 = 1.6 x 10~12N

(iii) From F = ma, acceleration of electron,

_ F _ 1.6 x IO-12 a m 9.1 x 10-31

1.76 x 1018m s 2

Example 37The capacitance o f a parallel plate capacitor when in air is 3qF and in the presence o f a dielectric material 6fiF. The dielectric constant isA. 3 B. 18 C. 2 D. 9 JAMB 200632

358

Solution

D ielectric co n sta n t = d Îelectric o f m ed ium = 6fxF d ie lec tr ic o f a ir 3fiF

2

EXERCISE 2!• The force of repulsion between two point positive charges SfiF and B/iC separatedat a distance of 0.02m apart is A. 1.8 x 10-10Af B. 9.0 x 10~BNC. 9.0 x 102N d . 4.5 x 103N f— = 9 x 109Nm*C~21

UtCo *JAMB 199836 Ans: 9.0 x 102M

A charge of 1.0 X 10_s coulombs experiences a force o f4QN at a certain point in space. What is the electric field intensity? WAEC 199042 Ans: 4 x 1 06AfC-1 3- A point charge of 1j0 X 10_7C experiences a force o f 0.01 AT in a uniform dectnc field. Calculate the magnitude of die strength of the field.

WAEC 1993“ Ans:! x lOsV m 1(N C'1}4. A charge of 1.0 X 10~SC experiences a force of 401V at a certain point in space. What is the magnitude of the electric field intensity at the point in Newton per coulom b?

WAEC 199738 Ans: 4.0 x 106^ A charge of 1.6 x 10_1°C is placed in a uniform electric field of intensity2.0 X 10 NC *. Calculate the magnitude o f the electric force exerted on the charge.

WAEC 200443 Ans:3.2 x 10 'SNd- A charge of 2.0 X 10_SC experiences a force of 80N in a uniform electric field. Calculate the magnitude o f the electric field intensity

NECO 200643 Ans:4 x 106N C '17. If the force on a charge of 0.2 coulomb in an electric field is 4N, then the electric field intensity of the field is A. 0.8 B. 0.8N/C C. 20.0N/C D. 4.2N/C E.20.0C/N

JAMB 198347 Ans:20.0N/C8. The electric field intensity in a place where a charge of 1 0 10C experiences aforce of 0.4N is A. 8.0 x 10_12AfC_1 B. 8.0 x 109WC-1C. 4.0 X W9NC l D. 4.0 x 10~11NC~1 JAMB 200523 Ans:4.0 X lO’ NC- 19. A point charge of magnitude 2fiC is moved through a distance of 0.2m apiind uniform field of intensity ISVmT1. Calculate the work done on the charge.

WAEC 199940 Ans:l x 10~SJ10. If 8 X 10 zy of work is required to move 100/tC of charge from a point X to apoint Y in an electrical circuit, the p. d between X and Y is A. 4.0 x 102kB. 4.0 x 104k C .a O x lO 2 D. 8.0 x 104k JAMB 199742 Ans:8.0 x 10*k11. Find the work done in moving a 2C charge between two points X and Y in an electric field if the p.dis 100 volts A. 50/ B. 100/ C. 200/ D. 400/

JAMB 199841 Ans:200/1 2 . The potential difference between two points A and B situated at a distanr*» d apart is k. Which o f the following expresses the magnitude of the electric field intensity between the two points assuming the field is uniform. A. Vd B. dV~l C. V d 1D. V2d E. d2V WAEC 199339 AnsYd" 113. What is the magnitude of the electric field intensity between two plates 30cm apart, if the potential difference between the plate is 4.2K?

NECO 200240 Ans: 14.0km-114. An electron o f charge 1.6 x 10_19C is accelerated between two métal plates. If the kinetic energy o f the electron is 4.8 x 10"17/, the potential difference between the plates is A. 30Y B. 40k C. 300k D. 400k JAMB 200333 Ans. 300k1 5 A battery is connected across the plates of a 2 .0nF uncharged parallel platecapacitor. The charge on the capacitor after a long time isA. 2 5fiC B 10.0fiC C. 20.0fiC D. 50.0pC JAMB 199239 Ans:10.0//C

359

*■ _A Pïate capachor bas a counnou plate area of 5 x lO^wf2 and prise^ » b o n ° f 2 x 10 3m. Assnnnug free spaee, whaf b die capacitance?A-¿25X 10 ” P B. 430 x 10"*7F C .2 2 5 x ir “ F D 4-50 x UT3*F Uo - 9.0 X 10 “ C2# "1»!-2] JA M B 199842 Aœ:Z25 x 1(T14 A17. Two 50ftC parai le] prias capaciton are connected ■ séries. The a VK** ™ then connected across a 100Y baOery. The charge on each prise ai the «P«tor » A. 5.00 x 10 5C R 230 x 10‘ 3C C . L25 x ÎO^C

100X 10 C JA M B 199444 Ans: 2-S X l<r3CVS her nn3 «vatiifi awiî ioq :r>q

_ + « |-< ? I I

V' *3i x ; ; •' ,• 10 i <*: oc; ci-jroc * j Adt- to JOTO* 4 «3tr h jq /.

Cl x f <n/, 4y f« i "JT?-'zri -yn/ :;T»! i p v i i AiO.O h --..or» :

fig . ¿18x f tf .

6gF connected in séria.the figure abovfc shows Aren capacitors'2/tF, 3fiF gnriarisl (fifkiChut ac¿ t S W V & W * * * IAMB to r t» Am: to

-Of v OA: r.A " Ti-ei ~3>Vrittffralni *lo blsiY iirrtanb irndiwc £ ni c,v >5 «i '(•! - 31

11J Lfl:> *5iir no D^no/J 'V\; Of x S F.znA.

¿fi.ii oinoata mndhru

1_ yA% i \'c icala 3ift mdl .VW- iW^O.Oy.i ' 3 \ / r..f-

Fig. 2.19 ! ;: ..»: sav!r.-.rT--'j/t* .What is the total capacitance in the represented by the diagram? A- Cj + -f- C3R ± < + ^ 4 o i C. j y . E . CtC2C3 " ‘0 V

vm- Mc rn? 0 to rf^uoir.i fca/r m WAEC1994“ A m C i+ C 2+C32*l Two capacitors of capacitance 0.4pF and 0.5//F are connected m parallel and charged to a potential difference Of 501A Determine the total charge acquired.

• <»i >' tn it q b mod sguab lo W A E C 200S34 Arts 45pCV 0 : > J f . 'i Y ban X ttjofcvi Sv .t* erfi «iiicsis t.rgsr^-, m r t .*.->321.

V ' 01 x 0.8:«:: A ' ' V *?.? f *J 01 x C

\00£t> 3 :

Til. % * \L>0XH \' T Q F -.<>■' sx>t rtb.c iilit■ ,1 l -k 1 _ on •, HI ;to-xs -o !*:* -z-'.j-f. c.-ri : Z \

.■ •;■ F l g r ; 2 i 2 Q : ' 1 - :d J t o

The effective capacitance beK\ n points "X and Y in the diagram is 1.0¿lF. What is the >alue of the capacitance C. mca ired in microfarad? W.4EC 19954S Ans. l.QuA; .V • . :i ■ n-j- . ; ; . n .' :i ^22.

s' . • : > n / )0S vo . w -------------- 1 jJ i i f lq iM f fn V.- ^ ^Y Jjfli-fc):J'..

•jYiiLi ist/trjtoq tjrii 0 .'aril n ixy*. 'sy -d '.»<

-JlKfq ¡jfier^q b:

n g. 2.21 r\q0.0 f . ' it/

/nA “ FOUS y r h — f■•.n.rru r lo !a 's'OOL

ru ,no| « : {,nol 1 t >‘t. OS T

360

^ , mmn — - o f t e cacmt-Using the diagram above, calculate dte « W * * ffly r i W 11 ^ rf..5 6

, • rX jiM e i e *ffl«**23. The diagram below iq w M » * « * " * cm* Lcapacitance m the cecboo.

In tte circuit shown above, calculate the effective capacitance between X and Y A B . 2 £ / iF C.12tiF D . 4 ^ E.13#iF JAMB 198435 Ans:2^/iF27. a. Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.b. Two parallel plate capacitors of capacitance 2fiF and 3pF are connected in parallel and the combination is connected to a 50K d . c source.Draw the circuit diagram of the arrangement and determine the

(i) Charge on either plate of each capacitor(ii) Potential difference across each capacitor (in) Combined energy of the capacitors

WAEC 1996“ Aits: a. W = £ b. (i)l x lO^C, 13 x 10 4C (ii) SOK(iii) 6.25 x 10"3/2 f a series arrangement of three capacitors of values 8pF, 12/if and 24//F is connected ia series with a 90P battery.

(i) Draw an open circuit diagram for this arrangement,(i) Calculate the effective capacitance in the circuit,(ii) On closed circuit, calculate the charge on each capacitor when fully charged.

(iii) Determine the potential difference across the 8pF capacitorWAEC 2001£14 Ans: (ir)4/iF (iii)360*rC (iv)45K

29. Three capacitors L, M, and N with capacitances C3, C2, and C3 respectively are to be combined. If C3 > C2 > Ct , which of the following arrangements will give maximum capacitance? >-

N M L

Fig. 2.23

30.

ÑECO 20 0742 Ans:B

2 pF

Fig. 2.28

- u

H \

2pF

2pF

Three 2pF capacitors are arranged as shown in the above circuit. The effective capacitance between E and F isA. 0.75pF B. 1.33pF C. 3.00pF D. 6.00pF JAMB 198939 Ans: 1.33pF

31.

FiV ?.29

• r ; ‘ capatiUi • e >n t:. . r ,*F B 6 .OpF ( 2.0 u '■ JA . 'B i r 9 t 39 Ans: 2.0pF

32. Which nl the following comhimilions of 2///* capacitors will G vc capacitance of 3fiF across terminal XYV

A*2/ih 2///'

H I----------I I -

2///'

A*-2tiF ¿OlL

2/-//*'

C.2 fiF 2uF '¿¡j F

x*---- 11---- 11---- 1 I---- ^

D.A*-

Fig. 2.27

• Y

JAMB 19 8 842 Ans: A

33. Given 3 capacitors 0.3fiF, 0.5fiF and 0.2/iF, the joint capacitance when arrange to give minimum capacitance is A. 03fiF B. 1.0[iF C. 0.1/tF D. 0.5[iF

JAMB 2 0 0 450 Ans: 0.1//F34. Calculate the energy stored in a 10/iF capacitor if the potential differencebetween the plates is 201C WAEC 198941 Ans: 2 x 10~3J35. Calculate the energy stored in a 20fiF capacitor if the potential difference betweenthe plates is 40K. WAEC 199249 Ans: 1.6 x 10"2/36. The energy stored in a capacitor of capacitor 5mF is 40J. Calculate the voltageapplied across its terminals? WAEC 199544 Ans: 4000T37.

Fig. 2.30

10 nF

H |_1 2 K

A capacitor o f 10jiF is connected across a cell of 12 volts as shown above. Calculate the energy stored in the capacitor. A. 7.5 x 10~4/ B. 1.4 x 10~3yC. 1.4 x l O " 4/ D. 1.2 x 10-2/ JAMB 199241 Ans: 7.2 x 10~4y38. A capacitor 8(iF, is charged to a potential difference of 1001C The energy stored by the capacitor is A. 1.0 x 104y B. 8.0 x 10y C. 1.25 x 10yD. 4.0 x 10"2y JAMB 199 443Ans: 4.0 x 10-2/

3 nF

Fig. 2.31 |------ -------------- 100lA

363

In the circuit above, the potential across each capacitor islOOK. The total energy stored in the nvo capacitor is A. 3.0 x 104/ B. 3.0 x 102J C. 2.5 x 1 0 'zy D. 6.0 x

] JAMB 199938 Ans: 2.5 x 10~2JThe energy stored in a capacitor o f capacitance 10/uC carrying a charge o f 100fiC

isA. 4 x l 0 -3/ B. 5 x 10_4/ C. 5 x 104/ D. 4 x 102JIT . JAMB 200230 Ans: 5 x 10-4/Use the information below to answer question 41 and 42.An isolated metal sphere of radius of radius R, carrying an electric charge, Q, is situated m a me ium of permittivity, er . A test charge is placed at a point P, distance r from the u ace o t e sphere. Let e0 represent the permittivity o f free space.

The electric potential at P is given by the expressionA.-

42A.

4Ti£0err B.-47t£0£r (R+r) c. 471 £„£r (R-r)D.-

47t£0f r RWAEC 200936 Ans: B

The magnitude of the electric field intensity at P is given by the expression JL n Q C Q n Q

4TT£0£r r 2 B.-47t£0£r (R+r)2 ' 47I£0£r (R -r)2 D.-

’ 4n£0£r R2WAEC 200937 Ans: B

43. The electric potential at a distance d from a point charge q in air of permittivity ea is expressed asA. 47T£0£Î B. 477 E n d 2

c. 4n£0d2 D. E.-477£02dNECO 200940 Ans: D

44. Calculate the work done in charging a capacitor of capacitance 20p f through a potential difference of 1000V. NECO 200941 Ans: 10.00J

364

3CURRENT ELECTRICITY

RESISTIVITY AND CONDUCTIVITYResistivity of a material is defined as the resistance per unit length per unit cross sectiona area of the material.

RA Rnr2 " _ R nd2Resistivity p = — or p =■ —-— or p — ^ ~

Where p = resistivity in ilm R = resistance in ilA = cross sectional area = n r 2 = in (m) l = length of wire in m

d = diameter of wire in m 2 r — radius of wire in m 2

'X

Electrical Conductivity is the reciprocal of resistivity. It is measured in (flin) 1 or

Electrical conductivity a = — = — = ---------- = — ——p /M R x n r 2 R x n d 2

Always make sure you do your calculation in uniform unit of measurement. Do not mix up m 2 and cm 2 or m and mm.Preferably, express all measurements in S.I unit.Note: 1mm = 0.001m or 1 x 10_3m ,

lc m = 0.01m or 1 x 10_2m 1mm2 = OD00001 or 1 x 10_6m2 lc m 2 = O.0QOT or 1 x lQ -4m2

Example 1Calculate the resistivity of a wire of length 2m and cross sectional area 0.004cm2 if itsresistance is 3.0 ohm. WAEC 198851SolutionLength l = 2m = 200cm; cross sectional area A = 0.004cm2; resistance R — 3.Oil

RA 3 x 0.004 0.012Resistivity, p = — = — — = --¿qq = 0.00006ficm = 6.0 x 10 5ficm

Resistivity in flcm can be converted to ilm by dividing by 100 (100cm = lm )

0 000Q- = 0.0000006 = 6.0 x 10"7ilm 100

Example 2A constantan wire has a cross sectional area of 4 X 10"8m 2 and a resistivity of 1.1 X 10_6ilm . If a resistor of resistance 1 l i l is to be made from this wire, calculate the length of the wire required. WAEC 200539Solution

A = 4 x 10_8m 2; Z = 1.1 x 10"6f im ; /? = l l i lR A RA 11 x 4 x 10~8

From p = — . length, l = ~ = 1-X x 10-6 = °-4m

365

Example 3A wire of length 100cm and cross sectional area 2.0 x 10 ’em lias •' 0.10ft. Calculate its electrical conductivity. NhCO 2002.SolutionLength l = 100cm ; cross sectional area = 2.0 x 10_3cjn2; ^ —

resistance3l)

o .io n

of

Electrical conductivity a = - = — y p RA

1 0 00.1 x 2.0 x IO“3

= 5.0 x iosn_1cm

Example 4A 2m wire o f resistivity 5.5 x 10_7ftm has a cross sectional area o f 0.50mm ■ C alcu ate its resistance. NECO 2004Solutionl - 2 m -, resistivity, p = 5.5 x 10~7ilm ;

Cross sectional area, A = 0.5m m 2 = 0.5 x 10~6m ( l t n m 2 = 1 x 1 0 6m)

n RAFrom p = — , piresistance R = —

A5.5 x 10~7 x 2

0.5 x IO "6

1.1 x 10-6 0.5 x 1 0 -6

2.2Ì1

Example 5A wire of length 15m made of a material of resistivity 1.8 x 10~6flm has a resistance of 0.27fl. Determine the area of the wire. A. 1.5 x 10~4m 2 B. 1.0 X 10-4 m 2 C. 2.7 x 10~5m 2 D.7.3 x lO “6??!2 JAMB 19 9 2 36Solutionl = 15m ; p = 1.8 x 10_6il?7t; R - 0.27H

RA pi 1.8 x 10-6 x 15 2.7 x 10“ 5From p = —~, a rea , A = — = ----------------------- = ----------------

l R 0.27 0.27

= 1.0 x 10~47n2

Example 6A wire o f 5H resistance is drawn out so that its new length is two times the original length. If the resistivity o f the wire remains the same and the cross sectional area is halved, the new resistance is A. 40fl B. 20ft C. 10ft D. 5ft JAMB 2 0 0 3 32 Solution

Original wire New wireResistance Rx = 5ft R2 =?

Length h = l l2 — 21 {two times the original length)

Resistance Pi = P p 2 = pjsam e resistivity)

Area A x = AA

A 2 = ~ (area is halved)

Ri A t /?2^2ResistivitiesP' = L

Pi - .l2

Equate p x and p 2 because they are equal

R2A 2

Cross multiplying, R2A¿l\ = R iA ^l2

R i A \ l 2T herefore, R2 = —

A 2‘1

366

Substituting, r 2 = 5 X A x 21 _ 5 x 2 x /t x 2/ _ 5 x 2 x 2 = 20fi4 x i

GALVANOMETER CONVERSION: SHUNT AND MULTIPLIER

I. Conversion o f Galvanometer to Ammeter. A galvanometer, also known as moving coil meter, can be converted to an ammeter by connecting a low resistance shunt in parallel with the galvanometer as shown below.

\

Fig 3.1

Galvanometer

Shunt/ = Ammeter reading of the converted galvanometer

= Current flowing through galvanometer at full scale deflection R = Shunt resistance r = Galvanometer resistanceRemember, potential difference (p. d) = current x resistance Current flowing through shunt = / — lg Current through galvanometer = lgPotential difference across shunt = shunt current x shunt resistance

Potential difference across galvanometer = galvanometer current x galvanometer resistance

= lg x rBecause r and R are in parallel, they have same potential difference

p. d across shunt = p .d across galvanometer

( / - lg )R = /s x r

I g X rShunt resistance, R - . _ .

1 ‘g

Example 8A galvanometer of resistance 5.Oil has full scale defection for a current of 100m A How would its range be extended to 1.0A? By placing a resistance of A. in parallel B. | i l

in series C. 45 il in parallel D. 45 il in series E. | i l in parallel JAMB 198449

SolutionGalvanometer resistance, r = 5.0fl Shunt resistance R =?Galvanometer current, lg = 100mA = 0.1A Ammeter reading I = 1.0A

I j X r 0.1 x 5 _ 0.5 _ 5 " R = / - lQ = 1.0— 0.1 “ 0 9 ~ 9

5Ans: A resistance of - fl in parallel

367

Example 7 i- resistance 2O to anThe diagram below illustrates the conversion ol a galvanome rtf 10mA. Calculateammeter. The galvanometer gives a full scale deflection for a currcn

b WAEC 199645the value of R.

V}0A

Fig 3.2

Solution

Fig 3.3

- 0 -Tg =20

V T10 mA

CDR

99A ' r = 20

____________ 1------1-----------------

Galvanometer current Ig = 10mA = 0.01/1

Galvanometer resistance r = 20

/„ x r 0.01 x 2

Ammeter reading / = 10A

Shunt resistance R =?

R =_ a 0.021 - L 10 - 0.01 9.99

= 2.0 x 10“3O

Example 9A galvanometer has a resistance of 50. By using a shunt wire of resistance 0.050, the galvanometer could be converted to an ammeter capable of reading 2A. What is the current through the galvanometer? A. 2mA B. 10mA C. 20mA D. 25mA JAMB 199941Solution

Galvanometer resistance, r = 50 Shunt resistance R = 0 .050Galvanometer current, lg =? Ammeter reading, 1 = 2

Substitute into R = 'a x rI - L

to obtain

Cross multiplying, 0.05(2 - lg) = 5lg 0.1 - 0.05/5 = 5lg

0.1 = 5.05 lg

0.05 1 a * 5

2 ~'a

368

/ = -2JL = 1 98 x 10~ 2A * 2.0 x 10'2A = 20mA n 5.05

IRNext time remember that galvanometer current, lq — ^

//. Conversion o f galvanometer to voltmeter. A galvanometer can be converted to a voltmeter by connecting a high resistance multiplier in scries with the galvanometer as shown below.

Fig 3.5

lg — Current through galvanometer at full scale defectionR = Resistance of multiplierr = Galvanometer resistanceV1 = Potential difference across multiplierV2 = Potential difference across galvanometerV = Voltmeter reading of the converted galvanometer (Total p.d across circui';

Potential difference across multiplier, V1 = lg x R

Potential difference across galvanometer V2 = lg x r

Voltmeter reading = p. d across multiplier + p .d across galvanometer

V = V1 + V2

V = lgR + lgr

V = lg (R + r )

Rearranging, lg (R + r ) = V

V VR + r = — ••• Multiplier resistance R = ----- r

‘g lg

Example 10A moving coil meter with an internal resistance of 100H has a full scale deflection when a current of \0mA flows through it. What value of resistance would convert it to read lOFat full scale deflection? WAEC 199555Solution

Fig 3.6

369

Galvanometer resistance r = 100ft; voltmeter reading, 7 — 107Current at full scale deflection lg = 10mA = 0.01/4

Multiplier resistance R = j - — r = ^ — 100 = 1000 — 100 = 900ft

example nThe maximum permissible current through a galvanometer G of internal resistance is 0.05/4. A resistance R is used to convert G into a voltmeter with a maximum rea mg o 1007. Find the value of R and how it is connected to G.A. 20,000 ohms in parallel B. 19,900 ohms in series ^C. 1990 ohms in series D. 100 ohms series. JAMB 1Solutionr = 1 0 ft; Jg = 0.05/4 ; 7 = 1007

vMultiplier resistance R = -— r

R = - 10 = 2000 - 10 = 1990ft in series0.05

POTENTIOMETER. WHEATSTONE BRIDGE. METER BRIDGEL Potentiometer A potentiometer is used to measure resistances of wire and e. m. f of cells as shown below.

Fig 3.6

E2

For the potentiometer above,

Where £ , - e. m. / o f cell A E2 = e . m . f of cell B lt = Balance point of cell A l2 = Balance point of cell B

e1 = ia

Ez lz

Example 12The balance length of a potentiometer wire for a cell of e. m . f 1.637 is 85cm. If die cell is replaced by another one of e . m . f 1.077, calculate the new balance length.

NECO 200744Solution

E2 = 1.637, = 85cm ; E2 = 1.077, l2 =?

370

l;or ,i potentiometer. a± = !±Ei h

1.63 85 1.07 x 85New balance length l2 = — — 55.80cm

II. H hcatstone Bridge. The Wheatstone bridge shown below is used to accurately measure resistance. ,

Fig 3.7

B ' At balance point. B and D arc ut the same potential, therefore the potential difference between B and D is zero For resistances Rv R2, R3, and /?4 connected as shown above,

Et _r 2 ~ r 4

Example 13

Fig 3.8

In the diagram above, the galvanometer indicates a null defection. Calculate(i) The value o f R(ii) The potential difference between P and Q NECO 200543Solution

P

371

1

... „ /?! R3(0 From the digram, —- = — becomes

R 2

4 2 3 X 2 6_ — _ o — ,------ = — = 1.5113 /? 4 4

(ii) The potential difference between P and Q is zero at null defection or balance pom

Example 14

Fig 3.10

SolutionAns: 0

In the diagram below, the galvanoirieter indicates a null deflection. What is the potential difference between P and Q! / WAEC 199347

III. Meter Bridge. The meter bridge is a special application of a wheatstone bridge and is also used for measuring resistance accurately.

Fig 3.10

For the meter bridge above at null defection,R t— = — If l i is know n, th e n lz = 100 - L"7 If. 1

Example 15In a meter bridge experiment, there is a zero defection of the galvanometer when Zt = 4 0 cm as shown in the diagram below. Calculate the value of the resistance R.

WAEC 1 9 8 9 45io n R

Fig 3.11

372

Solution

Given: /, = 40cm

. > . •, 1 MI1 •

/j = 100 - /i = 0 1 o II ON o n S

From the diagram, R> = h ' R2 lz

becomes

1 Oil 40cm « ' ■ > \ • ' 1

/? 60cmlOfl x 60cm 10n X 60

Therefore, R = 40cm— = 40

,1|'

isn

Example 16In a meter bridge experiment, two resistors 2fl and 3fl occupy the left and ng t gaps respectively. Find the balance point from the left side of the bridge. 43A. 20cm B. 40cm C. 60cm D. 80cm JAMB 1997Solution

2n 3n

Balance point from left side of bridge.

ZxFrom the diagram, — = — becomes

R2 lz

h3 1 0 0 - / !

Cross multiply, 2 x (100 — l1) = 3/x

200 - 2/a = 3Zi

200 = 3Za 4- 2/j

. i- 200 = 5/x■ ■ " ■ - ■ 200- ■ : .».«■'()•• ■ S " i ■ .i h = — ~ 40cm . ... .

■ ! . J ■ ; . < i".- I. to , \ . ■ ■ *. - *•i o <- i.' !: 1 iio i, . / .. . ■ . . • ' < •

EXERCISE 3 . , Vj1. Calculate the length pf a constant wire of cross sectional area 4tt x 10~8m 2 and resistivity 1.1 x 10_6n m required to construct a standard resistor of resistance 2 i n .( Take n as y ) W A E C 199244 Ans: 2.4m2. A given wire of resistance 1 On has a length of 5m and a cross sectional area of4.0 x 1 0 '3m 2. Calculate the conductivity of the wire. • , ;

W A E C 199550Ans: 1.25 x l O ^ n T 13. A wire, 1.0m long and with cross sectional area 2.0 x 10-7m 2 has a resistance of0. in. Calculate the electrical conductivity of the wire. ; ,

W A E C 200838Ans:5.0 X l O ^ l ^ m " 1

373

4.of 0.1 Oil. Calculate its electrical conductivity.

