hoUr s PHYSICS XI (Subjective) 1

Chapter 1

MEASUREMENTS

|Learning Objectives|Understand what is Physics?

Understand that all physical quantities consist of a numerical magnit

Recall the base quantities and their units.Describe and use base, supplementary and derived units.Understand and use the scientific notation.Use the standard prefixes and their symbols to inbase and derived units.

i.

2.

3-4-5-

sub-multiples or multiples to both6.

Understand and use the conventions for indicating units.Understand the distinction between systematic errors and random errors.

Understand and use the significant figures.

7.8.

9-Understand the distinction between precision and accuracy.

derived quantity by simple addition of actual, fractional or percentage10.

ii- Assess the uncertainty In auncertainties.

12. Quote answers with correctnumerical and prac* cal w ;rk.

13- Use dim. ' ;iona» ' to

*4* Der* _ rm

scientific notation, number of significant figures and units on all

check the homogeneity of physical equations.. ie in simple cases using dimensions.

CONCEPT MAPMeasurement

Ti i I1

ignincantPhysical Prcctt% or OConventions forScientificPhysics Prefixes FiguresQuantities and accuracyindicating UnitsNotation

1 •o• Importance ofPhysics &Random Rrror

zBase Units Derived Units oDimension of ^Physical quantity

Checking the Derivationhomogeneity of of possible formula

physical equation O

3

i J I l a

-L*J S(.hobr’* PHYSICS XI fSi»bJ« < tivi j 3

For Your InformationScience is the knowledgeof logical reasoning.

A m<»f> ha-, always wanted to observe, think and reason about the world aroundh,rri Man tried to find ways to organic the disorder in the observed facts aboutthe natural phenomena and material things in orderly manner which results forthe single discipline of toi*n< <>,railed natural philosophy

I

Q j How ran we classify the study of nature?

Areas of Physics

MechanicsHeat & ThermaElectromagnetsOpticsSoundHydrodynamicsGeneral relativityQuantum mechanicsAtomic physicsMolecular physicsNuclear physicsSolid-state physicsParticle physicsSuper conductivity

Classification of the Study of NatureThe study of nature may be clarified into two branches:1. Biological ScienceThe science of living things is called Biological science.

2. Physical ScienceThe science of non-living things is called Physical science

ynamics

Q,2 What it physics?n. ,4

MvPhysicsSKF' ''' nPhysics the branch of science concerned with the properties of matter and

energy and the relationships between them.In other words, physics is basically the study of howPhysics is an important and the basic part of physicalIt is the experimental science.

objects behave.I science.

Interdisciplinary Areasof Physics

•Astrophysics•B»opnys»cs•Chemical Physics•Engineering Physics• Geophysics•Medical Pnysics• Physical Oceanograpn>

0,3 Describe the main frontiers of fundamental science.

Main Frontiers of Fundamental ScienceThere are three mam frontiers c ! fundarm ntal s ence

1) The world of extremely larg e. universe2) The world of the ext <. efy sma.. . e particles such as electrons, protons,

neutrons, meson''ndo> «v3) The world of . lidr' * ed things (from molecule at one extreme to the

Earth 3t th® othe It i - tne world of complex matter

04 Dev H * som. new branches of physics. Also describe the role ofPbytfcs developing technology.

Bf»r. ,s hy'.ici&V the T of HT century many physicists started believing that every thing'^ &*f - " has beeo discovered However, about the beginning of the 20'*'n,irrwt " * w e /per .mental farrs revealed that the laws formulated by the<s i' vest .gators nv-ed modification

Corrpixer chips are made trom•vUws ot vw metakoxJ stacona senconduaor

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• l-'hyik * «1%, » pluy. anand angloiroilnH.

• •„ion, u and iruhnoloKV are a vital force for change in the outlook ol

mankind• the Information motfla and the fa^t munm « »l communication* have

brought all parti « »! tin- world in close enntac 1 wllh one anothurlivenh In one part of the world Immediately reverh»*rate round the globe

» Wrs > ir t» living the age of Information technology.

• the computer networks are product* of (. trips developed from the basicIdeas of phytic* Ihe ( hips are made of silicon Silicon can be obtainedfrom tandIt It up tO Ul whether we make .1 sandi astle of a ( ompuler out of It .

‘iMiit*ifluidity n ih" Hii Hunlnv* fl« > w mfi

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1 f.lilrtiu' " limps, With .1 lump down to /cm.Ihr HMinmMtiim whlili »l«u trk al rashtauce Is zero I*- i«lled (lie rrlttr.al lcniper»Ui»e ( (, ) end vnrle*. wllh the Individual matoii*!. lor

rtdiloved hy 1oollOB niHlorlaU wllh olthu, liquid helium or liquid nltrugan. IltK 4U%r thr -m

to ••Important role In the development of technology

SB-10»A/dUMirV' O< V«M*IM H > »*W *u,,U’* fHfUtiiel 1"iir- > rrltlirtl lemptudiure-. .,i «L

mulertub luiV« Hu uli-i tiltel 1 tnUttince, meenlng elm iroiv. can imvel Himuid, lliiun Ireuly, lhoy < ,m 1 any Ur'Ke aniuunt *. nl rlettrU.il current

fill long pe,luds ,,f II,nr without losing runny a*, heat. Superconducting loop* ol wlm have been shown to tarty electrical currents fur

si’Vi i (il yom •* wllh tin me.c.n, .ililu- luss ltd', pinpedy has Inlplli allnns In, oUsclil, al |iuWi*i trwnsrnlsfilun, II tiansmlsslon lines , an he made ofiii|in, 1 onduiHiip ,"Iainli and h >, till aI slur,,|{e dftvli CS.

It is til" Inmi\ h LI| phyvlts which Involves the betuivlur and pmperile* ol Unlit, Including Its Interaoloris with mallei and du-construction of Instruments that use or deled It Optics usually descrlbtl the behavior ol visible, ultraviolet, and Infrared light

ttvdlfillvilioikl Ih" I), am Ii of sclent " that deals with the dynamics of fluids, especially Incompressible fluids, In motion, it Is concernedwith (ho me, hmdi .il properties of fluids It inll . that how quickly 0 object tan imvel In 0 Hold (aemple a person swimming In water .

UcatVMhlKmalvm U 1» ,)ne I> f the four fundarnmUel Inlerectlons In nature. The other throe the strong Into,action, the weak intriaiuuu

and gravitation I lectromegnellsm 1» the force that causes the* Interaction between electrically charged panicles, the areas In which ihK

happi.Mis me , ..II. , 1 I , I.- , AHI ,.-iI, I,,-l,Is I I", tmnuKnnthm Is I,".|KHIMM lor |" m Hi ally .,11 H„- ph. -mmu- n., encountered In dally Ilf *.W 1\

the PHCeptlon of gravity. Ordinary matter takes Its form «s a result of inter molecular forces between Individual molecules In matte,I In, iromagiielltiTl Is also Ihd force whuh holds electrons end protons together Inilde atoms, which a,.* the building blocks of mo

I , Include!tht Itudy of 4 This governs il,. pro(MW Involved in chemistry, which .nlsr from IntiMactions between the HI^' I, .„ui KMMM

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dJpqptiyilil ( he hifliK I, ill physici r oou'fiiml wllh Urn physli id and , hmoli nl properties, origin, am' 1 solution of Ih hu.ila ioi f 1

liUgf&dJit II,e bifloch nl phyih s whir h deals will , the s. lerdlfl, study of biologic al prot • 1, 1 * «' »,, ol 'lie laws of l * f » yv i s Ie

I chulor etlpO In toils, stre»se» m ,d .Ooln. In cbeletal and muse uhu nlror lures,

Aeilrdvoarnik! Tfin hranth of physics whirl, deals with the study of the movement of *lr and otherIntarirtlOili of air with moving fd,)*i u, sur h ai, planes, and of Ihn elf". Is of moving air on slaflonat y olijnr U, ».u« h "*•b"!li

^ ^ ^Lywt'jlOr.tf fh» l.omch of physl. » wldi l, d* nls will, l) ,r hi lmvlo, ol Ih, luntnlai urdv, , < In ll« ' nlln lv II h. ihoone of widest so »

spaifrorn nl physicsthyikMl CilimflWV ff ,« brefrf I, III 11,,- rnlvlr y that ic i on, eined will, the physic a! stro, tore of r hemli al < urn|iO"mU, theth*y have, the way they react writ , other romponmls,, .„,d Ihr bonds that hold Iheli atoms t'ogrllmrMwliat ULUIUlMIMiU I * b ft,r Muriy nl irh /sl, ol loniHrlt,,.. „,d pliy - l1 > 1 / inutttn wllhln ihr o, min,proper ru- . nl ui Pun water s

MlJn.il Lh /iUi If Is Ih* appllf ,(lor, of phytic t lo medkln* It gmarally » one or » ,s phytlr s as applied to mndfrel end redlolhnrepy. . af uuhlrh laUU* of DhvilC!(onfollU II /.*» • ../ (A. ' W ,» lu<>. th..h.p.of th* ,.r,h, It, All thou- quantities In terms of which law* pnys

magnruf foldt>>* fl /narnlr -. hi M,e I mil, a , whottt and ul . romporn „t puts, the I wtlV* Inlemal IlfUtfure, CdfnpOlltlori j, plty.U t »l qUdntlllf '..th* gpiipminn of vtilrunlsm and > • , forn ion, tr .r» t / rirolf .t • >l ry» 1« Inrlorling snow end lie. wl* fl6P,u ,’‘ 1,1 ^ p(fv Types of Phyilcal Qugntltlil«1moH»t,*ro, I0004pt.*m, rn*g/,eto»ph«rr and o>lar rrrestrlal rti.idoM, and analogous problem* etfeOClfted Hh the mod" Phyilul quantities dtd divided Into

Wh.it art? physical quantities? Discuss its different types and way to

maasura tha bast quantltlai. —unit nl eneflmn,

Physical Quantitiesnpoclelly the motion* *"«! PhV 9,(

be described aru calledcan

Darlvad Quantities

Chapter 1 [Measurements]Scholar s PHYSICS - \1 (Subjective) 7

Base QuantitiesThe mini*ptopkmlfE«mD*es

^'' zr' “ ass t-“c etc.

:e - ice - = -;

fiuxir~a definitions are based on other physical quantities are

cJied denied qi+ma&s*

>£ cr

Measurement of Base Quantitiesnea»^e-e-.: rase Quantities -ivoK'es two steps:

The dv>ce of the standard~- e c ^ccedufe comcar -zrs quantity to ce measured with theStandard.

Properties o* an ce= Standard- - rea standard - as: , c princeal characteristics;

ft is invariable.

(t) Meter The distance traveledbv light in vacuum during atime of V299.7924.58 second,(h) Kilogram It is defined ss themass of 5 Platinum (.50%) and•re um (ioK) ailov cylinder,3_gcm in diameter and 3.9 cm inheightInternationalWeights and Measures fnPrance.Th-s mass standard wasestah' shed in root.

(5i) Second The duration inwhich the outer most electronof the cesium. 133 atoms make9. 2,631.770 vibrations.(*v) Kelvin It is the fractionV273.16 of the therimotfynam<temoer3ture of the triple pointof water.(v) Ampere The unit of electriccurrent is ampere. It is thatconstant- current which ifmaintained in two straightcars;'s ' conductors of infinitelength of regig c12 circularcross-section and placed 3

meter of length.(vi) Candela The unit of!u-:-.ous intervsrty is candela,itis defined as the (urncr«oc:intensity in the perpendiculardirection of surface of1/60000 scuere meter of ablack body 'adiator at thesolidification temperature ofplatinum under standardatmospheric pres s c e(vfi) Mole The mcle is theamount of substance of asystem which contains as manyelementary ert ties as thereare atoms in 0.012 kg if carbon12. Or.p mole of any substancecontains 6.022s * ion entities.

Temperature(v)number f i f ths physical quantities in terms of which other

be defined are called base quantities.T kelvin K

1Light intensity(vi)nttes can L candela cd

Amount of substance(\ ii) mole mol.n

(2) Supplementary Unitskept at theBureau of The units which are neither base units nor derived units are called

supplementary units and are oftenly used for geometrical quantities. They(ii) The solid angle

are:Plane angle

Standard Definitions of Supplementary Units

(i) Radian

(0accelerator -norserr-m force etc.

It is the plane angle between two radii of a circle which cut off on the

circumference an arc equal in length to the radius of the circle. It is shown in

figure.

(ii) SteradianIt is the solid angle (thby an area of surfacefigure.

Vi

/ree dimensional angle) subtended at the center of sphereequal to square the radius of the sphere. It is shown in

k

0 •s access:e~~cS£ Tv.: -e:- *e“ e'3 =‘£ often ncompatible and compromise has to be“ade between tr-em.

SymbolSI UnitPhysical Quantity

radRadianSteradian

Plane angleSolid angle sr

0,6 W is numatiora system of units? Discuss the units upon which it isbu?t up.

(3) Derived UnitsThe SI units for measuring all other physical quantities are derived from the base

and supplementary units., such units are called derived units.International System of Units1" ’roC a- ntem3dor’2 con-- nee agreed on a set of definitions and standardto describe:* e p- s cs quantities. The system that was established is called thesystem ntematiioral (S l""he system ntematjona- is formed from three kinds of units:("I Base un't (2) Derived units(1) Base UnitsTra'& 3re seven base units fior different physical quantities.'ength , ^ass. t me, temperature, electric current, luminous intensity, amount ofsubstance. 1 / / - ^

In terms of base unitsSymbolUnitPhysical Quantitykg ms'2

NNewtonForceNm = kg m2 s"2

JJouleWorkJs'l= kgmV’WWattPower

( 3) Supplementary units Nm ~ = kg m sPaPascalPressureAsCCoulombCharge

Q.7 Write down the conventions for indicating the units. What 3re scientific

notation? —No. Physical Quantity Symbol of unitSymbol of quantity SI Unit(»> Length l mmetre

Conventions for Indicating the UnitsUse of SI units requires special care, more particularly In writing

prefixes: .

(ii) Mass kgM kilogram(m) Time . second s

Electric Current)Aampere

Chapter 1 [Monturomonti8

Scholar ’* PHYSICS XI 'Subjects#) 9Full name of the unit does not being with .» capital lettereven If namedafter scientist O K ,newton.The symbol of units after a scientist has Initial capital suchas N for newton.The prefix should be* written before the unit without any

space, such as

1 x 10 ‘m is written as imm.A combination of base units Is written each with one space

I SI FACTOR FACTORSymbol JPowr) (Languago)l

SIThe uncertainty Is usually described as an error in measurement.Types of ErrorsThere are two major types of errors I

Random error2. Systematic errorRandom ErrorRandom error is said to take place when repeated measurements of thequantity, gives different values under the same conditions.CausesIt is due to some unknown reasons.Reduction of random ErrorThe ranuum error can be reduced by taking several readings of same quantityand then taking their mean value.

Prolixio- Wntlliio,,

WHlrWlontrillionbillionmillion

thoutsndbmulroii

l•id

10"1’petA

UO 10"Tl«fS1.10*GU‘U*»

m#Q.i M 10'(HI)kkilo 10'

10'hhttetotloca 10' Ionda

dd*cl tenth10Civ) N *10' hundredththouMndth

millionthbillionthtrllllonth

quidrllllonthqulntllllonlh

coiUl capart.For example, newton meter Is written as Nm.Compound prefixes are not allowed. For example ippFmay be written as ipF.A number such as 5.0 * 104 cm may be expressed inscientific notation as

5.0 x 10* m.When a multiple of a base unit is raised to a power, the power applies

'«

mllll 10 'in

micronano

10 *l>10’(v) n

pIco 10"Plomto 10 "t Systematic Error

Systematic error occurs when all the measurements of particular quantity areaffected equally, these give consistent difference in the readings.CausesThe systematic error may occur due to

Zero error in measuring instrumentPoor calibration of instrumentIncorrect calibration on the measuring instruments

L(Vi) alto 10"a

(vii) 0

to the whole multiple and not the base unit alone, Thus,1km2 =i(km)2 = 1 x io

(‘m2

(viii) Measurement In practical work should be recorded immediately in the most convenient unit, e.g., (•••)Micrometer screw gauges measurement in mm, and the mass of calorimeter in grams. But before Reduc tion of Systematic Error

calculation for the result, all measurements must be converted to the appropriate SI base unit. Systematic error can be reduced by comparing the instrument with another

(0(ii)

instrument which Is known to be more accurate. Thus, systematic error Isreduced by applying a correction factor to all the reading taken on aninstrument.

Q.8 What is scientific notation?. A m! Jt Scientific Notation

Numbers are expressed in standard form culled scientific notation, w hichemploys power of ten.

W The internationally accepted practice is that there should be only one non zero

* digit left of decimal. Thus the number 1347 should be written as 1.347 x 103 and.0023 should be expressed as 2.3 x to"3.Q.9

Q.10 What are scientific figures? How can we estimate the number ofsignificant figure In the physical measurement and explain the way torounding off data.

For Your Information Significant FiguresIn any measurement, the accurately known digits and the first doubtful digit arecalled significant figures. OR

3 2 x 10’ In other words, a significant figure is the one which is known to be reasonablyK.6 XI04H jc IQ'1

1 X 1 0’1 x 1 0«

I X 1 0 ”

Interval (s)

Ayfi of ihc universeApe of the earth( >ne year( >nc dayTime betweennormal heartbeatsPeriod of audiblesound wavesPeriod of typicalradio wavesPeriod of vibrationof nn atom In u solidPeriod of visiblelight waves

What are the sources of errors In the measurement of a physicalquantity? What are the types of errors? How can we reduce the errorany measurement? ^ reliable.

How to Increase the number of significant figuresWe can Increase the number of significant figures in a measurement by

Errors and UncertaintiesAll physical measurements are uncertain and imprecise to some limit. There arethree sources of errorssources of errors(Ij Negligence or Inexperience of . person,

hi) faulty apparatusInappropriate method or technlq

improving the quality of our measuring instrument.General Rules for deciding number of significant figures(1) Digits

All digits 1,2,3,4,5/6,7,8,9 are significant.Zeros may or not be significant.2 x IO'” W

HDApproximate Valuos of Some Rujes^ jef0S

Time Interval* (I) A zero between two significant figures is Itself significant.(Hi) UP.

Chapter 1 [Measurements10 Scholar s PHYSICS — XI (Subjective)Interesting Information 11

to the left of significant figures are not significant.Zeros

For example,

None of the zeros in 0.0046 or 02.59 is significant.

(Note: these zeros are used only to locate decimal position)

Zeros to the right of significant figure may or may not be significant.

In decimal fraction, zeros to the right of significant figure are significant.

tMmm (kg)

10-»(ii) In case (i) 72.1 have the smallest number of decimal

rounded off to the same position which is 75.5m.In case (ii) the number 4.10 has the smallest number of decimal places andhence, the answer is rounded off to the same decimal positions which Is then8.13 m.

places, thus the answer is- • E!®ctronO Proton

- Uranium atom

- ] DNA mciocui*- 9 Cell

10-»

10-»

(iii) 10-1«

(iv) Describe the rules for rounding of data up to appropriate precision.Q.1110-10

For exampleAll the zeros in 3.570 or 7.4000 are signifreant. However, in integers such as 8000

kg, the number of significant zeros is determined by the accuracy of themeasuring instrument.

If the measuring scale has a least count of 1 kg then there are foursignificant figures written in scientific notation as 8.000 * 103 kg.If the least count of the scale is 10 kg, then the number of significantfigures will be 3 written in scientific notation as 8.00 * 103 kg.If the least count of the scale is 100 kg, then the number of significantfigures will be 2 written in scientific notation as 8.0 x 103 kg.If the least count of the scale is 1000 kg, then the number of significantfigures will be 1written in scientific notation as 8 x 103 kg.

(v ) When the measurement is recorded in scientific notation or standard form,the figures other than the powers of ten are significant figures.

For example,A measurement recorded as 8.70 x 10*1 kg has three significant figures.(2) Multiplication and division of NumbersIn multiplying or dividing number, keep a number of significant figures in theproduct or quotient not more than that contained in the least accurate factor .

5.348x10:x 3.64x10*1.336 '

As the factor 3.64 x 104, the least accurate in the above calculation had threesignificant figures, the answer should be written to three significant figures only.(3) Addition or Subtraction of NumbersIn adding or subtracting number, the number of decimal places in the answershould be equal to the smallest number of decimal places in any of thequantities being added or subtracted,

m this case, the number of significant figures is not Important,

it is the position of decimal that matters.

For exampleSuppose we w sh to add the following quantities e *pr ssed in rr ters.

it10-«

10« iBookRounding Off DataThe non-significant figures should be deleted by using the following rules:

If the first digit dropped is less than 5, the last digit retained shouldremain unchanged.If the first digit dropped is more than 5, the digit to be retained isincreased by one.If the digit to be dropped is 5, the previous digit which is to be retained isincreased by one, if it is odd and retained as such if it is even. Forexample the following numbers are rounded off to three significantfigures as follows;43-7556.854673.65064.350

Cor10«

10'0 Oitonkor 1.10«

Moon* Evorwt2.10*»

- Earth10»3--io»

Order of magnitude of somemasses.

is rounded off as 43-8is rounded off asis rounded off asis rounded off as

56.9Do You Know?Mass can be thought of as aform of energy. In effect, themass is highly concentratedform of energy. Einstein'sfamou- equation.E = me1means

gy = mass x (speed ofY

According to this equution 1

kg mass is actually 9 * 1016 J.EXPLANATION!If m » lkg. thenEnergy me2

* lkg x (3 »,0* ms*1)* * 9 x 1016 J

73-664.4

Q.12 What do understand by the terms precision and accuracy?