WAEC 199646 Ans: 2.0 x

2.0 x 10' 3cm2 has a resistanceA wire of length 100cm and cross sectional area

I 0 ' 6i lo 7 crosT sectional area 0.7mm

cm - t^ ...... is 4.9 x

5. The resistivity o f a given wire of cross sectional 10_4(lmm. Calculate the resistance of a 2m length of the 'r^oQ /lPR3 Ans: 1.4(1

WAEL g o x 10” tti

6. A resistance wire of length 2m and o f uniform cr0Sq*34t' ^ lS. 5 5 x 1 0 _7flmhas a resistance of 2.2il. Calculate its resistivity. WAEC 19 resistivity o f7. A piece o f resistance wire of diameter 0.2mm and resistance

. ( 22\8.8 x 10 7(lm, calculate the length o f the wire — 7 )

WAEC 2002PB3Ans: 2.5i X8. A 0.6i2-resistor is made from a wire o f length 4m and resistivity • __s 2What is the cross-sectional area of the wire? NECO 2008 Ans. .9. A piece of resistance wire of diameter 0.2m and resistance 7 as r8.8 x 10~7(lm, Calculate the length of the wire. -

WAEC 2003PR3 Ans: 2.5 x 10sm10. The resistance of a piece of wire of length 20m and cross sectional area10-6m 2 is A. l.Oil B. lO.Ofi C. 0.5(1 B. 5 .00 [resistivity of the w i r e - 10_7(lm] JAMB 200518 Ans: l.O il _6 2 .11. The resistance of a 5m uniform wire of cross sectional area 0.2 X 10 m is0.425(1. What is the resistivity of the material of the wire?A. 1.10 x 10_6(lm B. 4.25 x 10_6(lm C. 2.40 x 10"7(lm D. 1.70 X 10 ilm

JAMB 198 935 Ans: 1.70 X 108(lm12. A conductor has a diameter of 1.00m and length 2.00m. If the resistance o f thematerial is 0.1(1, its resistivity is A. 2.55 x 102f1m B. 2.55 X 10s(lmC. 3.93 x 10_8(lm D. 3.93 x 1 0 -6n m JAMB 200735 Ans: 3.93 x 10_8(lm13. A milliammeter with full scale defection of 10mA has an internal resistance o f5il. It would be converted to an ammeter with a full scale deflection o f 1A by connectin areisstanceof A. —- ohm in series with it B. — ohm in parallel with it

9 9 9 9 K9 9 . . 99

C. — ohm in parallel with it D. — ohm in series with it E. 2 ohm in series with itJAMB 198320 Ans: B

14. An ammeter of resistance 0. l i l has a full scale deflection of 50mA. Determine the resultant full scale deflection of the meter when a shunt of 0.0111ft is connected across its terminals. A. 400mA B. 450mA C. 500mA D. 550mA

JAMB 198943 Ans: 500mA15. A 0-10mA galvanometer with a coil resistance of 3011 can be converted to a 0- 10A ammeter by using A. 0.03(1 series resistor B. 0.03(1 shunt resistorC. 9.99(1 shunt resistor D. 9.99(1 series resistor JAMB 199045 Ans* B16. A moving coil galvanometer of 300(1 resistance gives full scale deflection for 1.0mA. The resistance, R, o f the shunt that is required to convert the galvanometer intr, „ 3.0A ammeter is A. 899.70(1 B. 10.00(1 C. 0.10(1 D. 0.01(1

JAMB 199137 Ans: 0.10(117. An ammeter o f resistance 5(1 has a full scale deflection when a current o f 50mA flows in it. The value of the resistor required to adapt to measure a current o f 5 4 ‘

A. 19.80(1 B. 5.00(1 C. 0.05(1 D. 0.25(1 1SJAMB 2 0 0639 Ans- 0 05(1

18. A galvanometer with a full scale deflection of 1.5 x 10_3A has ao f 50(1. Determine the resistance required to convert it into a voltmeter readreS1StanCe1 S V ■ WAEC 200148 Ans: 9 S 0 lT 8 ^ t019. A milliameter of resistance 5(1 and full scale deflection o f 50mA is to be measure a potential difference of 50T. What should be the resistance o f th- ™ u r ^

A. 95(1 B. 110n C. 595(1 D. 995(1 E. 1005(1 multiplier?

374

J A M B 198028 Ans: 995ft

20. The galvanometer G in the diagram below has a resistance scale deflection for a current of 10mA

Fig 3.13

of 5ft and gives a full

If the galvanometer is to be used as an ammeter reading up to 2A, which of the followingequations is correct? A. 2R = 20 B. 1.99 x R = 0.01 x 5C. 2 x R = 0.01 x 5 D. 1.99 X 2 R = 0.01 x 5 E. 1.99 x 2 R = R - 0.01

NECO 2000so Ans: B21- The value of the series resistance required to convert the galvanometer below into a voltmeter with a full range deflection of 15F is

0.015Af-------- 0 —

5ft

Fig 3.14 *----------------- 15P------------------►

A. 985 ohms B. 990 ohms C. 995 ohms D. 1,000 ohms E. 1,005 ohmsJAMB 198230 Ans: 995ft

22. A galvanometer of internal resistance 50ft has a full scale deflection for a current of 5mA. What is the resistance required to convert it to a voltmeter with full scale deflection of 10F? A. 1750ft B. 1950ft C. 2000ft D. 2500ft

JAMB 199741 Ans: 1950ft23. A potentiometer wire carrying a steady current is lm long with a standard cell o f e .m .f 1.1V suitably connected, a balanced length of 44.0cm was obtained. What is the e. m. f o f a cell which gives a balance length of 68.0cm?

NECO 200048 Ans: 1.7F24. Calculate the value of R when G shows no deflection in the circuit illustrated below.

Fig 3.15

WAEC 198844 Ans: 36fl25. When a known standard resistor of 2.0ft is connected to the 0.0cm end of a wirebridge, the balance point is found to be at 55.0cm. What is the value of the unknown resistor? A. 1.10ft B. 1.64ft C. 2.44fi D. 27.50ft

JAMB 1990®8 Ans: 1.64ft

375

26. In (he diagram below! the galvanometer indicates a nu potential difference between X and Y?

W hat is the

Fig 3.16

27. What is the

WAEC 198 9 42 Ans: 0 value o f R when G shows no deflection in the circuit illustrated

Fig 3.17

28.

Fig 3.17

Calculate the resistance R required to balance the bridge in the circuit above

2d.

fig 3.18

R 2.Q

I he diagram above shows a meter bridge n which lw< of the arm-nid 2C. A balance point is obtained at 6 0 m i fiom thi . \ p . ,A. I (> U. 1.3iZ C 3.Oil IV .'>.00 JAMB L ' W

contain resi leu IAns: CU'.

R

7C

30.

'/Oil

Fig 3.186 0cm

--------1____ ] -----

Z---------- Il__________________________

. _ - ------— ivugui m i . inc vuiuc 01 n ui duiuhuu jtuiiii * 1»A. 3.012 B. 13.3ft C. 15.0H D. 30.012 JAMH 199442 Ans: 30.0i2 7 qn ^ 'V-4C reS‘Stivity ^ * 10-5^ cni has a cross-scctional area o f

x 0 cm Calculate the length of this wire that will be required to make a 4.0ft resistor. A c<r. „n,wl7o A , 0__WAEC 2009 Ans: 68.18cm

. 377

4ELECTROLYSIS, MAGNETIC AND

ELECTROMAGNETIC FIELDSELECTROLYSIS . , -dElectrolysis is defined as the chemical change (decomposition of compoun s; m a due to the flow of current.

Faraday’s first law of electrolysis states that the mass (M) of a substance librated or deposited during electrolysis is directly proportional to the quantity of e ectncity ( ) passing through the electrolyte.

That is, M oc Q Or M - Z Q

Remember, quantity of electricity Q = It

Therefore, M = Zlt

Where / = current in amperes (A)Z = Electrochemical equivalent kgC-1 or gC~l

Also, the mass of the substance deposited during electrolysis depend on the density ( p ) of the substance, the surface area (A) of the cathode and the thickness (d ) of the substance deposited on the cathode.

mass—------ Mass = density x area x thicknessvolume

M = p x A x d M = Zlt

p x A x d = Zlt

Faraday’s second law of electrolysis states that when the same current passes through different electrolytes for the same time, the masses of the elements deposited or librated are proportional to their chemical equivalent.

Mass of elment A Chemical equivalent of A Mass of elment B Chemical equivalent of B

Electrochemical equivalent (e. c. e) of a substance is defined as the mass of the substance deposited during electrolysis by one coulomb of electricity. (One ampere of current flowing for one second). Copper has an (e.c.e) of 0.00033pC_1 because when a current of 1A flows through a copper solution for lsec, it deposits 0.00033p of copper.

M M pAd

Example 1Copper of mass 0.33g is liberated by passing 100 coulombs of electric charge through a copper voltameter. Calculate the mass of copper deposited when a current of 2A is passed through the voltameter for 5minutes NECO 2007£7

378

Solutione.c. e ,Z = 033.9/1000C = 3.30 x \Q~Ag /C

C urrcnt / = 2A ; time = 5min = 5 x 60 = 3005 ;

Mass deposited M = Z l t = 3.30 x 10~4 x 2 x 300 = 0.198.9

Example 2 „A charge of one coulomb liberate 0.0033.9 of copper in an electrolytic process. How long will it take a current of 2A to liberate 1.98,9 ° f copper in such a process?A. 5m inu tes B. 30m inutes C. 50minutes D. 6Ominutes E. 120m m utes

JAMB 198535Solutione. c .e ,Z = 0.0033gC~l ; current / = 2A ; mass liberated M = 1.98#

M 198From M = Zlt, time t = — = ------ ---------= 300s = 5min

Zl 0.0033 x 2

Example 3In an electrolysis experiment, a cathode of, mass 5g is found to weigh 5.01 a after a current o f 5A flows for 50 second. What is the electrochemical equivaler >f the deposited substance? A. 0.00004^/C B. O.OOOO29/C C. 0 .02500^/6D. 0.05000,9/C E. 0.00001^/C JAMB 197930SolutionMass deposited M = 5.01,9 ~ 5,9 = 0.01 g; current / = SA ; time t = 50s

MFrom M = Zlt, electrochemical equivalent Z = —

0.01Z = - — — = 0.0004^/C or 4 x 10- V : -1

j X j U

Example 4Copper o f thickness d is plated on .the cathode of a copper voltameter. If the total surface area o f the cathode is 60cm2 and a steady current of 5.0i4 is maintained in the voltmeter for 1 hour, calculate the value of d. [density of copper = 8.9 x 103k g m ~ 3] [electrochemical equivalent of copper = 3.3 x 10~7kgC -1] WAEC 2007£S SolutionArea, A — 60cm2 = 60 x 10_4m2 ; current, / = 5/1;Time, t = lhour = 3600s; e .c .e .Z = 3.3 x 10~7/t£C-1Density p = 8.9 x 103k g m ~ 3 ; thickness d =?

From p x A X d = Z lt

Z ltThickness d ----------p x A

3.3 x 10"7 x 5 x 3600 d ~ 8.9 x 103 x 60 x 10-4

= 1.11cm

5.94 x 10~3 53A = 1.11 x 10“4m

Example 5When a current o f 1.0/1 flows in an electrolyte for 3.0/t, a mass of 1.0g is deposited. What mass will be deposited in 10.0/z if a current of 3.0/1 flows through the electrolyte?

NECO 200246

379

Solution _Firstly: 1 = 1.04 ; t = 3 hours = 10800s ; Secondly: t = 10 hours = 36000s, —

Firstly:M

Z = — =

Secondly: M = Z lt = 9.26 x 10"s x 36000 x 3 = 10#

example osteadv^cuiT^Ctr<!J^SIS °^copPer 0 0 tetraoxosulphate (vi) solution, an ammeter shows a Calculate the e ^ ^ m'nutes while 6.6 X 10~4kg o f copper is liberatedis 3 30 x i q— C ammeter reading. (The electrochemical equivalent o f copperSolution 9 } WAEC 1997”

m ™ = 6^ x 10- r ' *ime>i = 30 min = 3 0 x 6 0 = 1800s ; o-b X 10 Z = 3 .3 0 x l0 - 7kgC~'

M 6.6 x 10"From M = Zlt, current used / = - = 3,30 x 10-7 *1800- = 1.11A

Ammeter error = current used - ammeter reading = 1.11 — 1-0 — O .lllA

Example 7What is the mass of hydrogen liberated in an acidulated water voltameter if a current of 6A flows through the voltmeter in 1 hour?[electrochemical equivalent of silver = 0.001118gC_1][equivalent weight of silver = 108]SolutionCurrent / = 6A ; time t = 1 hour = 3600s

Quantity of electricity used Q = It = 6 x 3600 = 21600C

1C of electricity releases 0.0011180 of Ag

21600C of electricity will release x of Ag

x = 21600 x 0.001118 = 24.15,9 of Ag

Mass o f Ag _ Chemical equivalent o f Ag Mass o f H2 Chemical equivalent o f H2

24.15 _ 108mass o f H2 1

24.15Mass o f H2 = — — = 0.224#

E X E R C IS E 4A1. A cun-ent of 2A is passed through a copper voltameter for 5minutes. If theelectrochemical equivalent of copper is 3.27 x 10” kgC~l , determine the mass of the copper deposited. WAEC 2001*3 Ans: 0.19620 or 1.96 X 10~*kg2. During the electrolysis of copper (ii) tetraoxosulphate (vi) solution, a steady current of 4.0 X 102A flowing for one hour liberated 0.480 of copper. Calculate the mass of copper liberated by one coulomb of charge.

WAEC 2000£9Ans: 3.33 x 1O~703. In copper plating, a current of 0.54 is allowed for a cathode area of 100cm2. If this current is maintained constant for 100 minutes, the thickness of the copper deposited

380

Will he approximately [the electrochemical equivalent of copper - ° .° 0 0 > s /C dens,t> of copper = 1 0 \ g / a n 3] A. 10_2mm B. 10-3mm C. 10"4»mn D.lO rnm E. 1 0 -6mm JAMB 198231 Ans: 10 mm^ The electrochemical equivalent of silver is 0.0012^/C. If 36.0,9 ^ f silver i deposited by electrolysis on a surface by passing a steady current for . minu cs. current must be A. 6000/4 B. 100/4 C. 10/1 D. 1/4

JAMB 198944 Ans: 100/45. The electrochemical equivalent of platinum is 5.0 x 10 7kgC To plate out 1 0 kg o f platinum, a current of 100/1 must be passed through an appropriate vessels orA. 5.6 hours B. 56 hours C. 1.4 x 104 hours D. 2.0 x 104 hours

JAMB 199144 Ans: 5.6 hours6. The electrochemical equivalent o f a metal is 1.3 X 10 kgC C The mass ot the metal which 2.0 x 104C of electricity will deposit from a suitable electrolyte is _A. 6.5 x 10~2kg B. 2.6 x 10~2kg C. 6.5 x 10~3kg D. 2.6 x 10 k g

JAMB 199847 Ans: 2.6 x 10 3kg7. A current o f 0.5/1 flowing for 3hours deposits 2g o f a metal during electrolysis.The quantity o f the same metal that would be deposited by a current o f 1.5/1 flowing in lhouris A. lOg B. I8 9 C. 2g D. 6g JAMB 200240 Ans: 2g8. 2.1g o f silver is deposited when a current o f 0.5/4 is passed through a voltameterfor 60 minutes. Calculate the mass of silver that will be deposited when a curre"t ot 1.0/1 is passed through it for 15 minutes. NECO 2008E14 Ar 1.05g9. In the calibration of an ammeter using faraday’s law of electrolysis, the ammeter reading is kept constant at 1.20/1. If 0.990g of copper is deposited in 40 minutes, the correction to be applied to the ammeter is A. 0.03/1 B. 0.04/1 C. 0.05 D. 0.06/4 [e.c.e o f copper = 3.3 x l O ^ g C ' 1] JAMB 2 0 0 339 Ans: 0.05A10. During electrolysis, 2.0g o f a metal is deposited using a current o! 0.5/4 in 3hours. The mass of the same metal which can be deposited using a current o f 1.5/4 in 1 hour is A. 3.0,9 B.2.0g C. 0.5g D .l .Og JAMB 200649 Ans: 2.0#11. A cathode of mass 5g weighs 5.01,9 after a current of 5/1 had been passed through the voltameter for 50 seconds. The electrochemical equivalent o f the metal is

NECO 2 0 0 743 Ans: 4.0 x 10~sg /C12. 118.8cm2 surface of the copper cathode of a voltameter is to be coated with 10_6m thick copper of density 9 x 103k g l 0 ~ 3. How long will the process run with 10/4 constant current? A. 15.0 min B. 5.4 min C. 20.0 min D. 10.8 min[e.c. e o f copper = 3.3 x 10~7kgC~3] JAMB 200741 Ans: 32.4s or 0.54 min

13. In an electrolysis experiment, the ammeter records a steady current o f 1/4. The mass o f copper deposited in 30 minutes is 0.66,9. Calculate the error in the ammeter reading.[Electrochemical equivalent of copper = 0.00033^C _1]

WAEC 2008F2 Ans: 0.11/414. A steady current o f 1.2A is passed through a copper voltameter for 30s.Calculated the mass of the copper deposited. [Electrochemical equivalent o f copper 3.3 x l O - ^ C “ 1] NECO 200945 Ans: 1.19 x 10~2,915. An electric charge of 9.6 X 104C liberates 1 mole of a substance containinu6.0 X 1023 atoms. Determine the value o f the electronic charge.

WAEC 2009E6 Ans: 1.60 X 1 0 - iqC

381

M AGNETIC FIELDMagnetic field is a region surrounding a magnet within winch “ y jn ., magncticother magnetic substance. When a charge q moves with a Jfield B , the magnetic force F is given by

F = qVB Sin0

Where F = magnetic force in Newton (A/)V = Velocity in m /sB = Magnetic field in Tesla (7) or W eber/m2 6 = Angle between V and B

Note: 1. If V and B are in the same direction or parallel to each other, 6 — 0 or 180 Therefore, F = 0

2. If V and B arc perpendicular or at right angle to each other then 6 — 90, Therefore, F is maximum.

The force F on a conductor of length / carrying a current / in a magnetic field B is given by;

F = Bll SindF is zero when B and / are parallel to each other. F is maximum when B and / are perpendicular.

Example 1A particle of charge 5C moves perpendicularly to a magnetic field of magnitude 0.017'. II the velocity of the charge is 1.5ms-1, calculate the magnitude of the force exerted on the particle. WAE C 199341

SolutionCharge q = 5C ; magnetic field B = 0.017 6 = 90 (perpendicularly)Velocity V = 1.5ms-1 ; magnetic force F =?

F = qVBsin 6 = 5 x 1.5 x 0.01 x sin90 = 0.075/V

Example 2A proton moving with a speed of 1.0 x 106m s-1 through a magnetic field of 1 07 experiences a magnetic force of magnitude 8.0 x 10-4 /V. The angle between the proton’s velocity and the field is A. 45 B. 30 C. 60 D. 90°(proton charge = 1.6 X 10-19C)SolutionV = 1.0 x 106m s-1 B = 1.07 ; F = 8.0 X 10~147V,

F = qVBsind

F 8.0 x 10-14 8.0 x IO-14Sin0 = qVB = 1.6 X 10-19 x 1.0 X IO6 X 1 ~ 1.6 x 1 0 ^

sin# = 0.5 6 = sin-10.5 — 30”

JAMB 20056

q = 1.6 X 10 10C

Example 3An electron enters a magnetic field of flux density 1.57 with a speed of 2.0 x 107mv 1 at an angle o f 30° to the field. Calculate the magnitude of the force on ih ' electron.[Electron charge, e = 1.6 x 1019C] NEGO 2006£u

3 8 2

SolutionFlux density (magnetic field), B = 1.57; V = 2.0 X 1 0 7ms ,

6 = 30; cj = 1 . 6 x 10_19C

Magnetic force F = qVBsmQ = 1.6 x 1 0 '19 X 2.0 x 107 x 1.5 X sin30°N

= 2.40 X 10-12

Example 4A conductor of length 2m carries a current of 0.8/1 while kept in a magnetic field of magnetic flux density 0.57. The maximum force acting on it is r0A. 8.0/V B. 3.2N C. 0.8N D. 0.2N JAMB 2000SolutionLength l = 2m ; current / = 0.8/1; magnetic flux density B = 0.57\

Magnetic force F = Bll = 0.5 X 0.8 X 2 = 0.8N

EX ERC ISE 4B1. A particle of charge q and mass m moving with a velocity V enters a uniform magnetic field of strength B in the direction of the field. The force on the particle isA. qVB B. mqVB C. qVb/m D. mVB/q E. 0 WAEC 19 9 444 Ans: 02. A charge of 1.6 x 10~19C enters a magnetic field of flux density' 2.07 with a velocity of 2.5 x 107ms~1 at an angle of 30° with the field. Calculate the magnitude of the force exerted on the charge by the field. WAEC 2003£14 Ans: 4 X 1 0 "12/V3. If a charged ion goes through combined electric and magnetic fields, the resultantemergent velocity of the ion is A. B. EB C. ^/g D. E — B

JAMB 200032 Ans: A4. The diagram below shows a closed square box of side 0.5m in a uniform electric field E in the direction shown by the arrows. What is the flux </> for the boxA. 0.5E B. 2.0E C. 0.27 D. 0.07 JAMB 200130 Ans: A

5. The force on a charge moving with velocity V in a magnetic field B is half o f the maximum force when the angle between V and B is A. 90 B. 45 C. 30 D. 0

JAMB 2 0 0 3 43 Ans: 45°6. Two long parallel wires X and Y carry current 3A and 5A respectively. If the force experienced per unit length by X is 5 x 10“5yV, the force per unit length experienced by Y is A. 5 x 1 0 ~SN B. 5 x 1 0 '4/V C. 3 x 1 0 -6 7 8 9/V D. 3 x 10~S7V

JAMB 200738 Ans: D7. A particle o f charge q and mass m moving with a velocity V enters a uniform magnetic field B in the direction of the field. The force on the particle isA. BqV 13. BqV/m C. BqVm D. mVB/q E. mV/Bq NECO 200045 Ans: 08. Calculate the magnitude of the force due to a magnetic field of 0.407 on an electron moving perpendicular to the field with a speed of 4.00 x 106m s_1.[e = 1.60 x 10-19C] NECO 200548 Ans: 2.56 x 10~13/V9. A charge of 1.6 X 1 0 "19C moving with a velocity of 2.5 x 107m j_1 enters a magnetic field o f flux density 2.07 at an angle of 30° with the field. Calculate the magnitude o f the force on the charge. NECO 200845 Ans: 4.0 x 10~12N

3 8 3

10. A band of 500 rectangular loops of wire of area 20cm by ’ . ^of magnetic field which changes from 1.0T to 0.4T within 5 secon s, induced e.m.fi. * „ . nA. 5.60V B. 24.00V C. 0.24V D. 2.40V JAMB 200 ^ ^

ELECTROMAGNETIC FIELDElectromagnetic field is a field arising from the combined interaction of magnetic le and electric field. Electromagnetic induction is the generation of electric current or voltage in a conductor as a result of a relative motion between a conductor and a magnetic field.

Induced electromotive force (e .m .f ), E, in a straight conductor of length /, moving with a velocity V, perpendicular to a field of magnetic induction, B is given by;

E = BIVAlso the e. m. f generated by an a. c generator is given by

E = NABa)E = NABoosinG

Where B = Magnetic field strength (T or weber/m2) l = Length of conductor (m)

V = Velocity (m s-1)N = Number of turnsA = Area of coil (m2)co = Angular velocity (rad s -1)9 = Angle of inclination in degrees

Example 1A circuit has an area of 0.4m2 and consist of 50 loops of wire. If the loops are twisted and allowed to rotate at a constant angular velocity of 10 rad s -1 in a uniform magnetic field of 0.47*, the amplitude of the induced voltage is A. 8 V B. 16V C. 20V D. 80V JAMB 200341

SolutionNumber of turns N = 50; area of coil A = 0.4m2 ;

Angular velocity co = 10 rad s - 1 ; magnitude of magnetic field B = 0.4T

Induced e. m . f or voltage E = NAB to

E = 50 x 0.4 x 10 x 0.4 = 80V

TRANSFORMERSA transformer is a device for changing the voltage of an a.c supply. A step up transformer converts low voltage to high voltage. A step down transformer converts high voltage to low voltage.

The induced e .m .f in the secondary coil and in the primary coil of a transformer are related to the number of turns in the coil by the equation.

Secondary e. m . f (E5) Number o f turns in secondary coil (!VS)Prim ary e. m. f (Ep) Number o f turn in prim ary coil (Np)

384

Also, power in secondary coil = power in primary coil.

That is, Ps = PpOr ls x Es = Ip x Ev (from P = l ' 1')

Rearranging, ^ = y ............................S

From equation (1) and (2) above, we obtain

Ns_NP

Where IP - current in primary coil ls = current in secondary coil

(2 )

From, . power outputefficiency = -------------------x 100,

power inputwe obtain

. power in secondary coilTransformer efficiency, € = ---------------- ----------- r r x 100power in primary coil

£ = ^ - x 100Ppls x E s

e = —— ^ - x loo

e =

Ip x EPIs * N s/ p x Np

x 100

Turns ratio = —~NsNP

If Ns/N p >1, it’s a step up transformer.I f Ns/Np < 1, it’s a step down transformer

Example 2A transformer is required to give 120F from a 240V mains supply. If the primary has 5500 turns, how many turns has the secondary? NECO 200351SolutionEs = 120V, EP = 240V, NP = 5500, Ns =?

E1 = N± Ev Np

N c =Es x N„ 120 x 5500

240 = 2750

Example 3A transformer with 5500 turn in its primary is used between a 240V a. c supply and a 120V kettle. Calculate the number of turns in the secondary. WAEC 198842SolutionNp = 5500; Ec = 120V; Ep = 240V; Ns =?

N«F r o m ' T p = V p ’

Nc =E« x N„ 120 x 5500

240 = 2750

Example 4The current in the primary coil of a transformer is 2.5/4. If the coil has 50 turns and the secondary 250 turns, calculate the current in the secondary coil. (Neglect energy losses in the transformer). WAEC 199749SolutionIp = 2.5/4, Np = 50, Nt = 250, ls =?

385

From ^ L = '± N„ / , ' current in secondary coil is,

50 x 2.5 250

0.5/1

Example 5A 95% efficient transformer is used to operate a lamp rated 60W, 220V from a 44001/ a. c supply. Calculate the

(i) ratio of the number of turns in the primary coil to the number of turns in thesecondary coil of the transformer. £14

(ii) current taken from the mains circuit. WAEC 2007SolutionEfficiency 6 = 95%, Ps = 60W, Ep = 4400, Es = 220V

(i) ^2. = ÜJL . NpNs Es " N.

4400220

= 20

(ii) e = - f - x lo o PsIp x Ep

x 100

. , Ps x 100Main current /„ = —---------

P e x Ev60 X 100 6000

95 X 4400 ~ 4180000.01435/1

lp = 0.014/1 Or 14mA

Example 6A transformer supplies 12K when connected to a 2401/ mains. If the transformer takes0.120A from the mains when used to light a 12V, 241V lamp, what is its efficiency?

NECO 200250SolutionEs = 12V, E„ = 240V,

Efficiency, 6 =

lp = 0.120/1

24Ip x Ep

x 100 =

Ps = 24 W,

24000.120 x 240 x 100 = ■

28.8

6= ?

= 83.33%

Exam ple 7A transformer which can produce 8F from a 240F a.c supply has an efficiency of 80%. If the current in the secondary coil is 15/4, calculate the current in the primary coil. A.0.625/4 B. 1.600/1 C. 2.500/1 D. 6.250/1 JAMB 199846SolutionEfficiency G— 80%, Es = 8V, Ep = 240V, ls = ISA, Ip =?

Is x Es

Ip * Ppx 100

Current in prim ary coil, lpls x Es x 100

e x Ep15 x 8 x 100 _ 12000

240 x 80 ~ 19200 0.625/1

Exam ple 8A 40KW electric cable is used to transmit electricity through a resistor of resistance 2-0fi at 800F. The power loss as internal energy is A. 5.0 x 102W B. 4.0 x 103W C.5.0 X 102W D. 4.0 X 102W JAMB 200742SolutionPower transmitted, P = 40KW = 40,000M/; resistance R = 2.0ii,

voltage V = 8001/P 40000 _

From, P = ' V>current , = v = “^ 5 " ~ 5M

386

Power loss, I2R = 502 x 2 = 5001V = 5.0 X 103 W

E X E R C IS E 4C*• The magnetic flux in a coil having 200 turns changes at the time rate o f 0.08 Wb The induced e .m . / in the coil is A. 1.6V B. 16.0 V C. 25.0 V D. 250.0V

JAMB 200043 Ans: 16.0V2- A transformer with 5500 turns in its primary winding is used between a 240 a. c supply and 120 V kettle. Calculate the number of turns in the secondary winding.

WAEC 199451 Ans: 27503. A house is supplied with a 240 a. c mains. To operate a door bell rated at 8 V, a transformer is used. If the number of turns in the primary coil o f the transformer is 900 calculate the number o f turns in the secondary coil of the transformer.

WAEC 199149Ans: 304. If a transformer is used to light a lamp rated at 60W , 220V from a 4400V a .c supply, calculate the

(i) ratio o f the number of turns of primary coil to the secondary coil in the transformer.

(ii) current taken from the mains circuit if the efficiency of the transformer is 95%WAEC 1992s3 ( i )2 0 : l (ii)0.01AA

5. A transformer has 400 turns and 200 turns in the primary and secondary winding respectively. If the current in the primary and secondary winding are 3A and 5A respectively, calculate the efficiency of the transformer.

WAEC 200242 Ans: 83.33%6. A voltage and current in the primary of a transformer are 200V and 2A respectively. If the transformer is used to lighten 12V, 30W bulbs, calculate its efficiency.

WAEC 199936 Ans: 75% Hint: Ps = 10 x 30W = 300W7. A transformer is rated 240V. If the primary coil is 4000 turns and the secondaryvoltage is 12V, determine the number o f turns in the secondary coil. A. 250 B. 100 C. 150 D. 200 JAMB 20051 Ans: 2008. A transformer has 300 turns of wire in the primary coil and 30 turns in the secondary coil. I f the input voltage is 100 volts, the output voltage is A. 5 volts B. 10 volts C. 15 volts D. 20 volts E. 25 volts JAMB 198147 Ans: 10 volts9. A transformer has a primary coil with 500 turns and a secondary coil with 2500 turns. When the voltage input to the primary coil is 120V, the output isA. 6000V B. 600V C. 240V D. 60V E. 24V JAMB 198546 Ans: 600V10. The primary winding of a transformer has 400 turns and its secondary has 100turns. If a source o f e . m . f o f 12V is applied to the primary, the secondary e . m . f will be A. 3V B. 6V C. 24V D. 48V JAMB 199547 Ans: 3V11. The primary coil o f a transformer has N turns and is connected to a 120V a .c power l<ne If the secondary coil has 1,000 turns and a terminal voltage of 1,200 volts what is the value o f JV? A. 120 B. 100 C. 1000 D. 1200

JAMB 2 00143 Ans: 10012. A voltage o f 240V is connected to the primary coil o f a transformer. Calculate theratio o f the primary turns to the secondary turns if the voltage available at the secondary coi, is , 5V NECO 2005so Ans: 1:16 or 0.0613 A step down transformer has 300 turns in the primary coil and 30 turns in the secondary coil. If the input voltage is 100 volts, what is the output voltage?

387

NECO 20074“ Ans: 101/ (14. A transformer is required to supply 12Vrms to operate a toy train set from a2401^ ms If the number of turns in the secondary coil is 100, calculate the number of turns required in the primary coil. WAEC 2 0 0 843 Ans: 200015. A step down transformer has a power output of 50W and efficiency of 80%. If the mains supply voltage is 200K, calculate the primary current of the transformer.A. 0.31 A B.3.20A C. 3.40/1 D. 5.00/1

JAMB 200843 Ans: 0.31/t % W in t:e= ^-x too

16. A transformer has an efficiency of 92.5%. The ratio o f the numbers o f turns in the primary coil to that in the secondary coil is 128:45. If the current passing through the secondary coil is 9.0A, calculate the current passing through the primary coil.

' NECO 20CJ949 Ans: 3.42A17. A transformer with a primary coil of 800 turns and a secondary coil o f 50 turns has its primary coil connected to a 240V a.c mains. If the current passing through the primary coil is 0.5A, calculate the

(i) potential difference across the secondary ends.(ii) current passing through the secondary coil assuming no power losses.(iii) Power in the secondary coil if 10% of that in the primary coil is lost.

NECO 2009EI4 Ans: (i) 15V (ii) 8.0A (iii) 108W

■ . r.i >)f < . •''.ill >/ <! t ■ '!• > i . -i. •

)! . •. * i, i ,; ".‘ A;\ lirv/ iiw i i.mov r bin

, t.j><;i- . i '.000 ...ii/' O',, i,'J . r•j ,w ■

388

5 ->ril '■

VJOSS =

SIMPLE A.C CIRCUIT ^ ^ byAn alternating current (a .c ) changes its direction periodica y are reppSKfWJ* 3s alternating voltages. Alternating currents and alternating vo g jnoTiua or!!follows. i ri§u

/ = I0S m 2 n ft or / = /0Sinott / tnaTiuo odJ lo tnolBvmpu

V = K,Sin27r/t or V = F0SinoJt

Where / = Instantaneous current (j4)la — Peak or maximum cu rren t'll)/ = Freqtieniy ( ti'z )" '- t = Time (s)

OJt = 8 = Phase angle o f current or voltage V = Instantaneous voltage (V)V0 — Peak or maximum voltage (F)

PEAK AND ROOT MEAN SQUARE =r = zn'c tThe peak value o f an a. c is the maximum value of the current r£c6rde a • c CY The r .m .s current is defined as the steady or direct current (d .c ) which P ^ ^ jq ^ g /^ same heating effect per second in a given resistor. f , v b3ln3231q37 2[ o .n n B lI

Root m ean square current, T r ;my3rtv7^ =-0 .71/# ' / z .s" .t acri It truoiio fi n i l i dnonu-y

■VIM! -1'Iff

noiiuIoZ

uoni - - \ nortaupo rrovig or/J m oldi inula / 1upo (a .b ) inonuo toorib orfT

A = A = ;SV £V ^

•"Ibiji urn i>nG lolunirnortub yIqiiluM

v2noiiulof

A o .e i = »Peak current, 10 = y/2 X lr m s = 1.414/r m s

ni oiuni; laariqSimilarly, Root m ean square voltage, Vr ms = — 0.71Fo or \

,. v2 tn : u n >lE9q, rriot-i

Peak voltage V0 = y/2 X Vr m s = 1.414I£ m s>i ; - \ .tnanuo zuoonBJnElan,

E xam ple 1 •• .a. . i_The cuiTent, / in an a. c circuit is given by the equation: / = 30sin 1007rt where t is the tune in seconds. Deduce the following from this equation. 2(i) Frequency o f the current •' ' ' 1 (wVPeak value o f the'cilfriifiiit3^1 liuaiio :>.d ns nl( i i i ) r . m :S value o fIh e current >.i rb- .¡- > o n n b liib dift&iifeinBttniSolutionThe equation / = 30Sin 1007th can be qq^ipared with

/ = las in 2 n ft

(i) lOOnt - 2 n f t100 = 2 f

no ilu lo?,NJ081 = ogBlIov jIro'I

s b v -) e I o til - °0d£ 2i stay A

PI lu> slgn r 02Bi.

100f ---------= b0H21 2

di) Peak current l0 = 30/1

(iii; r .m s cu rren t, Ir.m.sjo_ = l 0 V2

M

IV, . \l

. > .niofiogiT mu

M ultiply the denom inator and numerator by \J2

\[2 x 30 30V 2 2 0 /2

,r2 X \ T “ ~ 7 T ~ ~ 2 ~.21.

3 8 9

Example 2Calculate the peak voltage of a mains supply of r. m. s value 220V. NECO 2000SolutionVrm J = 220V

Peak voltage, Va = 1.414 X Vr m j = 1.414 x 220 = 31 IK

Example 3The current through a resistor in an a. c circuit is given as

C. 2Aequivalent o f the current. A. ~ A B. 2^2Av2

2sino)t. Determine the d .c D.V2A

JAMB 200140SolutionFrom the given equation / = 2sino)t, peak current l„ =2AThe direct current (d. c) equivalent of a. c is the root mean square current lr.mj

r _ lo _ 2‘r j j u ---- ~j=- — —V2 yfl

Multiply denominator and numerator by V2

2 V2 2V2 2V2 ^= v ! x v ! = W = — = ^

Example 4I f an a .c is represented by / = /0sino>t. Calculate the instantaneous value o f such a current, if in a circuit it has r. m. s value of 15.0/1 when its phase angle is 30°.

WAEC 1989E3SolutionIrjnjs = 15.0/1, phase angle u)t = 30

From Irm_s = - p , peak current I0 = V2 x /r = V2 x 15 = 21.21 V2

Instantaneous current, / = I0smu)t

= 21.21 xsin30 = 21.21 X 0.5 = 10.61A

Example 5h an cue circuit the peak value of the potential difference is 180K. What is the instantaneous potential difference when it has reach th o f a cycle? WAEC 198951 SolationPeak voltage V„ = 180K, instantaneous potential difference, V =?

A cycle is 360° | th o f a cycle = ~ x 360 = 45°

-• Ppase angle (ot — 45°

V = VgSmojt

V = 180 X Sin45

From trigonom etry, sin45 = ~ V = 180 x — = 90V2K

Example 6The instantaneous value o f the induced e. m . f as a function o f time is e = r 0sinajt where

is the peak value o f the e. m . f . The instantaneous value o f the e . m . f , one quarter of a £e B. C O D. ^ IAMB 200749die period is A. ~

\\

390

Solutione = f0sin<«*t 8 = u)t, substitute to obtain e = fosin0

One period = 360°; .*. i x 360 = 90°

Substitute 0 = 90° into equation

£ = eosin90 = f „ x l = f„ Ans: B

R E A C T A N C E A N D IM PE D A N C E . ClthcrReactance is defined as the opposition offered to the passage o ^ ¡udmctn*

the inductor or capacitor or both. Reactance due to an inductor is ca reactance. Reactance due to a capacitor is called capacitive reactance. . rMit

Impedance is the overall opposition offered to the passage o f an a ' ^ mi5L ncc Mlj containing a resistor any one or both of an inductor and a capacitor, impedance are both measured in ohms (il).

amplitude of voltage across inductor or capacitor^Also, reactance = amp,itude of cutTCnt through inductor or capacitor

A C Circuit Containing Only Resistor

Fig 5.1

Ohm’s law (V - IR) can be applied to the a. c circuit with only resistance. Substitute V — (¿smalt and / = /-sinu>t

Resistance, R = — = ° mo>ti ioSinwt

Substitute V0 = V2 x I?r ms and l0 = y/2 x ir r

Vr r.m_s_ . „ K V 2 x v rResistance , / ? = — = —=.------to yf2 x lr t Irr

Example 7A voltage supply o f 12VrnLS and frequency 90Hz is connected to a 4(1 resistor Calthe peak value o f the current. NECO 2004SO «JculateSolutionVrms = 12V, resistance R = 4(1

r „ K-.mj , Vr.ms 12From R = ------ , lr ms = —— = — = 3A*r.ms ^

Peak current, /„ = y/2 x lr ms = y/2 X 3 = 4.2A

391

Example 8 .Calculate the amount of heat generated in an external load of resistance 8il it a11 a - c °peak value 5-4 is passed through it for 100s. NECO 2003SolutionResistance R = 8fi, time t = 100s, peak current Ia= 5A

1r.m.sJo_V2

5— = 3.54 AV2

Heat energy H = l2Rt = 3. 542 x 8 x 100 = 10025.28 = 1.0 x 10*J

A. C circuit containing only inductor.

O

Fig 5.2L = inductor Vr ~ s f

From ohm’s law (V — 1/?), voltage across inductor is VL = / x XL

V VInductive reactance XL = y- = r m s = coL = 2n fL

From VL = / x XL and XL = 2nfL , inductance L of an inductor is given by:

, VL VL1 = o—77 or L = —2n fl col

Exam ple 9

Fig 5.3

- o -

/ r m nL = 0.9H

If the frequency o f the e. m. / source in the a. c circuit illustrated above is — Hz.W hat is the reactance o f the inductor? WAEC 199141Solution -,... „Frequency / = H z , inductance L = 0.9H

Inductive reactance XL = 2n fL

2n x 500 x 0.9= --------------------- = 2 x 500 x 0.90 = 900Ün . . . 1 • ‘ " • -

7 . ' ': » ' . , ,.r 11 > . - . . I r‘ ‘ . r ' i,.' , . .. /< ,-r . , . /Exam ple 10 . , .Calculate the inductance o f the coil in the circuit diagram shown below.

L----------------m r m n —

-------- e —F ,g 5 .4 2 4 0 V,— Hz WAEC 200244

392

Solution looCurrent, / = 2.0/1, voltage, V = 240F, frequency, f - n

Inductance, J C - 240 - ™ = 0.6H2* f 2 ^ x i 5 2 x 2 400

I

Example 11A 220V, 60Hz a .c supply is connected to an inductor passing through the inductor? ( n = — )SolutionFrom V = \X L , current through inductor, / = y l

V 220 220

of 3.5//. What is the current NECO 2001

1 -2 n fL 3 x 3.142 x 6 0 x 3.5 1319.62

= 0.167 A

Exam ple 12At what frequency would a 1 OH inductor have a reactance of 20000?A . ^ t f z B TtoHz C V Wz D. \00nH z JAMB 199939 SolutionFrequency / =? inductance L = 10H, inductive reactance, A/, = 20000

xFrom XL — 2nfL , frequency / - —

2nL2000 100

f = -----------= ------Hz' 27T x 10 n

Example 13Calculate the energy stored in an inductor of inductance 0.2H through which a current of5.0A flows. NECO 200649SolutionCurrent, / = 5.0A; inductance, L — 0.2H

1 , 0.2 x 5.02 „Energy, E = - L I 2 = -------------- 2 .5 /

A. C Circuit Containing Only Capacitor.

From ohm’s law (V = 1R), voltage across capacitor, C, is Vc = I x Xc

v t'r.m .s * 1Capacitive reactance Xc - ^ ^ 2 n fc

From Vc = I x Xc and Xc = capacitance, C, of a capacitor is given by:

393

r - 1 r ___—C ~ 2nfV c ° T C ojVc

Example 14If the frequency of the a. c circuit illustrate below is H z. What would be the r in the circuit?

Fig 5.6WAEC 1992s2

SolutionFrequency f = ^ H z ; effective capacitance. C = 2/cF + 3fiF = 5//F = 5 x 1 0 F

Capacitive reactance Xc =1

2nfC 2ttx — x 5 x IO '6 n

10.005

= 20on

Example 15At what frequency would a capacitor of 2.5nF used in a radio circuit have a reactance o f 250fl? A. — Hz B. 200tt/ /z C. 2000?tHz D. — Hz JAM B 2 0 0 2 s0

n 800SolutionCapacitance C - 2.5fiF = 2.5 x 10_6F, capacitor reactance 2TC - 250fl

X' = 2 ^ C ••• / =

1 1 1 x 106 1000000' ~ 2n x 2.5 X 10-6 x 250 ~ n y . 1250 x 10~* ~ n x 1250 ~~ n x 1250

800= ----- Hz

n

A. C Circuit Containing Only Inductor And Resistor.

2 ut = J r 2 + x 2

394

Current in L - R circuit, L = ~rf v + xl

The applied voltage V also lead on the current / by an angle 8 given by.

ViT a „ » = - 'A

IR R

Effective or total circuit voltage, F = y/V^ + V*

Example 16For the diagram shown below,(i) Calculate die inductive reactance of the circuit.

0.25 H----------- m r m n —

Ra

Fig 5.8

(ii) If the current in the resistor R is 0.05/1, calculate the potential difference across the inductor. WAEC 199541SolutionInductance L = 0.25//, frequency f = ^ Hx , current / = 0.05/1

(i) Inductive reactance XL = 2 n fL

= 2n x — x 0.25 = 50fln(ii) Potential difference across inductor, VL = IXL = 0.05 x 50 = 2.5V

Example 17A circuit consist of a 10.011 resistor and a 0.8H inductor connected in series to a 240F, 60Hz a. c source.(i) Draw the circuit diagram of the arrangement

Calculate its:(i) Inductive reactance(ii) Inductive reactance(iii) Impedance circuit of the circuit(iv) Capacitance of a capacitor that must be connected in series with these

components to obtain maximum current [rr = 3.142] NECO 2005s14Solution

Fig 5.9

« I.]— i---------r r r r mlo .o n 0.8 H

240F.60 Ht

R = ion, L = 0.8H. Vrm s = 240V. f = 60Hz

395

(ii) Inductive reactance XL — 2 n fL = 2n x 60 x 0.8 - 3 0 1.63fl

(iii) Impedance ZLR — J R2 + X£ = VlO2 + 3 0 1 .632 = V 91080.66 — .->01.8 i

(v) Maximum current is obtained at resonance when XL = Xc

301.63 = X C

301.63 = —— 2 n fC

Capacitance, C =1

2 x 3.142 x 60 x 301.63= 8.8 x 10- bF or 8.8/iF

Exam ple 18

L-m r m .

V, = 8V V„ = 6V

Fig 5.10v

In the series a .c circuit shown above, the potential difference across the inductor is ^K-.m.s and that across the resistor is 6Vr m s. The voltage isA. 2V B. 10K C. 14F D. 48F / i4 M t f l9 9 9 i5Solution

VL = 8F, VR = 6V

For a L — R circuit, Vr m s = -JVR +

= V82 + 62

= V64 + 36

Vrm s = vToo = 10V

A. C Circuit Containing Only Resistor and Capacitor.

R. C Series circuit

Fig 5.11

For R. C circuit, the im pedance ZRC -i r .m s io

JRC = Ir 2 + X 23/

C urren t in the R. C curren t, l0 =J r ^ + x 2

396

The current 1 also leads on the applied voltage V by an angle 8 given by

Tan IXç1R

XcR

Effective or total circuit voltage, V = J v f

Example 19An a. c source is connected in series to a capacitor of capacity reactance of 103V3 and a resistor of resistance 103O. The impedance of the circuit is A. 2000V3H B. 20000 C. 1000V3 D. lOOOflSolutionCapacitive reactance Xc = 103V3, resistance R = 103O

Impedance ZRC = + X 2C

= A/ a 0 3)2 + ( l0 3V3)2

= J lO 6 + (103)2(V3)2

= V106 + 106 x 3

= V106 + 3 x lO 6

= V4 x 1 0 6

= 2000H

Example 20

Fig 5.12

3 on x = 4onH I—

>2001 50 Hz

In the a. c circuit above, the current value is A. 6.67/1 B. 4.00/1 D. 0.58/1 JAMB 200046SolutionResistance R = 30il, K-.ms = 2001/,Frequency / = 50Hz Capacitor reactance Xc = 40fl

C. 3.00A

VFor R. C circuit, current / = , ■ ~~

yj R + X c

200 _ 200 V302 + 402 ~ V25ÖÖ

200"50 = 4.00A

Example 21The resistance in a series R. C circuit is 50. If the impedance of the circuit is 130, calculate the reactance of the capacitor. WAEC 199654SolutionR = 50, Z = 130, capacitor reactance, Xc =?

For R. C, ZRC = y/R2 + Xc

397

13 = + X *

132 = .I2 + X£169 - 25 = Xf:

144 = X2 Xc = V T 44= 120

A 2.0nF cupacitor is connected in series with n 3000 resistor across an a potential difference 50l£ms and frequency ( alculatc the r .m .s value o

. . . . . . 2,1 N EC 0 2 0 0 6 51current in the circuit.SolutionCapucitancc, C = 2.0uF = 2 x 10-6F, resistance, R — 3000,

- - r _ 3 1 4 „Vr.mj = 501/,

, 314 „frequency, / = ~ n z

Capacitive reactance Xc = „ =L 2nfC2nf C 2 i r x ^ x 2 x 10"6 Zn

0.000628= 1 5 9 2 .3 6 0

From ZRC — —r.m.s

1 r.m.s... Vr.ms Vr.ms___________ 50 50

Z*C J r 2 + X l V3002 + 1592.362 V2625599.821

50= 0.031,4

Example 23A 40f/F capacitor in scries with a 400 resistor is connected to a 1001/, 50H z a. c supply.

(i) Draw a circuit diagram of the arrangement(ii) Calculate the

I. Impedance in the circuit;II. Current in the circuit;III. Potential difference across the capacitor.

SolutionC

WAEC 2008E14

Fig 5.13 100P, 50Hz

Capacitance C = 40fiF = 40 x 10-6F ; Frequency f = 50Hz

(i) Impedance ZHC = yjR2 + X2

1 1

resistance R = 4 0 A ; Vr.m.s = 100P ;

= 2n fC 2 x rr x 50 x 40 x 10-6r z = 79.5811

ZRC = V402 + 79.582 = 89.07A ’

(«) current, / =P Pvr,m.s __ vr.m.s

J r2 + x* ~ 2

398

10089.07

1 .12/1

(iii) Potential difference across capacitor V = I X Xc

V = 1.12 x 79.58 = 89.35K

A. c Circuit Containing Only Inductor and Capacitor

Zlc = V (* z .-* c )2 = XL ~ X C

Current in L - C circuit, L = 0 — = ------—V(*l - * c)2 Xl - X c

In the L. C circuit, the inductor voltage VL and the capacitor voltage Vc are out of phasebyl80°. ________

Effective voltage V = V( VL — Vc)2 = VL — VC

Example 24The frequency of the a. c circuit below is — Hz. What is the reactance in the circuit

Fig 5.15-rrm rv 0.9 H 2 nF

N.ECO 200252

Inductance L = 0.9//, capacitance C -2 fiF - 2 x 1 0 6 F Inductor reactance X, = 2rtfL = 2 n x 500/rr x 0.9 = 900Í1

1 1 1Capacitor reactance, * c ' 2nfC ~ 2n x §00 x 2 x 1Q_6 ~ ÔÔÔ2 ~ 500fi

Circuit reactance, ZLC = Jx* + X¿ = V^OO2 + 5002 = VlOÓOOOO= 1029.56 = 1030Í1

399

A C Circuit Containing Resistor, Inductor and Cap

Z = y/R2 + (XL - X c}¿

Current in RLC circuit, L = .y/R2 + (XL - x cy

V ,- V r IXL - I X C XL - X CPhase angle is given by; Tan 6 = ——--- = ------—--------p

VR IK «

Effective voltage, V = -JVr + (VL — Vc) 2

Example 25An a. c circuit o f e. m. / 12U has a resistor of resistance 8fl connected in series to an inductor of inductive reactance 16Q and a capacitor of capacitive reactance 10iL The current flow in die circuit is. IAA B. 14.0,4 C. 1.2A D. 12.0A JAMB 200429 SolutionVrjnjs = 12V. R = 80, XL = 16Q, Xc = 10O

VCurent in RLC circuit / = -

/ =12

y /R l+ (X L - X C¥

12 12V82 + (1 6 -1 0 > z VS2 + 62 V64 + 36

12 12Vïôô ~ io ~ 1ZA

Example 26A source of e .m ./of240K and frequency 50Bz is connected to a series arrangement of a resistor, an inductor and a capacitor. When 1he current in the capacitor is 10A the potential difference across the resistor is 140K and that across the induct™ 50P. Calculate the

(i) Potential difference across the(ii) Capacitance of the capacitor(iii) Inductance of the inductor WAEC 2002£l4

Solution

Fig 5.17

400

Potential difference across inductor VL = 50P Frequency f — z< 1 — \r\a Potential difference across resistor VR = 140V Current throug capaci or Potential difference across capacitor Vc = ?Potential difference across RLC circuit V = 240V

(1) V2 = V2 + (VL - VCY

2402 = 1402 - (50 - Vc) 2

38000 = (50 - Vc)2

V38000 = 50 - Vc

±194.9 = 50 - Vc

Vc = 50 + 194.9 or 5 0 - 194.9

Vc = 244.9V or -144 .9P

Potential difference across capacitor Vc = 245F

, „ 1 10 10 0 0 Capa0tanCe C - w v c= 2n X 50 X 245 = 7 6 9 3 7 ^0

C = 1.299 x 1 0 '4 = 130 x 10-6 F or 130/iF

From VL = IXL and XL = 2nfL

inductance,! = Vl2 T lf l

5027T x 50 x 10

127T X 10

= 0.01 S9ff

Exam ple 27A series circuit consisting of a 100H resistor, a coil o f 0.1 OW inductance and a 20nF capacitor, is connected across a 110F, 60Hz power source.(i) Draw the circuit diagram of the arrangement. Calculate tile(ii) Inductive resistance (iii) Capacitive reactance (iv) Impedance o f the circuit(v) Current in the circuit (vi) Power loss NECO 2002£14Solution

R 1 Co -------m n r i ----------- 11—10Oil 0.10H 20n F *

Q -----F ,g 5 -18 H OP, 60 Hz

R =10011, L =0.10H, C = 20/iF = 20 x 1 0 -6F, VTms = HOP, f = 60Hz

(ii) Inductive rea c ta n ce ^ = 2nfL = 2n x 60 x 0.10 = 37.7011

(iii) Capacitive reactance Xc =

(iv) Impedance for RLC circuit,

1 1_________2nfC 2tr x 60 x 20 x 10-6 - 132 63 ii

Zrlc = 'JR 2 ± (XL - Xc) 2

z RLC = VlOO2 + (37.70 - 132.63)2

= yJlOO2 + ( -9 4 .9 3 )2

= V10000 + 9011.70

401

= V Í9Ó IL7 = 137.88ft

(vi) Power loss, P = l 2R = 0.82 x 100 — 641VR _ 0.8 x 110 X 100 _ 63 g _ 64W.

or P = 1 Vcosd = IV x — = 137.88

resonance in a.c circuitA resonance circuit is an RLC circuit through which maximum alternating currents flows. The frequency at which resonance occurs is called resonance frequency C/o) occurs when XL = Xc . Also, at resonance VL = Vc

Resonance frequency, = — -—y J° 2nyfLC

Maximum current at resonance I = V /R

Example 28In a series L — C circuit, the inductance and the capacitance are 5H and 2fiF respectively.Calculate the resonance frequency o f the circuit NECO 2000s1SolutioaInductance L = 5H. capacitance C = 2/xF = 2 x 10~6F

Resonance frequency, f a = — -— = ________1_________2nVLC 2n x V5 x 2 x 10"6

1 _______ 1 _ 1° 2n x Vl X 10_s 2n x 3.16 X 10-3 1.987 X 10-5 50.3/Zz

Example 29

Fig 5.19

15fl 5 fiF 8 mHCD------11----- rmn>

------- Q ----------E = E0sm<j)t

In the a. c circuit diagram above, the resonance frequency is5000 B 2500

n HzC. 5000

i t H zA.nHz

SolutionInductance L = SmH = 8 x 10~3H,

D. JAMB 19984S

Resonance frequency f 0 =

fo =

capacitance C = 5¡iF = 5 X 10"6F 1

2nyfLC 2rr x V8 x 10-3 x 5 x 10~6

1 _ 1_______ 1.0000 25002rrV4 x IO"8 2n X 0.0002 4* T ~ Hz

E xam ple^) ^ ^ ^ 10-12F and an inductor are joined in series. The value of inductance that will give the circuit a resonant frequency of 200KHz is

l B.lH C.¿H D-S» JAMB 2 0 0 1 "16

the

402

Solution ,nnnnnU rC = 20 x 10-12F; Resonance frequency, fa = 200KHz - 200000 ,Inductance L =?