Wc use many devices tophysicalmeasure

quantities, such aslength, time andtemperature They illhave some limit of

enerPrecisionPrecision means how close the measured values are to each other.A precise measurement Is the one which has less absolute uncertainty. Theprecision of a measurement Is determined by the Instrument or device being

usedThe precision of a measurement depends upon the size of the unit you use to

make a measurement. The smaller the unit, the more precise the measurementThe precision of a measurement describes the units you used to measuresomething.For example, you might describe your height as 'about 6 feet’. That wou dn t 0«very precise. If however you said that you were '74 inches tall , that wou e

more precise.

light)

=l.45768982xl03

precision.EXPLANATION:Each device formeasurement hat somaleast count Theprecision of the dev,c«depends on m leastcount therefore* alldevices have some limitof precis100.

AccuracyAccuracy means how dose a measured value (result) Is to the actual (true)

0072.1 2.7543 value.mtMurtfnrrt I.th.dHtennc* b«w..n

f tha difference, the leal eccurata3 42 The accuracy of aand the accepted correct aniwtr. Tha bigg®

An accurate measurement

40. QQ3 im is one which has Its* fractional 0'75 523 your measurement.h 1273

Chnptor 1 [MnnuuromontajSiliolut Vn PHYSICS ^I (Nuhjetiivc)12

13iniM'.mi'lm'iit depends on t » » « • fr .u uon.d or

llw .K ( u i i i ( y ol .iunu*rt «ilnfy In tli.il nu',isurii*nu'nlpen iwit .iKi1 error

percentage 1

ExampleWhen the objft. t Is riMonled as 25.5 cm by using .i meter rod having smallest

division In millimeter, it t *. the dlffeieme of two ridding of the initial and

position The uncertainty In the single reading as discussed before is taken as

tO. OS ( m which is now double and i .died .ibsolule urn ertalnly equal to 10 I

Absolute ufucitalnty, In effect, Is equal to Ihe least count of the measuring

instrument , lids is c ailed precision:

#. 10.', I 0.1. in < nil ,ill p,Irillllji ii '.' i )u .1 hm,"lllill/ f.Vi ll"W "/id M"‘T In p/ < idu< *ili«' * fiiir * iunj/c nt lohmrn,All III' Clll'llll‘i III llllB IMMali.iv l,"ii made himi |u->llliuflo him i ol* »i*i "ICXI’LANAIJONiAny t olmif i .in l,n produ* «•»!Iiy it Ulllillilr 1 iiiII11111ii11< >ii uiod. M H ' I I .III«I Mil' minimI Wll toifIII!-*i Will' ll pioduicwhin* lijdil, when mixed nr*1' lin' d il'i inm|)lrilK'|||,H7"'limn. Ii h found IIMIyellow mid him*, or jm*cri midmagenta, or ted mid peacockbluei- olimif». IIn* appealail": tdcolours h ii procni olMibiimilvr nuluro. lied glaixnluiorbit »11 oilier culounisxceplmg Ilie it .l '1 lue. bluen wllllc inixiure ol green midred. The coloim obtainedwith paints nml inks resultfinrn II subtractive pmi es

.md 1 yin, i i i ' i )" n i j t

x, 26.8 I 0.1 c mIhi* difference x between them Is recorded as

x x, x,(>6.8 I 0.1) ( lO.yjn

•16.3 1 0.2 cm iIn Case of Multiplication and Division 1VPercentage uncertainties are added.Tor exampleI be maximum possible

< m

Case (I)Precision or absolute uiuertalnly (least t onnt ) 10.1 cm

0.1 cm 0.004 value of resistance R of conductorbtenti.il difference V and resulting

nd as follows;

Fractional uncertainty25.5 cm0 l t m 100 0 1

determined Irom the1 urrent How T by using

nts o.earncMinpIcmmUiiymeIK IsI) b .Percentage uncertainty

Case (ii)Another measurement taken by vernier calipers with least count as 0 0 A

recoded as 0.45 cm. It hasPrecision or absolute uncertainty ( least count ) ' 0.01 cm

25 5 cm 100 loo V0.8/1 I O.OjjA

for V B jOO5.2 100

3 ®h°ut 2%

ftQinty for / SPJJ5 1000.S4

X100

a about 6%

;e unceri

age0.01cm 0.02Fractional uncertainty - F8? Y8iir Informationrtalnty in the value of resistance R when V is divided by I is 8%

5.2V0.45cm

0.1cm 100 2.00.45cm 100 100

lie result Is thus given as R = ~ 6.19 V/APercentage uncertainty * 0.84 A= 6.19 ohms with %age uncertainty of 8%So the reading 25.5 cm taken by meter rule is although less precise but is more

accurate having less percentage uncertainty or error.Whereas the reading 0.45cm taken relative, measurement which important . Thesmaller a physical quantity, the more precise instrument should be used . Here

the measurement 0.45 cm demands that a more precise instrument, such asmicrometer screw gauge, with least count 0.001 cm, should have been used.

because % uncertainty for V is 2% and for I is 6%. So,Total uncertainty = 2% + 6% = 8%

Hence | ATOMIC C LOC Kflic cesium atomic frequencystandard at the Nations!Institute of Standard andTechnology in Colorado

(USA), .ii Is the primarystandard for the unit of timeEXPLANATION:To meet the n*n1 for n K»-n.crstandard of i'

'w, Jockhas been developed. In this

the frequenciesassociated with certain atomictransitions can be measuredprecisely. In l%7. the SI unitof time, the SECOND, wasredefined by IHh CiencrnlConference on Weights andMeasures.using characteristic frequencyof cesium-135 atomAccording to this standard

R= 6.2 ± 8% ohmsR= 6.2 ± 0.5 ohms [8% of 6.2 = 8/100 *6.2 = 0.5]

In Case of Power FactorMultiply the percentage uncertainty by that power .For exampleIn the calculation of the volume of a sphere using V=4/3(7tr )

% age uncertainty in V =3x% age uncertainty in radius r.When the uncertainty is multiplied by power factor, then it increases theprecision demand of measurement. If the radius of a small sphere is measure as2.25cm by a vernier calipers with least count 0.01 cm. then therecorded as r = 2.25 ± 0.01 cm

These .ire not decorationpieces ol glass bui ure ihecurliest known exquisiteand sensitivethermometers,built by theAccadomiu del Cincmonto( 1057- 1(,(,7). in .Florence.They contained alcohol,some limes,coloured redfor easier reading.

Q.13 How can you assess the total uncertainty In the final result ?

Assessment of Total Uncertainty in the Final ResultEvery measurement has a degree of uncertainty associated with it .

The uncertainty derives from the measuring device and from the skill of theperson doing the measuring. The total uncertainty in the final result can befound as followsIn Case of Addition and SubtractionAbsolute uncertainties are addedFor exampleThe distance Y found by the difference between two separate positionmeasurements

Absolute uncertainty in r = Least count = ±0.01 cmO.Olcm 100

^x — = 0.4%%age uncertainty in r

Total percentage uncertainty in V = 3 x 0.4 = 1- 2%2.25cm 100

Chapter 1 [Measuremp

^ Scholar 's PHYSICS - XI (Subjective)

14 -V = 4/3 ( nr } )

= 4/3 * (3-14) * (2.25)

= 47.689 cm 2 with 1.2% uncertainty

(i)Speed For Your Infro motionThus volume Length

As speed Dimensional variables— > Force, velocity etcNon - Dimensional van . es— > Plane angle,solid angl . etcDimensional constants-* g, G. k etcNon - Dimensional con « nts

— > Reflective index, dicla ic

constant etc

TimeDimension of length jlj _ rDimension of time " (T) =^' 1Dimension of speed = [v] =

( jj )Acceleration

As acceleration

the result should be recorded as

V = 47.7z 0.6 cm'In Case of Average value of Many Measurement

Find the average value of measured values.

Find deviation of each measured value from the average value.

The mean deviation is the uncertainty in the average value.

Hence

VelocityTime

(ij Dimension of velocityDimension of timeDimension of acceleration - ( a) =

(id For Your Information

^-MLTlinMLr2](Hi) Dimensional analysisdeals with the physicalquantities in theirqualitative meaning

,3l= [Tl fThe six reading of the micrometer screw gauge to measure the diameters of awire in mm are; 1.20, 1.22, 1.23, 1.19, 1.22, 1.21.

1.20+ 1.22 +1.23 +1.19 +1.22+1.21

INTRUSTING INFORMATION(iii)ForceA forces mass x acceleration

Dimension of force= dimension of mass x dimension of accelerationAs-cThen Average = 6 &out* part m Mn \ [F ] = [M][LT 2]

[F ] = [MLT 2]= 1.21 mm

The deviation of the readings, which are the differences with out regards to thesign, between each reading and average values are 0.01, 0.01, 0.02 0.02 , 0.01,0.00

M3000

Uses of Dimensionsj Using the method of dimensions called the dimensional analysis

the correctness of a given formula or an equation and can also derive it.\0T (i)Checking the homogeneity of the physical equation

In order to check the correctness of an equation, we are to show that dimensi'.— of the quantities on both sides if the equation is the same, irrespective of th»

form of the formulas. This is called the principle of homogeneity of dimensions.

(ii)Deriving the possible formulaThe success of this method for deriving a relation for a physical quantity depend^

on the correct guessing of various factors on which the physical quant ’

)>800 ; we can checktooo

0.01+0.01+0.02+0.02+0.01+0.00 •ooMean of deviation = •oo

foo

6VJO

= 0.01 mmIn Case of timing experimentThe uncertainty in the time period is found by dividing the least count of timingmeasurement instrument by the number of vibrations.For example

The time of 30 vibrations of a simple pendulum recorded by a stop watchaccurately up to one tenth of second is 54.65. Thus, the time period is given by

T = 54.6/30 = 1.82s

Uncertainty in time period -

Eacap«

*00

300

700PMnduKjm V(M»gulpnrtq cko-iowi

lOOTlid* (JiiVICO «nOf’f . «* I11 -irtuii -'* * ehv.- : ^

1»*.'.o

depends.

Ix'iisi count

Nuol vibrationsThus, time period T is written as T = (1.82 ± 0.003)s

Q.14 What do you understand by dimensions of physical quantities? Lxplainwith examples. Also write its uses.

For Your InformationDimensions of Physical QuantitiesI hi - (Itinelisions of „ physical quantity representsquantityL

,J,I bi,i,c phvslCt,l Quantity teprt .nrited is by .i ,, . ciflc symbol with in squarefcnackHsr ’hw dimensions ot length

I ho pull ill I | II;IIIIIIH,>

|| i \ r sill IU* illllli - |lMt ' H‘>nature of that physical • W Oils , ll »H|IU I ' llv III'

' mipiilr in.'inv nli'in\niii.ilm maim 1'ium

I'hliulvk * niiiHiiinlC. .ms . .ii,1 • I ' .iu linn'11

. mass and hm. an. |l |, |M| and | I | respei; lively i nit . ..i u> •l> M I l k '

r

MkSt holm W 1*1 1Y HIC.H \| (MulipuIlv < ) 1 /

IAN 1 1 1 II >IMNNONMH I HVSK AI Wmidcngll.SlrrwMlmin

IhiMliC nindnliuFocal lengthlipfnidlve

hide*

x ~v /r» -I'7A

», - AV/V1' - „/i;

HI

1/n.T |MI , Fl 7|Dlmcmloii l*» -Unittymb#) or

FormulaPNytfecil

OuontityN \ •!"

IIOlii Hour

|M1.11 7|I*# - N|MI “ r'l - i M i(M I I 1'!- H IIMHTTI - rn~

|M l l '|- [M 'l

am Fw i' .iH I

> Uivi1l eugth N flout

Mmint

secondtime (

m \>-A t\ ckKU> . speed|M"l \ | - |l I |. Multiple Choice Questionsin/s4 - AtXecvkraiKm

|M"l.3T“|- n’l'\V r v4

\\ olULUC Four possihh' answer,s to each statement are Riven below. Tick ( S) / lie correct answer:f [Ml. V| |M1, ’|kg< IIIp -uvAIVlINlty|MLT *|N = kgms *F»1114Force The main frontiers of fundamental sciences are

(a ) One(c) ThreeThe bianch of physics which deals with the ultimate particles of which the matter is composed is

called(a) Plasma(c) ParticleWhich one of the following is not a unit of energy?(a) Kilowatt(c) JouleSteradian is the unit of(a ) Plane angle(c) TemperatureWhich one of the following is a base quantity?(a ) Area(c) Current

1.[Ml. ' 1 |l*a N/m-¥ = K \Pressure (b) Two(d) FourNs "kgms 1 IMi.r'iP *niv

I »ApMomentum,

Impulse 2.|M L- r *|Nm

J = Nm- kgm\s *

i * IFW = Fd

Torque, \> ork.Fncrgy

physics(b) Nuclear

Solid statelMi/r'|W = J/sP = W/tPo\> cr (d)radian noneAngular

displacement0 3.

(b) Ergrnrad/sto •AO/AtAngular

\ elocitv

Kilowatt hour(d)

4.inrad/s’u = Au>/Al Solid angle

Intensity of light(b)Angular

acceleration (d)

Js =kg m“ s 5.Angularmomentum

L = mvr(b) Charge

ForceTivuTl1 = mr*(d)Kg mMoment of

inertia has base units identical to6. If p is the momentum of an object of mass m, the expressionNm:/kg:(.»' = Fr* /ni|ni2( »‘r;n itational

constant(b) Force(d) Velocity

By increasing the number of observations In an experiment, the error in measurement:(b) Decreases(d) None of these

(a) Energy(c) Power\v r -\m/s2Acceleration

due to gravityg =\\7m

7.(a) Increases(c) Remains sameHow many significant zeros are there in 0.0005010?

|TjTime period r seernf =1/T

to 53 2rcfFrequency,

angular.frequency

Hz = cydes/sec l.(b) 4(d) 6

Error In the measurement of radius of sphere Is 1* The error in the calculated value of its volume Is:

(b) 5%

(a) 3(c)|M L T T |Ns/rn kg 111 s'q «l'7(mnCoefficient of

viscosity (a) 7»

V

Chapter 1 [Measurements Scholar’s PHYSICS- XI (Subjective) 19

18

Short Questions of Exercise(d) 1%(c) 3% *

10. The percentage errorsmaximum error in the calculation of K.E. will be

in the measurement of mass and speed and 2% and 3%# respectively. Tfy

Q.1.1 Name the several repetitive phenomenon occurring in nature which serve as reasonable time

J ]( }\' jJ/ (Lhr 2005,Bwp 2006-2008,Mtn 2009,Lhr 2010,Lhr 2011)

Ans. The phenomenon that repeats itself after certain intervals of time is called repetitive phenomenon,

which can serve as reasonable time standard. .Examples:(i) lattice vibrations in a crystal.

(ii) the rotation moon around the earth.(iii ) the rotation of Earth about its own axis.

(b) 5%(a) 1% standard?

(d) 11%(c) 8%How many significant zeros are there in 5.00 x 10 3?11.

(b)(a) 3 4(d) 6(c) 2

12. Zero error is the example of(3 ) Personal error(c) Systematic error

13. Poor calibration is the example of(a) Personal error(c) Systematic error

(b) Random errorAll of these(d) (iv) sunrise and sunset.

(v) change of weathers.Give the drawbacks to use the period of pendulum as time standards?

(Grw 2003-2008, Rwp 2005,Mtn 2006, Bwp 2008, Lhr 2009,Grw 2010-11)

fVp/

(b) Random errorNone of these

Q.1.2

(d) As the time period of the simple pendulum can be expressed as;Ans.

thas the dimensions of14. T = 2R

I

but this time period can not served as reasonable time slandered due to several reasons.

Reasons:(i) time period of the simple pendulum varies with Q and g varies with altitude.(ii) The length of pendulum may change with temperature (in winter or summer)

Q 1 3 Whv do we find it useful to have two units for the amount of substance, the kilogram and mole?

(Mtn 2003, Mir Pur 2006,D.G.Xhan 2006,Mir Pur 2009

S

(a) velocity(c) MomentumThe unit of energy is(a) Joule, j(c) Joule, JThe dimension of light year are(a) [LT1](c) [ML2r2]Which one of the following is not dimensionally correct equation

(b) vf = V| + at

(b) AccelerationNone of these(d)

> •

UL*sl15. and its unit symbol is:

(b) joule, J(d) joule, j

16.(b) [M°L°T](d) [M°LT°]

Ans. Reason:When we are concerned with number of particles

^

then we use

mole of a substance contains the equal number of particles (i.e., NA = 6.022 -mole as the unit of amount o

substance because one1°23).But in case when we are concerned quantjty Lmatte^we use the unit kilogram.

Q.1.4 Three students measured a length of needle with a scale on which the minimum d.v.s.on .s 1 mm an17.

E = me2(a)

s =?t2(c) S = vt2 (d) recorded as(i) 0.2145m (ii) 0.21m (iii) 0.214m which record is correct and why.

Ans. 0.214 is correct record.

ThfTcorrect record is 0,214, because the least count of the scale is t mm or 0.001m

Q,, r;=i:;p-— — •you make regarding to experimental data used In computation?

Ans. Analogous statement:

Its analogous statement is

"A result of experimental data is only

data."

(Grw 2008]

18. The gravitational constant G has dimensions(a) [ML’Vj(c) [ML2r2]Which of the following has the same dimensions as that of momentum

(b) Energy(d) Impulse

(b ) [M°L°T](d) [M L*1T°]

So length can t

19.fa) Force(c) stressSI unit of light year is(a) candela(c) Second

20 .(b) Meter(d) None of these as much accurate as its least accurate reading in experimen

ANSWERS3.a 4. b 5. c2. c 6. 7. b 8. c 10. c9. c

14. a13.c I S. 16. d12. c 17. c 18. a 20. b^19.

Chapter 1 [Measurements]

Q 16 The period of simple pendulum is n^asured by a stop watch. What type of errors is possible in tin*Q.l.t* rnepenuuu r " (Mir Pur 2004,Lhr 2006,Fsd 2008)

20 Scholar’s PHYSICS — XI (Subjective) 21

;Solved Examplesperiod?Possible ErrorsAns. Example 1.1:|

There are two types of possible errors:

(i) Systematic error:Stop watch may be faulty ( zero

(ii) Random error: Negligence and inexperience of person (of the time to start or stop the stopwatch\

remains fixed for an instrument while random error changes for every

The length breadth and thickness of a sheet are 3.233m, 2.105m and 1.05 cm respectively. Calculatethe volume of the sheet correct upto the appropriate significant digits.

Given Data:error may present or the calibration is incorrect ) .

Note: Systematic errorobservation.

Does a dimensional analysis give any information on constant of proportionality that may appears in

algebraic expression? Explain.

Length of sheetBreadth of sheetThickness of sheet = h = 1.05cm

= 1.05 x 10“ 2m

= t = 3.233m= b = 2.105m

Q.1.7(Federal 2003-2005, Grw 2005-2008)

Dimensional analysis does not provide any sort of information about the value of constant of

proportionality, it can be determined by experiment. But dimensional analysis can provide the

information about the units of the dimensional constants like g (gravitational acceleration), G

(universal gravitational constant), k (spring constant) etc.For example:For time period of simple pendulum

To Find:

Ans. Volume of the sheet = V = ?Calculations:

Volume = (length) x (breadth) x (thickness)V = t x b x hV = 3.233 x 2.105 x 1.05 x 10” 2m3

V = 7.14573825 x 10~2m3

As the factor 1.05 cm has the minimum number of significant figures equal to three. Therefore volume

recorded upto 3 significant figures.V = 7.15 x 10"2m3

As the factor 1.05 cm has minimum number of significant figures equal to three, therefore, volume is

recorded upto 3 significant figures, hence, V = 7.15 m3.

T = 2n

where dimensional analysis provides no information about the constant 2rc.

Q.1.8 What are the dimensions of pressure and density?(Rwp 2003, Bwp 2004, Mtn 2004-2005, D.G.Khan 2005,

Grw 2005-2006-2009, Federal 2006, Lhr 2009, Lhr 2010-2011)

Ans. ( i ) Dimensions of pressure: Example 1.2

IF] [ma] [M ][ir2] The mass of a metal box measured by a lever balance is 2.2 kg. Two silver coins of masses 10.01 g

and 10.02 g measured by a beam balance are added to it. What is now the total mass of the box

correct upto the appropriate precision?Given Data:

= [ML_1r2](Pressure) = [P] =7M= [a] = (L*j

(ii) Dimensions of density:[m] (M)

= [ML-3]

.1.9 The wave length depends on speed V of wave and its frequency '/'• knowing that [X] = [L], [v] = [Lf[Density] = [p] = mi = 2.2 kg

= m2 = 10.01 g = 0.01001 kgMass of the metal boxMass of first silver coinMass of second silver coin = m3 = 10.02 g = 0.01002 kg

l ] and [f] = [T1]. Decide which one is correct (i) / = vA (ii) / =

(i) For f = v>.Dimensions of LH.S. = [f] = [T1]Dimensions of R.H.S. = [vX] = [LT

_1][L] = [CT1]As from (1) and (2)

(Mtn 2 >05, Grw 2009) To Find:Total mass of the box = m = ?

Calculation:Total mass when silver coins are added to box

m = mi + m2 + m3

m = 2.2 + 0.01001 + 0.01002m = 2.22003 kg

Since least precise is 2.2 kg, having one decimal place, hence total mass should be to one decimal place

which is the appropriate precision. Thus

..•(1)— (2)

LH.S.5= R.H.S.So. the equation is not dimensionally correct.

v(ii) For f =r-

Cimensions of LH.S.Total mass = m = 2.2 kg

=[f) = [r1!_M=

(LT ]~ W "

[L]

~(DThe diameter and length of a metal cylinder measured with the help of vernier calipers of least

count 0.01 cm arc 1.22 cm and 5.35 cm. Calculate the volume V of the cylinder and uncertainty inDimensions of R.H.S.

= (T1] .42) it.