1At resonance o r 2n fL = 2 nfC

Rearrange and substitute, 2n x 2 x 105 x L = — x 2 x 105 x 20 X 10_l2

1L = 2n x 2 x 10s x 27T x 2 x 10s x 20 x 10

Example 31Given that the voltage amplitude of the a. c source shown in figure 5.19 is 100V

(i) Determine the resonant frequency of the source.(ii) Calculate the maximum current passing through the resistor.

Fig 5.20

SolutionR = lOOfì; L = O.lff; C = OA/iF = 0.1 x 10~6F; VrnLS = 100V

1 1(i) Resonant frequency, fa =

fo =

(ii) Maximum current lr mj! =

2nVLC 2W 0.1 x 0.1 x 10"6

12n x 0.0001

V V¥r.ms yTjrus

= 1591.55Hz

2 V* 2 + (*r.-*c)2

If current is maximum, XL = Xc or XL — Xc = 0. , _ Knus 100 , ,

rm j R 100 1A

P O W E R IN A .C C IR C U ITAverage power in a. c circuit P = IVcosd or P = 12R Where I = root mean square current, Ir ms

V = root mean square voltage, Vr m j 6 = phase angle, angle o f lag or lead between / and y

resistance Rcos0 = power factor = :------ ------- = —

impedance Z

Example 32The power dissipated in an a. c circuit with an r. m. s. current o f 5A, r. m. s. voltage o f 10K and a phase angle of 60° is A.25W B. 70W C. 120W D. 125W

JAMB 199746

403

SolutionIr.m.s = 5A, Vrm j = 1OF, phase angle 9 = 60°

Power P = lVcosG = 5 x 10 x cosóO = 251V

Example 33 „When an a .c given by 1 = 10sin(1207r)t passing through a 120. resl® ^ 0 4 5 ^ dissipated in the resistor is A. 1200W B. 600W C. 120W D. 30W JAM B Solution

/ = 10sin(1207r)t From equation, peak current l0 = 10A, resistance/? I2 il

Root mean square current, Vr m v = 7.0 7AV2 V2

Power, P = /2ms/? = 7.072 x 12 = 599.8 = 600W

Example 34The diagram below illustrate an a.c source of 50V ( r .m .s ) , — H z connected in series71with an inductor o f inductance L and a resistor o f resistance R. The current in the circuit is 2A and the potential difference across L and R are 30V and 40V respectively. Calculate

(i) The power factor o f the circuit(ii) The average power dissipated in the circuit

R

I = 2A I

-rWfm--30V--------*** -40V-

- @ -

Fig 5.21

SolutionVr.m.s= 50V;

50 Vr m s ,^ -H .

f = ~ Hz; 1 = 2A; VL = 30V-

W AEC 1 9 9 8 49

VR = 40V

resistance R(i) Power factor, cos0 = :---- ------- = —

impedance Z

VR 40From, V = IR for a resistor, /? = — = — = 2 Oil

VL 30From, VL = IXL for an inductor,XL = — = — = 15fl

For L — R circuit, Z = y]R2 + XL

= V202 + 152

= V400 + 225 = V625 = 25il

„ « R 20 Power factor, cos0 = — = — = 0.8 Z 25

(ii) Average power P = I2R = 22 x 20 = 80W

EXERCISE 5.E Calculate the peak voltage of a mains supply of r .m .s value of 220V

WAEC 199451 Ans: 311V2- In an a. c circuit the peak value of the potential difference is 1801/. What is th instantaneous potential difference when the phase angle is 45°?

WAEC 1994S6Ans: 90V23- In <4. C circuit theory, the root mean square ( r .m .s ) current, Ir.m.s>an< Pea^(maximum) current lQ are related by A. l0 = /r m s /V 2 B. lr.ms = !o / ^C - 'r.m .s = I / / 0V2 D. I0 = l / / r m.sV2 JAMB 199246 Ans: B

4. The voltage of the domestic electric supply is represented by the equation V —31 lsin314.2t. Determine the frequency of the a .c supply. A. 50.0Hz B. 100.0HzC. 311.0Hz D. 314.2Hz (tt = 3.142) JAMB 199 747 Ans: 50.0Hz5. When a certain a. c supply is connected to a lamp, it lights with the samebrightness as it does with a 12V d .c battery. The r .m .s value of the a .c supply is A. 8.51/ B. 17.01/ C. 3.51/ D. 12.0K JAMB 200647 Ans: 12.0V6. Calculate the peak voltage that could be obtained from a 220 Vr m s main supply.

NECO 200551 Ans: 311K7.

Fig 5.22

Calculate the inductance L of the coil in the circuit shown above.WAEC 199538 Ans: 0.6H

8. An inductor o f inductance 10H carries a current of 0.2A. Calculate the energystored in the inductor. WAEC 20 0 244 Ans: 0.2/9. Calculate the inductance of an inductor whose reactance is one ohm at 50Hz.

WAEC 2 0 0245 Ans: 3.18 x 10~3H10. A 2H inductor has negligible resistance and is connected to a ^ H z a .c supply.

The reactance of the inductor is A. 200Ü B. 500 C. ~ C l D. — OJAMB 2 0 0 036 Ans: 2000

11. The energy stored in an inductor of inductance 5mH when a current o f 6A flows through it is A. 1.8 x 10 2J B. 9.0 x 10 3/ C. 1.4 X 10 2JD. 9.0 X 10~2J JAMB 2 0 0 427 Ans: 9.0 x 10~2J12. If two inductors o f inductances 3H and 6H are arranged in series, the total inductance is A. 18.0H B. 9.0H C. 2.0H D. 0.5H JAMB 200515 Ans: 9.0H13. Calculate the energy stored in an inductor of inductance 0.5H when a current of3A flows through it. NECO 200449 Ans: 2.25/14. A direct current o f 5A flows through a 0.2H inductor. Calculate the energy storedin the inductor. WAEC 20084S Ans: 2.5/15. Calculate the energy stored in an inductor of inductance 0.5H when a current of5A flows through it. NECO 2 0 0 847 Ans: 6.25/16. Calculate the energy stored in an inductor o f inductance 0.5H when a current of5a flows through it. NECO 200847 Ans: 6.25/17.

5 mH 10 mH 20 mHFig 5.23 o— i m m — f m m \ — i m m — o

405

th rr G'VCn threc 'nduc‘ors of inductances 5mH, lOmH and 20mH connected in series,n% Cc nntivc inductancc is A. 0.35mH B. 3.50mH C. 2.90mH ¿5.00m// JAMB 200844 Ans; 35.00m//

A ^ 0 0 C' t0r caPac' tancc 25//F is connected to an a. c power source o ffrequency ~ / / z . Calculate the reactance of the capacitor.WAEC 199543 Ans.lOOil

* . _ ^ /. DOwer source o f19. A capacitor o f capacitance 25fiF is connectc ’ Q9740 Ans:frequency — Hz. Calculate the reactance20.

of the capacitor. WAEC 1997 ioon

Fig 5.24

240l/rm_s-o o—

H 4n *"-o rrm -X. = 3Í1

In the diagram above, determine the r. m. s current. A. 3 1A B. 48A C. 6(M D.80AJAM B 2 00138 Ans: 48A

21.

Fig 5.25

0.1H-rmrn-

-G>-7 S V .-H z

1.5A

From the diagram above, the inductive reactance and the resistance /? are respectively A. 2 5 ila n d 5 lf l B. 20fi and 50ft C. 1 0 ila n d 5 0 n D. 5 0 iland45 il

JAMB 20 0739 Ans: lO il and 5 0 il

22. Calculate the reactance of the inductor in the circuit diagram shown, (nr = — )0.7 H SOU

i i 1 t

L------------------- © -------------------- 1

Fig 5.26 230F ,60Hz WAEC 200 142 Ans: 264Í1

23. In a series R — C circuit, the resistance o f the resistor is 4fl and the capacitive reactance is 3fl. Calculate the impedance o f the circuit WAEC 19933s Ans: 5fi24. Calculate the following in the series circuit shown below

(i) Reactance of the capacitor.(ii) Impedance of the circuit(iii) Current through fhe circuit(iv) Voltage across the capacitor(v) Average power used in the circuit

406

1 <)///•' i non

~ 0 -Flg 5.27 9 0 /, 25//z r f i i v

WAEC 1988i4 (i) 636.6212 (ii) 1105.412 (iiij O.O^^ the circuit is25. The resistance of u series R — C circuit is 2003s0 Ans: 1^121312, calculate the reactance of the capacitor.26.

theFig 5.28 0.9// 2pF

If the frequency of the a .c circuit illustrated is — Hz what would be the reactance creuit? " WAEC 199043 Ans: 4 0 0 027. A source of e . m . f 2 4 0 / and frequency 50Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the p entia difference across the resistor is 140/ and that across the inductor is 5 0 / . Draw the vector diagram of the potential differences across the inductor, capacitor and the resistor. Calculate the(i) Potential difference across the capacitor.(ii) Capacitance o f the capacitor.

(iii) Inductance o f the inductor.!/AEC 1991£3 Ans:(i) 2 4 5 / (ii) 130¡iF (iii) 0 .0 lo9H28. In the diagram below the resistor has a resistance o f 812 while the reactance o f the inductor and the capacitor are 1012 and 1612 respectively. Calculate the current in the circuit.

L C R

Fig 5.29 WAEC 199 6 53 Ans: 13.0A

29. A series RLC circuit comprises a 100-12 resistor, a 3-H inductor and a 4 -g F capacitor. The a. c source o f the circuit has an e. m . f o f 1 00 / and frequency o f — H z.

(i) Draw the circuit diagram o f the arrangement. Calculate the(ii) Capacitive reactance(iii) Inductive inductance(iv) Impedance o f the circuit(v) Current in the circuit(vi) Average power dissipated in the circuit. W AEC 1999t4Ans: (i) 781.12512 (ii) 96012 (iii) 204.8212 (iv) 0.488A (vi) 23 .81 /

30. In a scries L — C circuit, the inductance and the capacitance arc 0.5H and 20uFrespectively. Calculate the resonant frequency o f the circuit ^

WAEC 19984BAns: 50 .31 tfz

* S T i i f “ ‘ . " T * ,hC V°" aBCS thc « « * 0* and theinductor are 3 0 / and 4 0 / respectively. What is the voltage across the capacitor?A 3 0 / B. 4 0 / C. 5 0 / D. 7 0 / JA M g 19g9P48 ^ w

407

32.

A.

a series RLC circuit is expressed asThe resonance frequency of - An• B. 2W ZC C. s û D. E. 2M NECO 2007« Ans: A

¿n\[ÍZ, , . , i - i sinoJt and the voltage, V =33. In a purely resistive a. c circuit, the current, / '«•l^sincut. Calculate the instantaneous power dissipated in the circuit in im

C. i„v„

34. resonance

I) I V * in2ait E I 2V„2 WAEC 199337 Ans:/0K0sm cut In the circuit diagram below, calculate the energy stored in t e i

O.lfiF100ÎÎ 0.1//

Fig 5.30 100F, 50HzWAEC 200141 Ans: 0.05J

35. An alternating current with a peak value of 5A passes through a resistor of resistance 10.Oil. Calculate the rate at which energy is dissipated in the resistor.

WAEC 199 736 Ans: 2 5 0 W36. What is the direct current equivalent of an alternating current 5sincut?

NECO 2 0 0 750 Ans: 3.54A37. A 1201/, 60W lamp is to be operated on 220F a.c supply mains. Calculate the value of non-inductive resistance that would be required to ensure that the lamp is run on correct value A, 2000 B. 3000 C. 5000 D. 1000 JAMB 2 0 0 746 Ans: 200038. Calculate the average power dissipated in an a.c circuit with an r. m .s current of SA, r .m . s voltage of 10F and phase angle of 60°. NECO 200849 Ans: 25W39.

Fig. 531

/• C- r r m r v

V.-40V , l 4 VC=110V_ ,__ y„=8o_

■0 -

From the diagram above, if the potential difference across the resistor, capacitor and conductor are 80V, 110V and 40V respectively, the effective potential difference isA. 116.3V B. 50.0V C. 230.0V D. 106.3V JAMB 200944 Ans: D40. An inductor of inductance 1 .OH is connected in series with a capacitor o fcapacitance 2 . 0 in an a.c. circuit. Calculate the value of frequency that will make the circuit to resonate. WAEC 200944 Ans: 112 5Hz41. An ammeter connected to an a.c. circuit records 5.5A. What is the peak current inthe circuit? WAEC 200945 Ans: 7.8A42. A source of e.m.f. 110V and frequency 60Hz is connected to a resistor an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential difference across the resistor is 80V and that across the inductor is 40V.

Draw the vector diagram of the potential difference across the inductor the capacitor and the resistor. Calculate the:

(i) Potential difference across the capacitor;(ii) Capacitance of the capacitor;(iii) Inductance of the inductor [n = 3.14]

WAEC 2009hl4 Ans: (i) 115.5V f ii» 45.9 x K,-<>r nr4rViF (iii) q.05 |.

408

6ENERGY QUANTIZATION AND WAVE-PARTICLE PARADOX

ENERGY QUANTIZATION ,According to the quantum theory suggested by Max Planck, heat or e ectromagn radiation is emitted in fixed, discrete or separate amounts known as quanta.The energy £ o f a photon or quantum is given by:

E = h f

Where h = Planck’s constant = 6.6 x 10_34/ sc = Velocity o f electromagnetic waves in free space = 3.0 X 10 ms X = Wavelength o f electromagnetic wave (m) f = Frequency of electromagnetic wave (Hz)

Electrons in an atom have definite energy levels with definite values like Eot , E2, E3...e. t.c. The energy of the electron could change from one level to another, for

example, the energy of the electron could increase form lowest energy level i ground state E0 to Ex or decrease from E3 to a lower energy level E2 or Et

Energy change, AE = En - E„ = h f

Or A E = En - E 0 = y

Where En = Energy in excited state, n = 1, 2 ,3 --------Ea = Energy in ground state.

The greater the energy change, the greater will be the frequency of the emitted radiation.

Example 1An electron makes a transition from a certain energy level Ek to the ground state E I f the frequency of emission is 8.0 X 1014//z , the energy emitted is A. 8 .2 5 x 1 0 ~19/ B. 5 .2 8 x 1 0 ~19J C. 5 .2 8 x 1 0 19J D. 8 .2 5 x 1 0 19J [h = 6.6 x 1 0 -347s] JAMB 200348Solution

Frequency / = 8.0 x 1014//z, h = 6.6 x 10_34ysEnergy emitted E = h f

E = 6.6 x 10-34 x 8.0 x 1014 = 5.28 x 10~19/

Example 2Calculate the energy carried by an X-ray o f wavelength 6.0 x 10_1°m [Plank 's constant = 6.6 x 10 34/s , velocity of light = 3.0 x 108m s_1]

WAEC 1995s8SolutionWavelength X = 6.0 x 10_1°m, velocity o f light c = 3.0 x 108m s_1h = 6.6 x 10"34/ s

heEnergy E = h f or E = —

409

6.6 x 1(T34 x 3.0 x 108 _ 3 3 x l 0 - ^ j6.0 X 10-»

Example 3 -1 3 .6 e V energy level.An electron makes a transition from — 3.4eV energy 1c

NECO 2003 £15

Calculate the(i) value of the loss o f energy due to transition in joules.(ii) frequency of the emitted radiation

[leV = 1.6 x 10'19y ; h = 6.6 x 10'34ys]So,ution , i p - - 1 3 6eKInitial energy level E, = —3.4eV; final energy level 6/leK = 1.6 x 10"9/, h = 6.6 x 10_34/5(i) Loss o f energy, AE = E* — E -

= - 3 . 4 - ( - 1 3 . 6 )= -3 .4 + 13.6 = 10.2eP= 10.2 x 1.6 x 1(T19 = 1.63 x 10"18/

p 1 fio y i n-18(ii) From E = h f , frequency f = — = — — ^q_34 = 2.5 x 101SHz

Exam ple 4An atom radiates 1.5 x 10-19/ when an electron jumps from one level to another. What is the wavelength of the emittedradiation?[Planck's constant = 6.6 x 10_34ys] [speed of light in vacuum = 3.0 x108m s_1 ] NECO 2005s8SolutionEnergy radiated, E = 1.5 x 10~19J, h = 6.6 x 10_34ys, c = 3 X 108m s-1

EheT

hC 6.6 x 10~34 x 3 x 108 ravelength, A = — = ------- ; 4 0-1 O = 1.32 x 10~6i

Exam ple 5A photon o f wavelength A0 is emitted when an electron in an atom makes a transition from a level o f energy 2Ek to that o f energy Ek. If the electron transits from - Ek to Ek level, determine the wavelength of the photon that would be emitted. WAEC 200144 Solution

heEnergy charge, AE = Ex — E2 = —

1st instance AE = 2 Ek — Ek = Ek

he•• Ek = T .................................................................................... ( 1 )

Ao 2

2nd instance AE = ^ Ek — Ek = ^E k

Making the constants he subjects o f equation 1 and 2 we have

1.

2.

he = A0Ek

Equate he in both equations to obtain

410

A-o^k *Ek3

New wavelength, A 3A0Efc

Ey= 3A„

Example 6In a model of the hydrogen atom, the energy levels Wn are given by the formu ^

= ~ r7 where n is an integer and R is a constant. Determine the energy re e transition from n = 3 to n = 2 W A B C 2001'SolutionEnergy released E =

W3 =w 3 - w 2

R~ W ~

R 9 '

R R - + -

_ R _ R ~ 4 9

9R - 4K _36 “ 36

Electron Volt and Kinetic EnergyIn atomic theory, the energy of an electron is expressed in joules (/) or electron volt (eV). An electron volt is the energy acquired by an electron in falling freely through a potential difference of 1 volt Therefore, leV = 1.6 x 10-19/-

Emitted electrons are accelerated through a potential V and acquire a potential energy eV. The potential energy is converted into kinetic energy and the electrons finally acquired a velocity v. This process is expressed as follows.

eV = ^ m ev 2

eV = EK

eV = h f

Where e = electronic charge = 1.6 x 10_19CV = accelerating voltage or potential difference (V) me = mass of electron 9.1 x 10 ~z lkg v = velocity of electron ms-1 Ek = kinetic energy of electron ( / ) h = Planck’s constant = 6.63 x 10-34/s f = frequency of electron (Hz)

Example 7An electron of charge 1.60 x 10”19C is accelerated under a potential difference of1.0 x 105P. Calculate the energy of the electron in joules. WAEC 2008£7SolutionElectron charge e = 1.60 x 10"19C ; accelerating p. d , V = 1.0 x 10SK

Kinetic Energy EK = eV= 1.60 x 10-19 x 1.0 x 10s

411

1,60 x 10 '♦/

Kxnmplc 8 •locitv of theA M)QKV is applied across an X~ray tube, calculate the rn|1*'™l n7nflfi47 electrons produced.|m0 m 0.1 x 10 *xkq ,e = 1.6 x 1 0 “ l‘*C) W SolutionPotential di(Terence V - 500KV ■ 500000K = 5 X 105V Mass of electron m , ■ 9,1 x 10"3,fc,</ ; e = 1.6 x 10 1 ?C

t> =

1 7eV * -m .tr

2eV = mev 2

[2 x 1.60 x IQ-19 x 5 x 10s yj 9.1 X I Q " 31

yj 1.76 x 1017

4.2 x l O ^ s ' 1

Kxnmplc 9II the frequency of an emitted X ray is 1.6 x 10 '* ///, the accelerating potential is A. 6.6V B. 66.3V C. 663.0V I). 6630.0V|e = 1.6 x 10_,9C, h = 6.63 x 10“3Vs] JAMB 200244Solutionf = 1.6 x l0 lft//7, e = 1.6 x 10~19C, h = 6.63 x 10-34/ s

From eV = hf ,h f

accelerating potential, V = — e

6.63 x 10"34 x 1.6 x 1016 " = ■--------- i s l T i o ^ ----------= 6« 1'

Kxnmplc 10The potential difference between the cathode and turget of an X-ray tube is 5.00 X 104V and the current in the tube is 2.00 x 1 0 '2/1. Given that only one percent of the. total energy supplied is emitted ns X-radiation, determine the(i) maximum frequency o f the emitted radiation.(ii) rate at which heat is removed from the target in order to keep it at a steady temperature.(Planck's constant, h = 6.63 x 10"34Js; electronic charge e = 1.60 x 10_19C]

Solutionh = 6.63 x 10~34/s; V = 5.00 x 104V;

WAEC 2004E1S

e = 1.60 x 10~19C; Current, / = 2.0 x 10-2i4

eV(I) From h f = eV ,maximum frequency,/ = —

f1 % o f t'V 0.01 x 1.60 x 10 '19 x 5.00 x 104

/i 6.63 x 10'34 1.207 x 107Wz

(ii) Rate o f heat energy removal power, P = 1 V( 100 — 1%) or 99% of heat has to be removed./» = 99% x / x V

« 0.99 x 2.00 x 10 2 x 5.00 x 104 • 990 W Or 990 Js '

412

£ ¡ £ £ 2 * ! " « « ■ E 'n « c .n E, u „ . o „frequency falls\> ' roin 11 n,c,nl surfucc when electromagnetic radiation o f sufficientr«dintion is less r/ *' ," cla* ^ nielnl emits elcetrons provided that the wavelength o f the hc emitted no m'!!' " ccr,uin va,uc. If the wavelength is too long, the electrons will not ultraviolet radiatio Uf ^ Intcns',y and duration o f the radiation. For example, when emitted by the zinc'am ** ^nr*'cu nr frequency and wavelength falls on zinc, electrons arc

*s illuminated by 1*/* Ji 1'*V ,C*°Ctr*01 is the emission o f electrons from a metal plate when it °nly on the frequen^ max'mum kinetic energy o f the emitted electrons depends beam. ° r Wuvclength of the incident light and not the intensity o f the light

function(Wa) o f the1™ °i1Cr nccdcd to pull out an electron is called the work frequency o°r wavelength*.», ,<hrcsho,iJ frequency ( f0) or wavelength (A0) is the minimum

When a photon 3 mUSt ^ cxcceded f°r electron emission to occur.*his energy known nc .^ t*ua" ,um ° f light energy, E or h f , is incident on a metal, part o f remaining e n e r n , W°rk f,,,Kt'on W ) A re te s ">« electron from the metal. The

represented by the C'“ 'r0n * kine'iC e"CrSy o f ™ s P '°ceSS is

h"Cr8y °f Ph0'°" = W° * function + kinetic energyE = K + ek

^ f ~ hfo + ~m ev 2

hf = hf0 + eV

h f hco f ~ ~ h eV Ao

W here W0 = h fa = — = work function ^

K . f 0 = Threshold wavelength (m), threshold frequency (Hz)

k ~m ev - eV = maximum kinetic energy (/)

Depending on the question given anv nr a .•app|ied 1 y’ cotnbtnatton o f the above eqnations can be

Example 11The work function o f a metal is 4 65eV a o,

- p WlM, i s , e k i „ e n c e „ e ^ o f m e S ^ S K r S ^ 1fSolution W AEC 199J59 '

Work function Wa = 4 65eV pnpm ,„ f .„ u, ” « W . energy of,adtahon , £ ( / . / ) = 6 .86e(,

b = W 0 + EkKinetic energy, EK = E - WQ

= 6.86 - 4.65 = 2.21 e l '

Example 12The work function of a metal is 8.6 x 10-19/. Calculate the wavelength of its threshold frequency.[speed of light in vacuum = 3 x 108 m s~ \P lanck 's constant = 6.6 x 10~34/s ]

WAEC 199460SolutionWQ = 8.6 x 10_19/, c - 3 x 108m s_1, h = 6.6 x 10"34/s

4 1 3

Work function W0 = hf0 or Wa = —X0

heThreshold wavelength, A0 = —

. = 2.3 x 10'

Example 13Calculate the frequency of the photon whose energy is re' with a kinetic energy o f 1.9 X 10~16eV. If the work function l e - 1.6 x 10 19y, Planck's constant, h = 6.60 x 10-34/s]

quired to eject a surface electron q 1.33 X \0 ~ 1(,eV.