From equations (1) and (2LH.S. = R.H.S.

So, tr.e equation is dimensional ^ correct.

Given Data:Least count of vernier caliper = 0.01 cm

Chapter 1 [Measureme^22Scholar’s PHYSICS - XI (Subjective) 23

* d = 1.22 cm

= 5 35 cmDiameter of metal cylinderlength of metal cylinder i

To Find:Volume of the cylinder * V » VUncertainly in the volume = ?

Calculations:Absolute uncertainty in length 7 0.01 cm

0.01tin dimension ofL.H.S.^dimensions .H.S.As '100 0 2%% age uncertainty in length x So that equation is dimensionally Correatoo100100

5.53cm0.01cm1.22cm

ndU4'

Total uncertainly in volume 2(% age uncertainty in d* m. 1 r 1 • t . i - w . »n U npth )

- 2 (0.8%) f 0 2%-1.8%

^ faic period of sample pendulum (Fjg. 1.2) using dimensional analysis.

The various possible facton on which the lime period T may depend arc:(liven Data:

or* Derive a relation for the timAbsolute uncertainty in diameter a

. .< )V -As volume (i) Ungihof the pendulum (f).(ii) M*wVllicb.,h(i»).(in) Angle0 wtudi ihStfuc*) makes with the vertical,(iv) Aetderation doc to gravity (g).

A

/ / / / / // / / / / /To Find:Putting these values in cqu ( 1 ) urn T 7KjcloUonJbr the tune penod of simple pendul

Calculation,:US reUfcon for the time penod T will be of the form,

•x f x t f x g*V 1 14 ^ ’ 6 , Ji. i • . oUint*4

0 \Tl H(as 1.8% of 6.2509070 a 6 7509070 0 I .

constant m* t 0‘gd . (I)100Taking dimensions on both sides, we getV•(6.2i0.1) (cm)1

Where 6.2 (cm)1 is calculated volume .uni 0 I m 1 . < > ; in i*lbus

S - r OA s

e - -(WTnHTWn Orr, \ * t

0 - ll.L 1!where % i* he speed of mCheck the correctness of the relation v Thus ASo equationI becomes,

(T) * constant (M)*|L]b (I L 1 )c (FT Vxml in-Is > iu instretched string of tensiun F, length (

Given Data: Comparing the dimensions on both sides

m* im“I’.quationIM|° - (Ml*

b » c - 6 * 4IF)° -It)To Find:To check the eoneelness of that equation I- quoting powei* on both sides, we gel

Culcuhilions: 1-2d- 1 or d-- -a- 1

IM d-0 o r b - ~ d o r b ' -

'Y iL°r i0 - Fl-uidmid d in cqu.( 1 )Substituting the value* of a.b,c

ii

x m° * f 1 * 1 * 8 1constantii

I - constant K ( J xg *

/

Chapter 1 [Measurements] Scholar's PHYSICS- X I (Subjective)24 25

Or S = cttT = constant putting values, we getorg • 3.0 X 10RX 3.1536 X 10

9.5 xl 0'5 mHow many seconds are there inHow many nano seconds inHow many year in 1 second

S =The numerical value of the constant cannot be determined by dimensional analysis, however it can be S =found by experiments. 1.2 (a) a

(b) r?Find the dimensions and hence, the SI units of coefficient of viscosity' r\ in the relation of Stokes lavfor the drag forces F for a spherical object of radius r moving with velocity v given as F = 6rcr|rv.

Given Data:

(c)Given Data:

1 year = 365 daysStokes lawWhere

F = 671 r|rvF = drag force,

r = radius,r\ = co-efficient of viscosity.

To find;(a) Seconds in 1 year(b) Nano seconds in 1(c) Years in 1 second

U T y

999

AndCalculations:To Find:

(a) As 365days1 ye;(i) Dimension of co-efficient of viscosity = ?(ii) SI unit of co-efficient of viscosity = ? 365 x 24 hours1

1 year =1 year =1 year =1 year =

As 1 yearand 1 secondSo 1 year

1 year1 year

365 x 24 x 60 minutes365 x 24 x 60 x 60 sec31536000 sec3.1536x107 sec= 3.1536 x 107 sec=109 nano second= 3.1536 x I 07 x 109 ns= 3.1536 x IO16 ns= 3.1536 x 107 sec

Calculations:F = 6Tv Tjrv

6TI is a number having no dimension, so it is not taken in dimensional analysis.Thus

CO As

[F] = [1J1V]IF ]M =or DIM

Putting the dimensions of F, r and v in R.H.S., we get

1MLT ~2 ]

MILT’1]

hi = [ML-'r1 ](ii) Thus Sf unit of co-efficient of viscosity is kg m-1s 1

11 second

= 3.17 x 10'8 yearsThe length and width of rectangular plate are measured to be 15.3 cm and 12.80 cm, respectively.Find the area of the plate.

In) = T year3.1536 x 10

1 secondOr1.3

Given Data:

Exercise Problems Length of rectangular plate - £ - 15.3 cm= W =12.80 cmWidth of rectangular plate

To find:A light year b the distance light travels in one year. How many meters are there?Given Data:

= A = ?Area of rectangular plateCalculations:

Time * t = 1 yeart “ 365 dayst * 365 x £4 hourst 365 x 24 x 60 minutest 365 x 24 x 60 x 60 second *,

t B 31536000 sccor it 3.1536 x JO7 second

Speed of light v c 3.0 y 10* /nj 1

Area = length * widthA = £ x W

As

putting values, we get= 15.3 x 12.80= 195.84 (cm)2

- 196 (cm)2

AA

Or AAdd the following musses given in kg upto appropriate precision. 2.189, 0.089, 11.8 and 5.32.

1.4Given Data:

2.189kgTo find: 0.089kg

S 7Distance( akufotion*:

mmmmm mA

r’‘_ Eri- W ' fMMChapter 1 [Measurement*

Scholar’s PHYSICS- XI (Subjective) 27m3 = 11.8kgm4 = 5.32kg

least count 0.1Uncertainty in time measurementTo ftnd: = 0.005 .v= —Add these masses up to appropriate precision no. of vibrations 200.005 1Calculations: Percentage uncertainty in time

Total uncertainty in g = % uncertainty in Jen= 0.1%+2x0.25%= 0.1%10.5%= 0.6%g = 9.76 cm 2 with

0.25%Total mass = m = mi + m2 + m3 + ITM

m = 2.189+0.089+11.8+5.32m = 19.398kg

As in the given masses, 11.8kg is the mass of least precision, having one'decimal place, which is tlappropriate precision.

Hence Total mass = 19.4kg

2 0 1:c uncertainty in time period)

So

Find the value of g and its uncertainly using T = 2n — from the following (As 0.6% of 9.76 *

g = (9.76 ±1.5 g

Hence

1.6 What are the

Given Data:

Measurements made during an experimentLength of simple pendulum = t = 100 cmTime for 20 vibrations = t = 40.2 sLength w as measured by a metre scale of accuracy upto 1 mm and time by stop watch of accuracupto 0.1 s.

— .. \ ^dimensions and is of gravitational constant G in the formula F = G

m , m 1.Gravitational force = F -= Gr2

Given Data: To find:Length of simple pendulum = t = 100 cm = lmTime for 20 vibration s

Dimension of GSI unit of G

= ?= t = 42.2 s = ?

42.2tI Calculation:Time period = 2.01s20 20 m,m 2F = GLeast count of meter rod = 1 mm = 0.001m

Least count of stop watch = 0.1sTo find:

fTijITl -jAcceleration due to gravity = g = ?2_ (Dimensions of force ) xjDimension of length)Calculations:

Dimension of GAs time period of simple pendulum is given by (Dimension of ma.\s) x ( Dimension ol mass)

fMLT-lL-]T = 2n

Squaring both sides , we get ML3!"2

=[tvrVT~:]As SI units of force, length anJ mass are Newton, meter and kilogram respectively. Sog = 4*‘ -7{ T 2

Or

Nm 2Putting values . we have -2— or Nm2 kgSI units of G == 4 x (3. l4)2 xl kg

is dimensionally correct, where vi is the velocity at / - 0 , a 1gShow' that the expression v, - v» 1 0 is

acceleration and Vfis the velocity *« ( ri 11 1(201)2

g = 9.76 ms'2

Given Data:Calculation for uncertaintv: v.= v -1-atFirst equation of nation is

To show:The equation v, -v, i-al is din.eniioaally cart***-

^%Msm" -rjilt. u•X V'l / ggrasHfigMawg.

Chapter 1 [Measurement Scholar’s PHYSICS XI (Subjective)2829

Proof: [ M f = [[Mr*[i]' =[r3o'‘][rr1 =[7T2‘

Dimensions of L.H.S of the equation [vr ]=[LT'’ ]Dimension of R.H.S of the equation = [v( +at]

R.H.S =[v.]+[a][t]R.H.S =[LT-1]+[LT"2 ][T]R.H.S =[LT"1]+[LT1]R.HS = 2[LT"' ]

As 2 has no dimension being a number, soR.H.S. = [LT1]

Thus R.H.S. = L.H.SHence the equation is dimensionally correct.The speed v of sound waves through a medium may be assumed to depend on (a ) the density p ithe medium and (b) its modulus of elasticity E which is the ratio of stress to slain. Deduce by llmethod of dimensions, the formula for the speed of sound.

i

A-2b =-1 or b = 5 L - (4)

Ration (2) we get

Now pulling values of ‘a’ and ‘b’ in equation ( 1 ), wc get

ant

Equation powers on bolh sides, we geta + b = 0 or a = -b...(2)-3a- b = 1 ...(3)and

1andsubslituting value of b in cqi

1a2

1.8v = constant p 2 E 2

i

fitGiven Data: v = constantSpeed of sound depends onDensity of medium = pModulus of elasticity = E

Show that the famous “ Einstein equation” E = me2 is dimensionally consistent.

E - me2

To find: 1.9Formula for speed of sound v dimensionally Given Data :

Calculations: Einstein equation

Einstein equation E = me2 is dimensionally consistentAs speed of sound depends on the following factorsv cc p° and vec EbOn combining, we getv oc p* Ebv = constant p° Eb ( 1 )

Where wc have to Find the values of powers a and bAs the dimensions ofVelocity = v = [LT‘]

mass

To prove:

Calculations:E = me2

Where E is the energy in joulesDimensions of L.H.S of equation = E = ( ML2!2 ]

(As E =W = Fd)Dimensions of R!LS of equation = me2 = [M][LT )

[ML2r2]Since the dimensions of both sides ol equations arc same

L.H.S.=R.H.S.

Hence, the equation is dimensionally consistent or correct .radio, r with uniform

1,0 Suppose, wc arc told dial the acceleration of a particlc^ovvnt^inj^cirHvv^ say

'v'/d'clcrminc the

speed v is proportional to some po*cr of r } > 1

powers of r and v ?

Acceleration of a purticlc moving in circle depend on

Radius of the circleThe uniform speed

powers of r and v i .c., n

As

Or ...0 )

-.( I )

. . .(2)

= [ML*3 ]Density = p =volume Thus

stressElastic modulus -= E *= - [ML"*T 2 ]strain

Because,

dimensions of stress =|ML'1!2 )

Strain is dimension less, because it is rai ioNow, writing the dimensions of bolh sides equation ( 1 )

( LT" 1 ]*. Constant [ML HML r2 ]b

3a *comparing the dimensions on both sides, /^c get

F(Asslress -A ^ Given Data:

•rv

To find:in 2kiILJfr ) =|M*^][L « ?, m ?Or

I

Chapter 1 [Measurers.30 ; Scholar's PHYSICS - XI (Subjective) 31( alculations:

As givena * r"a « vm

on combining, wc geta a. r" \'n

a = constant r" vm

Chapter 2

VECTORS AND EQUILIBRIUMwriting the dimension of both sides, we gel

ILT2] = Constant * [L]n * [LT']m

IL]'[T]'2 = Constant * [L]", m[T]'mcomparing the dimensions on both sides, wc get

ing Objectives

[ L]' = [L]n* m

IT] 2 =nr Understand and use rectangular coordinate system.Understand the idea of unit vector,null vector and position vector.Represent a vector as two perpendicular components known as rectangular components.Understand the rule of vector addition and extend It to add vectors using rectangular components.Understand multiplication of vectors and solve problems.

6. Define the moment of force or torque.7. Appreciate the use of the torque due to a force.

Show an understanding that when there is no resultant force and no resultant torque, a system Is In

equilibrium.9. Appreciate the applications of the principle of moments.10. Apply the knowledge gained to solve problems on statics.

1.Equation powers on both sides , wc get

n + m *1-m =-2m = 2

putting value of 4 m’ in equation ( 1 ) , we ge tn + 2 = 1n = -1

2..-(I)> 3-

. ..(2) 4-5.

Or

8Scholar’s PHYSICS (Objective) F.Sc. Part Iare available in market

By the Same AuthorsConceptual M.C.Q’s Answers with Hints, Past Papers,

Short Questions & MCQ’s

U*Scholar’s PHYSICS — XI (Subjective) 33

Physical QuantitiesQuantities which can be observed as well as measured are called physical quantities.

ORQuantities in terms of which all the laws of physics can be expressed are called physical quantities.Types of Physical quantities

(i) Scalars (ii) Vectors

A physical quantity which has magnitude only is called scalar quantity.(i)Scalar

Examples'S

Time, distance, mass, temperature, speed, energy, work, volume, area, electric charge etc.(ii)Vector

A physical quantity which has both magnitude and direction is called vector quantity.Examples

Force, velocity, displacement, torque, momentum, acceleration, weight, angular velocity, electricintensity etc.

How is a vector represented?Q.1

Vector RepresentationA vector is represented in two ways.(i) Symbolic representation (ii) Graphical representation

Symbolic RepresentationIt is represented by bold face letter such as A, d, r and v etc. It can be also berepresented by a letter with an arrow placed above or below the letter such

-*as A or A

Graphical RepresentationIt is represented by a straight line with an arrow head at its one end. Thelength of line represents magnitude of vector (according to suitablescale).Arrow head represents the direction of vector.

Note:Representation of magnitude of vector

The magnitude of vector is represented by light face letter such as A, d, r

and v etc or by the modulus of a vector such as A , v etc.

Q-2 What is rectangular coordinate system?

‘Rectangular Coordinate SYStem/(Cartesian Co-ordinate System)

The set of two or three mutually perpendicular lines intersecting

point is called rectangular coordinate system.The lines are called coordinate axes. Onehorizontal axis) The other is called y-axis or ve

fto both * and y-axes Is called.,- The point of Intersection Is called on&n

7.O 'at a

of these lines is called X-axis (or

rtical axis. The line perpendicular

Chapter 2 [Vectors And Egad,Scholjr s PHYSICS — \ 1 tSuhjectixc )U 35

Two dimensional coordinate system (Plane)

consists of fn-o perpendicular lines then it » s called two dimension* * This Ime gives the magnitude of resultant vector R The direction of

resultant vector is from tail of vector A to the head of vectorB'in pu* «»1, ^ Commutative Propertyrequired to represent a ve*vhik throe anglesreared lo reprevw a \cca space

if the systemcoordinate system.

Direction of a vector in planeIt is represented by the angle which the vector makes with po$ , iive x -

axis in anti-clock wise direction.

For Your Informath,

from figure, it is dear that either we add \ to R or resultant is same ie,I

\ - B B- AIt mfins that when vectors are added;the fLutt is thlsa e forHence, vector addition is commutative

Y order of addition.

_Eiplain the following terms:Resultant vector (jj)

*x X Q 4 For Your InformationO Vector subtraction KID ulijfH » *l iw <. ' w i l ihe rnj \imum when ihc> jrc

.il* »ny flu.* .imc dirtvtuut jruiAIII he minimum when lhc>arc in dirccii< >n

(iii) Multiplication of vector by a scalar(iv) Unit vector(vi) Equal vectors

(v) Null vector+v

Three dimensional co-ordinate systems (Space)if tne system consists of three perpendicular lines, then it »s ca ed tnree6 r- ensjona) co-ordinate systems.Direction of a vector In spate

tt is represented by three angles which the vector ma« es w.thi *, and :axei

(i) Resultant Vectorl > 1 1 tor M hu h liu % thi %ume rffret u\ Ihr cumhiinti iffret of nil the Melon to

hr uihh ai% t ailed rnultaot vector.(ii) Vectors SubtractionThe subtraction of a vector »s equivalent to the addition of \nnn 1 vt i" > »* nh if\

dire* tott mrru JExplanation

Consider two vector \ and B To subtract B from vector A First , take the

negative of vector B Add ( B ) into vector A graphically as shown in fig So

A B « A M B )

(iii) Multiplication of a vector by a scalarA vector can be multiplied by

1 a positive number2 a negative number1 a scalar with dimension

1. Multiplication with positive number

When a vector \ »s multiplied by a positive number fl (I* n > 0) Then the*

product vector will h#ve magnitude equal ionA and direction as th.it of

0, nVVcv’Mf. .uhll.lvUoM 1.1' v. IV »t

».ibo IHI* oroiuui.iim itw1

Ffir Vvur Idlur lu.iliundl Oevcribe the addition of vectors by head to tail rule. r**t reverse pftx.ru of verieraddition h called resolutionof vector

Head to tail rule»t is a graphical method to add two or more vectors,

Explanation' fm *•0raw ttuf representative lines of vectors A Mid li Job ‘he tall of

<

vector B , with the head of vector A In* * • tail ot »ector A with headof vector B ,

II vi.iiuf \ i. multiplied i

inimbi i rf . iilui |*UI>IUM I «I

rKgiimv. il.,' Muyniiudi nl

new \ cv t < •< K

B\

II \

2. Multiplication with negative number

Whon ,i vector A is multiplied by a negative* number n (i

to that of A .0). then th** product vector will havee n <

magnitude equal ton V but direction »'/7 *‘ " <

Chapter 2 [Vectors And Equity36 Scholar .s PHYSICS — XI (Subjective )

3. Multiplication with scalar quantity

When a vector A is multiplied by a scalar quantity n, then the product vector will be a new physical quant,

equal to the product of dimensions of n and A

Q 5 What is meant by component of vector and what are rectangular components?(i) Find rectangular components of vector.

How can we determine a vector from its rectangular components?(H)whose dimensions are

Examples . yyv is momentum! p = m v ] Components of a vector

I The effective value of a vector in a Kiva,direction is colled component of a vector

A vector may split up into two or more than two parts these parts are known ascomponents ol vector.

Rectangular Components of Vector

V Product of mass m and velocity v— » _

> Product of mass m and acceleration a is force!F = ma )

> Product of force F time t impulse [ I = Fx t ]

For Your InformationResultant of unit vectors iand j is N/2

(iv) Unit vector 'A vector whose magnitude is equal to one with no units in a given direction iscalled unit vector.It is represented by a letter with a cap or hat on it.

Mathematical Form t

— * — A

If A is a vector with magnitude A, then A = A A

Jhe components off vector which ore perpendicular to each oilier are called

f trectangular camponruts.

Explanation

Consider a vector A which makes an angle 0 with x-axis as shown in figure.Y

\OM is projection of vec tor A on x - axis and ON is projection of vector A on y -

axis MBy head to tail rule* A kA=OR \\A [As ON=MP]OP = OM + MPo.Examples

A A A

> Direction along x, y and z-axes are represented by unit vectors i, j,krespectively.

> unit vector r represents the direction of r .A

> unit vector n represents the direction of normal drawn on a certain surface

J1 H

X \ !( > \ 1(1)A = AJ + A J •

Thus Avi and Aj are the components of vectorA . SinceAj and Aj are at

right angle to each other, so they are called rectangular components.

X- Component of AIn right angled triangle OPM,

= cos0

ORFIQ. 2 S<»)1 IU :.< >An

(v) Null or Zero VectorA vector whose magnitude is zero and direction arbitrary is called a null vector.

— ¥

It is represented by 0Example

A Sum of a vectorA and its negative vector (- A ) is a null vector , i.e.

OMFor Your Information

OPPosiiion vector of origin is

— = cos0null vector.The acceleration of a bod)

moving with uniform velocit)is null vector.Verticalprojectile at thepoint is null vector

ORA

(2 )Ax = A cost)ORvelocity °*A +*(- A ) = O

^ Sum of vectors taken head to tail along the sides of a closed polygon isnull vector

highest Y - Opponent of AIn right angled triangle OPM,— '

= sin(>( M\ ) Equal Vectors OPFor Your InformationTie vectors are said to be equal vectors if y Pave same magnitude and same

direction regardless of the position of their initial points.Example Hie*

A ,Two like parallel vectors ofequal magnitudes arc equal

— = sinUORA

A = A sin 0vectors.— *Two parallel vector A and of sa ne mag. ide add direction are equal vectors. (3)

Ay = A sin()

Chapter 2 [Vector* And

Scholar s PHYSICS — XI (Subjecthe) 39Vector A i ro~ its rectangular componentsQ.7 Describe the method of addition of vectors by r. ;ular components.W =r twde of A

r * = err ?e>M

*V-V Vector Addition by Rectangular Components

Consider two vectors A 3nd B representedb / tHe lines CM an

m figure.By head to tail rule the resuJtjflPaf these two vectors is#=A + B

their rectangular components.