NECO 2002E1°

SolutionKinetic energy, EK = 1.9 * iQ-u>eV _ i g x 10- i6 x j 6 x 10-9 _ 3 04 x 10-35/

k function, W0 = 1.33 x i 0-ieel, _ j 33 x 10_i6 x j 6 x 10- i 9 _ 3.13 x 10“35/

c r Ek + W0Frequency, / = — — -

h f = Ek +

3.04 x 10~35 + 2.13 x 10~35 = ? 83 x 10- 2Hz 6.60 x IO '34

Caesium has a work function of 3 x 10-19/- The maximum energy o f liberated el when it is illuminated by light of frequency 6.7 x 1014Wz is A. 1.42 x 10_19y B. 3.00 x 10“19y C. 4.42 x 10-19/ D. 7.42 X 10 J

[h = 6.6 x 10-34/s] JAMB 200847SolutionW0 = 3 x 10"19y ; / = 6.7 x 1014Wz; h = 6.6 x 10_34/s

h f = Ek +

Maximum kinetic energy, EK = h f — W0

Ek = 6.6 x 10-34 x 6.7 x 1014 - 3 x 10-19

= 4.422 x 10"19 - 3 x 10"19

= 1.422 X 10"19y

Example 15A photo emissive surface has a threshold frequency of 4.0 X 1014Wz. If the surface is illuminated by light of frequency 5.0 X 1015Hz, Calculate the(i) Threshold wavelength(ii) Work function(iii) Kinetic energy o f the emitted photo electrons[c = 3.0 x 108m s-1,h = 6.63 x 10- 34/s] WAEC 2008E1SSolutionThreshold frequency f a = 4.02 x 1014Hz Radiation frequency / = 5.0 x 1015Hz c = 3.0 x 108m s- 1 ; h = 6.63 x 10-34/s

(i) From c = X f, threshold wavelength A0 = —JO

¿0 =

(ii) Work function W0 = h f0

3.0 x 108

4.02 x 104 = 7.46 x 10~7m

= 6.63 x 10"34 x 4.02 x 1014

= 2.67 x 10~19y

414

(iii) Kinetic energy EK = h f W0 19= 6.63 X 10-« X 5.0X 10«- 2.67X10= 3.315 x 10'18 - 2.67 x 10"19= 3.048 x 10'187

The grapl/below represents the result of a certain photoelectric experiment where Ihe stopping potential Vs is plotted against the frequency /Determine from the graph, the(i) threshold frequency(ii) threshold wavelength(iii) work function of the material r\-™nNECO 2006£1S(iv) value of Planck’s constant [c = 3.0 x 10 ms , e — 1-

Vs<V)

Solution(i) Threshold frequency = intercept on f-axis

... f 0 = 4.7 x 1014Hz

(ii) From c = Xf, threshold wavelength isc 3.0 x lO 8 _ _

Xo = To = 4 ^ T o * = 6-38x10 m

(iii) Work function W0 = e x intercept on Fs-axis

= 1.6 x 1 0 '19 x 1.9 = 3.04 x 10"19y

(iv) Planck’s constant (h) = e x slopeM 0 .8 -0 0.8 _ 15

From graph, slope - ^ 6 .6 - 4 .7 1.9 x 1014

., h = 1.6 x 10"19 x 4.2 x 1 0 '15 = 6.7 x 1 0 '34ys

415

S topping PotentialThe stopping potential of a photoelectron emission is given y,

eVs = Ek

eVs = h f - W a

h f - W 0 _ EkTherefore, stopping potential = --------------~~

Where Vs = stopping potential in volt (V)e = electronic charge = 1.6 X 10-19C

Ek = maximum kinetic energy (J)Wa = work function (J)

The maximum kinetic energy of the photoelectrons emitted from a meta su ac 1 0.34eT. If the work function of the metal surface is 1.83eK, find the stopping poten a . A. 2.17T B. 1.4917 C. 1.0917 D. 0.3417 JAMB 200344SolutionMaximum kinetic energy EK = 0.34eT ; WQ = 1.83eV

h f = W0 + Ek

= 1.83 + 0.34

= 2.17eV

Stopping potential,h f - W0 (2.17 - 1.83)eT

e e0.34eV

e= 0.3417

Alternatively, we could have solved it quickly by using, eVs = EK

Ek 0.34eKStopping potential, = — = ----------

e e0.3417

Example 18Light o f wavelength 5.00 x 10_7m is incident on a material of work function 1.90el7. Calculate the(i) Photon energy(ii) Kinetic energy of the most energetic photo electron(iii) Stopping potential[Planck's constant h = 6.6 x 10~34/ s ; c = 3.0 x 108m s_1, le V = 1.6 x 10-19/]

WAEC 2001“SolutionX = 5.00 x 10_7m; Wa = 1.90eV = 1.90 x 1.6 x 10"19 = 3 04 x 10“19/- h = 6.6 x 10 '34/s; c = 3.0 x lO ^ s" 1 ‘ J '

he(i) Photo energy, E - h f = —

6.6 x 10"34 x 3.0 x 108 E = --------5 .0 0 X 1 0 -’-------- --- 3 9 6 * 10" 9'

(ii) Maximum kinetic energy EK = h f — W„

= 3.96 x 10-19 - 3.04 x 10~19

= 9.2 x 10~2OJ

416

(iii) Stopping potential Vs =h f — WQ

e9.2 x 10 20

= 1.6 x 10-1*0.5757

d u a l i t yin another it behaves '\'V nature ^ecause in one instant it behaves like a wave, whileA, of a particle wave is give^b*™ °f paitic*e- According to De Broglie, the wavelength,

Where h = Planck’s constant = 6.63 x 10~3AJs m = mass of wave particle (kg) v = velocity of wave particle (ms-1)P = momentum of wave particle (kgms *) , . 1 ««A And

An electron of mass, me,with accelerating voltage, V, has a kinetic energy, 2 e momentum, P = mev

Therefore, eV = ^mev1 2

Combining this equation and De Broglie’s, the following equations can be derivedI. eV = ^m ev2

or 2eV = mev22eV

Rearranging, —— = mev

2eV , hSubstitute mev = -----into A = ——v mev

hA ~ 2eV

vhv

II. eV 1 2 = - m evz2 e

_ 2eV~ rne

v =

Substitute the above equation into P — me vP = mev

P - m e2eVm,

417

2 eVmj

J me

Substitute P = y]2eVme into A = —

yf2eV rne

All symbols and constants maintain their usual meanings and values.

Example 19 ... velocityCalculate the wavelength o f an electron of mass 9.1 x 10 kg moving wiof 2.0 x 106m s_1. [plank's constant = 6.63 x 10~3*Js ]Solution 34.Mass o f electron me = 9.1 x 10~31kg; v = 2.0 X 106ms l ; h — 6.6 x

From De Broglie,mev

6.63 x 1 0 '34 9.1 x 10"34 x 2.0 X 106

= 3.64 x 10"10m

Example 20If electrons are accelerated from rest through a potential difference of 1 OK V; what is the wavelength of the associated electrons?A. 1.22 x 10_11m B. 3.87 x 1 0 -lom C. 2.27 X 10u m D. 2.27 X 1014m [me = 9.1 x 10~31kg, e = 1.6 x 10"19C, h = 6.6 x lO "34;«,] JAMB 20084S SolutionAccelerating potential difference V = 10KV = 10000K = 1 x 104V Mass o f electron, m e = 9.1 x 10~31kg wavelength,A =?Electronic charge, e = 1.6 x 10-19C Planck’s constant, h = 6.6 x 10_34/ s

A =6.6 x IQ '34

j2 e V m e V2 x 1.6 x 10“19

Example 21The mass and wavelength of a moving electron are 9.0 x 10_31fc^ and 1.0 X 10- lo m respectively. Calculate the kinetic energy o f the electron, [/i = 6.6 x 10-34/s ]

WAEC 2001*®Solutionm e = 9.0 x lO -^kg-, A = 1.0 x 1 0 -10m; h = 6.6 x 10"34/5

h hFrom, A = ----- , velocity v = — -

m ev m eA

Kinetic energy KE = ^ m ev 2

, h 1Substitute v = — - into KE = - m .v L

m eA 2 e

1 / h \ 2 KE = - x mc x (— -

2 \meA/

418

1 m e h2 ~ 2 X rfe X V

hZ = _____ (6.6 x 1(T34)z 4.356 X 10~672meA.2 2 x 9.0 x lO“31 x (1.0 x n r 10)2 _ 1.8 X 10"iU

= 2.42 X 10~17J

Example 22Calculate the minimum wavelength of X-rays when a voltage of 60KV is applied to an X- ray tube, [e = 1.60 X 10“19C; h = 6.6 x 10“34/s; c = 3.0 X 108ms_1]

NECO 2000£1SSolutionV = 60XV = 6 x 10'4!/’; e = 1.60 x 10-19C;h = 6.6 x 10_34ys Velocity V = 3.0 X 108ms_1

hv 6.6 x 10-34 x 3.0 x 10----- ------------------------------------- 1 X 102eV 2 x 1.60 x 10“19 x 6 x 104

H eisenberg U ncertainty PrincipleAccording to Heisenberg, it is not possible to get an exact measurement of both the position and momentum of an atomic particle at the same time. If the electron position x is measured to a high degree of accuracy, then the simultaneous measurement of the momentum will be to a very low degree of accuracy.

The Heisenberg uncertainty principle states that if Ax is the uncertainty in measuring x, and AP is the uncertainty in measuring P, then their product is equal to or greater than h, the Planck’s constant

That is, Ax. AP > h

Therefore, Ax or AP > —AP Ax

The uncertainty principle can be applied to the measurement of the energy E of an object at a particular time t. If the uncertainty in E is AE and the uncertainty in t is At the principle is represented as follows

AE.A t> h or h .

AE.At > —2n

In all the above equations, Ax.AP.AE and At are the uncertainties in the position, momentum, energy and time respectively.

Remember that momentum, (P) = mass(m) X velocity (v).

Example 23If the uncertainty in the measurement of the position of a particle is 5 x 10_1°m, theuncertainty in the momentum of the particle is A. 1.32 x l()-442VsB. 3.30 x 10~**Ns C. 1.32 x 10' 2*Ns D. 3.30 x 10"24/Vs[h = 6.6 x 10- 34/s] JAMB 200342Solution 0Uncertainty in position Ax = 5 x^lO mPlanck’s constant h = 6.6 x 10 Js

419

h 6.6 x 10 _ m " 24A/sUncertainty in momentum, AP > - — > —— . n_iF" — A AAx 5 X 1U

From uncertainty principle Ax. AP > h

Example 24The uncertainty in determining the duration during which an particular energy level before returning to the ground state is 2.0 uncertainty in determining its energy at that level.[Take ^ = /i = 1.054 x l ( T 34;s]Solution

electron remains in a X 1 0 '9s. Calculate the

WAEC 2006£8

Uncertainty in time, At = 2.0 x 1 0 '9s; — = 1.054 x K T 34/ 2n

From uncertainty principle, AE. At > —2n

h i 1Uncertainty in energy level, AE > — x — > 1.054 x 10 34 x 2 q x 10-9

> 5.27 x 1(T26;

Example 25The uncertainty in the velocity v of a moving electron of mass 10~30kg is 3 x 106m s Calculate the uncertainty of the simultaneous measurement of its position X.[h = 6.62 x lO"34;*] NECO 2005E1°

SolutionMass of electron, m = 10~30 kg; Velocity of electron, v = 3 x 106m s-1

Uncertainty in momentum, AP = A(m x v)Ax. AP > h

h h „ 6.62 x 10-34Ax > — or Ax > —------- r ••• Uncertainty in momentum. Ax > ----^ --------------z

AP A(m x v ) J 10_3° X 3 X 106

Ax > 2.21 x 10- lo m

E X E R C IS E S1. An electron jumps from one energy level to another in atom radiating 4.5 x 10_19joules. If Planck’s constant is 6.6 x 10-34/s , what is the wavelength o f the radiation?[Take velocity of light = 3 x 108m s-1] WAEC 198957Ans: 4.4 X 10-7m2. An atom radiates 1.5 X 10~19J of energy when an electron jumps from one energy level to another. What is the wavelength of the emitted radiation?[Plank's constant = 6.6 x 10~34/s; speed of light in air = 3 x 108m s-1 ]

WAEC 199739 Ans: 1.32 x 1 0 -6m3.

Fig 6.2

l?f

0

p

h fII, , I , , V, r

O.OeV-2 e V—5.7eV-12 .0eV

The diagram above illustrates the energy transition of five electrons of an atom.(i) Which of the transition will produce the emission of longest wavelength?(ii) Which of the transitions will produce emission of highest frequency?

WAEC 199942 43 Ans: (i) 1 (ii) v

420

-27

4. In which of the following transitions is the largest quantum of energy I'ban hydrogen atom when the electron changes energy levels.' _ _[n is the quantum number] A. n = 2 to n = 1 B n = 1 t 0 " 47 Anu. AC. n = 2 to n = 3 D. n == 3 to tt = 2 WAEC 200 « * u ,5. The energy associated with the photon of a radio transmission atA. 1.30 x 10~29/ B. 2.00 X 10_29y c. 1.30 x 1 0 '28/ D. 2 .0 0 x 1 0 ' 2[h = 6.63 X 1 0 '34/s] JAMB 200342 Ans: 1.32 X 10 ^6. The energy associated with the emitted photon when a mercury atom c ange from one state to another is 3.3eT. Calculate the frequency of the photon.A. 8.0 x 1014Mz B. 3.1 x 1052Wz C. 1.3 x 10 ' 15Hz D. 3.2 x 10 J Hz [e = 1.6 x 1 0 '19C ; / t = 6 .6 x 1 0 '34ys] JAMB 20068 Ans: 8.0 x 10 Hz

7. 1 he energy E of a photon and its wavelength are related by EX = X. The numerical value of X is A. 6.60 X 10-26 B. 1.99 X 10"25 C. 1.99 X 10D. 6.60 x 1 0 '28 [/r = 6.63 x 1 0 '34ys; c = 3 x 108m s_1]

JAMB 200726 Ans: 1.99 X 1 0 '8. An electron makes a transition from - \.60eV energy level to -lOAeV energy level.Calculate the energy loss due to transition. NECO 2004s6 Ans: 8.8eV9. An electron of charge 1.6 x 10_19C is accelerated in vacuum from rest at zero volt towards a plate of A0KV. Calculate the kinetic energy of the electron.

WAEC 198858 Ans: 6.4 x 1 0 "15/10. An electron of charge 1.6 x 1 0 '19C is accelerated in vacuum from rest at zero volt towards a plate of 40KV. Calculate the kinetic energy of the electron.

WAEC 1991s7 Ans: 6.4 x 10_15y11. An electron is accelerated from rest through a potential difference o f 10KV in a vacuum. Calculate the maximum speed acquired by the electron, [electronic charge = — 1.6 x 10~19C; mass of an electron = 9.1 x 10"31k^]

WAEC 1998s3Ans: 1.57 x1087ns_112. A metal is illustrated with a radiation of energy 6.88eT. If the kinetic energy o f the emitted electrons is 1.50eT, calculate the work function of the metal.

WAEC 199 056 Ans:5.3eK13. A metal has a work function of 4.375eT. Calculate its threshold frequency.[/i = 6.6 x 10'34ys, leV = 1.6 x 10'19y] WAEC 199945 Ans: 1.06 x 1015Hz14. If light with photon energy 2eV is incident suitably on the surface o f a metal with work function 3eV, thenA. No electron will be emittedB. The few electrons emitted will have a maximum kinetic energy of leVC. The few electrons emitted will have a maximum kinetic energy of 3eV.D. Many electrons will be emitted JAMB 199450 Ans- A15. A light o f energy 5eV falls on a metal and the electrons with a maximum kinetic energy o f 2eV are ejected. The work function of the metal isA. OAeV B. 2.5eV C. 3.0eV D. 7.0eV JAMB 1995s0 Ans- 3 Oe^16. The work function of a metal is 2.1eV. Which of the following pairs correspond to the threshold frequency and wavelength of the metal respectively'' f/t = 6 6 X 10-34/c = 3.0 x 108m s_1, leV = 1.6 x 10"19y ] ' JS’A. 4.6 x 10~26Hz, 6.5 x 1034mB. 4.6 x 1 0 '7Hz, 6.5 x 1014mC. 6.5 x 1014Hz, 4.6 x 1 0 '7?nD. 6.5 x 1014Wz, 4.6 x 107mE. 6.5 x 1034Wz, 4.6 x 1 0 '26m NEC0 2006S5 ^17. A light of wavelength 5.0 x 10 m is incident on a metal resulting inphotoemission of electrons. II the work function of the metal is 3.04 x 10_19y Calculate the ’ C(i) frequency of the light

421

(ii) energy of the incident photon(iii) maximum kinetic energy of the photoelectrons _34

[speed of light = 3.00 x 108m s-1; Planck's constant = 6.6 x 10 Js]WAEC 1998£4 Ans: 0)6 .0 x 1014Wz ( it)3.96 x 10-19/ (iit)9.2 x 10 /18. Light o f energy 5eV falls on a metal of work function 3eV and electrons areliberated. The stopping potential is A. 15.OP B. 8.0P C. 2.OP D. 1.7P

JAMB 200041 Ans: 2.0P19. Light of energy 5.0eP falls on a metal of work function 3.0eP and electrons are emitted, determine the stopping potential. [Electronic charge, e = 1.60 X 10 J

WAEC 20 0 548 Ans: 2.0P ^20. An electron of mass 9.1 x 10~31kg moves with a velocity of 4.2 X 10 m s between the cathode and anode of an x — ray tube. Calculate the wavelength. ^[Take Planck's constant h = 6.6 x 10-34/s ] WAEC 2000£1° Ans: 1.73 x 10 m21. A photon of wavelength 6.0 x 10-7m behaves like a particle o f a certain mass. The value of that mass isA. 1.1 x 10~3Skg B. 3.5 x 10~36kg C. 2.2 x 10~27kg D. 2.2 x 10 k g[h = 6.63 x 10-34/s , c = 3 x 108m s-1] JAMB 2 0 0 724 Ans: 3.5 x 1 0 '36kg22. If Ax is the uncertainty in the measurement o f the position along the x-axis and APX is the uncertainty in the measurement of the linear momentum along the x-axis then the uncertainty principle relation is given asA. APxAx > h B. APxAx = h C. APxAx = 2n D. APXAx = 2 n /h E. APxAx = 1

NECO 200760 Ans: A23. What is the uncertainty in the measurement o f time, if the uncertainty in measuring the energy of an electron of mass 10~30kg is 2.45 x 10-19/?[h = 6.6 x 10-34/] NECO 2 0 0 460 Ans: 2.69 x 10-15s24. The uncertainty in the energy of a particle is 1.0 x 10 ~10J determines its uncertainty in the time measurement.[Planck's constant = 6.63 x 10-24s]

NECO 2 0 0 860 A n s : 6.63 x 10-24sThe table below shows the energy distribution for various levels o f an atom. Use itto answers question 25 and 26.Energy level (n) 1 2 3 4Energy (eV) -13.6 -3.39 -1.51 -0.85

[h = 6.6 X 10 3*Js; e = 1.6 x 10 19C; c = 3.0 x 108m s-1 ]25. Calculate the first excitation energy of the atom.

WAEC 200946 Ans:1.60 X 10-1826. If the atom de-excites from n = 2, what is the wavelength o f the emitted radiation?

WAEC 200947 Ans: 1.2 x 1 0 -ym27. An electron of mass 9.1 x 10-31 kg moves with a speed o f 107ms '. Calculate the wavelength o f the associated wave, [h = 6.6 X 10~34Js]

WAEC 200950 Ans:7.25 x 10-11m28. An electron jumps from one energy level o f -1.6 eP to one of-10.4 eV in an atom Calculate the energy and wavelength of the emitted radiation.[h = 6.6 x 1 0 '34Js; eP = 1.6 x 10 "19J; c = 3.0 x 108m s " 1]

WAEC 2009EI5 Ans: AE = 8.8 eV o r 1.41 x 1 0 -8J, A = 1.4 x 10~7m29. The work function o f a metal is 1.06 x 10_18J. What is the threshold wavelength o f the metal? [Planck’s constant h = 6.6 x 10_18J, c = 3.0 x 10m s_1]

NECO 200956 Ans: 1.9 x 10-7 m30. An electron makes a transition from an energy level o f -1 .5 1 e P to that o f — 13.6eP in an atom. Calculate the

(i) loss o f energy due to the transition in joules,(ii) frequency o f the emitted radiation.

[ le P = 1.6 x 10-19], h = 6.6 x 10-34JS]NECO 2009EI5 Ans: (i) 1.93 x 10-18j (ii) 2.92 x 10l s Hz

422

There is no democracy in physics. We can’t say that some second-rate guy has as much right to an opinion as Fermi.

Luis Alvarez (1911 - 1988)U.S. physicist.Referring to Enrico Fermi, who achieved the first controlled nuclear reaction.

423

7RADIOACTIVITY AND NUCLEAR REACTION

RADIOACTIVITY {Radioactiv n\ is defined as the spontaneous disintegration ol an unstable atomic t »• with the omission of u. p’ or y - rudtwUtws atnl roloaso ol energy. I in > nl 11 ,Kelement has its unique decay constant anil half life. ___

I'lie disintegration or decay o f radioactive element or atom is a ratuotn proots, that is totally independent of ohemioal ootnhination, temperature and pressurt. t occurs when an unstable nucleus usually with atomic number greater than So undergoes n.itura decay in a bid to achieve a stable condition. . .f ig 7,1 shows a typical decay curie for a disintegrating radioactive atom o tame > plotting the number of atoms present against half-life. If the initial number ol atom is i at time 7\ and falls to "} at time 7\, then the halt-life (T) is given by; T — T2 ~ Tx . In uk same manner, the half-life is also given by; T = 7'< — T2. if the number ol atom reduces from -> to -k

Fig 7.1 Half life and Decay•

Half Life: Equations of Radioactive DisintegrationI'he half life o f a radioactive clement is the time taken for ha lf the atoms o f the element to decay. Half life could also be defined as the time taken for a given mass o f radioactive substance to disintegrate to lull/ its initial mass.

Assume that a radioactive element with half life of 5seconds contains 192 atoms initially.

Aller the first 5seconds (I half-lie), % atoms would have decayed and % atoms will be left.

After 10 seconds (2 half-lives), 144 atoms would have decayed and 48 atoms willremain.

After 15 seconds (.1 half-lives), IhS atoms would have decayed and 24 atoms willremain.

After 20 seconds (4 half-lives), ISO atoms would have disintegrated and 12 atoms will be left.

After 25 seconds (5 half lives), ISO atoms would have disintegrated and 0 atoms will remain.

424

The following new equations can be used in calculations involving half-life- It is fundamentally different from the conventional method you are familiar with.

1. H alf life, T = t _ t X log2 .............7 ilog2 R ~ logR

^Where R = 2n = — and n — j,J

2. Number of atoms decayed, Nd = Nr - N2(R ~ 1\

= "■(— )

3. Fraction remaining undecayed, / r = - ^ = —

7.2

7.3

4. Fraction of atoms decayed, f d = = —- — ................................... 7 .4

WhereT = half life of radioactive element t = time taken for radioactive element to decay

n = number of half livesNl = initial mass or initial number of atoms present/initial count rate N2 = final mass or final number of atom remaining undecayed/final count rate Nd = number of atom or mass of atom that has decayed or disintegrated. f t = disintegrating ratiof r = fraction of initial number of atoms remaining undecayed. f d = fraction of initial number of atoms that has decayed

Though the term, “ disintegrating ratio” is a newly coined expression, it is NOT a new

or additional concept in physics. It does not contradict any term or concept in

radioactivity. It is simply a coined name for an established relationship [—■ = 2" j and

its modification, R = = 2” and simplified application in solving radioactive decay

problems.

For purpose of identification and reference, equation 7.1 should henceforth be called, Zhepwo radioactive equation. Other derived equations (equations 7.2, 7.3 & 7.4) should be referred to as Zhepwo derivative(s), so as to differentiate between the two groups of equations.

Details of how these equations are derived; their relationship with the decay constant and decay curve, and proof of (heir applicability to WAEC, JAMB and NECO questions and examples and exercises from fifteen physics textbooks are contained in a ground-breaking book by Solomon Dauda Yak wo: Can These New Equations Significantly Simplify Half-Life Calculation? (ISBN 978-978-49081 -5-3)

A good knowledge of the theory of logarithm is necessary for using some of the above equations nevertheless, like the multiplication table, it is advisable to commit th following to memory.