1'’ < as Sro //r

A =,A>A/ (4)“

'D 'ec! on o* A Resolve the vectorsA , B and R i' ' 'er * «* r ec •' 3'ge OPV

"SB £ 3 rf' -n pr — — % «»« S«m » >*T>< proceM by *h*h a vector iat>c obtained from iu recurtptbcofT^onenu i» called compotHmof 9 vector

OMA,

C* ",• A

*« U* -4 (S)rAA, /

0. 6 bef r* poprhon vector.FI«. ? o

X- Component of Resultant•9

OQ is ' ' Magnitude of x component of vector A , MS is the magnitude of * -Portion VectorThe vector whkh represent* the positton of a point or a particle with respect to

*«fir id origin n tailed position vet tor. It is denoted by rIxpUnJtiOO

f f ' - o dimere nmol inordinate system (plane), tho position of a point P (a, b) isfipr «/ *,* f »U'd by

component of vector |i and OK is the magnitude of x - component of vector Kf r o m figure

OR 00 «mOr OR v 00» MS ( since OR MS)

| r a* bj ] ( DR, A, * B,

This shows that the sum of magnitudes of x components of A and H is equalIfh« mtgnitud* ol tint position v « » t o r i s

Je i b' I ho Own * 1' .u iukils ill

llm incredible b.ihukiw•id art- in equilibrium.i.xn ANARION:Ihe ('him- e <iuobuh .ucnt Htablt equilibriumlu » .iii'v *. the Imr throughthe ventre of nuw.prnsc*Ihrough rtic twvo ,nv.t Ititic Ittv (alts in orouiviUc tin- IMU- .iivathen the <utoh.iL\ 'MII notbe m.qylUlmum

r to the magnitude of x component of resultant vet lot U .Y Component of Resultant

ve<..b)

OM Is the magnitude of y component of vector A ,5P Is the magnitude of y

component of VIM. tor It and UP is the magnitude of y component of vet lor 1<I rum figure

RP AS+SPHP QM SP IvInccRS • QM|

[ R»‘A> + B> 1This shows that the sum of magnitudes of y components of \ and It is equal

o X

»* II a *(•»In thru dimensional inordinate system (ipat* ), th» - position • *f .« point P (a, b, c)U rtprttonud by U )

• * » m,i it i i b | ck

to the magnitude of y component of resultant vcvtoi Kand Ms magnitude iss/a * i h; u 'i

Chapter 2 [Vectors Ami Equity40 Scholar’s PHYSICS - XI (Subjective)41

m m.

Since Rxi and Rxj are the rectangular components of resultant vector R ,hence

R = R,i + RyjPutting values of R» and Rv from equation (1) and (2) we get,

R =(A,+ )i+(Ay +By)j— *

Magnitude of Resultant Vector R

The magnitude of resultant vector R is

R = V'R «!+R >

i

Y0U determine the angle 0 of the vector R by its rectangularHow cancomponents

Q.8

Table 2.1Determination of Angle (9)

1) First find <f> by the following relation.

4> = tan'1 [ —

II Y

R. -R +

R. +\ AR, *iK )

Where ^ = the angle which R makes with nearest x-axis

2) By the signs of Rxand Ry find the quadrant in which R lies as follows:If both R, and Ry are positive, the resultant lies in the 1st quadrantand its direction is

o = t

X XR. -R. ~

R.* ~

/III Y' IV|ft = (3)1st quadrant

Y 0 =*Fo** Your informationDirection of Resultant Vector R

— »

The direction of the resultant vector R is given byIf R is ihc resultant of vectors A

If R . is - ve and Ry is +ve, the resultant lies in 2nd quadrant and itsdirection is

and B then us magnitude isR yj\* •n:4 2AUcos0Special Cases:If 0 - 0°

xtan0= — 0 = 180’- <J). 2nd quadrant

Or R , * A + B 0 =180 -*If 0 * 180°Rv If both R* and Ry are -ve , the resultant lies in 3°d quadrant and itsdirection is

- 1 00 = tan Rm», A-D ' * (x 1 XIf 0 °0°0 = 180*+ 4>.R-VA * +BJ-t (4)0 » tan

A, + B, If Rv is + ve and Ry is -ve, the resultant lies in 4th quadrant and itsdirection isIn General

Do You Know?Ihc sum .> f vector " Inch forms the 0 = 360* - <j>

MULTIPLICATION OF TWO VECTORS*

For any number of coplanar vectors A,B.C,D, we can writesides of open polyi on is not ero.

)!N K +B.+ C X + )J +LA,+ BV +C,+ y'R = „ Vectors can be multiplied in two ways:

'.c III

^ *Scalar ProductAndVector ProductA > 4 By + C\ +

,Aj + fC,+-10 - tan

AQ-9 Define scalar product of two vectors.Give examples.Summary

find the x and y-components of all given vectors.(•)Add x-componcnts of all the vectors to find the x-component Rx of the resultant vector.00 Scalar Product (Dot Product)Add y-components of all the vectors to find the y-component Rv of the resultant vector.(hi)

Find the magnitude of resultant vector R by usinj;(lv)

Chapter 2 [Vectors Anti Kquili42SthoUr’s PHYSICS - XI fSuhi,-,B

Where A and B are the magnitudes of vectors A 43

B. A = ABcosOFrom equations (1) and (2)

and B and 0 is the angle between them.Physical meaningDot product of two vectors is equals to product of magnitude of one vector andthe component of the second vector in the direction of first vector.

From figure

(2 ) (since AB = BA] B

,° 8 Cos 0 . N<6A .B =B.A \

V* NFig. 2.10 (a) (ii) Perpendicular vectorsIf two vectors are mutually perpendiculascalar product is zero. i.e.

\0 \

B *-* -* ^ -»

A .B = A (projection of B on A )

-• -«A.B = A (magnitude of component of B along A )

Fig. 2.10 (b) A

B\ -*-1<5> N A .B- ABcos90°\

Or sA .B = AB (o)s

0* NA .B = A (B cost))A .B = 0

lits veSimilarly Fig 2.10 (b) A

AIn case of un ctors,

' j = ( ») (1) cos- - kSimilarly j . k = 0 and k • i = 0— — — — —i . j = j . k = k i = 0

B. A = B (Projection of A on B )-* -4 -t

B. A = B (magnitude of component of A along B )

B. A = B (A cosO)

0 = 90"= o

AThus

BExamples

1) Work is scalar product of force and displacement

[ W = Fd ]2) Power Is scalar product of force and velocity

(P- F v ]3) Electric flux is scalar product of electric intensity and vector area

E- A ]4) Magnetic flux Is scalar product of magnetic field strength and vector area

»(ill) Parallel and anti-vectorsIf two vectors are parallel (0 = 0°) to each other then their scalar product is equalto the product of their magnitudes. I.e.

A ,D= ABcosO0 = AB (1) = ABThis Is the positive maximum value of scalar product.If two vectors are anti-parallel (0 = i8o°) then theirscalar product Is negative.

0 = 0e

A*

B«e = 180°

A.B= ABcos180° * AB (-1) = -ABThis is the negative maximum value of scalar product.(Iv) Self Scalar productThe self product of a vector is equal to square of Its magnitude I.e.,

A .A = AA cosO0 = A2 (1) = A2

In case of unit vectors,

i.i = (i) (i) cosO0 = (1) (1) (1) = 1

Do You Know?

Magnitude of a vector A is v A . A(4,a l* A ]

&1Q Write down the characteristics of scalar product of two vectors. For Your InformationWhy K.E is scalar quantity eventhrough velocity is a vector.

1 .KB - -mv *

1— m ( V . v)2Characteristics of Scalar Product

(i) Commutative propertyScalar product of two vectors is commutati t

If A and B be two vectors and 0 is the angle be veen them. Then

B Similarlyj . j=land k . k = 1

Thus h , jA A A

. i= j. j=k .k = 1

eand A.B= ABcobO B CoseMF»Q. 2.10 (a)

OP B A = ABAcosG

44

Scholars PHYSICS - XI (Subjective)45. Sea s' proc jet n terras rectan.g-yiar components

fingers o* the right nsnd in the director of"epre>s"^ the c 'ectior tr AxBExamples

~orc-e ‘ts the vector cro6ixt of ooscs[ r = r*F]

2) Force or = n*-:ng charged csrt*de -3 .f*: is. e odtj of changed partide ard -eghetiefie-Aike.[F

I3. Angu s' mor-ecrur. ,s . ^ r.c' product ofmomentum p e- (L = rujp )

'ttati- Erect thumjCy-: oer two .’erto' A and B in scace. then

A = A^l+Ajj*A.k ad B = BJi-B„j-B.k

- 3 = -. — A. j — A.i B, -B, _ -B.i

= A.B.Gi)-A¥B,(Lj>-A.BlfLk)

A,B,(^i -

ArBt<ki>-AJB,(k-i)-AA<ki)

= A B.P)-AA<«A.B, c -A/5. -A.B, •

- -. 5 ' - 5 - - r

5': -'orcer e

tor:y..r.r

VXB; ]Yew fedTag off the edge

positicr sector r r: '-.rV*tf >:>- 3a 10 i'< id~\n: -FYPLAVATIOV- v

^ a hoAag a *3.7teak as n r

- .0,12 Wrrte down the characteristics o' vector c^oc-c:of two vector*.

A*i ic£ 3'A sar- B = A.B - v 3 - .VE; CGi-r Characteristics of Vector Product(1)Voituc - of Commutative law" e cr: : s croc : : vector A =": 3 is 'ct sc— JfcstTve

• A =':: :t >.c r:: • r - rr -:5> r*' r — ' e*

£G«I SET- «!» A. ffiulI** IStK £TC2.badl

rr=cr*:r s fflcTsr* — •<; 5>£sr^

»5 3C£I**cas? = V.B,*A^(

AA-AJB.-AAZ5r& jeLSJSii =E*

*3A,3 - A 3. - \J,- :i: - = A_- m r r.AB w

ns acusecr or re -sec ri r~_

r»c JZ:-*: A = -A.sashr '' ~ y&XH: irc prw.-;* r- 3a*r

3' A =A3s«8 -aj 2 c -:S i3 = = i

r~- ecjatons r:(2]

At* 3 - -3 *A. ed:':*OCJC2 >:<i : ::.-Iiicr tfl usaxrr a- X

0® A*3 * B x A1) Perpendicular Vectors'A cross product of two perpeiwiciPar 8 = 90' 3 ~ ~

~ 2grc-ce. ie.

AxB=A3s3nW c

A c r»: •** . - ~ i ; - . i— *- - 3 <- t^ZZZrS A 3TC i S I*''A A1

L _* = l3sia

' -7' •err: icc 2 .: T IGC 1 is — e _r ' -ecccr

7T43

: ^ 'c r=rc2.r : - r ire 12.A* 3= A3 3 aAx 3= A3 n

*-• case & ts vectors

Ii =c:' of prxc -»ct'e. lirecsicr cr , '-— -r:r;c,c A* 3 C3r 1 : _

*c cy ' " = ~c__-e.

; grt 'arc-Ci- *s. is zr -e r -c ^ ecicrs c* ier “ e = - - 2 ;* - ecrc 'i. zc .2’ i

h*c:zr K r- rc 2 r" rc^gr 5ns cr

i x j= (i) {iJsin 90' i= W W W k- k

A A JB

j x k = i a o d k < i = js -* any.

t--: pant e -«-ges. Cun

/Chapter 2 [Vectors A46 EquiiM

Scholar’s PHYSICS — XI (Subjective)

+ AyB»(j*i)+ AyBy(jxj)+ AyB,(jxk)

+ AxBx(kxi)+ A 2By (kx])+ AxBx(kxk)

= AxB,(0)+ AxBy (k)+ AxB,(-j)

+ AyBx(-k)+ AyBy(0)+ AyB2(i)+ A 2 Bx ( j) + A2 By (-i) + A 2B

2 (0)

=(AyBi -A2By )i + (A 2B„-AxB2 ) j+ (AxBy-AyBx )k

AHence 47

i X j = k j x k = i k x i = j BNote:-j x i =-k k x j =- i i x k =- j 0 *i(iii) Parallel and Anti-parallel Vectors

The cross product of two parallel (0 = o°) or two anti parallel (0 =i8o°) vectors is•a null vector, i.e.In case of parallel vectors

AxB = ABsin0° n = AB (o) n = o n = O x

In case of anti -parallel vectors

A

B<r

Why do you keep your legs farapart when you have to stand itthe aisle of a bumpy-ridingbus?EXPLANATION:When you stand in the aisle of abumpy-riding bus, you are inunstable position and you mayfall. To make you stable, youkeep your legs far apart, so thatyour base area may increase andthe line from C.G may passwithin the base area, so no torqueis produced.

Ax B= (AyBx - AxBy)i+(AXBX -AxB2)j+(AxBy - AyBx )kAxB = ABsin180° n = AB (o) n = o n = 0

This result can be written in determinant form as

j kAx A, Ax

By

(iv) Self Vector product

The self product of a vector A is null vector. i

AxB=Ax A = AA sin0° n = AA (o) n = 0 n = O

In case of unit vectors,i xi= (1) (1) sin 0° n =(i) (1) (o) n = 6

B,For Your Information

Cl.-J3 Define and explain the term torque or moment of force.Pivot Point or Axis ofRotation is the point oraxis about which anobject rotates. Torque isthe tendency of force toproduce a rotation

Similarly TorqueDefinitionThe turning effect of force produced in a body about an axis is calledtorque.

ORThe product of magnitude of force and the perpendicular distance fromaxis of rotation to line of action of force is called torque.

ORThe moment of e force can also be defined as the vector product of theradius vector from the axis of rotation to the point of application of theforce and the force vector.Mathematical Form

j xj= kxk = OSo,

i xi =0(v) Area of Parade ogramThe magnitude of cross product of two vectors represents the area ofparallelogram formed with these vectors taken as its two adjacent sides.

Area of parallelogram = (length) (height)

* (A) (B sln0)AB sin 0

j * j=o kxk = 0turningpoint

Ths nut Is stay to turn with « spanner.O

3-n7br- tF* magnitude of ( AxB) •s It Is easier still « the spanner has a

long handle.to0JAXBJ Where £ >= perpendicular distance.And F * Magnitude of applied forceDependence of torqueTorque depends on the following factors;

1 ) Magnitude of fora2) Perpendicular distance from

called moment arm.

Area of parallelogram T* %[v\) Vector product in Rectangular component Form

Cornier two vector A and B In space, tn*- r

A ' A,i A, j+ A,k and B•Bfi 4 By j + BfkAxB * ( A 4 1 4 A j t A/ j/ f l S f : B/j B,k)

- + A,B/i/j; Atha0* i )

axis of rotation to line of action of force

Unit

Chapter 2 [Vectors And Kgu^48 Scholar * PHYSICS- XIFor Your Informal!^Torque is important;

the operation of electrvmotor which isvacuumdishwashers,

unit of torque is newton meter (Nm) and its dimension is j^ML T JExamples of torque

• Tightening of a nut with a spanner (wrench).and off the ground due to torque imbalance

Calculate the torque due force acting on a rigid body.

The SI - t r = (r sinO)Fr = r FsinOused ,

cleaner, in vector form,compi^printers, videocassot

recorders, **—• A seesaw rotates on r = r FsinO nwatt

Q.14 pumping stations etc.r =rxF

Torque in a rigid body

Consider a rigid body as shown in figure. Let Fis the force acting on the

body at point P. r is position vector of P with respect to pivot 0 and 0 is

the angle between Fand r .

Resolution of force FResolving the force into its rectangular components we have

—FsinO = component of force perpendicular to r

Direction of torque

Where n represents— a —

' containing r and F. It can be foImportant Note

Torque is the counter pcTorque plays the same role in angular motion as force plays in linearmotion.Torque determines angular acceleration as force determines linearacceleration.Anti-dock wise torque is taken positive and clockwise torque is takennegative.

title Of UK poo* pickedAguiaii * w»U Con >vuihe other leg ude ways? If not.then why not?EXPLAIN ATW)*:You cannot raise the other leglick wayv betauje m O&wgso, your bait ur J willdecrease and you will he to

mutable equilibruim ind > O«J

rce for rotational motion.

FcosO =• component of force along the direction of r .

Torque due to force about 0

As the line of action of FcosB passes through point 0, so the torque due

to this component is zero. Hence torque due to force F is equal to the

torque produced due to FsinO and is given by

r = r (Fsin0)

z - r Fsm9

may Call down. be*. toline from C.H vnj| fillthe hue area.

Interesting ApplicationQ.15 What is value of torque if the body is at rest or rotating with uniformangular velocity?

OR Torque acting on the body will be zero,

/ ReasonIn this case angular acceleration is zero, so torque will be

(2nd law for rotational motion)

In vector form,

r = r FsinO n

r =IaOR r = r * F

r =l(0)= 0

- pS /Alternatively

Resolving me position vector r m r0 fts components we have

Qi6 What Is equilibrium? Give Its types.nt

am ba mEquilibrium

r «,«nO & component of r perpendicular toF

r cosO - component of r along the

Torque due to force FIn this case the torque is given by

r* / F

Where t * r sinU = moment a*Fus

EXPLANATIONA body is said to be in equilibrium if it is ai rest or moving wuk umfivelocity under the action of a number offeree*.

Types of equilibriumThere are two types of equilibrium.

1. Static equilibriumIf a body is at rest , It Is said to be in static

r®CtlO* f i 1Tto m *•»tw . abe by

1Examples to

Book lying on a table

- *• *> - 1

BE SN »

1 ' * mChapter 2 [Vectors And E Hquilibri^Scholar’s PHYSICS- XI (Subjective2. Dynamic equilibrium

If a body is moving with uniform velocity. it is said to be in dynamic51Noteequilibrium. 1. We will apply the conditions of equilibriforces are coplanar.

2. To calculate to torque we choose3. A most suitable place ispass.

ExamplesA car moving with uniform linear velocity

• A body is rotating with uniform angular velocityMotion of a paratrooper

um to situations in which all thean axis .The position of axis is arbitrary,one through which line of action of many forces

Point to ponderQ.17 State the two conditions of equilibrium.Do you think the rider in the figure is really in dangeT? What ifpersons below were removed?EXPLANATION:The rider in the above figure is not in danger because the line'through the C.G passes through the base area. The rider will bein danger when the people below are removed because, now,the line from C.G may fall out of the base area.Alternate:If the persons are removed then effectiveraised so the raider is i

First condition of equilibrium (equilibrium of forces)The vector sum of all the forces acting on a body must be null vector.

L? =oIn case of coplanar force, 1st condition can be expressed as:

z?.-3£h =o

(1)j.e.center of gravity willm more unstable condition.

(2 )Can You Do?

And (3)AWhere FORMULAE

y FK = sum of x-directed forces— *£F? = sum of y-directed forces

Commutative law for vectoraddition

Subtraction of vectors

Vector representation

NoteIf the rightward forces are taken as positive then leftward forcestaken as negative.If upward forces are taken as positive, then downward forces are taker?

as negative.Forces which lie in a common plane are said to be coplanar.

Second Condition of equilibrium (equilibrium of torques)The vector sum of all the torques acting on the body about an axis must be nullvector.

With your nose touching theend of the door, put yourfeet astride the door and tryto rise up on your toes.EXPLANATION:When you try to rise up onyour toes the line of C.Gwill fall outside the basearea and you may fall.Also in this case, line ofaction of our weight passesthrough the axis of rotation.Moment arm is zero, torquewill also be zero. So we cannot rise.

Unit vector4

5 Null vector

Vector in terms ofrectangular components

6

5>-oi.e.7 x-component of a vector AQ.18 Under what conditions the body Is said to be in complete equilibrium?8 *y-component of a vector A •

Magnitude of vectorA9Translational equilibriumWhen first condition is satisfied, the linea ccelerabon 0' body is zero and thebody is said to be in translational equilibriumRotational equilibriumWhen second condition is satisfied, angular srcelerc ten of body is zero and theoody is said to be in rotational c ’libri n.Thus for a body to be in complete « ibrium, be conditions must be satisfied,i.e. both linear acceleration and angu. accelera on must be zero.

— *Direction of vector A

Position vector of a pointP(a, b) In planePosition vector of a pointP(a, b, c) In space

Chapter 2 [Vector* And K iSibnl Scholar'* PHY81C3 - XI (Suhjecth't) 53

R =.J(A^ +B,)J +(A,+ Ry)2Magnitude of resultant ( R )13 -I of vectors A and B iSelf Vector product of unit

vectors i. jandk31 * . *•i*i - O l * i m° k n k - OAy 4* By

Direction of resultant ( Rl 0 = tan l( Vector product In terms ofrectangular components

Ai'B - rAyB/ * A,B,)U(A,B. - A,B,)j * (A,B, - A y B t )i14 Ax -fBx 32of vectors A and B

15 I Scalar product of two vectors AB = ABcosG ?

A* BAngle between two vectorsl Scalar product of two| perpendicular vectors

-isin33AB = 0 A andB AB

Scalar product of unit vectorsi * j = 0 j k = 0 k i =017 i. jandk Area of a parallelogram and

vector productArea of a parallelogram A • Bj * AHs.nO34

^I Scalar product of two

paraHel vectors AB = AB TWFTorque35 t •r * F T •rFfinOnScalar product of two anti-

^ 1parallel vectors I?.-0 - IF,-0AB =-AB IF =01" condition of equilibrium36

A A = A2 I* -®20 Self dot product of vector A 2nd condition of equilibriumA = 37

Self scalar product of unit* •i.jandk

k - k = ljj= ii i = l21vectors

Scalar product in^erms ofI rectangular components A - B = A ^B, 4 AyBy + AjpBj

)' A «B„4 AyBy 4 AZB /Angle between two vectors -iAB 0 ® cos0* cos“‘ AB23 ABA andB

A cos0 - A BAB14 | Projection of vector A on B AcosO ®

B

AB *25 Projection of vector B on A B cos 0 » A •BBcosO ®

AVector product of twovector*

*Ax b AllsinBn

Vector product of two1 1 B ABner rrVector product of unit

vectors l. jandhVector product of twoparallel or anti paraHel

k * i j\* )* k * < *H -iM M

*9 A x B -O

cross product vectr»0 • iAn A 4*0A

Chapim J [\ m x n And \M

s4 holm '« PNYAICA \| (Si»h|i « Uvr ,Multiple Choice Questions 55

( t ) pit»Hn« l MI fuui % h\V \

If iTt«i»liinl*» of scalar and v#,!of pfftduHl o|t(-)(t ) 60*Reverse proton of vector addition 1% called(,« ) negative of A vector(c) resolution of a vectorWhen a voctor It multiplied by a(a) Does not » hang#

(c) Change by 270°Tho minimum number of vectors of un aqual magnitudes, whose vector sum can ba «aro it

(b) 2(d) 4

tha magnitude of thalr ratultant It maximum whan the angla

Id) M »* A»nata» t ol pal•lUitog» antwo vectors ara tema. than angla between themItIt.