425

!f R = T Then log2 R = n

" I f 2 — 21 Then4 = 22 8 = 23

16 = 2*32 = 25 64 = 26

128 = 27 256 = 28 512 = 29

1024 = 210

Example 1In 24 days, a radioactive isotope decrease in mass from 64g to 2g. What is the half life of the radioactive material? WAEC 199458

log2 2 = 1 log2 4 = 2 log2 8 = 3 log2 16 = 4 log2 32 = 5 log2 64 = 6 log2 128 = 7 log2 256 = 8 l°g2 512 = 9 log2 1024 = 10

Example 1 .In 24 days, a radioactive isotope decrease in mass from 64g to 2g- What is the half life of the radioactive material? WAEC 1994s8

Solution: Conventional method64g —► 32g = 1 half-life

32g —► 16g = 2 half-life

16 —* 8g = 3 half-life

8g —► 4g = 4 half-life

4g —> 2g = 5 half-life

If 5 half-life is equal to 24 days

Then 1 half-life will be T

/. T x 5 half-life =1 half-life x 24

Solution: Zhepwo methodDecay time, t =24 days; Half-life, T — ? initial mass, = 64g;Final mass remaining, N2 = 2g

Nx 64R = t = T = 32T =

log2Ä24 24

1----- r r = — = 4.8 dayslog2 32 5

days^ 1 half life x 24 daysT = 5 half life

4.8 days

Example 2The half life of a radioactive material is 6 hours. What quantity of 1 kg of the material would decay in 24 hours? WAEC 199752 Solution: Conventional method After 6hrs, ^ kg decays, ^ kg remains

After 6hrs, ^ of^ kg decays, ^ kg remains

1 1 1After 6hrs, - o f- kg decays, - kg remains

After 6hrs, ^ of^ kg decays, kg remains

Therefore, the material decayed would be:

Example 2The half life of a radioactive material is 6 hours. What quantity of 1 kg o f the material would decay in 24 hours? WAEC 199752 Solution: Zhepwo method Half life T = 6 h; initial mass Nx = lk g -, mass of material decayed, Nd = ?

R = 2n = 24 = 16

Na= Wl(^ T i ) = u s ( i i ^ )

426

I k g - i ^ K g IS= 1 6 *«

Example 3A radioactive sample initially contains N atoms. After three half-lives the number of atoms that have disintegrated is A.% S .H c D. —

JAMB 1985S0

Solution: Conventional method After 1 half-life, j decay, ~ remain

After 2 half-life, ^ of j decay, ^ remain

After 3 half-life, ^ o f^ decay, j remain

Example 3A radioactive sample initially contains N atoms. After three half-lives the number of atoms that have disintegrated is

»■ T5N

' 8JAMB 19855

Solution: Zhepwo methodInitial number of atoms Ni = N; number of half lives,n = 3; Numbers of atoms decayed, Nd =?

R = 2n = 23 = 8Number of atom disintegrated = sum of

disintegrated atoms or fractions

N N N IN

Nd

Example 4After three half -lives, the fraction of a radioactive material that has decayed is

B. i C .fa - ; D .-8

JAMB 19924

Example 4After three half -lives, the fraction of a radioactive material that has decayed is

B.Aj i c . -3 3 D. - 8

JAMB 199249

Solution: Conventional method1 half-life => \ decays, i remains

2 half-life => J decays, J remains

3 half-life => j decays, J remains

Fraction decayed = original fraction - remaining fraction

1 7= 1 ~ 8 ” 8

Note that original fraction is 1 or -

Solution: Zhepwo methodNumber of half-lives, n = 3;

fraction decayed f d =?

Disintegrating ratio, R = 2n = 23 = 8

f R ~ 1 8 - 1 7f d ~ R ~ 8 ~ 8

Example 5 .The half - life of a radioactive element is 24 hours, calculate the fraction of the original dament that disintegrated in 96 hour 2006

Example 5The half - life of a radioactive element is 24 hours, calculate the fraction of the original element that would have disintegrated in 96 hours NECO 2006s8

427

Solution: Conventional methodAfter 24hrs, j disintegrate, j remain

After another 24hrs, ^ decay, j remain

After a further 24hrs, - decay, 7 remain

After another 24hrs, 7- decay, — remain 16 16

Fraction disintegrated = sum of decayed fractions

1 1 1 J _ _ 15“ 2 + 4 + 8 + 16 " 16

Solution: Zhepwo methodHalf life, T = 24h; decay time.t = 96h\

fraction disintegrated, fa =?t t 96h A

T ~ n * ~ T ~ 2 4 h ~ *R = 2n = 24 = 16

R - 1 1 6 - 1 15~ R ~ 16 ” 16

Example 6A radioactive isotope has a half life of 8 days. What fraction of the atoms will remain after 72 days? NECO2004s9 Solution: Conventional method Let n be the original number of nuclei. After 8 days, j disintegrate, — remain

After 8days, ^ of j decay, j remain

After 8days, ~ ° f ~ decay, j remain

After 8days, “ of “ decay, remain

After 8days, ~ ° f ~ decay, remain

After 8days, - of — decay, — remain

After 8days, ~ of decay, remain

After 8days, \ of decay, remain2 128 266

After 8days, 7 of -7- decay, -7- remain

••• After 72 days, fraction of atoms that will remain is -7- or 1.953 x 10-3

Example 6A radioactive isotope has a half life o f 8 days. What fraction o f the atoms will remain after 72 days? NECO2004s9 Solution: Zhepwo method half-life, T = 8days; decay time, t = 72 days; fraction remaining f r =?

R = 2n = 29 = 512

1Fraction remaining, f r = —

R1

5 Ï2= 1.953 x n r 3

Example 7A radioactive substance has a half life of 20 hours. What fraction o f the original radioactive nuclei will remain after 80 hours? WAEC 1990ss Solution: Conventional method Let n be the original number of nuclei.

After 20 hours, ^ disintegrates and ^ remains.

After a further 20 hours, '/2 o f ^

Example 7A radioactive substance has a half life of 20 hours. What fraction o f the original radioactive nuclei will remain after 80 hours? WAEC 199055Solution: Zhepwo method Half life, T = 20hrs;

decay time, t = 80hrs

t 80Number of half lives, n = - - — = 4

428

disintegrates and ^ remains.

After another 20 hours, Vi of ^ disintegrates and ^ remains.

After another 20 hours, Vi of - 8

disintegrates and remains.

Therefore, after 80 hours, fraction of the original number remaining would be —

Disintegrating ratio,/? = 2" = 2* = 16

Fraction remaining, fr = 1 = 1

Alternative direct method T = 20; t = 80

1 1Fraction remaining, fr = ^ = ^ 7

- 1 - 1.

~ 2* ~ 16

Example 8Two radioactive elements A and B has half-lives of 100 and 50 years respectively. Samples of A and B initially contain equal number of atoms. What is the ratio of remaining atoms of A to that of B after 200 years? WAEC 198860 Solution: Conventional method Let n be the original number of nuclei Sample A: half-life = lOOyrs After 100 years, ^ disintegrates and 7 remains.After another 100 years, ^ disintegrates and ^ remains.Therefore, after 200 years, fraction of the original number remaining would be -

Sample B: half-life = 50yrs After 50yrs, j disintegrate, j remain After 50yrs, ^ decay, ^ remain After 50yrs, j- decay, ^ remain After 50yrs, decay, ^ remain

After 200yrs, fraction of atoms that will remain is ^

Ratio of A: B = ~ (multiply both sides by 16)

Ratio of A: B = 16 Q ) : 16 ( ^ ) = 4:1

Example 8Two radioactive elements A and B has half-lives of 100 and 50 years Samples of A and B initially contain equal number of atoms. What is remaining atoms of A to that of B after 200 years? WAEC 198860 Solution: Zhepwo method

Half life T

Decay time, tt

Number of half lives n - -

Disintegrating ratio R = 2"

A B100 50

200 200

2 4

4 16

respectively, the ratio of

429

41

16

N

Atoms remaining N2 = —

(initial no of atoms /V, = l)

Multiply by 16

Ratio of A to B

16 x -4 16 x A1

Example 9A radioactive substance has a half life of 80 days. If the initial number of atoms in the sample is 6.00 x lO10, how many atoms would remain at the end of 320 days? A. 3.75 x 109 B. 7.50 x 109

C. 3.00 x 1010 D. 5.63 x 1010 JAMB 199848

Solution: Conventional method

After 80 days, ~ (6 x 1010 atoms) decay, 3 x 1010 atoms remain

After 80 days, | ( 3 x 1010 atoms) decay, 1.5 x 1010 atoms remain

After 80 days, ^ (1.5 x 1010 atoms) decay, 7.5 x 109 atoms remain

After 80 days, ^ (7.5 x 109 atoms) decay, 3.75 x 109 atoms remain

after a total of (80 + 80 + 80 + 80)Or 320 days 3.75 x 109 atoms remain.

Example 10The percentage of the original nuclei of a sample of a radioactive substance left after 5 half lives is A. 1% B. 8%C. 5% D. 3% JAMB 200245Solution: Conventional methodPercentage of original left

amount left (fraction remaining)

original amount

Example 9A radioactive substance has a half hie ol 80 days. If the initial number of atoms inthe sample is 6.00 x lO 10, how manyatoms would remain at the end of 320 days? A. 3.75 x 109 » .7 .5 0 x 1 0 *

C 3.00 x 1010 D. 5.63 x 1010 JAMB 199848

Half life 7 = 80 days; 1Qinitial no. of atoms = 6.00 X 10 decay time, t = 320 days;Final number of atoms remaining N2

_ t t 320T ~ n ” " ~ T ~ 80 "Disintegrating ratio, R = 2” = 24 — 16

Nx 6.00 x 1010NxR = W N2

N? = —r =R 16

= 3.75 x 109atoms

Alternative direct method

T = 80; t = 320; JVt = 6.00 x 1010

NrN? = 2 t/T

6.00 x 1010 6.00 x 1010— 2320/80 2*

6.00 x 1010------—------= 3.75 x 109 atoms

Example 10The percentage of the original nuclei of a sample of a radioactive substance left after 5 half lives is A. 1% B. 8%C. 5% D. 3% JAMB 200245Solution: Zhepwo methodPercentage of original left

amount left (fraction remaining)----------------7-7— ;------------------ — x 100original amountNumber of half lives, n = 5

430

1 half-life

2 half-life

3 half-life

4 half-life

5 half-life

i decays, ^ remains

i decays, ^ remains

2

l 4

^ decays, ^ remains

7- decays, — remains

~ decays, — remains

after 5 half-lives, — o f the original

nuclei is left. Note that initial fraction

present is 1 o r -l

Percentage of original left

1 /3 21

x 100

= — x Klo = 3.125 = 3%

Exam ple 11A radioactive substance o f mass 7680 has a half life of 3years. After how many years does this substance leave only 6g undecayed? WAEC 200648Solution: Conventional method

After 3 years,7 6 8 g _

2384g

After 3 years,3 8 4 g _

2192g

After 3 years,192g_

2: 96g

After 3 years,9 6 g _2

48g

After 3 years,4 8 g _2

24g

After 3 years,24g _2

12g

After 3 years,i 2g _2

: 6g

Thus making a total of

(3+3+3+3+3+3 +3 )yrs or 21yrs

Exam ple 12 .An element whose half-life is 10 days ,s o f mass 120. Calculate the time during which 11.250 o f the element would have

Original am ount or fraction N\ = 1

Disintegrating ratio, R = 2n = 2s = 32

1 1Fraction remaining, f r = — = —

R 3Zf r

••• percentage left = y x 100

= y ^ x l 0 0 = y X l 0 0

= 3.125 = 3%

Example 11A radioactive substance o f mass 768g has a half life of 3years. After how many years does this substance leave only 6g undecayed? WAEC 200648Solution: Zhepwo method

Half life T = 3years; initial mass present, N1 = 768g; final mass remaining, N2 = 6g; decay time, t =?

. Nt 768Disintegrating ratio, R = — = -----

N2 6

= 128

t = T x log2R

t — 3 x Zo02128 — 3 x 7 = 21 years

Example 12An element whose half-life is 10 days is of mass 120. Calculate the time during which 11.250 o f the element would have

431

decayed? NECO 2005s7Solution: Conventional method10 days => 6g decay, 6g remain

10 days => 3g decay, 3g remain

10 days => 1,5g decay, 1.5g remain

10 days =* 0.75g decay, 0.75g remain

••• after 40 days, (6 + 3 + 1.5 + 0.75)g or

11.25g would have decayed

decayed?Solution: Zhepwo

NECO 2005s7method

Half life T ~ 10 - 1 2 g;

Nd = Nl - N 2 > _ Nfinal mass remaining, N2 1

N = 1 2 -1 1 .2 5 = 0.750

/? - — = — = 16 * ' JV2 0.75

T ____L _ ... t = r io g 2 fllog2Æ

... t = io X log2 16= 10 x 4 = 40 days

; from 1000 to 150 in 6 days. What is the half hfeExample 13A radioactive element decrease in mass : of the radioactive material?Solution _ 1 -Initial mass present, Nt = 1000; final mass remaining, 2 0»Decay time, t = 6 days

* i 100Disintegrating ratio, R = — = -yg- - 6.66/

t 6 daysHalf life, T = = ,og2 6 667

While 4,8,16,32 can easily be expressed as 22,2 8, 24, 2s respectively, it is very clear that expressing 6.667 as 2 raised to the power of a given number will be a very, very difficult task. So, when you encounter such a situation, you are advised to use the variant of the Zhepwo radioactive equation, equation 7.1 on page 118

t x log2 6 days x log2 _ 6 days x 0.30103 logR “ log6.667 " 0.824 2.2 days

log 2 and log 6.667 can be obtained or solved directly from your scientific calculator

Example 14The time it will take a certain radioactive material with a half-life of 50 days to reduce to 1/32 of its original number is A. 300 days B. 150 days C. 200 days D. 250 days JAMB 20059Solution: Conventional method

1 half-life, 50 days => reduces to -2

2 half-life, 100 days =* reduces to -4

3 half-life, 150 days => reduces to -

Example 14The time it will take a certain radioactive material with a half-life of 50 days to reduce to 1/32 of its original number is A. 300 days B. 150 days C. 200 days D. 250 days JAMB 200S9Solution: Zhepwo method

Half life T — SO days; decay time t =?

fraction remaining f r =

432

4 half-life, 200 days =* reduces to ^

5 half-life, 250 days =* reduces to

time taken to reduce to ^ o f original

number is 250 days

Rearranging,t

T ~ log2R

R = 32

t = T X log2 R

t = 50 x log2 32

= 50 x 5 = 250 days

Example 15A radioactive substance has a half-life of 3 minutes. After 9 minutes, the count rate was observed to be 200, what was the count rate at zero time? NECO 2000s8 Solution: Conventional method

Example 15A radioactive substance has a half-life of 3 minutes. After 9 minutes, the count rate was observed to be 200, what was the count rate at zero time? NECO 2000 Solution: Zhepwo method

Note that count rate at zero time is the same as initial count rate.

9 min => count rate 200

6 min => count rate (200 X 2) = 400

3 min => count rate (400 x 2) = 800

0 min => count rate (800 X 2) = 1600

count rate at zero time is 1600

Final count rate N2 = 200;half life T = 3min; decay time t = 9 minInitial count rate N2 =?Note that count rate at zero time is the same as initial count rate

Disintegrating ratio R = 2n = 23 = 8

R = T T •■ Ni = N2 x RN2

Nx = 200 x 8 = 1600

Example 16The count rate of a radioactive material is 800 count/min. If the half life of the material is 4 days, what would the count rate bel6 days later? A. 200 count/min B. 100 count/min C. 50 count/min D. 25 count/min JAMB 200340 Solution: Conventional method

Example 16The count rate of a radioactive material is 800 count/min. If the half life of the material is 4 days, what would the count rate bel6 days later? A. 200 count/min B. 100 count/min C. 50 count/min D. 25 count/min JAMB 20034° Solution: Zhepwo method

Initial count rate = 800 count/min, half-life = 4 daysafter 4days => ^(800) = 400 count/min

after 8days => ^(400) = 200 count/min

after 12days => ^(200) = 100 count/min

after 4days=> ^(100) = 50 count/min

a the count rate 16 days later is 50

Initial count rate, N2 = 800; Half life, T = 4 days;Final count rate N2 —1 decay time, t = 16 days

Disintegrating ratio, R = 2n = 24 = 16

R Ni . ' = ^ i = 800N2 " 2 R 16

count/min - 50 count/minAlternatively,

N2 800 800 800~ 2tfT “ 2i6/4 “ 24 “ Ì 6"

= 50 count/min

433

E xam ple 17The half life o f a radioactive source is 1 minute. If a rate meter connected to the source r e n te r s 200pA at a given time, what would be its reading after 3 minutes? WAEC 199857Solution: Conventional method

A rate meter measures the count rate of radioactive substance.

Half-life = lm in

After 1 min, rate meter reads - (200//) = lOOp

After 2 min, rate meter reads - (100//) = 50pAfter 3 min, rate meter reads - (50//) = 25p••• rate meter reading after 3min is 25//

Exam ple 18In 90 seconds, the mass of a radioactive element reduces to — of its original values. Determine the half-life of the element. WAEC 199853Solution: Conventional method

Let n be the original mass of radioactive elementAfter 1 half-life element reduces to -

After 2 half-life element reduces to ^

After 3 half-life element reduces to ^

After 4 half-life element reduces to ^

After 5 half-life element reduces to ^

If 5 half-life takes 90 sec

Then 1 half-life takes T1 halflife x 90 sec _ 90sec

Hence T = -------5 t e m e ” 5= 18sec

S ^ o f a r . d . o a c . v e ^ e i s l

minute. If a rate meter connecsource registers 200/r/l at a g. vent,m e,

what would be its reading a eminutes?Solution: Zhepwo method

A rate meter measures the count rate o f radioactive substance.

Half life T = lm in ; decay tim et = 3m m initial count rate /Vi = 200pA;

final count rate N2 = ?

T = t — — = — = 3n n ~ T 1

R = 2" = 23 = 8Disintegrating ratio,_ /V 1 = 200pA = ” R 8- *n 2

Alternative Zhepwo m ethodNx 200 200/M _ 200pA

Nz = 2 t/T ~ 23/ i - 23 _ 8

= 25 pA

Example 18In 90 seconds, the mass o f a radioactive element reduces to ^ o f its original values. Determine the half-life o f the element. WAEC 1 9 9 8 53Solution: Zhepwo m ethod

Decay time t = 90s;

fraction of initial mass remaining, f r

J _ _ 1

32 ~ R

32

Rearranging, disintegrating ratio, R = 32

Half life, T = _____ = 90 _ 2 2log2 R log2 32 5

= 18s

half-life = 18sec

434

Example 19A radioactive substance has a half-life of 3 days. If a mass of 1.55 <7 of this substance is left after decaying for 15 days, determine the original value of the mass. WAEC 200449Solution: Conventional method

Half-life = 3 days

15 days = or 5 half-life, 1.55g remain

12 days = y or 4 half-life, 2 x 1.55g =

3.1g remain

9 days = | or 3 half-life, 2 X 3.1g =

6.2g remain

6 days = | or 2 half-life, 2 x 6.2g =

12.4g remain

3 days = ^ or 1 half-life, 2 x 12.4g =

24.8g remain

0 day = ^ or 0 half-life, 2 x 24.8g =

49.6g remain

original value of the mass = 49.6g

Exam ple 20In 90 seconds, the mass of a radioactive element reduces to 1 /16 of its original value. Determine the half-life of the element. NECO 200356 Solution: Conventional method

Let n be the original mass of radioactive elementAfter 1 half-life element reduces to ^

After 2 half-life element reduces to -4

After 3 half-life element reduces to -8

After 4 half-life element reduces to —16

If 4 half-life takes 90 sec

Then 1 half-life takes T

1 halflife x 90 secHence T =

4 halflife90sec

422.5sec

A^adioactive substance has a half-life o f 3 days. If a mass o f 1.55# o f this substance is left after decaying for 15 days, determine the original value of themass. WAEC 200449Solution: Zhepwo method

decay time t — 15 days;

final mass remaining, N2 = 1-55*7

t t 15 _T ~ n " n ~ T ~ 3

Disintegrating ratio, R = 2n = 25 = 32

Nx = R N 2n 2

■■■ original mass, = 32 x 1.55

= 49.6 g

Alternative Zhepwo methodN2 = 2t/T x N2 = 21S/3 x 1.55

= 2s x 1.55 = 32 x 1.55 = 49.6 g

Example 20In 90 seconds, the mass of a radioactive element reduces to 1/16 of its original value. Determine the half-life o f the element. NECO 200356 Solution: Zhepwo method

Decay time t = 90s;

fraction of mass remaining f r

J _ _ 116 ~ R

_i_ _ 1 6 ’

Rearranging, disintegrating ratio, R = 32

Half life, T = — = — 90 =log2 R log2 16 4

= 22.5s

435

••• half-life = 22.5sec

Decay Constant ,he,„SUnun»usr,K of decayThe decay constant of a radioactive atom is tn dN •of the radioactive element. disintegrates seCOn< ’ ~dt ' 1S

For a radioactive atom, the number of atoms ^ instantdirectly proportional to the number, N, of atoms presen .............y 5

T hais. ..... .. ........................................................................ _• . . . . ca lled the decay constant. The

Where, A is a constant peculiar to the radioactive a creases. negative sign shows that N becomes smaller as time, < m Rearranging 7.5, we obtain decay constant;

- 4 ® ................................................ 7 6number of disintegrating atoms per second

^ number of atoms present at that instantBy integrating equation 7.5, the following is obtained

N = ...................................................................................Where N„ = number of atoms present at time t = 0 , .

The value of half-life T, can be derived considering that, at half-life the number ot decaying atoms is half its initial value.

That is, N = -y and t = T

eqn 7.7, N = Nae -Ai becomes

Y = »„*-"•

Dividing both side by Na we obtain;

Taking logarithm of both side to base e we have;

•oge- = log^e- -1*

Applying subtraction law and power law in the theory of logarithm, we get

l°ge 1 ~ logc 2 = — AT log,, eRemembering that Ioge 1 = 0 and loge e = 1, the equation becomes

— loge 2 = - AT

In 2 = AT 0.693 = AT

••• Decay constant, A = — 9- 7

or Half — life, T = 2^21A .................. 7.8

436

Decay constant is measured in per second, s -1 while half-life is measure m rsejfThe value of In 2 can be obtained from your scientific calculator. Don t o t r y with the process of deriving the decay constant. Equation 7.8 and equation ar

, . , , 0.693 , d N ___ i «/you need, so commit them to memory. A = —— ana — — /u’

Example 21 . , . „A piece of radioactive material contains 1020atoms. If the half-life of the mat isseconds, the number of disintegrations in the first second isA. 3.47 x 1018 B. 6.93 x 102° C. 3.47 x 102° D. 6.93 x 10

JAMB 200947SolutionNumber of atoms initially present, N = 102°, half-life, T = 20s

A =0.693 0.693

20= 0.03465

dNThe num ber of atoms disintegrating the 1st second, = — AN

— = 0.03465 X 1020 2 3.47 x 101® d t

Example 22A radioactive element has a decay constant of 0.077s-1. Calculate its half-life.

WAEC 2008s®Solution

, 0.693 0.693Decay constant A = 0.077s 1 .*. Half life, 7 = —-— = ^ - = 9sJ A 0.077

Example 23The half-life of a radioactive substance is 2 seconds. Calculate the decay constant.

NECO 200060Solution

Half life, T = 2s0.693 0.693

Decay constant, A = ——— = —-— = 0.347s 1

Example 24 Example 24A radioactive element decays to one eight A radioactive element decays to one o f its original quantity in 9 seconds. eight of its original quantity in 9 seconds.Calculate its decay constant. Calculate its decay constant.

WAEC 199857 WAEC 199857Solution! Conventional method Solution: Zhepwo method

Let n be the original mass of radioactive fraction remaining, fr = i ;element . ' s '

n r'fAfter I half-fife element reduces to - decay time t * 9s

K f . i . . fn 1 v tvVfter 2 half-fife element reduces to - f = — .*'• v-

a R^ Aftef 3 half-fife element reduces to ,^ ^ RearTangit#,^isinte

" ---- “ . f

] *> Vt

If 3 half-life takes 9 sec

Then 1 half-life takes T1 halflife x 9 sec

Half life, t =log 2 R Iog28 3

Hence T =Decay constant, A =

0.693 0.693

3 halflife

437

= 0.231s_ 9sec-----3 ~ = 3SCC

Decay constant, A = = ?:69-T 3

= 0.231s'1

- l

Relationship between Decay Constant(A) and Disintegrating R atio(ft)

foUow^^3^11 15 related to the established equation for decayconstant (A) as

A = 2.303 logRt

.........7 .9

Where R = —So N2KiverL^^]10™ ^ C3n ^ uset* to die decay constant even when the half-life is not

time (t^are number of atoms(Ari), final number of atoms (/V2) and decay

Example 25A radioactive element decays to one eight of its original quantity in 9 seconds. Calculate its decay constant

WAEC 199857Solation: Conventional methodLet n be the original mass of radioactive elementAfter 1 half-life element reduces to ^

After 2 half-life element reduces to -4

After 3 half-life element reduces to -8

Example 25A radioactive element decays to one eight of its original quantity in 9 seconds. Calculate its decay constant.

WAEC 199857 Solution: Zhepwo method

Fraction remaining, fr =

decay time t = 9s

Rearranging, disintegrating ratio R = 8

If 3 half-life takes 9 sec

Then 1 half-life takes T1 halflife x 9 sec 9sec

Hence T = ----- - ■ ------ = ——3 halflife 3= 3sec

0.693 0.693Decay constant, A = —- — = —-—

decay constant, A = 2.303

= 2.303

logRt

log89

2.303 x 0.90309 9 = 0.231s-

= 0 .231s'1

Example 26If | | th of the radioactive atoms of an element decay in 8min, find the decay constant of the element.Solation: Conventional method

Example 26If ^ th of the radioactive atoms of anelement decay in 8min, find the decay constant of the element.Solution: Zhepwo method

438

If — atoms have decayed, then the 16

fraction of atom remaining is

After 1 half-life, ^ o f atoms remain

After 2 half-life, - of atoms remain4

After 3 half-life, ^ of atoms remain

After 4 half-life, — of atoms remain 16

— of atoms remain after 4 half-life. 16

This means that after 4 half-life, the

number of atoms left is — th of the 16

Fraction decayed ,fa — ;

decay time t = 8min

_ R - 1 f a - R

15 R - 116 “ R

16 R - 16 = 15 R

1 6 R - 1 S R = 16

R = 16lo g R

decay constant, A = 2.303 f

2.303 x log!6

8min

15

original.