,»*•* M**\ IM ( ) /A « cVrrwr uw.iHvr:» »>*r awtan h» nn 4 rei arr7 (b )a

(d)

» 7 .ftdk out the walat quantity* t‘ .

(6) torque

(d) ImpulseBCt'un of a victor

* >!» • ation of a vector<•» / . ^ (d)negative numl»*r, lt * iKJ momentum

Add«t*n of VWCtOri Ob#V*<*) commutative law

(0 avKKMttva law

Mat m mbe ol components of a vector may be;

Ia) onetc) three

II diraction(b) CharggibvW

Change \ by 1*0'> I

(b) distributive law(d) .01 of these

V

(d)

14.iI. (a) l

3(0(bl two

(d) infiniteTwo Kvcof act together on an ob|*ct . Tha magnitude of their resultant force Is minimum when th#

act attaj 1BQ'tc) 45*

15. Two forces act togctlbatwean the forces Is(a) 0°(c) iso'

* 1 (b) 904(d) 270°

(b) 90**

(d) 0°Megmtucie of resultant vector of t>N and 8N which are perpendicular to each other Is:

(b) 10 N

(d) 2 N

6. If AxR points along * aids then the vector A and B must be In(b) yi * plana(d) All of these

xylano(c) <4planeA vector In space has(.1) one(c) three

(a)V S.fa) 14 N

k) 20 Nrectangular components

(b) two(d) Infinite*(V are A »3i - 2j- k.B * - A t - i j a k, thenA. Two

18. What Is tha angle that the given vector makes with y-aali? A 2i • v l:Jr| (a) 30°

(c ) 90°

19. If for two non-tero vectors A and B, A , B 0* the vectors will be:

(b ) perpendicular(d) ft •60°

a

(b) B Is negative vector of \(a) B ©Hipyillgl to A (b) 60°(d) 120“A

*(d) B is perpendicular to A

•f a force trlSNb applied parallel to moment arm of 5m, then torque Is(a) 25 Nm

(C ) 10 Nm

(c| Both a 4 b

(a ) parallel(c) anti-parallel

20. Which pair contains one vector and one scalar?(I) Displacement; Acceleration (II) force. Kinetic energy

(b) 5 Nm(d) 0 Nm

g lf A B * 0 then A> B *(Iv ) Power; Speed(b) only (t»)(d) only (Iv )

(Hi) Momentum, Velocity(a) only (l)(c ) both (II) and (ill)

(4) 0(c) Ari

(b)

(d )

^ 9. W H' l - P i - . than angle between vectors Is: ANSW1Mla) 0*(c) 120*

The magnitude of product vtdor f Lt^ A> L r(a) sum of adjacent U<k >f p# aeiog.

ik AA3 d(b) 7.4ii5.b4.a1.a JLd2.d19 AIk A(d| lb.b II?.C15. A14.cII. b 13. d12. c

^ 10. Ii equal to:*area of parallelogram

Scholar*!PHYSICS - XI (Suhjn,jve ) 57

Short Questions of Exercise-s ^ X- r— Ax0 dQ 2 \

Ant.Unit VectorA vector who%tffpft>fnitd bv * lctte» with * cap or hat on it.

Mathematical Form-*

-* *it \ uvector * th magnitude A. then A A A

IDrngnitud* i\ ont with no units In .1 given direction Is called unit vector. * (II) If the vector lies In i— QIJ."' quadrant , bqth ofIt;rectangular comoonents will hjv« opposite signs.

I [ >

1 x

AY"

Aguiar components of a vector Is not zero, can Its magnitude be iero? Upleln(O.G- Khjn 2004.MV Pur 2004,Uv 200*.IV WO, Grw ion)

r-#1

Y '0* Q.2.4 If one of the recta

No. its magnitude can not be zero.(n) Aov^tion Vector'He vectv> w*ch rec*esents the ooiiton of point or a particle with respect to fixed origin is cal

-*A is demoted by t .

Jitwo c*-er»uona< ctxyc'nate system (plane), the position of a point P (a.b) is represented by

r »n*bjm three dawensicn*; cooro natt system ispace) . the position of a point P (a, b, c) is represented by

Ans.MReason

The magnitude of a A is given by

yjAt: + A/ + A*This equation shows that magnitude of the vector will be zero only wh**n i i of u ".‘angular

A =

components are zero.Q.2.5 Can a vector has components greater than the vector's magnitude?\ x « ai-bj^ck|

(MI) Components of a vector(Rwp 200«.Lhr 2004.Mir Pur»0*. fin* *OOfi.kV *tml

Ans. The rectangular component of a vector can never be greater than the vector' s magn. r ;, :•* * mffitXft* &Pwn the- • 3 .ector - a g »^ direction is called component of a vector. A vector iay

to vector's magnitude. The component of a vector other than maybe

*c magnitude of the vector.Explanationof three sectors fives a xero resultant What can be the possible or entation of

(D.G ^an :006,hd 2

are "eprvsenttd by the sides of tnonoie joined by head to tail rule, their sum *

CU-2 TH* *

The magnitude of the vector A is given by

A2 - A, VA =

OR

AJ Ay*AiA,

Theshows that themagnitude of rectangular component canbeequalorQ 2-6 Can the magnitude of a vector have a negative value?Ans. No, It can never be negative.

ReasonBy definition the magnitude of J vector quantity «s a K*er|

{As it is measured by the length of the vector which can Hf» *tr be oegatnr*

=> A2 A,3 OR=> AiA, OR, B ar#d C as srown »n figure, it is e’ear tha: sum of the vectors is *

tai c' the'rst vector coincides erth the head of the la sf vector.

A B * C « 0

* Aa

>> it on*nu >n will both th* rect*nful»r comportIti </v ponenunew opposite signs?C . »>e vector +* tr r . '/ its rt erguler components win be negative

li,Oxi .edor AA »

In this case, we always take the

ScholarPHYSICS ~ XI (Subjectiw-i59»ay about the components of two v*cto«7

Explanation

Ans, Consider two vectors A and B

Q.2.7 If A * B* 0 ,What can you

Sum of their respective components will also be null vtCtQf

ExplanationIf A = Ati+ Ayj ancj B*B.i Brj

(0JlJSAns.

as shiA= B and angle betwec

in figureo vector Is 90°. [Given]

AsA +B-0

(A,i + Atj) >(fl.i + Brj)«0i 4 oj(A. fB,)i+(A, 4 Bf )j.OUOj

ThenBy using head to tall rule.

H = A and R ' = A - BORProof

Comparing the coefficient of iand j on both sides, we haveMagnitude of RA,+ B, * o| and

In vector form

A. + B» »0Q.2.8 Under what circumstances would a vector has components that ere equal in magnitude ?

and A, B,«0

( 2), it is clear/* K .

Ans. It is possible only when the vector makes an angle ofi^ndlhionProof (3)

45

B45" "?\ ALet A* and Av be the rectangular components of vector A

Ay * A,A sln0 * A cos0sin0 » cos0sinOcosOtan0 1

8-Un ' (l)6 A5l

Q. 2.9 Is It possible to add a vector quantity to a scalar quantity? Explain.Ans. Ate,It isnot possible.

ReasonBoth physical quantities are dlffinnt In their physical nature, scalars have magnitude onlyvectors have magnitudes as well as direction.Scalars can be added by simple arithmetic rules while the vectors can be added by special rules (*algebra).

Qxio,Can you add zero to a null vector?Ans. £te/ it is not possible.

ReasonBoth lero and null vector are two vent hysica quan ibes one is scalar and the other is vectc

•we can not add zero Into a null vector.

if oFrom figureZLOM * ZNOM = 45°

45“OROR

4> -BTherefore,OR - 1 ZION «Z10M ZNOM » 45* 45° *90°OR

NSo R and R *

Hence proved.are perpendicular to each other.OR

=0Alternate Method<w 1005,•* »OOt.MWl

«-q -4#Angle between R end R‘

(b) ( A c B).(A - B)-4 -4 -4 - -• -* -* “.A. A- A . B B.A - B , B

. A2- A . 5 * A . B - B J

t

I'.' A , B 3 B,A ]

(vA.B=0 as AIB)

[V A.B]

(Grw 200$,»wv loot,Lhr loot,thr 200% Grv

= A3 - B3

A3 - A3

« 0' I -4

R is perpendicular to RSo02.11 Two vectors have unequal magnitudes,u their sum ^e equal to zero? Explain.(Sfdaoo&Mfrftjr joot-aoo&MmaooS' lhrao'O-Jon,*Ans. flte/ their sum can not bt ro.

Vector Ami

Scholar * PHYBIC& A [ 1 'bjattrvfj

9$' '*> ' « i4rt ,o*ifanfcc),

0,2.15 SMPPOM! the iUUn of d .j ^vectors?Am. The sum of theve /ectors wdJ te

ReasonIn this case, the head of the \cSithe resultant is zero. P

IdMagnitude of K and K

• ; W,.,1,5 v**« »'% •: **,'(•) r'"I*»* * •'** v* .-* jr,«-.i VWM,

'• • ' : -V"> •:• > *» • „•,• ;«R .RAs R *

W '*&r ' ***£$ v. *:*ae .; Y* o.- «*^(A +B).(A +B)

/-. - -* -* - -* •*

^A.A + A.B+B.A +B.B

R = ^A 2 +2 A.B+B 2

R * ^/A 2 +B2

JR' .R'

R'= \/(A -B).(A-B)

R'= J A - A.B-B . A+B1

R’= >JA2 -2A.B+B2

R' =>2 +JBi £From (1) and (2) it is clear that

R =R'

R =

R =

* tfci o '." e i's* .esr.v a:vow it |„r* y..Or r , •’ir, t'

(!) (vA.B=0 orAlB) D+E+D

And R' * E C

F B

Aaz.16 (a) traveling in different directions at equal speeds. The actual direction o f

motlo “i of X is due north but to an observer on Y, the apparent direction of motion of X lies north-east.The actual direction of motion of Y as observed from the shore will be(a) East (b) West (c) South-East (d) South-West.

Ans. (b) The actual direction of motion will be due west.Explanation

(vA.B=0 05 AJ_ B)(2)

* Let v * = velocity of ship X— fvy = velocity,of ship Y

v* - vy = velocity of ship X relative to ship Y

„ 0A +AB=0~

BOr

V, -hAB S V i- V y

AB =a — Vy=a =*This shows that vy is directed opposite to AB

l.e. due west

Q.a.16 (b)A horliontal forct F Is applied to a small object 9 of mass m at rest on tha smooth plant Indlnad

at an angle 0 to tha horliontal as shown In figure.Tha magnitude of tha resultant force acting up and

along tha aurfaca of tha plana,on tha object Is

(a) Fcoa8-mgiln0(b) F sln0-mg cos0(c) Fcoa0 + mgcos0(d) FilnB - mgilnB(a) mgtanO

(Federal 2003,D.G.Khan 2005, Fsd*Q .13 How would the two vectors of the same magnitude have to be oriented, if the were to be combi’

to give the resultant equal to a vector of the same magnitude?

Ans. It is possible only when the angle between two vectors is 120°ExplanationIf the two vectors are represented by two sides of an equilateral

triangle, then the third side represents their resultant such thatA=B=R as shown in figure. In this case the angle between two

vectors is 120°.

BA

\7 - \Tnral 2£ 0 EW -

6(T

sBROr

. 6 0' ( )C^ 120160’

A

Q.2-14 The two vectors to be combined have magnitudes 60N and 35N. Pick the correct answer from tb

given below and tell why it is the only one of the three that is correct.(I) 100N (II) 70N (iii) 20N

Ans. the correct answer is 70 N

jfeason‘JJjm of two vectors is maximum, if they are parallel to each other.

j.e 60 N 35 N = 95ftSAn of two vectors is minimum, if they are opposite to each other.

i.e 6oN + (- 35) N = 25N

?

Fig. 2 22

Scholar’s PHYSICS ~ XJ (SubjectiveAn* la) Fcos 9 •mg sin6.Explanation

Resolving F and W into rectangular components along perpendicular the inclined plane, we haVeF cos0- mg sinG = net force acting up along the plane.

/ J k

AiX A,y A,zThen, ~ty x X2 =

A*x A2y A2Zwhen components are reversed i e

Q.2.17 If all the components of the vectors A , and A 2 were reversed, how would this alter A , x ?(Mir Pur 200S, Lhr 2006, Bwp 2007,Grw it

Ans. / It would not change in this case.\ Explanation

\ we know that direction of A, x A ? is perpendicular to the plane containing A , and A , as showfigure (a )

A, x A, (A) * (A)

A, x A, * A,' x A,'•+ -* -*Q.2.18 Name the three different conditions that could make A , xA2 »0 ?

Ans.

AT(Mtn 2006,Bwp 2006)

0If A , and A, are two vectors then

x Aj AiA2 sin On .Figure (a) A. Figure ))

Conditions-• *

A , x A , Is zero if

1) A , orA 2 is a null vector.•* •*2) A , and A . are parallel. |U*., 0 = 0UJ

B) A , and Aa are anti-parallel

Now, If all the components of vectors A , and A; are reversed (l.e. If we take negative of vfd

A , and A ), then again the direction of (- A , ) x (- A ,) remains the same as shown In figure (b). J-> . -K .

A, X A 2 a (-At ) X (-A, )Alternate Method

• :l.e[ i.e., 0 = i8o°]

It would not change In this caseExplanation Q.2.19 Identify true or false statements and explain the reason

(•*) A body In equilibrium Implies that It Is neither moving nor rotating.body form a closed polygon, then the body Is said to be In equilibrium.I A A A

A, A,0 / A,yj • A,|k

^ A A AA 1 AJIMI A ^yj 1 AJ# k

(h) If coplanar forces acting on aAns. (a) This statement Isfeist,

let

and

-

Chapter 2 [Vector And Equjiih,

Scholar’5 PHYSICS - XI (Subjective)

Solved ExamplesReasonBecause in dynamic equilibrium body

(b) The second statement is true

Reasonin th s case lr' condition cf equilibrium is

equ ibr' jm

Q.2-20 A picture is

for which the tension in the string will be minimum.suspended from wall by two strings, as shown in figure. Resolve the tension in*

v mnvr or rotate with uniform velocitymaIF %simple 2.1

fromPan'originCKn'kn] as showi'aia ^tan' are represented by two points A (2,3,4)

(i) What are their position vectors?(ii) Calculate the dista

A (2,3,4).B (5,6,7),

asatisfied and the body is said to be in transla^

n'een the tv' o aeroplanes.

is suspended from a wall by two strings. Show by diagram the configuration of the * Givcn Data:-• .

Let the p cture srectargui3,‘components

= Ogives

TsmO TsmO = W

2T sm 0 =

4To Find:Ans.... AT of first aeroplaneT 0) on = FA = ?t

if! <r Position vector of second aeroplane = r5 = ?tween the two aeroplane = r = ?

T co*l T CO* •*

(ii)Calculation:

-v

(i) As position vector r in three dimensional space is given by

r = a/ + bj + ck

Thus position vector of first aeroplane A is

OA =rA = 2i* 3 j T 4k

and position vector of second aeroplane B is

OB = r.= 5i +6j+ 7k

(ii) According to head-to tail rule,

OA + AB = OB— > — *AB = OB - OA

tarn

\v- ;2sm0 *

K* T”ersor A I be mi*— urn fsinOis max mum

r LS

Ti

S r 7 = 1

e = 9cr iASc

A" = T or

4 SV ngj a'e vertical

T i l: I s i i • :s center c' g'a Try --se' the action o Its we » *?

Hen 2X3 V- MC4 2005 2XS.§wp 20C4, l*r 2306

' r = AB = rs- rA

7 = (5i -6j- 7k )- (2i -3 j - 4k )

r = 5 / - 6 j - 7k -2 / - 3 j

r = 3 / - 3 j ^ 3k

o r. Fid 2001, Lhr 2009.Gr*l

^ t rr. 3«»: t

ftusori through Divot OOlnt (Cfrtttr of gravity). 10*

itions of two acropUne, ihereto-e.i t' z 3- y ‘o'M R * • ?** *- f i un :***»giiiiudc of vector AB is the disianct between the pos

r = A3 = jof -Of -Of

r = 49+9+9

Pfit

« v 27: t f » 0 )t * : rSo

*wdirection* making angU Wt •o0*

i£ jd- brt ofon a body in

and ION »cl

llani force.forces of magnitude 10 N

r*'pectivcly with x-axi *. Find (he rc^ u-rtf tn# tysvl * MO o« i

- 49

’‘•s ts'T, V* fadL paMeJ Vto+p* f' '0*

i '

r> ar

fo'Y-* "0^

Chapter 2 [ VtUor And Fguititir^M Scholar',PHYS»C^ci (Su^v^y< #iv«rn Data:

Magnitude* of In si farceMagnitude of second force =Angle of find forceAngle of c< ond force - 0, 60''

670 tan 1 f ~ ^ *^p, ION

P2 = 20N 18 660 ctan''(l.l96)

* 0, » 30" b = 5oH

To Find: Find the* angle between two force, of equal magnialio equal to the magnitude of either of thee forces.Given D«*»:

*Resultant force l‘ 7-> « when the magnitude of their resultant is

( akululions:Ut F

-;and d arc the given force, and R j, their resultant.Step (I):x-component* i F . I - I F J J p t f f M

Let angle ofI* force *Jk*0°Angle oft* force « 02 ^ 0

* Pi* = Fi cot 0,Ihe x component of first force

= 10 co» 30” * 10 x 0866 « 8.66 NThe x -component of second force I I * co\ 0

To Find: —* 20 x co,60° a 20 x 0.5 * 10 N F

Angle between forces *0 = 7.ion:The x-com

y-componentsCalcula

The y- component of the first force *Fi,* Fi inn 0ponent of the resultant is:

F.-Fn+Fz,= 10 X tin 30° = 10 x 0.5 = 5NThe y component of the tecond force l;jy - I vm 0. ?%

m FjCOs0| + F2CO502or Ft = F|CosO° + F2cos0= 20 x sin 60° = 20,0 866 = 17 ^ 2 N

Step (Ilf: or F,- F| F2co$8The magnitude of x component I , of the resultant force F Similarly y-componcnt of the resultant is:

Fy - F|y + F2?I . I 1 . . I . Fy =" F,sin0| + F2 sin bF.*= 8.66 10 = 18.66N Fy = FjsinO0 FisinOStep (Hi): As sinO° = 0Fy •F2sin0

The magnitude of y component l\ of the resultant force Now the magnitude of the resultant is:

Fy * F|y Fjy R - y/ Fx 1 Fy3

F,= 5 17.32 = 22 32 N putting valuesStep (lv ): R - V(F^F:C0S°)2 + (FjSinO) 2

As R = Fi = F2 - FThe magnitude 1 of the resultant force K is given by. * (F + F cos0)2 +(F sin0)2

. - F2 + F2 cos20 + 2F2 cos0 + F2sin20F2 * F2 + 2F2cos0+ F2 (sin20 + cos20)F2OrF2

IAs sin20 + cos20 - 1

F 1 « F2 +2F2cos0 + F2ThusI* = y[a&M)2 (22 32)a

0 - 2FJcos0 + F2

I* - y/'M8.2 498 2 2F2COS0 * - F 2Or1

I V848 4 Or cose - - -[f;- 2W\ Or cos0 0 (- 0.5)

Step ( v >: 0 = cos 0.5)Or-4

Direction of the resultant ton given by L P^\20°\^us angle between two equal force.1 is 120°.

Chapter 2 [Vector And Equilibn,$8Scholar *8 PHYSICS - XI (Subjective!

69T-> IT*A . B = 26A force F* =2 i + 3 j units, has its point of application moved from point A( l, 3) to the point 8(5,7,

Magnitude of vector Q is:Find the work done.Given Data: B = V(3)2 +(-4)2 +(-12)2

B = i/9 + 16 + 144B = -/169

"

B = 13^ ^Putting values of A . B and B is equ. (1 ), we get,

y ^Projection of A on B = A cos 0 = —

= F* = 2*

i + 3 j unitsForceFirst point = A(1,3)Second point = B(5,7)

To Find:= W = ?Work done

13Calculation:

|A cos e = 2IThe position vector for point A is: rA = i + 3 j

The position vector for point B is: rB = 5 i + 7 j

d = rB “ rA

?= (5 i + 7 j )-( i + 3 j )

d = 5 i + 7 j — i - 3 j

d = 4 i + 4 j

The line of action 01 a force F passes through a point P of a body whose position vector in meters» A ^ — * * A

is i-2 j + l . I f F = 2i-3j + 4k (newton), determine the torque about the point ‘A’ whose position» A A

vector ( meter) is 2 i - j + k .Now,

Given Data:

Force = F = 21 - 3 j + 4 kThe position vector of point A = 7^ = 2 i + j + kThe position vector of point P = r 2 = i - 2 j + k

Thus,

Work done =W = F* . d To Find:W = (2 i + 3 j ). (4 i + 4 j )

W = 8 + 12Torque about point A = r = ?

Calculation:/ The position vector of P relative to A isr*AP = r = r2 - r,

7* = ( i - 2 j + k ) - (2 i + j + k )

7* = i + 2 j + k - 2 i - j - k

r* =-i -3 j

W = 20 unitsmiFind the projection of vector A = 2i — 8 j + k in the direction of the vector B = 3i - 4 j-1 2k .

Given Data:

A* = 2 i - 8 j + k

i? = 3 i -4 j - 12 kNow,

The torque about point A is:

= r* * FTo Find:

putting values, we getProjection of A* on B = A cos0 = *?7 = (-i -3 j ) x (2 i -3 j + 4 k )Calculation:

j k1Let 0 is the angle between A* and B* then,

A*. B* = AB cos0A . B

-1 -3 012 -3 4

7 = (-12 -0) 1 + (0 + 4) j -*- (3 + 6) k

7 = (-12 i + 4 j + 9k ) N m

I =

A c o s 0 =O r ( i )B

A*• B — (2 i 8 j + k ). (3 i - 4 j - 1 2 k )

A*. B* = 6 + 3 2- 1 2

N o w

Chapter 2 [Vector And Equin^70 i Scholar’s PHYSICS - XI (SubierH,,^71To Find:A load is suspended by two cords as shown in Fig. Determine the

maximum load that can be suspended at P, if maximum breakingstress of the cord used is 50N.

Distance covered by the man = d = 9Calculation:

Let beam can rotate about point A.

By applying second condit *^ahou'tpo.nt Ai .e. ST - 06 x 400 - 400 x d - 200 x 3 = 02400 = 400 d + 600400 d = 2400 - 600

400 d = 1800

Given Data:Maximum breaking stress of the cord = Ti = 50NAngles made by the cord are

0i = 60° and 02 = 20°w

Fig. 2.15 400 NATo Find:Maximum load at P - W - ?

Calculation:A1800Resolve Tj and T2 into rectangular components.

Applying the first condition of equilibrium for x-components. d - 1— 3.0m ••••>400d = 4.5 mIFjr= 0 gives ‘

T2 cos20° - TI cos 60° = 0Putting values, we getT2 * 0.94 - T, x 0.5 = 0

0.5 T|= 0.94 T20.94 T2

<The man can walk a distance 4.5 m fn point A.

A boy weighing 300N is standing at the edge of a uniform divingboard 4m in length. The weight of the board is 200N (Fig). Find theforces exerted by pedestals on the board.t *400 NT,

Example 2.9T,T, Sin 60* I, Sin 20*

60*, 20*

T, Cos 60* T,Cos 20*

Tl = Given Data:0.5A DWeight of the boy = 300 N

Weight of the board = 200 NLength of the diving board = 4 m

T, = 1.88 T2

As Ti = 50 N, then from equation (1)

T2 =1.88

T2 = —1.88

T2 = 26.6 N .Now applying First condition of equilibrium for y-component.ZFT = ° 8>ves>

Ti sin 60° + T2 sin20°- W = 0Putting values, we get

50 x 0.866 + 26.6 x 0.34 = WOr 43.3 4- 9.04 = WOr W = 52.34 NOr jW = 52 N1,

(1)w

1' .

Fig. 220(a)To Find:

Force exerted by the pedestals on the board = ?Calculations:

Let R|and R2 are the reaction forces exerted by the pedestals.C is the centre of gravity at which the weight of the board acts.Applying first condition of equilibriumAs IFX = 0 gives no information because there is no force acting along x-axisNow EFy = 0 gives

Ri + R2-200-300 = 0or R, + R2 = 500 N

A ".

,4 so

300 N ^(1) SOONaJOfU)

Applying second condition of equilibrium consider the point D as an axis of rotation. ]T-R > x AD - 300 x DB-200 x DC = 0-Ri x 1-300 x 3 - 200 x 1 = 0-Ri - 900- 200 = 0~Ri - 1100 = 0Ri =- 1100 N

Putting the value of Ri inequ. ( 1 )- 1100 + R2 = 500

^2 = 1600 N|Negative sign of R, shows that it is directed downward.

t = 0MBBBBI A uniform beam of 200N is supported horizontally as show n. Ifthe breaking tension of the rope is 400N, how far c^n the man ofweight 400N walk from point A on the beam as shown in Fig.?Given Data:

Weight of the man = Wj = 400 NWeight f the beam = W2 = 200 N

Breaking stress of the rope = 400 NLength of the beam =•t - 6 m

6.0m

wi— i2 rV'cctor And Equilibria

grholar s PHYSICS - XI (Subjective! 73

W hat is the unit vector in the direction of the vector A =4 i +3 j?n ExerciseProblenns 2.3Given Data:

\ has its tail ai

Given Data.

A-4 i +3 j

To find:U Unit vector = A* ?

Calculations:Point P (-2,-3)

Point Q (3,9) '

between points P and Q r

- * *

= r, = -2i -3J

* r2 — 3i + 9 j

By definition

A AA = —To find:ADistance

Calculation: .which is given byWhere A is the mai Vl

Position vector of point P

Position vector of point 0

laCement vector from point P to point 0

! 7 = (3*i + 9 j )-(~2i'

7 = 3/ + 9}+ 2/ + 3}-4 * ~r = 5i + I 2y

Distance between points P and Q

r = N^fHi2y

pulling v+ 9 = V25 = 5* r * r2 • ri

l DispThus

.4 r*o particles•4 -* x-axIs.-Magnitude of r d its orientation with respect to the

vector r, - r> anGiven data:

So,Location of first particle -\*3i +7 j

* rj-«2l+3 jr = N/ 25 + \ 44r = N/169

(T=T3 unitsJ Location of second panicleKan inaccin metervrdinatc system

Of the insect from this corner of the rocm

cooTo find:

<4 **Magnitude of (rx -0" •2.2

is sitting on anwhat is distancei - 0 - 7and orientation (direction)

Given Data: Calculations!- p(2, nThe position of insect from cornerCoordinate of origin = 0(0, 03 As relative position vector

To find:Distance of insect from corner r - ?

putting value, we get

7 « (-2i + v)-tf + 75)Calculations: of insect with respect to origin isAs the position vector

7.-21+ 3J -iM?r = 2i + jr

_— f ^ 2Magnitude of r = r = V* + )'

r = J(2 f +

= %/4 + 1= V5

pT24 1Hence

h

rhapter 2[Vgctor AnJgrholar’s PHYSICS - XI (Subjectiver 75

74 B = 3i-4 j

r =,/25+16

r = >/41

E53Orientation (direction) «

( ycp = tan' 1 -

To find:

(a)C = A+ BMagnitude and direction of(6)D = 3 A-2

Calculations:C = A+ BPutting the value, we get

C = (2/ + 3/C = 5/-}

(a)x4

tp = tan 1 "7

= tan-1 (0.8)cp = 38.6° . .

Hant lies in 3rd quadrant so direction is

9 =180° + <P9 = \ 80° + 38.6°0 = 218.6°f7~

2T9° Appr°*lIf a vector B is added to vector

_4i + 7]. What is magnitude of vector A ?

A + B = 6i+ j •

A-5 =-4i+7 j

q>

Now magnitude of C is C = -J(5)2 + (-l)2

' = V26

As the resu

=5.09

is 6 - + ]. if B is subtracted from A the ren C = 5.1 unitsA , the result is

2.5 Direction of Cf \

cp = tan'1 -v*

<p = 11°

As x-component is +ive and y-component is

0 =36O°- (p

0 =360°-11°

0 =349°

[Q~=~

349^

Given Data:(0

vector lies in 4th quadrant, sois-ive, so(2)

To find:Magnitude of A - A 7

Adding equation (1) and (2)

(A +B)+(A-B) = 6/ +}-4i + 7J

2 A = 2i +8 j

A =i+ 4;

Calculations;Hence

D = 3 A-25(b)putting the value, we get

D = 3(2/ + 3})-2(3/-4;)

A A_

D = 6?+ 9 j - fy +Magnitude 0 f A — A— Y ( 1) (4)

= y f r T I e D = 17;= Jn is D =Now magnitude of DA= 4.1units

f = A t B a n d l W E3Give*« X- li.W-B-M.-*•— — “ d ” 8" * ‘ -axis

the vector lies along >As x-component is zero so

s w.M x-axi5»

y.axis i .c - 0

Chapter 2 [Vector And Egllt|.76 Scholars PHYSICS - xijSubject^ 77

, A =5 i + j and B = 2i + 4j.Find the angle between the two vectors2.7 d = /•*->•„

d = (6i + 4 j )~( 2i- j )

d = 6i +4y-2i +) '

3= 4/ 4- 5 j A

Now work done = W = F.dPutting value, we get

W=(3i+2j).(4i+5j)W=(3)(4)+(2)(5)\V=12+10

W=22units|

Show that the three vector /+ j+ k , 2 /-3 j+ k and 4 /+ j-5 k are mutually perpendicular.

Given Data:4A = Si + j

B = 2/ + 4/Calculation:

As A.B = AB cos# -> ->

A .B (1)Thus cos6 -AB

Now

A.B = (5/ + /).(2; + 4 j )

A . B =(5)(2) + (1)(4)

2.9A .B = 10 + 4Given Da

A.B = 14The magnitude of vector A is

A = J( 5 )2 +( l )2 = V25 +1 = V26 = 5.09

5 = V(2)2 + (4)2 = V4 +16 = V20 =4.47Putting value is *equ\ ...( l)

cos 0=(

cos 0-0.164

A = / + j+k

B= 2i-3 j+kA A A

C = 4/ + j -5kTo find:

To prove, these three vector are mutually perpendicular.14 Criteria:

5.09x4.47These three vectors will be mutually perpendicular if A . B = 0,5.C = Oand C.A =*X)

Calculations:© cos*1(0.614)0-52° A.5= (/ +}+£).(2/-3}+ £)

A.5= (l)(2)+(1)(-3)+(1)(1)

A.5-2-3+1

HenceFind the work done when the point of application of the force 3/ + 2 j moves in a straight line1

the point (2,-l) to the point (6, 4).» 2.8

Given Dalis A . 5- 0Force F* 3/ + 2J

Point A (2,- 1 )und Point B (6,4)

This shows that A and B are mutually perpendicular

B.C-(2l-3j+k).(4i+ j-5k)

B . C -(2)(4)+(-3)(l )+(l)(*5)

B . C = 8-3-5-> -*B . C- 0

To find;

work done- W« ?Calculations;

The position vector of point A(2,- ! ) » rA » 2/ -)'t * tk

Thi position vector of point 1; (6,4) r„ « 6/ + 4 j

If displacement between point, and 1 is d , them

V

'

1

Chapter 2[Vector And Equine

Cchol»r,s PHYSICS ~ XI (Subjective!78

79•4 - « « » * " »

r.A « (4/ + y-5A ).(/ +;+ *)

C.A-(4)(l) + (l)(l)+ (-5)(l)

C.A = 4 + I-5

C = 8 units eastN

To find:A

(a) A x B (b) Axe (c) B * CB CWCalculation: -EC.A = 0

This shows that C and A are mutually perpendicular (a) Ax B = ABsin0n

As angle between A and B (i.c noi

AxB=:ADsin90°nputting values, we get

Kcault:Thus all these three vector arc mutually perpendicular, because their mutual scalar products arczero.

s

^ » _•B- 31— 4 A: , find the projection of A on B .Given that A » I- 2 > -» 3 A and B2.10

Ax B = 4 x3 (1 ) n

According to right Hand rule, direction of Ax B (i.e. n ) is vertically upward.

|AXB=12 units upward

1° =!Given Data:* * * *A = /- 2 j +3A

a A

B*3i-4kDirection:

To find:

Projection of A on B -- A cost) c?

^ ^Let 0 is the angle between A and B .then,

A> . B* •AB cosO

A cos 0 *

Calculation:(b) Ax C = ACsin6n

As angle between A and C is 90°

AxC=4 x8sin90° n

AxC=32( I )n

AxC =32n

A . B (1)orB

Now

A.B = (/-2}+ 3*).(3/ -0y-4*)

Ai= (l)(3) + (-2)(0) + (3)(-4)

A.B = 3+0-12

A .B =-9And B = 7(3)J +(4)2 = >/9 +16 = V25 =5Putting values in eq (1), we get

Direction:

According to right hand rule, direction of AxC (i.e. n ) is vertically downward.I -4 —•I AxC=32units downward

l (c) BxC=BCsin0n

As angle between B and C is 180°

BxC =3*8sinl80° n

BxC=3x8(0)n-91 -y This is projection of A on B .

Vectors A , B a n d C are 4 units north, 3 units we t and 8 units east, respectively. Describe care

Acos0=

fullyI

2.11 Bx C =0n(a) A ^ B (b) A x C (c) B x C BxC=0Given Data: Direction:

binary direction.As Bx C results into a null vector so it has arA = 4 units north

B = 3 units west

Chapter 2[Vector And Eg^ ^ ^^pHYSlCS- XI (Subjective)81

ini' effect of force about a given point is given by rx F where r is jk

-r -*A A „ ^

of application of F . Consider a force F = -3/+ j+Sk ^in N m about the origin ?

i j k0 -1 11 -2 0

tThe torque or turning _ _

from the given point to the pointA A A

acting on the point

r =2.12

7 ,+3i+ * (m). What is the torque in

r = i (0 + 2)-}(0 -l )+ iGiven Data:

F = (-3;+ j+5k ) (newton)

r = (7i-3j+k) (meter)

r = 2i + j+k

(b) in this case

r, - i + kTo find:

r = rxF = V

Calculation^:

As r = rxFputting values, we get

r = (7i+3 j+i)x(-3i -t- j+5k)

In determinant form we can write the cross product as,

i jt = 7 3 1|

H 1I “

r = i(l5- 1)-}(35 + 3)+ £(7+9)

r = 14/-38}>16fc

The line of action of force, F = i -2 j , passes

\ - k; Find *'aj the moment of F about the origin, fb)A- *

wkxh the position vector is i * k

r = r2-r{ = -i- j

Now using the formula

r' = rxF

? = (-/-;)x(/-2;)

& ) *-1 -1 01 -2 0

it '

r' =

?= /(0-0)-}(0-0)+F(2 +1)

through the point whose position vetf

the moment of F about the poir- 2.13 r' = 3* 6>/3 and 6 respectively. Find the angleoducts of two vector are

2-14 The magnitude of dot and cross pr

between the vectors.Given data:

Let A and B be two given vectors.r - i -2 j

• # A

?3fl&oa vector - : - ^ c Magnitude of dot product of two vectors

of cross product of two vectorsMagnitude

cf F v* or.gm r - ?

(bjxnnm&l ? MoutfeepoiaJ <rf » nthepoBi/.n vector it i k •f'"" To find:tv/o vectors

The angle betweenCalculation:

% VXZM fix t oi r j bout v gin

r * r / F

Chapter 2 Rector Aadr.U

Scholar’s PHYSICS - xi fSub,jectivc)<2)AB sin # = 683Given data:

!> riding equ. f2; by cqua'ion (1) we gelABsin 6 6ABcot0 6V3

Weight of the tractor = W = 15000 MWeight of the Bridge-Length of Bridge * 21.0 mLength of Bridge span between SWeight, from wheels =Wr =

Weight on the rear wheel*=Distance between front v4k,

1stndcoid

O - t a o V-^2.15 A toad «f 104*« suspended from a ctothea line. Thu diatom the linethe h'.rr/Aotal at each cad. Find the teuton in the etothea line.

uPP°rting ends*20 mj

3 * W*^sW-5000 N

A = W-W,= 1SOOWOOO =a and rear

!

,000 N•*-= 3 m1 To find:

Forces on the Br;= 7 F2

Ki that it make* an i» Calculi tiom: J— FfeZ22Orvea data;condition of equiJibri jm

/rce along x-axisff.Jf /wd - W - IQK

*- l f 5'I?.-o,i

Mow applying £F, - 0

7* *«

• doth i » , 7e * T ? z»

r vV —F, -F3-W.-W,.W( -0

F, *Fj-1000-1OMlC-5000 - 0F, *F3 - 23000 -0F, +F, - 23000 (I;

Now applying second condition of equilibrium Zr * 0I-et A i* the axis of rotation.'*> F,fABj- W,fAC)-W,fAC;- W, fAD> +F/0r 0

f-,' 20-»000»10-10000*10-3000«7*F/0 » 0Fj*20-*0000-100000-35000*0-0

F,*20-215000„,215000

* 20p, -10750 N

or F, -|0 750 * 10* Npj-io.mfl

Putting value of P* in equation fl >F, 10750- 23000F, - 23000 - 10750F, - 12250 NF, - 12.250 * 10* N\ j ~ 12.250

ffev/towsg 'Kr.i.s.r, into rta fectangnfar torr.prynenuApja/,«g #K firtt un/iAxm of eqv.l,bru*n - 0

T i//*; j* - 7 c/.*IJ* -0T <'*» 5*- 7 iv/sIS*

*•» IF> -0

7 %mH** 7 tm i f - W -027 am i f •w7- *

2ml?

i.-iL-2 /0 2/.T. *L

0 52Tz3Ef%

«i J s t n c s s a e— *— ^,k*r r->-^.upperta -U to- r«a, w ,„U »„TOl4,lk*^ "J

•SflilM

>'/*

V

'/r

2 U

Chapter 2 [Vector And Equi );^ Scholar^ PHYSICS - XI (Subj84jective)

A spherical hall of weight SON is to be lifted over the step as shown in

the Fig. 2.23. Calculate the minimum force needed just to lift it abovethe floor.

852.17Apply f, rsl condition ot equilibrium alonsj \ -

F=TSin.Waxisi .e.. VF = 0t— 4 V

Given data: T 'in 30F=T(>< )= W = 50 N- h = 5 cm

Weight of spherical Hal!Height of' the spherical

i^S an?5.

Re- 2,23 : jCb : 8bApply first condition of equilibnum alonRadius of the spherical ball = r = 20 cm

Tcos30°- W = 0

T= — — —cos30°T =-0.866T = 11.55;

To find:Minimum force required to lift the balls F=?

Calculation: FFrom diagramDE = 5cmCO = CE = 20 cmCD = CF.- DECD = 2 0- 5 = 1 5 c m

From the right angle triangle OCD(OC)2 = (CD)* + (OD)‘ (Pythagorean Theorem)(OD)2 = (OC)2 - (CD)2

OD = yltnc?-(CD )7

OD = V(20)J -(!5)J

T sin 30"

and|T = \ r. jiPutting th

Ort ucw=11.

= 5.77* 3f *** x *:* x x ****>:*

Scholar's PHYSICS (Objective)Part I & IIi OD = V400- 225

00 =^75OD = 13.2 cmOD = 13 cm (approx)

Take point ‘O’ as axis of rotation and apply second condition ofequilibrium i.c..£ r = 0

F x AO - w x OD -0F x 25 -50 x 13 = 0F x 25 = 657)

ATP FOR Part - 1or

Alternative To PracticalOnW= 50 N

650F ••=PHYSICS, CHEMISTRY,

BIOLOGY25

|F-?6 N|\ uniform sphere of weight I0.0N is held by a string attachedto a frictionlcss wall

2.1R

so that the string makes an angle ofuV’with the wall as shown in Fig. 2.24. Find the tens,on in thestring and the force exerted on the sphere by the wall.

&rnMPUTER SCIENCESGiven data:

Iweight of the sphere = W = 10NAngle between string and wall = 0 = 30°

tfiimcvNew Paper Patten I

To find:Tension in the string = T = ?

Calculation:Resolving T into two rectangular components T, and Tv .

Chapter 3 [Motion <fc p

^ Scholar’s PHYSICS

AccelerationThe time rate of change of vetoed nrUnit

' *Q body i

SIunit of acceleration is nt/sec> Tf)e

DirectionAcceleration is also a vector quantathe direction of change in velocity

V^the dl

Average AccelerationThe ratio of the total change in velar,*average acceleration. iT to the total tin

-*me taken is called

Let v, is the Initial velocity of the body which 1t,me' then;he «* "«* (VelocityM

Changes t0 the final velocity v"

v,-v,4?= change injj&city

-J Then' theavera«e ac«!£^o£js|ivenby

to

91Limitation of average velocityAverage velocity docr. not tell US

path is straight or curvedIf a squash ball comes Itimes, its total displacement is zero and also its average velocity is zero.Unit and DirectionSI unit of velocity is ni/.scc (i.e. ms'1). Its dimensions are [LT 1 J .It is

quantity and its direction is along the direction of displacement.Instantaneous Velocity

The limiting value of ,as time interval At following the time t, approachesAt

to zero is called instantaneous velocity.Mathematically

that the motion is steady or variable and the

called Acceleration.back to its starting point after bouncing of the wall several

mensions of acceleration arenymma M i

300.000.0X1

(m vtrjwn)210.000 f«Ot-3un Wiv«„„•"**«» u*«*.r».•00 f Mtti •»„.1.000 Uwn WUU.MI f ,MB

a vector rection of acceleration is alongFor Your Information

A'>w«a L_

33.1 r unUt^ Aj,)

•>- < i.«4

•.'*« IT* Inn,IIf f eBui «i tty*

c

^I

HI too

a ivnti4 Munwn n«>ng

001 (Mlkinu w»1

AA measurement ofmassindependentof gravity. The unknown mass m anda calibrated mass me are mounted ona light weight rod. If the masses areequal, the rod will rotate withoutwobbleabout itscentre.