If 4 half-life takes 8 min

Then 1 half-life takes T

1 halflife x 8 min 8minHence T = --------- -— ------- = — -—

4 halflife 4= 2min

2.303 x 1.204 8

0.347m in'1

Decay constant, A =0.693 0.693

T 2min = 0.347min" 1

Example 27In 24 days, a radioactive isotope decreases in mass from 128g to 2g. What is the decay constant of the radioactive material?Solution: Conventional method

128g => 64g —> 1 halflife

64g => 32g —» 2 halflife

32g =? 16g —► 3 halflife

16g => 8g —* 4 halflife

8g => 4g —» 5 halflife

4g => 2g —> 6 halflife

If 6 half-life is equal to 24 days

ITtcn 1 half-life takes T

T x 6 halflife = lhalflife x 24days

1 halflife x 24daysHence, T =6 halflife

Example 27In 24 days, a radioactive isotope decreases in mass from 128g to 2g. W hat is the decay constant of the radioactive material?Solution: Zhepwo method

Decay time t = 24days;

initial mass, = 128 g;

final mass N2 = 2g

p _ " i _ 1 2 8 gN2 2g

decay constant, A = 2.303 — Rt

_ 2.303 x log64 24days

_ 2.303 x 1.806 24

= 0.1733day-1

439

__ 24daysg = 4dnys

Decay constant, A = 0-693 _ 0 693 T 4 days

= 0 .17 3 3 d ay -1

The Hic^1****"®^Ratio a,id G raphical Solutionsin egration or decay o f a radioactive substance can be represent by a decay c

en t e number o f atoms remaining is plotted against the time t. The concept o f integrating ratio could also be used to solve graphical problems as the following

examp es illustrates. Answers that should normally be obtained from the plotted grap i •can be obtained accurately by calculation.

E xam ple 28The half-life o f uranium Xx is 2 4 days. Calculate the mass remaining u n c h a n g e d o b o f the substance after(i) 24 days;(ii) 48 days(iii) 72 days(iv) 96 days(v) 120 daysPlot your answer on a graph and hence determine:(vi) the mass remaining unchanged after 84 days(vii) after how many days there will be exactly 0.25g unchanged.(A.E.B) Chapter 47, page 579. exercise 4(d) .ABBO TA.F (1999). Physics. Fifth Edition.Heiiteittann

Solution: Zhepwo m ethodHalf-life, T = 24 days ; initial mass present, Nx = 0 .64g final mass remaining, N2 =?W e are asked to find N2 for decay time, t = 24, 48 ,7 2 ,9 6 , and 120 days.

NxFrom d isin tegrating ratio, — = 2"

twe substitute, n = — to obt

(i) t = 24 days T = 24 days;

Nx 2C/rn 2

Nx• n 2

~ 2 Vr

Nx =: 0.64g;JVa

n 2 “ 2f/ r :. _ 0 . 6 4 g _ 0 . 6 4 g 0 . 6 4 g

O t / T — - 1 1 -------n — 0.21

(ii) t = 48 days; T = 24 days; Nx = 0.64gw _ 3 _ - 0 M 9 _ 0 M 9 0 .64g

2 2c/ t 248/ 24 22 ~ 4 =

(iii) t = 72 days; T = 24 days; Nx = 0 .64gN _ = ° _ 0 .64g _ 0 .64g

2 2 ^ t 272/24

(iv) t = 96 days; T = 24 days; Nx = 0 .64g

3 2 g

0-16 g

23 - ~ ^ — = 0 .0 8 g

4 4 0

A/, 0 .64//^ T / r ~ ■¿'"•/¿'i

0.64// __ 0 M ^ _ o o 4/7

2 4 16

(v) t — 120 days; T = 24 days; TV, = 0.64///Vj 0.64//

~ ¿77? ~ 21Z0/Z4

0.64// 0-64// _ 0 02 g2s 32

Solution: Conventional m ethod(i) After 24 days, (0 .64«) decay, - (0 .64«) i.c. 0.32« remain

(ii) After 24 days, 0.32« decay, 0.32« remainAfter 48 days, i (0 .32«) decay, £ (0 .32g) i.c. 0.16« remain

('ii) After 24 days, 0.32« decay, 0.32« remain After 48 days, 0.16« decay, 0.16« remain

After 72 days, ^ (0 .16g) decay, - (0 .16«) i.c. 0.08« remain

(iv) After 24 days, 0.32« decay, 0.32« remain After 48 days, 0.16« decay, 0.16« remain

After 72days, 0.08« decay, 0.08« remain After 96 days, ^ (0.08g) decay, ^ (0.08g) i.e. 0.04g remain

(v) After 24 days, 0.32g decay, 0.32« remain After 48 days, 0.16g decay, 0.16g remain

After 72days, 0.08g decay, 0.08« remain After 96days, 0.04g decay, 0.04« remain

After 120 days, 4 (0 .04g) decay, ^ (0 .04g) i.e. 0.02g remain

Though the answers to (vi) and (vii) are to be determined from the graph, equations could also be used as shown below. Students are advised to plot the graph and com pare the ir answers with those obtained through calculation.

(vi) “. ..m ass remaining unchanged after 84 days... " t = 84 days; T = 24 days Nx = 0.64

Nt 0 .64 g 0 .64 g 0 .6 4 g~ 2 f ^ ~ 2 84/<24 _ ~ 2 f r ~ 11 .314 = 0 '0 5 6 ^

(vi) From, .. .after how many days there will be exactly 0.25g unchanged ... we obtain: Final mass remaining, N2 = 0 .25g; decay tim e t = ?

T = 24 days /V1 = 0 .64g

Ni 0 .64//D isin teg rating ratio R = — = — -g-~ = 2.56

T x logR _ 24 days x log2.56 24 days X 0 .4 0 8 2 4log2 log2 0.301

= 32.5 days

t x log2logR

441

Transformation of Elements through Radioacti' hen ¡t emits a - In natural radioactivity, a radioactive element u n d e rg o e s c . ef A and atomic nu ß-particle or y- rays. If a radioactive element X with mass n“ ^ eicnicnt Y, the nuc ca Z emits an alpha particle (Helium nucleus, 2 e) to orrn an equation is given by

AZX \H e + Az Z \Y

Similarly, if a radioactive element emits a beta particle (electron, element Y, the nuclear equation is given by

oe) and forms another

Emission of gamma rays (y) results in no loss of mass number and kombard radioactivity, high energy a- particle ( |He) and neutron (¿n ) are ^se 0 element thereby changing them from one form to another. Examp e o cc bombardment is;

1AN + \H e - * 1780 + \H

Example of neutron bombardment is given by

\\M g + ¿n * 2ANa + \H

Note: \H represents a proton\H represents a deuterium \H represent a tritium In represents a neutron _°e represents s beta particle 2He represents an alpha particle

Example 29A certain radioisotope of U emits four alpha particles and three beta particles. The mass number and the atomic number of the resulting element respectively are A. 219 and 87 B. 84 and 223 C. 223 and 87 D. 219 and 81 JAMB 199448 Solutiona-particle = \He ■■ 4a = 4\B e /^-particle = _°e 3/? = 3_?e

Let A and Z be the mass number and atomic number respectively of the resulting element Y.

2l \ U 4 | / / e + 3_?e + ^y

Writing a balanced equation for mass number we obtain 235 = (4 x 4) + (3 x 0) + A 235 = 16 + 0 + A 235 = 16 + A 235 - 16 = A

Mass number, A = 219

Writing a balanced equation for atomic numbers we obtain 92 = (4 x 2) + (3 x —1) + Z 92 = 8 + ( - 3 ) + Z 92 = 8 - 3 + Z 92 = 5 + Z 9 2 - 5 = Z

Atomic number Z = 87Ans = 219 and 87 or

442

Example 30A nuclide X is produced by bombarding a nitrogen (A) nucleus with an alpha (,<- particle with the release o f heavy hydrogen (D) nucleus as shown by the fo llo w ins nuclear equation.

a + + \ D wDetermine the values of P and q in the equation V/AEC tOP-Solution

a-particle = \H ea + ' l N — >pqX + I d

\ H e + X* N — > pX + jD

Writing a balanced equation for mass (nucleon) number, we obtain 4 + 14 = P + 2

18 = P + 2 P = 1 8 - 2 P = 16

Writing a balanced equation for atomic (proton) number we obtain,2 + 7 = q + 1

9 = <7 + 1 q = 9 - l<7 = 8

Ans: P = 16, <7 = 8 Or 'JX

Example 31

X$N + \H e —» 170O + X

In the equation above, the particle X isA. a proton B. a neutron C. an a-particle D. a -particle JAM B 20084®SolutionLet A represent mass number and Z atomic number

X$N + \H e —► 17b0 + i X

Writing a balanced equation for mass numbers we obtain 14 + 4 = 17 + A

18 = 17 + A

A = 1 8 - 1 7

A = 1Writing a balance equation for atomic numbers we obtain

7 + 2 = 8 + Z

9 = 8 + Z

Z = 9 — 8

Z = 1^ X become \X which is a proton ( \ t l ) Ans: A (a proton)

N U C L E A R R E A C T IO NNuclear reaction is divided into nuclear fission and nuclear fusion.Nuclear fission is the splitting up of a heavy atomic nucleus by neutron

bombardment. It results in the formation of two approximately equal lighter nuclei, move

443

Hcnirons and t],c .,ibombardment -is C ,SC vc, y *nrfic amount of energy. Uranium

s town below to forni Barium and krypton nuclei.¿JJ UUU'.'C- -

2UU -H In - » litB a + liK r + 2jn + energy

Pressure, 0f Jig|lt Is ,hc combination, under extremely liigb temperature andcutcriuni and tritium °i ,-°mi l,cav,cr ni|dcus with release ofhugc amount o f energy.

IC 01 nrc btsed to form helium nucleus as shown below.

+ \li —> 4n ,2n c + on + energy

Nuclear reaction ({“' *amount o f cncrcv T h-T ^ fusion) occllr with a loss in mass and the release ofhugc oss in mass, also known as mass defect, is given by

M a ss d e fe e t = of original substance in a. m. n - mass o f products in a. m. u

Note, a.m.u. means atomic mass unit and is denoted by the symbol, u.

The difference in mass that happens in a nuclear reaction is a measure of the binding energy o f the particular nucleus or nuclei.

Binding energy is the quantity of energy that must be put into a nucleus in order to break it into its constituent particles. Therefore, binding energy of a nucleus is proportional to the difference between the total mass o f the individual nucleons and the mass o f the nucleus. Usually, the total mass of the stable nucleus or nuclide is less than the sum of the masses of its constituent nucleons.

The energy released during nuclear reaction (binding energy) is given by Einstein’s energy equation.

E = m e2

Where c = velocity of electromagnetic wave (light wave) = 3 X 10Bm s-1 m = Mass defect in kilogram where la . m. u = 1.66 x 10~27kg E = Nuclear energy

The unit q f energy in nuclear reaction can be expressed in unified atomic mass unit (u) orthe electron volt (eK) or joules (/), as followsle V = 1.6 x 10“ 19y1 MeV = 1.6 x 10"13ylu = 931 MeV = 1.490 x 10_1°y

Example 32In a nuclear reaction the mass defect is 2.0 x 10-6 that the velocity o f light is 3.0 x 10°m s-1. SolutionMass defect m = 2.0 x 10-6# = 2 x 10^ k g Velocity o f light c = 3.0 x 10°m s-1

g. Calculate the energy released, given WAEC 199757

Energy released E — m e2= 2.0 x 10"9 x (3 x 10n)2 = 2.0 x 10-9 x 9 x 1016= 1.8 x i o uy

444

example . r- jIn a thermonuclear reaction, the total initial mass is 5.02 x 10~z7kg and th e tota mass is 5.01 x 10~27kg. The energy released in the process is A . 9.0 X 10 J B. 9.0 x lO "11; C. 9.0 x 10~12/ D. 9.0 x 10' 13J [c = 3 x 108ms ]

JAMB 200042SolutionInitial mass= 5.02 x 10~27kg\ final mass= 5.01 x 10~27kg; c = 3 x 108ms

Mass defect = initial mass— final mass m = 5.02 x 10~27 - 5.01 x 10"27 m = 1.0 x 10~29kg Energy release E = m e2

= 1.0 x 10~29 x (3 x 108)2 = 1.0 x 10“29 x 9 x 1016

E = 9.0 x 10“13y

Example 34In the fusion of hydrogen isotopes into helium, the decrease in mass is about 0.65 /o- Calculate the energy obtainable when l.Og. of hydrogen is used.[c = 3.0 x 108m s-1] NECO 2006E1°Solution

1.0g = 0.001 kg; c = 3.0 x 108m s_1

0.65Mass defect, m = —— x 0.001

100m = 6.5 X 106kg

Energy obtained E = m e2

= 6.5 x 10"6 x (3 x 108)2 = 5.85 x lO ^y

Exam ple 35A possible fusion reaction is \H + \H —> \H + \H + Q, where Q is the energy released

as a result of the reaction. If Q = 4.03MeV, Calculate the atomic mass of \H in atomic mass units. [\H = 2.01410u; \H = 1.00783u; lu = 931MeV] WAEC 2000£1S Solution\H + \H -> \H + \H + Q

Let x be the atomic mass of \H in u. lu = 931MeV

4.03 uTherefore, A.03MeV = = 0.004329u

Substitute given values into the nuclear equation to obtain:

2.01410 + 2.01410 —> x + 1.00783 + 0.004329

4.0282 = x + 1.01216

x = 4.0282 - 1.01216 = 3.01604u

Example 36The radioactive nuclei 2\°Po emits an a-particle to produce 2g|Pb. Calculate the energy, in MeV, released in each disintegration.(Take the masses o f 2\°Po =209.936730u; 2gfPb = 205.929421u; \He = 4.001504n, and that l u = 931Aiek)

WAEC 2006flsSolutiona-particle =

445

2l°Po —* \He + 2g\Pb 4- Energy (<?) lu = 931 MeV

Substituting given values into the nuclear equation, we o b ta in .

(209.936730)931 = (4.001504 + 205.929421)931 + Q

195451.0956 = 209.930925 X 931 + Q195451.0956 = 195445.6912 + Q

Q = 195451.0956 - 195445.6912

Q = 5.404MeV

Example S I n n qqnuThe binding energy of helium is A. 2.017u B. 0.033u C.4.03 u ■ • '[atomic mass of proton = 1.00783u, atomic mass of neutron = 100 «1

]AMB 200421Solution

Atomic mass of proton = 1.00783u

Atomic mass of neutron = 1.00867u

Atomic mass of \He nucleus = 4u

Number of protons = 2

Number o f neutrons = 4 —2 = 2

Mass ofprotons = 2(L00783u) = 2.01566u

Mass of neutrons = 2(L00867u) = 2.01734u

Total mass neutron = 2.01566u + 2.01734u = 4.033u

Binding energy = difference between mass of nucleon and that of nucleus

= mass of nucleon—mass of nucleus

= 4.033u - 4u

= 0.033u

Example 38The mass of a proton is 1.0074u and that o f a neutron is 1.0089u. Determine the energy evolved in stabilizing the nucleus of nitrogen o f a mass number 14 with 7 protons and 7 neutrons, [speed o f ligh t = 3.0 x 108m s-1; l u = 1.67 X 10- 27fc^]

WAEC 200549Solutionmass of proton = 1.0074u; mass o f neutron = 1.0089u mass o f nucleus = 14u-speed of light c - 3 x 108m s-1 number of proton = 7; num b» o f neutron = 7 lit = 1.67 x 10~27kg

Mass of protons = 7(1.0074u) = 7.0518tr

Mass of neutrons = 7(1.0089u) = 7.0623u

Total mass of nucleon = mass of proton + mass of neutron

= 7.0518 + 7.0623 = 14.1141u

Mass of nucleon in kg = 14.1141 X 1.67 x 10_27fc^

446

= 2.3570547 x 10~26kg

Mass of nucleus in kg = 14 x 1.67 X 10_27A:

= 2.338 x 1 0 -26kg

Mass defect = mass of nucleon — mass of nucleus

m = 2.3570547 x 10~26 - 2.338 x 10-26

m = 1.90547 x 10~2akg

Energy evolved E = m e2

= 1.9 0 5 47 X 10-28 x (3 x 108)2

= 1.714923 x 10“11/ « 1.715 X 10"11/Example 39Calculate, in joules, the binding energy for 9£?e.[ Atomic mass of %Be = 9.01219it; mass of proton = 1.00783u, mass o n r= 1.00867u, unified atomic mass unit,tt = 931 MeV, leV = 1-6 X 10 ] \ El5

NECO 2007SolutionMass of \B e = mass of nucleus = 9.01219tt Mass o f proton = 1.00783u; u = 931 MeV Mass o f neutron = 1.00867u; leV = 1.6 x 10-19/

Number of protons = 4Number of neutrons = 9 — 4 = 5Mass o f protons = 4(1.00783ti) = 4.03132uMass of neutrons = 5(1.00867u) = 5.04335u

Mass o f nucleon = mass of proton + mass of neutron = 4.03132u + 5.04335U = 9.07467u

Mass defect (binding energy) = mass of nucleon - mass of nucleus= 9.07467u — 9.01219u = 0.06248u

Binding energy = 0.06248 x 931 X 106 x 1.6 x 10“19y = 9.3 x 10~12/

Exam ple 40Deuteron and tritium fused to form a helium nucleus according to the equation

\H + \ H t + + Jn + QCalculate, in joules, the energy released.(\H = 3.01605t/; \H = 2.01410u; \H e = 4 .00260u ;)

( In = 1.00867; l i t = 931 MeV; le V = 1.6 x 10~19/ ) NECO 2004£15 SolutionTotal mass o f reactants = 2.01410u + 3.01605u

= 5.03015u

Total mass o f products = 4.00260u + 1.00867u

= 5.01127u

Mass defect = mass o f reactant - mass o f product

= 5 .0 3 0 1 5 -5 .0 1 1 2 7

= 0.01888u

447

Energy released = 0.01888 X 931M eV

= 17.57728 MeV

= 17.57728 x 106 x 1.6 X 10"19

Q = 2.812 x 10_12y

EXERCISE 7. toms would have1. An element whose half life is 3 years has N atoms. How many a

iqq?56 Ans:-Aatomsdecayed after 9 years? W A E C 1992 82. The half life of a radioactive element is 9 days. What fraction of atom jS

i B i c i D. H )A M B 199 549 Ans: ^

A radioactive element has a half life of 4 days. The fraction that has dec ydecayed in 36 days? A. — 3.day is A. - B. —J 4 16

C. 7“ D .i JA M B 20064 Ans: ^4. A radioactive substance has a half life of 20 hours. What fraction^of the ong radioactive nuclide will remain after 80 hours? W A E C 199355 Ans.5. A sample of radioactive material has a half life of 35 days. Calculate the fraction of the original quantity that that will remain after 105 days.

W A EC 2001£1S Ans: ^6. A radioactive substance has a half life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?A. - B. - C. — D. — JA M B 20072 Ans:-^

4 8 16 32 . . . . i7. A radioactive substance has a half life of 2 years. If the initial mass is 40,7, which

2years 3years 4years 5yearsA 20 ,7 10.7B 30 a 20,(7 10.7 M __________C 30.q 20,(7 13,7D 20 ,7 10.7 0 9

JA M B 198946 Ans:A8. A radioactive nuclide of mass 6.0g has a half life of 8 days. Calculate the time during which 5.25g of the nuclide would have decayed. W A E C 199554 Ans: 24days9. The half life of a radioactive substance is 14 days. If 48g of this substance is stored, after how many days will 1.5g of the original substance remain?

W A E C 2 0 0 048Ans: 70days10. 4g of a radioactive material of half life 10 days is spilled on a laboratory floor. How long would it take to disintegrate 3.5g of the material?A. 1 days B. 8^ days C. 30 days D. 80 days. JA M B 199 1 47 Ans: 30 days11. The count rate of an alpha particle source is 400 per minute. If the half life of the source is 5 days, what would be the count rate per minute after 15 days?

W A E C 199660 Ans: 5012. The half-life of a radioactive substance is 3 years. How long, in years, will it take 384 of the substance to decay to a mass of 6g? N E C O 200857 Ans: 18 years13. A substance has a half-life of 3 seconds. After 6 seconds, the count rate was observed to be 400. What was its count rate at zero time?A. 200 B. 1200 C. 1600 D. 2400 JA M B 19 9 048 Ans: 1600

448

14. If the fraction of the atoms of a radioactive material left after 120 yearsis the half life of the material? A. 2 years B. 10 years C. 20 years arsn. 24 y o n , 2000« Ans: 20 ^ ^

15. In 50 clays, a radioactive isotope decreases in mass from #half life of the radioactive material? NECO 2001 Ans. 1 W* What is the16. In 24 days, a radioactive isotope decreases in mass from 128 (J tohalf life of the radioactive material? NECO 2001 ns* c o n s ta n t .17. The half life of a radioactive substance is 2 seconds. Calculate t ic

WAEC199150 Ans: 0 .3465s' seconds?18. What is the decay constant of a radioactive element whose hal t c

WAEC 199655 Ans: 0.231 s e c '1 _4 t CalcU|ate its19. The decay constant of the isotope of a nuclide is 1.36 x 10 ,talf-lifc. NECO 2008£1° Ans: 5.096 *20. The half life o f a radioactive element is 5 seconds. Calculate its ccay

WAEC 199947 Ans: 0 .1386s ' ,f Hffe21. A radioactive element has a decay constant of 0.077s 1- Calculate i s

2 0 0 3 47 Ans: 9 s e c o n d s WAL, z y „.¡.i, 36 electrons22. Which of the following representation is correct for an atom A

and 40 neutrons? A. 38X B. C ~

23.

76Jf D . % X E.N E C O 200856 Ans: D

40IVCLit/ *----

•¿J. A radioisotope has a decay constant of 1 0 '7s _1. The average life of the radioisotope is A. 6.93 X 10°s B. 1.00 X 10"4s C. 1.00 x 107s D. 6.93 x 107

JA M B 200419 Ans: 6.93 X 106s24. Eight or-particlcs and six /7-particlc are emitted from an atom of 2|8fl before it achieves stability. What is the nucleon number of the final product in the chain reaction?

W A EC 199 3 58 Ans: 20625. Uranium of atomic number 92 and mass number 238 emits an alpha particle from its nucleus. The new nucleus formed has, respectively, atomic number and mass number.

W A E C 2005SO Ans: 90 and 23426. Eight alpha decays and six beta decays are necessary before an atom of 2gffl achieves stability. The final product in the chain has an atomic number ofA. 70 B. 78 C. 82 D. 90 JA M B 198949 Ans: 8227. An clement X of atomic number 88 and mass number 226 decays to form an element Z by emitting two beta particles and an alpha particle. Z is represented byA. 222Z B. 2H Z C. 2\%Z D. JA M B 199 349 Ans: B28. The nucleon number and the proton number of an neutral atom of an element are 23 and 11 respectively. How many neutrons are present in the atom?