AdVln* = lim —

At-»0 At V ? - V, AV3.v =At At Do You Know?/ Instantaneous Acceleration

|15i A vr/if /imams' va/«e of — as the lime intervals , following the time t, f 10

x approaches to zero is called instantaneous acceleration. J 0Mathematically

B ?ExplanationConsider a body is moving along the curve as shown in figure. The body moves— »from point A to point B in time A/ . Let r, is the position vector of point A and

r2 position vector of point B then displacement of the body during this shorttime interval is

*4 r 2o

o1 2 3 4

time (•)

How the displacement of a verticallythrown ball varies with time?A vauu = lim —51-0 AtArf=r,-r,

If value of A/ is small, the value of Ad will also become smaller. WhenEXPLANATION:The variation of displacementwith time Is parabolic.Positive acceleration

If the velocity of a body Is increasingjts acceleration Is positive.The Change n velocity miy Negative accelerationbe due to change In

for Your InformationAdA/ approaches to zero the point A approaches point B. In this case — -At

approaches to a limiting value called instantaneous velocity.NoteThe average velocity of the body may be zero even though its instantaneousvelocity Is not zero.Uniform velocityIf the body covers equal displacements in equal intervals of time, the body issaid to be moving with uniform velocity.Non-uniform velocityIf the body covers unequal displacements in equal intervals ( time, body issaid to be moving with noti-uniform velocity.Condition for uniform velocityif the average and instantaneous velocity of a body are equal the body is movingwith uniform velocity.

the velocity of a body is decreasing, its acceleration is negative. The„negatlve federation is also called retardation or deceleration..Edition for uniform acceleration

^or a body moving with uniform acceleration, Its average and Instantaneous acceleration are equal

I) magnitudeil) directionIII) In both of above

ll What are velocity- time graphs? Discuss.Velocity Time Graph

r».,nritv with time is called velocity-timeThe graph which represents the variation of velocitygraph.Let us

1time graph Is a horizontalWhen the body moves with uniform velocity, v

straight lin

*t ••

consider the case of a body moving along a straight path. \nfi

t

nee ref Sh0Wn ln figure -ered by the body = Area of rectangle^ista Flo.3.4OK

^instantaneous velocity does not change, the body is saiettb be moving with>T

Chapter 3 [M„linH92 SeboUr’* PHYSIO - XI (SMbjettiv.>

= (length] (wdth) . A native sign j,,pptt^^^oppos-te to that of inrtial vetocitv'

• The above equations canbodies by replacing a by g.

About acceleration due to Gravity (g)m the absence of air all bodiesfallfrufy

'

the action of gravity with uniform accelrr Zgravity It is denoted by g. ^ caAl<*occeler

Its average value near the surface of SL L

measure of strength of gravitationJf,eid * 9Jm/u4^ *

those quantities

also be applied

= vt Do You Know?direction Isv= s t to free motion ofCase II

When the body movesaetherat*on,,. the vfc .oot* time

shown m f gjre

Distance covered by the oody « Area of triangle

=i(base) (height)

V

With uniformly .releasing velocity (i.e. with uniform

Cr9c h is a inclined straight line with time axis as*

widerFKH5due to

fis a-II 0 v The value of g decreases with height as well* I v~ v

maximum value at the pofKJ n.1«,twper unit mass fl.c. gravitational field strength %* — )

2 22i sF‘«n

Al of m= vy , r m

* S fW

Case HIWhen r - «. body moves wit* r ; ^r.iforrni / mcreas- ng velocity (i.e, with

form ecoilenfoo i •* ' vt- 'ooty nm. < - graph r. <i < ///w as shown

DO VOU KNOW?Velocity time graph »never perpendicular to ,axis an n represent ir.fiaccelcniUnn

04 Discuss Newton's lav, of motion briefly.tXFLANATMJrt\txwOm% to,KBR> *•!««

*o f«g vrt

Note* ' « m «»;' • '. f ir .* 4' \# * o > s d' <A- - T.jtiori s e #iu- j to TI.I- slop* - of the?

tangent *•tf *f pOtf *

Vifr hcanc* of veiouty - time graphs.• ' v«dor * / * rr ,e gt pi • *. tou. to average act elerution

at trwAbout Newton's Law* of MotionSir lsa;»c Newton published his empirical laws of motion in his famous

' book "prlndplo* in i$87.These laws hold food for bodies moving withimaller spe - W as compare to the speed of lightFor fast moving objects such as atomic particles in accelerators,

relat vistic mechanics developed by Ablert tInstein is applicable

Newton's First Law of Motion / Law of InertiaStatementA body at rest wilt remain al rest and a body moving with unformvelocity will continue to do to, unless unbalanced external force eu*i w»

Ulkbodyrw% •e

m c«lM alfkmUMh.

g**14«I

§pamby.20 W

w10

LA (tauo

\i• ; +< ’ ’ i ! gr . .p» . to distance co /« - r *.-d h /the body

i

j Write Q'jdtf . t ' • vatlom of motion for uniformly accelerated bodies? 20l A*I' . . i '.- of a yfi

IhroWflI / *l,»i llm#>V*Iwily la ii|.w«rd«

Au

4«It

^lht\ Is alto known as law of inertia

Equation of Mot ,on for Uniformly Accelerated BodiesCwdtv •r.ody n mov.r|with ur form et':<deration along n straight line if ItsIr • # v* 0' *7 s ^ *n4 *h*r t re. # interval its final /elocity become wthgn

^ 4-- KM ,-tCil

tiruiunoNtoed

EXPLANATION:i"t (he fini half offvclotlly goo« on decf^and thun bttomci ie

hi nfht frame of reference, for example earth Is approitmatefy an ji fir*m#

a>eInertial frame of referenceThe frame of reference In which Newton’s first law holds * ca"ed *****

or «f(•) Mv* » r, §? vl * a•a We0 0

II maximum

I the hall move# Inillrecllon withvilotlly, till K

| tinning point JlglffcftI (tarfi/ig point, th# vr’| •'•"'inea oguai w JU upward Initial

*Nea *«eu(hi) /as *v/ - V,1 lie**IsM of reference r>2 Newton'iSecond Law of Motion

ttittmentWen a force is applied on a bod)1, U prod****direction. The magnitude of aueUmium fc directIQfx* and inversely proportional to its *****

Mathematically, it is expressed asF = m a

F - applied force on bodym = mass of bodya = acceleration produced

How to apply Iheat equations In problems• These eeiuatons are

acceleration• When tha ebjact moves a‘org a ilralghl line, the direction ofn^t on does nut change in suci t <isei all v« » r « * an be treatedKkiacaiars

usef^i o' v <'> linear r eu- jn with uniform

• lit prupian,! vecto- cen be tn tied like scalers, the0< initial velocity poslll** Where

GbapUt 3 * y^ V3I C S- X I (Subjective )'4*95

1— - P UJI

— C l j r i y, when two bodies of different massesdifficult to stop the maxslve one.

that the moving body has a q

"1Do you wear seat belt

J 1Newton's Third Law of MotionStatement

f( lion and reaction are eExplanationWh.n two bodies inttr.ct with each other then, action and reaction

Irxcc. art for the «urn length oftirr . e They neve^c^nth^am^gjbut always act on different bodies

hen it is! 1(hrswr Ny •* I.4MM In*tr *U**«* t»<4* ixmt Kr fX**• -•iM» • ' •! i« ...ijuul m magnitude and opposite in dire non ' < mm p4rii

moving with

to which it e

same :ities1then it »*This showsjn/thing that tnes to stop it. The qujfoty 0f movri, , t y of motion is called momentum.

.EXPLANAllON:A forw* I ll feppltal Ail (he f*.tigfey flUl ft ' JB ( A Use f « Md Afvari « vihnr tv A> A t<i» iiti|i, ,i i*im- 1 »,ijId Uft I** IT. ** in4M«cttkW U ttu ir.m n tftvhr**>*.h lh< N.-tfi, f'/ftt - f H(fwrfcncd H‘*b# b< »( tod Uf« bmitttuv r *.uu?- *»rJ fiwti *ht«t

When a moving car stopsquickly, the passengersmove forward toward thewindscreen. Seat belts'change the forces ofmotion and prevent thepassengers from movingThus the chance of injury isgreatly reduced.EXPLANATION:When you fall towardsforward direction, youapply force on the belt. Asa reaction the belt appliesforce on you in backdirection and thus you aresaved from injury.

. exerts a force ony which describes the

Q.7 Show that N s is equiv. msQ,3 What is inertia ? Explain.

f \Proof•— •' In » kg. * *• kg ms ’ ( v W =i kgUL

3N S = A#Inertia

The property of a hody due to which it tends to maintain its state of restor uniform motion Is called inertia.ExplanationIt Is a natural resistance to acceleration that all objects have. The greaterthe object' s mass, the greater this resistance . So,The mass of the object is u quantitative measure of its inertiaExamples

When you make a turn while driving a car, you move the oppositeway in which the car turns the corner .

• You get pressed back In your seat when an airplane takes off.Your face Is smashed against the windshield if your car suddenlystops against a brick wall.

1For Your Informulinr^Aci.cn ArtJ re actionb*l«kcc each ixhei Q.8 How force ar d momentum are related to each other? State Newton's

second law in tirmj of momentumJ.1‘uint InI’.ifi.lc

4I

A car «ccckr4*» Along a KMVO.** (art. *CIMU> ) Momentum and Newton's Second Law of MotionEXPLA % A TKW*

f ' t , \ttuc ro«d > Consider a >ody of mass m moving with velocity v , . A force F is applied on

CA* 0« of \he .v jimr 7 * :H°0Gt i,o r time t and its velocity changes to vfbut does n move ii. Acceleration produced by the force is

-* -4- v f -v,a = — 1

Point to Ponder

ZX& MFiril . . of motion gives tHdefinition of force w He secor * According to Newton's second law of motiontaw give the measurement c'

tns

HQ.6 Define and explain linear momentum.(2)F-m a

U,in* «Quation (1) in (2), we haveJforce.

POINT TO PONDEKWhat is the effefl on ihc speed of AMomentum

F = m^- V,fighter plane chising another* wh*it opens fire? What happens to ll*speed of pursued plane *hCI\

The product of mass and velocity of moving body is called linear momentum. tMathematically

OK mv.-mv Which hurt you In the(3)returns the fire? F =:above situations (a) or (b)EXPLANATION: tfire. Wherep =mvand think why?When the fighter plane opens

will be in **ck EXPLANATION:Where p = the momentum of the body its momentum » "* v, = •n'tial momentum of body The time of collision at Indirection due to reaction forcebackward direction and therefore !® ”?

ti* * vr = finalm = mass of the body

case (a) is smaller andtherefore, the impulsivespeed will decrease When

pursued plane opens fire in b* 2nd

direction, die momentum will &clor‘ ' ,n terms of momentumter "zsxssfr h *>+‘^omentum of bodyvelocity of the bodyv =

force F will be greaterDirectionwhich may hurt you.

V* \Since v is a vector quantity, so momenta n is alsohaving same direction is same as that of veioc ?/.a vector quantity

therefore, its speed will increase

PHYS|C^ ~ XI (Subjective)

servatlon of MomentumChapter 3

97f ConThis i« morr K«*IUMJI fo« vn of Nowton's second law of motion. Because it can

family be applicable for the rises when mass Is changing.for tMampleAs rocket acceletales, It loses mass because Its fuel is burnt and ejected to

provide greater thrust.Q.9 Define Impulse how It Is related to momentum?

*

isolate^ system of two smooth

moving with velocities v, and

[Poo You Know7]p. E is more ;Itcncrji Which will be moreeffective in knocking s bearclown.

Alform of force than F k "proof _ AlV. » tl»an, „h„m v k»«.<“ The“» «**« n J B n;

i.Opn^er an

and

a rubber bullet ora lead bullet of theii. * r1 same momentum

EXPLANATION:For knocking the bearimpulse = Fxt acts on thebear. In using the rubberbullet, the time of collision twill be smaller (due tobouncing) and impulsiveforce F will be larger hut incase of lead bullet, the timeof collision t will be greater(due to penetration) andthe impulsive force F will besmaller. Therefore, therubber bullet will be moreeffective, to knock down thebear.

r Your hair acts likecrumple (soft) rone on y^skull A force of «,N mightbe enough to fracture you,naked skull (craniumI *bony part), but withcovering of skin and hak.jforce of soN would btneeded Why It Is so?EXPLANATION:The hair on the skull act <$

a soft tone. For naked skuSa force of 5 N is sufficientto fracture because in thacase time of collision at asmaller and the Impulsiveforce F will be moreeffective, rn case of thecovered skull and hair,atbgreater and the force F

effective. Tcproduce the fracture, *| force SON instead of $*will be required.

m J1

/, respectively. Let

f = force exerted on m,by m2

j:= force exerted on rr\2 by m,According to Newton's second law of motion

-# -»- m, vj-m, v,

ft*.UtopolstWhen large force acts on a hotly for a very short internal of time, then theproduct of force and time for which the force acts, is called impulse offorce.Mathematically

jnd vi»

a

i •1_ (U1 =F*lUnit

St unit Of impulse is kg-m/scc or N-acc. It is same as that of linear momentum.The dimensions of impulse are (MIT It Is a vector quantity.Relation between Impulse and momentumAccording to Newton's second law of motion

/I

So the change in momentum of mass m,is-» - -*m, vj-m, v,=Fx t (1)OR

Similarlyr *-* —P _ m,v'.-m,v,p _ mv,-mv,

(2) Do You Know?t t 1Using equation (2) in (1), we have-» -*

T m v r -m v i

So the change in momentum of mass m2 is

m2 v'j-m,v2 =FxtAdding equations (1) and (2), we have

-* -* -* ^mi vj-m, v,+m2 v'-m2 v2 = F x t+F x t

v^m,v^-m,V, +pjxt.Wording to Newton's third law action and reaction are always equal but

°PPosite in direction.

r<tnot so OR <(2)l = x t

t

OR I =m Vf -m Vi (3)1Thus, impulse = change in momentum o the body

Instantaneous change in momentum of body due to impulsive force is calledimpulse. *A motor bicycle's :safetyhelmet is padded so as toextend the time of anycollision to prevent serious

OR (3)Does a moving object haveimpulse?

Concept of ImpulseSome times the applied force is not constant and it acts for short time. e.g. when. a hat hits a cricket ball the ball force varies from instant to instant during

* collision. In such cases it is more suitable to deal with force andimpulse) instead of either quantity alone.

• 1injury.EXPLANATION:The impulse is the product

No,When the body ismoving uniform velocity So,

of F and t. 'Hie paddedhelmet will increase the

then F = 0 so impulse Istime (i.e. F=-F2ero.time of collision t and thusWhen the body Is moving -*F+F* =0

=5^ Impulsive ForceForce acting on a body for very short internal of time is called impulsive force

the impulsive force F is(4)with variable velocity then Wns equati decreased.force is acting continuously- 00 (4) in (3), we haveb.10 What is an isolated system? State and explain the law of Hence no impulse isconservation produced because no fort*oMinear momentum. =0acts for short Interval of OR\4time. mi v,+m2 v2 =m,vj+m2 v2abated system \

= finalmomentumof system>sth6System on which no external agency exerts any force. at:vfii1

T Chapter 3 r», 1

»fl PHYSICS — XI (Subjective)98 Scho|^ 99

nd Inelastic collision?(V, - V',Kv, + V',) _ (v'i ~ V;)(V';+ V > )

(v',- v,)Q.11 Define elastic ai (v.- v',)

Commons(otd\mcns\oua\e\coWtsion*SmoothawAV^.*>iou-TOtatm^* \mpactparang(Headon^

V,+ V 1= V 2+Vj

V - v2 = v /2-v '1I \ OR

asijc ORElastic and Inelastic Collisionr

Elastic CollisionThe collision, in whichelastic collision.ExampleBouncing back of a hard ball from a

collision.Inelastic CollisionThe collision isinelastic collision.Example

Bouncing back of a hard ball from sandy floor

• Collision of two tennis balls.

,4(3)

V - V2 = ~ (v 1~ v 2)|is conserved, is called A)OR»/'*' Where(v,.V ) = magnitude ^ velocity of first bail reiative to second baH before the col.ision.(v',-v'i) = magnitude of velocity of second ball relative to fin

Relative speed of approach = Relative speed of separation

zero.

marble floor is approximately an elastic * ball after the collision;t

I Hence

which the kinetic energy’ of the system is not conserved is calledFind the expressions for the velocities;of two bodies m,and ma after

elastic collision in one dimension.Q.13

9

Determination of velocities after collision(l|

Momentum and total energy• are conserved in ail types of collision. We can calculate the velocities of the masses after collision by solving equations

Perfect elastic collision can not be possible. (1)/ (2) and (3). *IQ.12 Show that relative speed of approach is equal to relative speed of separation for elastic col Velocity of miss m,(i.e. v/ )

one dimension.

Note

IFrom equation (3 ),- v v= v, - v;+ v ', (5)

Elastic Collision in One DimensionConsider two smooth, non-rotating hard balls of masses m, and m2 moving in such a way so that theirane along the same straight line with initial velocities v, and v2 respectively. When they make head oncoiwith each other their velocities becomes v/ and v2' respectively as shown in figure.According to law of conservation of linear momentum

mfvT*m2v2= m,vi+m2V 2m,v,-m,Vt= m,v 2-m,v2

m,(v,- v',) = m2(v2'-vj (1)As collision is perfectly elastic, so K.E. is alsoconserved, i.e.

Using equation (5) in (1), we havem, (v,- v ',) = m2 [(v, - v2 + v 1) - v2)m,v,-m,v = m2 v*-m2 v2 + m2 v -m2 v2

m, v ', + m3 v ', = m, v, -m2 v,+ m2 v2 + m2 v2

(m, + m7) Vi, > ( m,- m2 ) v, + 2m2 v2

OROROR

OP (m, - m7 )(m, + m2 )

2m2OP v. = v, + (6 )1 V 21 (m, + m;)

Before cOlll»,on Velocity of massUsing equation (6) in (5), we have

(fflt ~ m2)(m, +in2)

1 tj ,I AJ >1— m,v, -m2v2 =r m,v , +-m2v 2. 2 2 21

/ 2 1 2

2m*v a *

2m2Vl

m 2(i.e. v/)

J 1- -m,v C7= 2m,1 v(m, +m,)

2mfr .1(m. + m,) J

= v,- v2 + v. +1 1-m.fv,1- v ',J)= V,

2 v',« V. Im^v,1- v',J) * m2(v j'-v,1)

Diaiding equation (2) by (1)nyv/ )

m,(v,- v',)

iL On, + m2 ) Jv 2 a lmi + m,/ ) + (m| ~ m 3 )

( m, m j)

| n > U ' 4ffli- (rn, m )(m, F m;)

2mm,

®i(v 2 -olll*i°n

FIB 3 9

iv, i viAfter c i

I;

w

.9'*

Chapter2S*. PHYSICS- XI (Subjective) 101100

’s

2m2Vj(m, +m2)

-m2 )2m, , ("LV‘^ (m, + nl2 )case (li)

v = 0V.+- v 2(m, + m2 )

v, +v a * i(m, +m2 )

o o2m(7)

(m, +<nj)(0)2m, v. +vi= ; r

(m, +m2 )iv, + (m + m)v‘= (m + m)

vI - O + O

v' = 0

mm

Before collision

v = 0 v = v.

. Q.14 and o o(m2 -m,) v,

(m, +m2)(m-m)

(m+ m)

2m, v. +!' <® m m

After collision2m (0)V, +v 2 " (m+ m)Case I• When m,and m2 are equal i.e.m,= m2 = m (say)

In this case, equations (6) and (7) become,

F.g. 3.11f

\

22Lv. +oV =2 2m i

2m2-m2)

ResultAfter collision,mass mi comes to rest

iv:= ' = VV2V, + V 2I II (m, +m2)(m,+m2)-

and m2 moves with the velocity of mi.2mv,

=(m-m)(m+m)

case (I)v.+ v2(m+m)i V, V, Case IIIWhen lighter mass mi

t.e.mi«m2 or mi*0 & v2=0.in this case,equations (6) and (7) become,

2m2 Vi(m, +m2)

2m2

9 9 collides with a massive body m2 at rest0 2mV. = V.+ V,1 2m 1 2m m, m,

Before collisionv| = 0+v2 case (iii)

y. _ (m,-m2 )1 (m,+m,)

v,=

(0-raj)(0+m2)

v^-^v.+O

v; = v, = 0V. +V2 I

9oandm.(0) mV. +9 9 li (0+m2)v--

2m' (m2-m,) v(m,+m2)

2

(m-m) v(m+m) 2-

Before collisionV, +!(m, +m2) <m,

After collislcFle-l-IS

v = 0v = - vm22m 9Vi- V, + 9v;=-V|;(m+m)

mand

After collision

2 2m 1 2m 22m

(m,+rrij)2

(m2-0)(0+m2)

2m, Fig 312V,= V. +(m,+m2)

i

v;= v, + ov-2 = V, 2x0 (0)2 (O+n^)

v'2 = 0 + 0v'2 = 0

v. +I

ResultThe masses m,and m2 exchange their velocities aftfir collisicCase IIWhen mt and m,are equal while the tar et mas.i.e. v2=0, In this case, e atir s (6) at. 7) bee >me,

ResultAfter collision mass m2 remains at rest while mibounces back with the same spaed as before collision.

at rest

Ldue!)102 pHVSICS XI (Subjcclivc)

103in..Case IV

^ change in momentumcollides with a lighter body nr.,at rest' ' j —. ( »o rn.)

When a massive body m,acO & va=o. time

i.e. m1»m7 or maIn this case, equations (6) and ( 7) become, '“ ' tin, \

* ' 0 Icase (|V)

- m vF =

t

Mewton's third law of motion, a forcef -» N

2m,(m, -m,)’ on the wall is equal but opposite. So

V2 From\\ = v. +i (m, + m2)

2m;

v, =(m, +m5)

v; *(g'S(m,+ 0)

Q( in(0)v, + t mii (m, +0) ni t

Before col||6|0niHii

t.0) -m, A— L v, +0vj * So,(m t- ci ) 11 v> vm.d te product of mass of water striking normally per second and change inThus force can be ca

velocity.cul0.1. „fVn, Example

Suppose wat

v 5: \'I I

and(m2 -m,) v(m, +m2)

22m

a pipe at 3 kgs"1 and its velocity changes from 5 ms 1 to zero on striking the wall,<63 • -V* - I *

After colinF'3 - 3.13

V. +v2 = i

(m, +m2 )I then

- 0) = 15 Nbne.vm OJ cornoo.»*1' n *iv. /om , r' l 1

2m .n yiil. ipujan•QL -V, +V, « I (m,+0) Q.16 Is mor ?ntum conserved, when momentum changes are produced byE anol is sai&S'n

(m, + 0), 2m,. * L V, + 0

explosive forces? Explain. F*C*s <m ybca sviwiwr i. Hfiw * bilioa.