W A E C 200146 Ans: 1229. \ \ N a + X —* 2° F + \H e What particle is X in the react on above?A. Beta B. Gamma C. Alpha D. Neutron. JA M B 200249 Ans: Neutron30. In a nuclear fusion experiment the loss of mass amounts to 1.0 x Theamount of energy obtained from the fusion isA. 3.0 X lO 'V B. 3.0 x 10'V C. 9.0 x 104/ D. 9.0 X 1010y [speed of light = 3.0 x 108m s '1] JA M B 199 8 49 Ans: 9.0 X 1010/30 The amount of energy released when 0.5 of uranium is burnt completely isA.*4 .5 x 1016y B. 1.5 x 1016y C. 1.5 x 108y D. 4.5 X 1 0 QJ( c -3 x 1 0 °m s '1) JA M B 200512 Ans: 4.5 X 106y

32. A substance of mass 2.0 x 10 kg undergoes a fission process which decreasesits mass by 0.3%. What amount of energy is released in the process? (c = 3 0 x 1 0 °m s '1) N E C 0 200859 Ans: 5.4 X 109y3 3 . A material of mass 1.0 x 10 ~3kgundergoes a fission process which decreases itsmass by 0.02 percent. Calculate the amount of energy released in the process. 3 x 1 0 8ms_1] W A E C 2 0 0 049 Ans: 1.8 X 1010/

449

.14.(,l„ • . .. .........me iiecreoKc n( ■ r i Z " ,C ,iM(io" o f 1 ■ «// o f Hie mal ‘ ■r.Jxio"/ | , .9.0X10»/

In n fission process, die decrease in iiium is 0.01%. How cnLrJi{uu................................... » . W > X , 0 /

|c * 3.0 x 10”m.v I ic,7/ ¡AMU 2003*° Ana: 0.0 / 1

•IS. In <i nuclear fusion process, four proton* each of mass M,, wc/c In -c< 1 Ia nucleus X of mattii M*. Which o f the following equation is correct/ ... . .A.4AÎ,. > Mx M .4 M ,» M * C. 4M,, < Mx I). Mr ~ Mx J A M i n W ” * " * -A •10. Calculate in joule«, the binding energy for

Aiom/c m uss o f ^ C „ = 5fl.9332«l (Mass of proton = 1.00703«][Mass of neutron = 1.00067uJ [Unified atomic mass unit u = 0 3 îMeV\¡U V = l.f, x 1 0 -” yj W/1/:'/.' 1999*5 An«: 0.20 X 10 /37. Calculate, in joules, the binding energy for J/,/ _[ Atomic mass of £/., = 7.01600«) [Mass of neutron = 1.00067«J lMaM (»ProU)n - 1.00703«) [Unified atomic mass unit,« = f)31McVj \1eV = 1.0 X 10 /J

/V//6YJ 2002m An«: 6.20 x 1 O '12/ . .18. A nucleus lias a proton number of 84. It emits an «-particle and then a /¿-partie c to achieve stability. What is the proton number of the product?

tVA/iC 1997s'1 Ans: 0339.respectively in the equation above?A. 10 and 20 13. 12 and 24 C. 20 and 10 D. 24 and 12 JAMB 199944 Ans: C40. If the decay constant of a radioactive substance is 0.23 Is , the half-life isA. 3.00s 13.0.12s C. 0.33s I). 1.50s JAM13 200ÎT' Ans: A41. A piece of radioactive material contains 102° atoms. If the half-life o f the materialis 20 seconds, the number of disintegrations in the first second is A. 3.47 x 10,n 13. 6.93 x l O 20 C. 3.47 x 1020 D. 6.93 X 101;;

JAMB 200947 Ans: A42. The half-life o f a radioactive substance is 5hours. If 5g o f the substance is left after 20hours, determine the original mass of the substance. NECO 2009'' Ans: 80g43. Determine the value of x from the nuclear reaction below.

jjA/a + proton —> '¡X + alpha particle. What are the values o f P and ry

I^Rn —* ngRa —» 2/? + energy NECO 20Q9W Ans: 22244. Calculate the decay constant o f a radioactive substance which has a half-life o f 25days. NECO 2009fe’ Ans:2.77 x 1 0 '2day_1

450

-

TABLES

Table 1: Quantities, Formulae and Units

QUANTITY DEFINITION FORMULA S.I.UNIT "

s y m b o l d i m e n s i o n

S peed d is tan c etim e

s metreper

secondm /s L T '1

V elocity d isp lac em en ttim e

s metreper

secondm /s L T '1

A ccelera tion change in velocity tim e

v -ua =—

meterper

secondsauared

m / s 2 ' L T '2

E quations o f m otion , rectilinear acce le ra to r

v = u + a t , s = u t 4 E quations o f m otion, gravitational acceleratio

v = u ± g t , h = u t -i

i:a t2, V 2 = t

n:

: \ g t 2, v 2 = t

i 2 + 2 a s

i 2 ± 2g h

Force mass x acceleration F = m a newton N M L T "2

F ric tioncoeffic ien t

fric tional force no rm al reaction

F

g = tarxG

Tensilestress

forcea rea

F F— ° r ---TA nrz

newtonper

metersquared

N /m 2 ML_1T " 2

T ensilestress

change in length orig inal length

eT

E lastic force force constant x extension F = k e newton N M L T '2

Y oungm odulus

s tre s ss tra in

y = FJ l 1\ e/l

newtonper

metersquared

N /m 2 ML_1T ~ 2

W ork done in elastic spring/string

average force x extensionW = \ Fe

1 , = - k e 2joule J ML2T ' 2

L inearexpansiv ity

in c rease in leng th a - h ' 1'

PerKelvin

orPer

Celsius

K "1or

° c _1orig in a l leng th x te m p e ra tu re rise

A reaexpansiv ity

increase in area a - A*~AlPer

KelvinorPer

Celsius

K“ 1or

°c - i

p M O 2-O 1 )

= 2 aoriginal area x tem pera tu re rise

C ubicexpansiv ity

increase in volumePerKelvinorPerCelsius

K " 1or

oC- i

y -

= 3Uoriginal volume x tem peratu re rise

A pparen tvo lum eexpansiv ity

_ tnass(vo lum e) o f liqu id expelled PerKelvin

orPer

Celsius

K " 1or

oC- i

m ass(vo lum e) of liq rem ain in g x te m p rise _ a p p a re n t in c rease in volum e

o rig ina l volum e x le m p e ra tu re rise

452

Real cubic expansivity

Actual increase in volume per unit volume per degree rise in temperature, when vessel expansion is considered

Yr = Ya + Y

PerKelvinorPerCelsius

K '1or

° c _1

W ork force x d is tan ceW = F x s

= m g x s

Joule or newton meter

Jor

Nm

ML2T ' 2

W ork done (horizontal) force x cosine of angle x distance = FcosO x s

= mgcosQ x s

Joule or newton meter

Jor

Nm

ML2T ' 2

W ork done (vertical) force x sine of angle x distance

Fsind X S

mgsinO x s

Joule or newton meter

Ior

Nm

ML2T " 2

K ineticenergy energy due to motion KE = ^

Joule or newton meter

Jor

Nm

ML2T " 2

Potentialenergy

energy by virtue of position or height PE = mgh

Joule or newton meter

Ior

Nm

ML2T ' 2

Pow er work donefenergy expended) time

p _ F*s t

_ m g x s t

= F x s

Watt or joule per

second

WOr

JS'1m l 2t -3

Efficiency(<0

power output v ^ qq power input V- E x 100

Pi

QUANTITY DEFINITION FORMULA S.I.UNIT SYMBOL DIM ENSION

Electriccurrent

quantity of charge time *—« II

<t |<

o

ampere A

Potentialdifference

c u rre n t x resis tance V = IR volt V

R esistancep o ten tia l d ifference

c u rre n t

Vs = 7

ohm n

R esistance in parallel

ohm n

Tw o resistor in parallel

RiRiR =Rl + /?2

ohm n

R esistance in series

r = r 1 + r 2 ohm n

E.m .f.(cell) in series

E = E-i + E2 volt V

E.m .f.(cell) in parallel

E = E1 = E2 volt V

E.m.f.Potential difference between cell terminals at open circuit

E= l (R + r ) E = V + v

volt V

453

p.d. across external resistor .. ef = J x 1 0 °

£f= - ^ - x 100R+r

Efficiency (electric cell)

e.m.f. of cell external resistance

extresistance+intresistance x 100

Electricalenergy

Charge quantity x po t difference

W = QV = IVt

= I2Rt = ?Joule J

Electrical electrical energy transfered

chIIa.

watt wpower tim e taken= i2r = v4

Electrical heat energy

H = l 2Rt = Pt

V2t= — = I V t

Joule J

QUANTITY DEFINITION FORM ULA S.I.UNIT

SYM BOL d i m e n s i o n

Resultant of twoperpendicularvectors

fi = J r ? + F¡

e = t a n "i [ 8

Resultant of two non­perpendicular vectors

R

= J f12+F}-2F1F2cos$

. _ i (F\ s in 0 \ “ = s,n l R )

Resolved horizontal component of force

force X cosine of angle given

F cos 61

newton N M L T- 2

Resolved vertical component of force

force x cosine of angle given F sin 6 newton N M L T- 2

Projectile time of flight

time taken to return to same projection level

2U sin0T = ----------

Qsecond s T

Projectilemaximumheight

highest vertical distance from projection plane

U2s\n20 H 2 g

meter m L

Projectilerange

horizontal distance from projection point to projection plane

U2 sin 29R = -----------

9meter m L

Equations o f motion under gravity: v = u ± g t , h = u t ± \gt v 2 = u 2 + 2 gh

Moment of force

force x perpendicular F x s newtonmeter Nm ML2T " 2

Densitymass

volume

SI*IIa

kilogramper

metercubed

kg/m3 ML-3

Relativedensity

mass of substance mass of equal vol H20

Upthrustdensity of fluid X

volume X qUF = p x v x g newton N M LT” 2

Centripetalacceleration

V 2 a = — r

meterper

secondsquared

m /s2 LT-2

Centripetalforce

mass x centripetal acceleration

m V 2 2 F = ----------- m ojr

r_------------------newton N M LT"2

! i

454

Aiitztalar angle turned 0(,) a - o r 2 n f

radianper r a d s '1

time* second

i / r 1Linear velocity angular speed x radius (amplitude)

V =3 6)V =a (,\A metreper

secondm / s

v = o)\!A2 - x 2

Linearacceleration

square o f angular speed x radius;

angular acceleration x amplitude

a = d )2 r

= (ü 2A a = n r

meterper

secondsquared

m / s 2 L T ' 2

TPeriod tim e taken t '¿7T 17* =* — « --- -3 —

n nt tsecond s

num ber of <>•;< ill.itinnsT - i

Frequency num ber of oscillations tim e taken

ri ut 1

7 t 2rt T

Persecond

or Hertz

s " 1 o r H e r t z

Period ofsimplependulum

Time for one complete cycle

¡7T a 2 IT -

i "

r t » r 2 If 2

second s T

Energy of simple harmonic motion

T = 2 n f i ; F = kc; r=2,rS : " ” Æ î ;T *•

Momentum M ass x velocity m x vNewtonsecond

N s o r

k g . m / g _M L T ' 1

M L T " 1Impulse Force x time or Mass X acceleration

1 = F x t = m x ( v - u )

Newtonsecond

N s o r k g m / s

Force Mass x acceleration

F = m am ( v - u )

F = — ------------t

newton N MLT " 2

MechanicalAdvantage

load out put force e ffo rt' input force

LM.A.= -

J

Velocity Ratiodistance moved by effort distance moved by load

V ' R 7 _ sinO _ 2 n r _ /?‘ V ~ r

Efficiencyworkoutp.it v 1(,0%

M. A£ = — x 100%

l/ fE = -¿y* x 100%h

work Input

5 (°p _ 32)C elsius(°C ) = ------— — , fah rcnh i

, , 9°c îit(°F ) = — + 32, K = °C + 2 7 3

Absolutetemperature

temperature measured from absolute zero r = 2 7 3 + °C kelvin K

Resistancethermometertemperature

0H - R 0

= _ X 100T 'MOO Ko Celsius °C

Heat capacityheat required to raise

temperature of substanceby 1°C

C = m ejoules

perkelvin

IK" 1o r

_ ]0C"X

Specific heat capacity

heat required to raise temperature of unit mass

o f substance by 1°C

<?m (0 2 ~ Oi)

joule per kilogram

per kelvin

J k g - 1K -1

o rJ k g ~ l o ç - i

455

Specific latent heat of fusion

heat required to change unit mass of substance at

melting point without temperature change

i = !m

IVtm

joule per kilogram

J k g ' 1

Specific latent heat of vapourization

heat required to change unit mass of substance at boiling point to vapour

without temperature change

L = — m

IVt

m

joule per kilogram

J k g '1

Relativehumidity

= 77 X 100%M

= — x 1 0 0 %S.V.P

S.V.P a t d e w p o in t v f|fW n ~ S.V.P at a ir t e m p e r a tu r e

Pressure force per unit area p = pgh

newtonper

metersquared

N / m 2 ML- 1 T -2

Work done by expanding gas

Pressure x change in volume

w= P(V7 - V i )

joules J ML2T " 2

V\ V2 , Pi p2 Boyle's law: = P2V2, Charles law:—* = —-, Pressure law: — = —

¡1 T2 ¡ i ¡2Pi Pi P2V2

General gas law: — — = — —1 i in

Cubicexpansivity

change in volum e from 0°C voi a t 0°C xchange in temp.

Vo - Vo1 V0 X 9

l o - l o Y L x e

per kelvin or per Celsius

K “ 1or

°c_1

Pressureexpansivity

change in p ressu re from 0°C voi a t 0°C xchange in temp.

pe - pop0 x 8

per kelvin or per Celsius

K " 1or

°c_1

Pressureforcearea

Fp = - , p =

pgh

newtonper

metersquared

N /m 2 ML- 1 T -2

Pi F2 Pi ^2 Pascal principle: — = — , Hare'sapparatus: — = —

Al A2 P2 n i

Frequency

n um ber of cycles tim e taken

nu m b er of w avelength tim e taken

1f - j hertz Hz T _ i

WavelengthDistance between successive

crest o r trough meter m M

Wave velocity

Wavelength X freq

wavelength period

IIH

I >*

II

meter1 Pcr

secondm / s L ' r 1

j

Mathematical wave equation: Y - Asi2 nx 2 tt /X

i — , Y = A sin — (x - v t ) , Y = Asin2n - f t )

Magnification

image distance object distance’

image height object height

VM = -

u

Angle of deviation

Angle through which incident ray is deviated

d = 100 - (t + r)

456

Equations are more important to me, because politics

the present, but an equation something for eternity.

Albert Einstein (1879 - 1955)German-born U.S. physicist.

The aim of research is the discovery of the equations which subsist between the elements of phenomena.

Ernst Mach (1838 - 1916)Austrian physicist and philosopher.

Q U A N TITY D EFINITIO N FORM ULA S.I.UNIT

SY M BOL d i m e n s i o n

Image formed by inclined mirror

3 6 0

n= 1

Mirror or lens formula

1 1 1

/ u V

f = - ^ ~u + v

Refractiveindex

sin of angle of incidence sin of angle of refraction

speed in air speed In glass

wavelength in air wavelength in glass

real depth apparent depth

s in i n = - —

s m r

o v*a * " va

na n g - , Á9

R

aD g

_ sin\(A+Dml„) sin |/t

Critical angle Angle of incidence when angle of refraction is 90°

1n = — —

s i n C

Wave speed

Wavelength x ffeq

wavelengthperiod

A

V ~ T

V = X f

v = V e

meterper

secondm / s L T - 1

Velocity of sound

2 x V = —

t

meterper

secondm / s L T " 1

Vibrating string & vibration in open open

Fundamental freq

First overtone

Second overtone

Third overtone

V

f ° ~ T i

A = 2 / 0 = y

, 2 vA — 4 /o = ~

v 3 v c , , V ib r a t io n in c lo s e d p ip e : f 0 = - A = 3 / 0 = — , f 2 = 5f 0 = — ,

7 v 4 1 ‘

___________________________________

457

S onom ete r: velocity of wave along string, v =

1 1Tfundamental frequency, f0 = — 1— ,

21J M

3 Tsecond overtone, f2 = 3fn = — — , t 72 70 2 ZjAf

j ï , frequency f ~ ¿ J

1first overtone, 2 f 0 - J

_ 2 ibird overtone, / 3 — 4f 0 — ¿ I

TM

¡ L¡M

TM

Velocity of sound In resonance tube

v = 2 f ( l 2 - l 1)meter

persecond

m /s L T '1

Beatfrequency II ir» i >

persecond or hertz

s - 1 T " 1

QUANTITY DEFINITION FORMULA S.I. UNIT SYMBOL DIM ENSION

Gravitationalforce

force of attraction

between two bodies

r Gml m2 r 2 newton N MLT"2

Gravitationalpotential

work done by gravitational field in taking unit mass from infinity to a

point

II -is

joulesper

kilogram

Jkg-1L2T -2

Escapevelocity

maximum velocity with which object

escape from gravitational field

V'=j2¿R

12GM

~ J R

meterper

secondm /s LT"1

Electric forceforce between two

charges1 <h<l2

4 ne0 r 2newton N MLT"2

Electric field intensity

electric force per unit charge

E = ■4 ne0r2

newtonper

coulombN c r1

Electricpotential

work done by electric field to

bring unit positive charge from

infinity

1 Ív = ----- •4 a f0 r

volt V

Capacitancecharge on plate

c 4eA farad F

1p.d. between plates

Capacitance in series

1 1 1 J_C Ci C2 c3 farad F

Capacitance in parallel

c = cl + c2 + cJ farad F

458

QUANTITY DEFINITION FORMULA S.I. UNIT SYMBOL d im e n s io n

Energy stored in capacitor

1 1 , W = - q V = - C V 2 joule

Resistivityresistance per unit

length per unit cross sectional area

RAP = -

Rnr1l

ohm-meter firn

ElectricalConductivity

reciprocal of resistivity

_ t _ / ° ~ p ~ RA

per ohm- meter (firn)-1

Shunt resistant R - ' » X r 1 l9

ohm fi

Multiplierresistant

VR = ~i r'g

ohm fi

Electrochemica 1 equivalent

mass deposited per coulomb during

electrolysisM pAd

Z = T t = ~ fT

kilogramper

coulomb kg/C

Magnetic forceF = qVBSinO F = BllSinO

newton N MLT-2

Induced e.m.f. inelectromagneticfield

E = BIV = NABcosinO

volt V

Transformerequations

Es Ns Ev Np Es Ip Ep Is

Transformerefficiency

Is x Ese = - r — rr-x 100 Ip x Ep

Is x Ns= , „ ■ X 100 Ip x NP

Alternatingcurrent

1 = l0Sin2nft ampere A

Alternating voltage >

V = V0S\n2nft volt V

Root mean f . square cu^tntf' lrms = v f ampere A

Roo&pearil*- / squ&ae voiltage

7 * ' Vrms V2 ■

■%volt

&L,---------V

---------^ ---1---—'+ .

A.C. resistance > < ¡"*fr *

Yr.ms

; V; ^ — h

• - r oh^i

at t ' A _ 4 _Inductive reactance vj| * t*

, j %ls — ciL = 2 nfL

X •• . . ohmÌ !

$ ■

InductanceVL _ VL

L 2nfl col henry H

459

QUANTITY

Voltage across inductor

Voltage across capacitor

d e f in it io n

Capacitivereactance

Capacitance

L-R circuit impedance

L-R circuit current

L-R circuit total voltage

fo rm u la

VL = / X XL

Ve « I * Xe

Xc = — toC

1= 2 nfC

C -2 rtfVc

I<oY,

■ J ^ + x î

/. =y/R2+Xl

v ^ j v s + vl

S.I. UNIT

volt

volt

ohm

farad

ohm

ampere

volt

SYMBOL d im e n sio n

n

n

R-C circuit impedance Z„c = jR t+ X ¿ ohm n

R-C circuit current U -

JR 1 + X,ampere

R-C circuit total voltage v = j v i + v i volt

L-C circuit impedance ZLC= X L - X C ohm n

L-C circuit current

/„ = Xr - Xr ampere

L-C circuit total voltage

R-L-C circuit impedance

R-L-C circuit current

R-L-C circuit total voltage

Resonancefrequency

Power in A.C circuit

Power factor

v = vL-v c volt V

JR* + (Xl ~Xc)¿ ohm n

jR* + (xL- x cy ampere

Iv* + (vL-v cy volt

fo = InjLChertz Hz

P = IVcosO

(“s0=i)watt W

Cosine of phase angle

cosO=resistance r Impedance i

460

QUANTITY DEFINITION FORMULA S.I. UNIT SYMBOL DIMENSION

Energy o f photon

E = hf he

e =tjoule J

Electron volt/ kinetic energy

eV = j m ( v2

eV = hf

Energy of photon(Einsteinequation)

work function + kinetic energy

E = Wa -h^mev2

hf = hfo + ev, , he h f = — + ev

A-o

joule J

Stoppingpotential

v _ h f - w a 5 e

volt V

Wave-particle duality equation (Dc Broglie)

ha = f

hX = — mvhv

X = 2eV h

yj2 eVme

Heisenbergprinciple

hAx > —APh

A P> —Axh

Ax > —------- rA (m x v)h

A E.At> — 2nAE.At > h

D isintegrationratio

NiR = 2n= l t

(n = t /T )

Zhepw o h a lf life equation

r = log 2Rt

l°32 ®

t log2 logR

N um ber o f atoms decayed

Nd = Nt - N 2 (R — 1\

.)= Nt (R - 1)

461

Fraction o f atomundecayed

ÌT Ni 1

~ R

Fraction of atoms decayed

f ~ Ndt d ~ KR - 1

R

Decay constant

A— I ®N\dtJ _ 0.693

= 2 . 3 0 3 ^ t

Table 2: Greek AlphabetUpperCase

LowerCase

Name UpperCase

LowerCase

Name UpperCase

LowerCase

Name UpperCase

LowerCase

Name

A a alpha H V eta N V nu T T tauB p beta 0 e theta s ( xi T V upsilonr Y gamma I L iota 0 0 omicron 0 <P phiA 5 delta K K kappa n 71 pi X X chiE € epsilon A X lambda p p rho 0 4> psiZ < zeta M P mu z <T sigma it CO omega

Table 3: Physical Constants

C onstan t Symbol V alueSpeed of light in vacuum c 3.0 x 108ms-1Gravitational constant G 6.67 x 10-11Nm2kg-2Electron rest mass me 9.11 x 10_31kgProton rest mass mp 1.67 x 10_27kgNeutron rest mass mn 1.67 x 10~27kgAtomic mass unit u 1.66 x 10_27kgAvogadro constant Na 6.023 x 1023molr1Planck constant h 6.63 x 10_34JsFaraday constant F 9.65 x 104Cmol-1Electron charge e 1.60 x 10"19CElectron-volt eV 1.6 x 10~19JElectron charge-to-mass ratio e /m e 1.76 x 1011Ckg-1Proton charge-to-mass ratio e / m p 9.6 x 107Ckg-1

Permittivity of vacuum O 8.85 x 10~12Fm-1Molar gas constant R 8.31447JK_1mol_1Acceleration due to gravity 9 9.807ms"2Standard atmospheric pressure P 1.01325 x 105Nm"2

462

Table 4: Fundamental Quantities and Units

Quantity Symbol S.I Unit SymbolLength / meter mmass m kilogramme kg___Time t second 5Thermodynamictemperature e ,T

kelvin K

Luminous intensity iv candela cdElectric current i ampere A

Amount o f substance n mole mol

Table 5: Submultiples and Multiples of Metric Prefixes

Submultiples Prefix Symbol Multiples Prefix Symbol

K T 1 deci d IO1 deca da1(T2 centi c IO2 hecto h1(T3 milli m IO3 kilo kK T 6 mi ero g IO6 mega MK T 9 nano n IO9 giga G1 0 "12 pico P IO12 tera T1(T 1S femto f IO15 peta P10-18 atto a IO18 exa EK T 21 zepto z IO21 zetta Z1(T24 yocto y IO24 yotta Y

463

UNIT CONVERSION FACTORS

MASS1 kg = 1000 g 1 g = 10“3 kg 1 mg = 10-3 g = 10-6 kg l Hg = 10-6 g = 10“9 kg

LENGTH1 m = 100 cm = 102cm 1 m = 1000 mm = 103 mm 1 m = 1000000 n m = 106 pm 1 m = 1000000000 nm = 109 nm 1 m = 3.281 ft l m = 39.37 in 1 cm = 0.3937 in 1 in = 2.540 cm 1 ft = 30.48 cm 1 yd = 91.44 cm 1 mi = 5280 ft 1 mi = 1.609 km 1 light year = 9.461 x 10IS m

ACCELERATION1 ms-2 = 100 cm-2 1 ms“2 = 3.281 ft/s2 1 cm-2 = 0.01 ms-2 1 cm-2 = 0.03281 ft/s2

SPEED1 km /h = 0.2778 m s '1 1 mile/h = 0.447 ms-1 1 mile/h = 1.609 km /h 1 m s"1 = 3.281 ft/s 1 mile/min = 60 mile/h = 88 ft/s 1 in /s = 2.54 cm/s

FORCE1 N = 1 kgms-2 1 N = 10s dyn 1 N = 0.2248 lb 1 lbft2s_1 = 0.0421 kgm2s-2 IN = 23.75 lbft2s—1

ENERGY 1 J = 1 Nm 1 kWh = 3.6 x 106 J 1 eV = 1.602 x 10“19 J

POWER1 W = l j s -1 1 hp = 746W

TIME1 min = 60 s1 millisecond = 1 ms = 10-3 s 1 microsecond = 1 ps = 10-6 s 1 nanosecond = 1 ns = 10-9 s l h = 3600 s 1 d = 86400 s l y = 365.24 d l y = 3.156 x 107 s

AREA1 m2 = 10000 cm2 = 104 cm2 1 m2 = 1000000 mm2 = 106 mm2 1 mm2 = 10-6 mz 1 cm2 = 10-4 m2 1 m2 = 10.76 ft2 1 cm2 = 0.155 in2

VOLUME1 m3 = 106 cm3 lm 3 = 109 mm3 1 cm3 — 10"6 m3 1 mm3 = 10-9 m3 1 litre = 1000 cm3 1 litre — 10-3 m3 lm 3 = 35.31 ft3 1 m3 = 61.02 x 103 in3

ANGLE1° = 0.01745 rad 1° = tt/ 180 rad 1 revolution - 360° = 2tt rad 1 rad = 57.30° = 180°/n 1 rev/min(rpm) = 0.1047 rad/s 1 rpm = 0.0167 rev/s

PRESSURE1 Pa = 1 Nm-2 1 bar = 10s Pa 1 mmHg = 1333 N m '2 1 bar = 10s Nm-2 1 mmH20 = 9.807 Nm-2 1 atm = 1D1.325 KNm-2

MASS-ENERGY EQUIVALENCE1 kg ♦-» 8.988 X 1016 J l u«-» 931.5 MeV leV <—» 1.074 x 10-9 u

464\

By academic freedom I understand the right to search for truth and to publish and teach what one holds to be true. This right implies also a duty: one must not conceal any part of what one has recognized to be true.

Albert Einstein (1879 - 1955)German-born U.S. physicist.

465

A LL-IN C LU S IV E CALCULATIO NS IN PHYSICS

NEW EDITIONThis book is creatively unique, highly simplified and very comprehensive in its concep , content and arrangement. If calculations in physics is one major reason stiii cuts nu | ■difficult and perform poorly in examinations, then this book is the complete, de rnitiv v an solution.Every concept with calculations in the WAEC, NECO and JAMB syllabuses is contained in this book and as much as possible, the arrangement is in sequence and in line witi t e c Secondary School Physics Curriculum.For every chapter, concept, topic and subtopic, A L L the calculation questions in WAEC SSC E (1988-2009), NECO/SSCE (2000-2009) and JAMB/UME (1978-2009) are included either as typical examples or as exercises (with answers).Some topics have been simplified by the formulation of new equations and methods that enhances the speed and accuracy with which a student can solve calculation problems.Students of preliminary physics course in universities, polytechnic and colleges ot education wi find this book a necessary tool in building a solid foundation for Advanced Level physics calculations.As a compendium of all past calculation questions in NECO, WAEC and JAMB classified into chapters, concepts and topics, A L L -IN C L U S IV E C A L C U L A T IO N S IN P H Y S IC S is a handy reference book for physics teachers and can be used to facilitate teaching and continuous assessment.

ABOUT THE AUTHORA graduate of education and engineering, Solom on Dauda Yakwo is a dynam ic and fledgling Educator, Author and Innovator. He has taught physics at the seco n d a ry school level for alm ost a decade and is the author of "What They Don't Teach You in School" and "Can These New Equations Significantly Simplify Half-life Calculations?"

The author has done a good deal of work in compiling this book for WAEC, NECO, JAMB, NABTEB and NDA examinations. This book may also meet the requirements for the Screening Tests now conducted for admission to most universities. The work is adequate and commendable. In summary. 1 can say the contents are relevant and adequate to the tests and examination needs of the senior secondary school students in physics.

Engr. M.D. Abdullahi, BSctPhysIcs),MSctEng).FIET,FNSE,FAEng.MFR.Chairman,National Business and Technical Examination Board(NABTEB)

be the best in its class. Thus, I strongly advise that you should have this book and at leastextbook. However, if you can't afford many but one, let that one be THIS ONE '

This book seems to one theory physics textbook ALL-INCL USIVE CALCULA TIONS IN PHYSICS

Dr. Jacob Tsado B.Eng„ M.F.ng.. Ph.Du „m t „, Electrical and Computer Engineering Department Federal University ofTechnology. Minna. Nigeria

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