0 — r v «Ii" v 9mO03dr • 19

v:Explosion is due to theresult of internal forces

m!

Momentum and Explosive ForcesTotal momentum remains same when momentum changes occur due to explosive forces within an isolatedsystem.Examples1- Explosion of a shell or bomb

v2 = 2v,r:Result

After collision, mass m, moves with saute velocity but the mass m2 moves withthe velocity double the speed of m,.

1

Q.15 Find the force due to water flow. When a movingc Suppose a bomb is falling, in its way it explodes into two pieces. The momentum of the oomb fragmentsquickly the P- combined by vector addition is equal to the original momentum of falling bomb, as shown in • gu'ejiiuve forward Wwindscreen. Sc:

’change the f®3 , -*motion and pfl* wnen a bu,let of mass m fired is from a rifle of mass M with a velocity \ .passengers iw® Momentum before fireThus the chance # ,n{tja|is greatly reduce-EXPLANATION Momentum after fire

2- Firing of a rifleForce Due to Water FlowSuppose water from a horizontal pipe strikes a wall normally,

hen it exerts 3 force on the wall. Let initially the velocity ofwater flow is v and on striking the wall, it comes to atrest so final velocity becomes zero. Thus,

initial velocity of water = v

final velocity of water

cO . J =>

GJ^momentum is zero as both bullet and rifle are initially at rest. X' V-

611rn When you .fonvard IS the velocity of recoil of the rifle.

55SS;Then- f*na. momentapphes p

-According to the law of conservation of momentum,

you 310 ** momer>tjm before fire = momentum after fireinjury.

= v1

= vf = 0 41 urn = m v + M \ 'change in velocity of water * A v=^-v = o - v = _ vmass of the water that hits the wall in tine t = mchange m momei. -> of xter = ^ =mA v=-my

Then according to Newton's st d law of r t;nn t ?fon, the force F exerted on the water

qrijfl0 = M v‘ \ )

OR M \ ' = -rn vis given bv

. PHYSICS - XJ Subjcctnt)

v or the rocket t. *.ne gases s= fan*M y nfie «$ *et> " The

*- cr 5 tr*e y reo> of ^s^ F = M a 2)

50 — e .eocr. r* ~eco r «s •en -:~ s ,^ ccs'ts'ec-s masse*' 'OC«e* s i = accelerator cr *trse*^ ;? v- expresso' for attention

Q.rj-c = •hy -oateL V 1 = “

m •© 3' = — ~M0' - e "2S i !*' r the’

"- ret tereases to me — ? r=locket Propulsion &ne passesPS***otkmg Principle '«ta izzzerz&r i — ease:

y certsesvsoor cr momentum ant fee ir Derive eepreaion kr the' -' ozzr zr 'XR? 3 r sec cr re s» 1?s c'ce-cc"e '

re od^y s { threwr a*. 5r angfe'lector! zlkirs 2, cr near.^Vopuisiec* ocr.ec moves r ejecting 3L— .£ gsssec frorr its 'csir ts *t -' engine. A ' r * w~ _

>e zr o-mec c ZJR~s cr rg- prsseje gases nese ^ser c r ~ e t-t *'c ~ t- -il Proj«ctacgji&ET the convene aettfertfi— due * p^nr* c*ier%me »~r .er. - g- i-rcr. "'he nxxet gains rr&tner.~_^ ~ e:-- to */«

'r.-BTrcr sr eoei-ec gases r-r r opposite iredoc .xt?:contir.jes to gair r*«21mere 5': ~<re ~ ^ hr»g ac engne of me rocket contrives to expe

girseo Dc re cc ar of -nc ec goes cr /versasing as long 2 re engines ie Afoccos to.ee c 5 ce r»voersrong E. at fees £'c" gr« - c-siie f me ~r~ 5 mn± r g terr me rocket *-* w* ssr*.* ,, 1«)». se c- v c rare'."4 Mt ' re t-— r «r a^r o/. er of re y re

, ^ L/ t - nation of pfojectiie motion, . ..

JS Oj^ier'.. ~.y. of s t»d» *'o-' Horaortak, fcorc a ?-o.« tue -aimerim. se »ei as '- -«

'3 err-sr:c r.« on eno^a ^p«art vm&Lto orerro^e re g^.-ty, 5S' «M

ocAet cEr*jmer aw. 1 .W4 kgw cr ^ Foc/r. e err re err; cektix". The ooc .gese:at tpeeo: J c -r

3=i atoo*•* •ass K * axter y-'5£T- tV^aona c redipr e^ =« » •^^** ^ ' 6~c m?roe r c mo: se e^a 'or.eu ere rr.ee flgtr/iet r/ngr me of

fare**?*<*$.V- re ooc. 'as * ^ vx&r* ' * -J-constat* ro«p*M( -w - 5l' - ,J3r*re^ t vc// fifcsr it »: c sce'C-e: 0>e tt ^Tucr r^ # >cr.e ro mx > "«c* # erJJ -ritr g'eeur s>eec

asceer cr E,r « net'/ : ce ert rter e . -A«^»erat#m of ^4ct Trajectory rne W?K* «f «

^>e c*r foto-vec c . re mtJjett e t ca ec n r^orn-* v re e/tscet ^ r * *<

>* ^fose# rf gxec'e^r.-efc rextr e nss*? >*• r t

' ^ .o

1Chapter 3 [\|„ti„n & f.106 pUYSICS - XI (Subjective) 107

So equ ( 2 ) becomes12

Equation ;i represents the horizontal distance while equation ( 2) represents theof tne body

Q.19 Oerlve expression for the instantaneous velocity of a projectile thrownat an angle 0with horizontal.

jal vertical velocity of projectile = v* = v,sinO

vertical acceleration = a¥ = - g

Lrtiral velocity at highest point = vN = oheight = y = H =?

• mu(

( 2)and V - o ''!

I 0maximum

ding to equation of motion,2aS = Vf

2“ vi

2

2ayy = V “ V'Yfio* 3CC0

1OR

2 l -gt H = o - (v, sinO)2

-2gH = sinO2

VA

rVinstantaneous Velocity (v.jSuppose a proper* >s Ved with ir.inai velocity at an angle i> witn horizontal.

v 2 sin OV -ir HPr - ig'VLet ^ r/.‘ I V * Tjme

i t v t e taken by body to cover rhe distance frto the place where it hits the ground is called the time

Worijorua! compo*1«nt 0* mrfjal vekK * ry»vt = v cmff

of nrtia velocit / r v(. - v.sinO

Horfao-ntal component of velocityV ce f ere >s rfo horizontal force along horizontal axis so acceleration a, = 0

* of /e oc*ty a* ar. / instant t is

e place ofB

projectiondf flight.As the body goes up and

*rtjca! distance, i.e .

y = 0r t j i i m i c a } velocity

acceleration due to gravity = a,= -gtime of fi ght - 1 =?

V

es back totaame level so it covers noim*rm*t,nq InformationJO.

f of projectile = v,* = v, sinOi si titadi1 i tun

''. a 0)v * x

r-: » For Your InformationV c V> i..c i\ jiizith.(1) . i?ill* . b» y

1./#r>ws* o* r«lodty/e' 'a 'O"oor - ' of /* v. / v a' /

* vw* a.t

H= — RS = v,t +^y * v*to « (v, ainG)i-~ gt ?

* gt 2 - (vj 8in 0) t

4 mJ,x

1 ,“art| r<1 1 n\O wfl2) V g ar.<f y - v *i«0 y

' / /•rxu* y* f * r At/ ,d* of /** V. * / 4 4' / rflsunt ‘ri

For Your Information0'

The factor which remains constantdufHlg the projectile motion arc* horizontal velocity•both x and y components pf

acceleration

* If /MU lath

% <*trtOf > tA vaVrfy** '*'**&* ***** m#4t angfe * „ fhfr .n 2v

(yin0m (hit t - (2)drop jrn*« of projectile

a/irr,tjrn distance which a projectile covers In the horizontal direction is calledfor,K* of pro)

*•Vh, t 4

' \ r t a* 0M

t / / ( / »„*>/ . M%*"" If,/cl. i c , (tuif ectlvt.' i di-.p(4/^rrt<enh -»f« ly-fr, I

~ a,t' bdeomes

2v. sinO

r, A * rt . .,• , m*kir>t *o * r.0Ai 0/ftb t' t ’ 4 *\

fr,) I fAtUgMi/ 4 /^v r#of#

f. /PI A 'l A l K;*4;

, ; - / - I- ' . v» -»»• / hi. v ’l.cii r4ttir.air»fV . . 4f4 *^ <-1 tlMI**0>r iJffi ,v)K affi 4» Ihr UH#

i ' Uifir viftK,-*l

, W V, COftO r

R „y/(2>inOcoiO)

Ui Mft. r+ v x 4 0K

OR

»/,y '/t f>H»U+" »,«* ,» .<„,M >***) II lirjfcHIHil tt<« rnnivnum

'« g/ * '/ r r <•

J+nt ' of r «• »« / *

(3) [v 2 sin 0 cosO sin 20|I II ' 1

108 Chapter 3 [Moti PHYSICS - XI (Subjective)5choiar s 109

For Your InformationMaximum RangeTne ^ange of me projectile will be maximum when sin20 has maximum value. for spherical earth

Ballistic trajectory will be elliptical- e*--*•— «o4«&B„,w>oee* pan

i.e.sm 20 =1

20 = sin^ (i)20 = 90*

0 = 45’

Uses oThe ballistic missiles are useful only for short range,For large ranges and greater precision powered ann

missiles are used.d rerr'oteifontrol guided

At high speed and for long trajectories the airfriction is not n

times the force of air friction is more than gravih/ it , .

The air friction creates difficulties for this purpose So tho a„.i £

needs a high degree of precision.f 6 a"g'e°f pr°>ectio"

Actual pathiOR 1OR

In the presence of airfriction, the trajectoryof a high speedprojectile fall short of aparabolic pathEXPLANATION:Due to air friction, thevertical and horizontalvelocities decrease and

ORe and somev,:sin28 Water is projected froent*^pips at the same speed-foe *an angle of 30° and

at 60°. Why arc the rangese^EXPLANATION:The range of a projectile isp*,

V1

R = — sin20

So, equation R = becomesg

V;2 sin2(450)R

* =g

v 2 sin90° therefore, ( the heightOR RM = gand range both decreaseg As sin 20 = sin 2 which make fall shortsin 60° = 0.866E v.2 parabolicof aAnd sin 20 = sin2 * 60°=a

120°* 0.866Therefore,R is same for boftikangles.

(4) [v sin 90° =1] trajectory, as shown ing the Figure.Note: FORMULAEWe can express the range of the projectile in terms of maximum range as

R=RDXJtsm20 For an angle less than 45*. the faqreached by the projectile arid ibc ®both will be less. When the iqkiprojection is larger than 45°, the Wattained will be more but the rr;again less.EXPLANATION:

IThe range is given by,R = .J_ sini:

Q.2i Define ballistic flight, ballistic missile and ballistic trajectory.AdBallistic Flight

When a projectile is given an initial push and is then allowed to movefreely due to inertia and under the action of gravityt then such a flight iscalled a ballistic flightBallistic MissileThe un-powered and unguided missile is called ballistic missile.Ballistic TrajectoryThe path followed by the ballistic missile is called ballistic trajectory.ExplanationBallistic missile moves due to the super position of two independent,motions:

a straight line inertial flight along the direction of launch• vertical gravity fall.• The gravity and inertia are responsible for the parabolic path of

projectile.Effect of InertiaDue to inertia an object should move straight off in the direction in which it isthrown, at constant speed equal to Its Initial su °x particu! ' I / In empty space.Effect of gravityDue to gravity straight path changes Into a curvedi ectory.for flat earth (For short ranges)The trajectory of projectile Ispar oli

Average velocity2 Vav =At

g

v -Limn —v a,-*° AtThe range depends on sin 20.When 0 = 45°, the range becomesmaximum that is

v2 v2

„„- — sin 90* mZLg g

The values sin 20 for all other anglor greater lhan 45° are less than I.therefore, < > » range, regardingangles is smaller lhan the range # 4-The height h is given as,

h_ v f jin1?

Instantaneous velocity3

V 2- VI A va»v 3Average acceleration4 At At

7 -Limit AV«M-*05 Instantaneous acceleration At

g8.The height depends on angle-renter values of 0, the height **

greater and smaller for smaller^ 2nd law of motion

Unaar momentum

2nd law of motion In terms ofmomentum

6 F m ao. 7 P m v

m v i -m v i8 F - t*4 r* “*1 o Fwf m v r- m v-*9 Impulse I - F x t •

, nijVj - n v|* m4 v,Law of consarvatlon of linearmomentum

10 mlvl

Relation between relative velocityof approach and relative velocity ofseparation

Multiple Choice Questions11 V|- v2 — (vi-vi)2m2Velocity of mass m,after collision in

one dimensional elastic collision, m| m jv , = — ! -

m, +m2four possible answers to each statement are given below. I ck ( S) the correct answer:12 V, + V 2I m, +m2

2m,Velocity of mass m2 after collision inone dimensional elastic collision

m2 m,V2= Which of the following can be zero if a body is in motion for some time?

(b) Displacement(d) None

13 v, + V 2Im, + m2 m, +m2 1.(a) Speed(c) Distance coveredIf the displacement covered by body is zero, then what can you say about its distance?

(b) It may and may not be zero

v mF = — vForce due to water flow14t

2.mv (a) It is negative(c) It mustbe zeroThe slope of tje velocit y time graph for retarded motion is:

(b) Positive

Recoil velocity of a rifle15 v = (d) It cannot be zeroMmv16 Acceleration of rocket 3-M (a) Zero m

(cl Negative

Two bodies are moving in opposite direction with velocity v. What is relative velocity between them

(a) \[iv(c) - .Area under velocity time graph represents.(a) Force

(d) NeutralHorizontal distance of an objectthrown horizontally from height h17 x = vttt

4-(b) 2v(d) Zero

Vertical distance of an objectthrown horizontally from height h

1ie,:18 y =

x-component of instantaneousvelocity of a projectile

5Vf* = V« = V, COS019 (b) Displacement(d) Accelerationy-component of instantaneous

velocity of a projectile(c) DistanceInstantaneous and average velocities becomes equal if body has •

(b) Uniform acceleration(d) Moves in a circle

vf> = vi sin0 — gt206.

Instantaneous velocity of aprojectile

(a ) Zero acceleration(c ) Variable accelerationInertia of an object is measure by its(a ) Volume(c ) Mass2nd law of motion defines

, (a) Inertia

When a body moves in a straight line then its displacement coincides with

(a) Distance <b> Force

lc) Vel0‘ity' (dlTThe rate of change in momentum of a body falling freely is equal to

(a) K.E. (b) Momentum(d) Weight

V = T/V i + V f t21

v? sin 2 9Height of projectile H = (b) Density(d) Temperature

222g

2v, sin 0 8.T =Time flight of projectile23 g (b) Acceleration(d) Both a and b

R _ vf sin 26Range of projectile24I 9 -g

v2Maximum range of projectile R25 max 1 10.

(c) PowerA small Which vehicle experiences the greater

h,collides head-on with a massive truck.

sports carlrnPact force (in magnitude)?(a) The car(ci They experience the same forcein ab°ve question, which vehicle experiences the grea er

^J3) The car tA\ None of these'c) They experience the same acceleration

(b) The truck(d) None of these

ceieration?U.

L xli

.pHYSICS XI (Subjective)CjjJjOlJJ112 Chapter 3 113

peflnitlon of acceleration:

The time rate of a change of velocity of a body i,called acceleration.Mathematically,A base ball ol mass m Is thrown upward with some Initial speed. If air resistance is neglectforce acting on the ball when it reaches Its peak is

13.(a) mg and upward(c) Zero

(b) mg and downward(d) None of these

A body is moving in a straight line such that the distance covered by it in time t is proportisquare of the time t. The acceleration of the body is:

* Av v, - v14. onal toth.

= initial velocity of body

= final velocity of the body

where v,(a ) Constant (b) Zero(d) Decreasing -(c ) increasing

SI unit of impulse is equivalent to(3) Force(c ) AccelerationTaking off rocket can be explained by(a) 1st law of motion(c) 3rd law of motion

vr15. — »

Av = change in velocity during tirm At

S.l units of Acceleration & velocity;

SI unit of velocity is m/sec.

SI units of acceleration is jft/sec*object is thrown vertically upward. Discuss the sign of acceleration due

velocity,while the object is in air?

(b) Momentum(d) None

16.(b) 2nd law of motion(d) None to gravity, relative toQ.3.2 An

Which component of acceleration is zero in projectile motion?( a ) Horizontal

17.(b) Vertical(d) None

A fighter plane drops a bomb when it is at the top of enemies target. Bomb misses the target duetc(b) Due to action of gravity(d) All of above

(Bwp 2003,Mir Pur 2004. Fsd 2008)

(c ) Both Ans. Sign of acceleration due to gravity relative to velocity is negative.

Explanation: JAll those quantities are assigned to be negative sign whose direction is opposite to the direction ofinitial velocity.So when the object is thrown upward, the direction of gravitational pull is opposite to the direction of

imtir:, velocity so the sign of acceleration due to gravity relative to velocity is negative.Q-33 Can the velocity of an object reverse direction when acceleration is constant? If so, give an example.

(Federal 2003-2005, Lhr 2005,Sgd 2005, Grw 2005-2010, Fsd 2008, Mir Pur 2009, Lhr 2010-2011)

18.(a ) Due to bad weather(c ’ Due to horizontal component of velocityA ball is thrown horizontally from the top of tower. What happens to the horizontal componer'its velocity?

19.

(a Firs: ncreased then decrease( c) Remains same

(b) Increases(d) None of these

What is angle of projection, so that the horizontal range is equal to maximum height?(a) Tan~1(4)

20. Ans. Yes, it can be possible.Example:When a body is thrown vertically upward its velocity goes on decreasing due to gravity and becomes

zero at the maximum height. After that it will reverse its direction of velocity, hut the acce.erat.on

remains constant during whole f ight [ i.e 9.8 m/s2 ).Q-3 -4 Specify the correct statement:

a* An object can have a constant velocity even its speed is changing,

b. An object can have a constant speed even its velocity is changing.

(b) Tam’(3)(d) Tan-i(1)(c) Tan- ’(2)

ANSWERSlO. dl. b 2. b 3. c 9. u '4.2) 8. b5. c 6, a 7. c20.111. c 12. a 13. b 19. c14. a 18. c15. b 17, a16. c

c- An object can have a zero velocity even its acceleration is not zero.d* An object subjected to a constant acceleration can reverse its velocity.

Statements (b),(c ) and (d) are correct.Explanation:

Short Questions of Exercises Ans.of virii*10.3 1 What ii the difference between uniform and variable velocity? From the explanation

velocity, define acceleration, Give SI ur ‘s velocity and acceleration? As when speed changes, velocity also changes so it s not^ n e - J ^ fcr)

direction. So it is a true statementwhen a moving object is stopped oy apply*r'fithethat instant but acceleration is not So it .s tru£

(a)(b)*» wp 2004, D.G.Khan 2005-2006,Lhr 20'*

A *i. Difference between uniform and variable velocity :

• In cast of uniform va' ocity the body covari taual dltolacemtnti in eauat Inttrvali of tim* ^case 3' to e re oc ty 'c in unta^' * dliDiacemtnti in equal Intervals of Uma,

• r case o' variable vt oc»ty, * magnitude or direction of velocity or both may chanatf©'m /e oc t / coth the mag de - J d » recnon" mains the fame.

C TOvelocity become* 1sudden hard bi .‘V

statementte)

\

si •

Chapter 3 [Motion\ ,1 (( PHYSICS - XI (Subjective)- 11 115

W \ UM \ A\ \ ub|i« ct I?* thrown veilU ally up, After leaching It maximum height, lt % velocity r^V(l

u\ ,u « r|iM.itlon iamAln* » onstnnt (I v 9.8 So 11 is also fruct statement .S A w\<% w standing on the top ol a tower throws .1 b.il! straight up with

tnltlii) velocity Vi and at the same time throws .1 second ball straightdownward with the same speed. Which boll will have large speed whenIt strikes the ground? Ignore air friction.

*% doth the balls hit the ground with H?/2HL.'»peoclt MpUrtetlon:the hall whit h is thrown vertically up with velocity v, will have samevoloi lly v, when It reaches hat k to the top of tower . So the two halls havejonn;downwaid vtloctty at top ol towei , Ilence they hit the ground with

same final volot ny.

jt 1«, true statement.. #*-Af"

h WM" *,bo* mo... w»b .nlio th„„m, „,«» « ' l t * m «**» «,„« , *..constant (l.e., Om/sec ) throughout tho motion of the fcft.dy

find the change In momentum for an object tuWjrted to a given force for a given time and state law

0|motion In terms of momentum?

•llower 2 Vp V,% I I V,Q.38

% (Bwp 2004, Federal 200s, Grw 2008)

= Initial.% m.mass of the body ^ v, a

J*, m final velocity 1 ti

velocity%Z I- - applied force

An.- LctZeintervalv /

- ft

Xi.6 f Kplaln tho < Ircurmlances In which tho velocity v and acceleration a of

(I) Parallel(II) Anti parallel(ill) Perpendicular to one another

.1 car are; Then

f motion,According, to Newtonr- *I- m a

- Jk(Iv) v I* zero but a Is not zero .1ni

ft(v ) a Is zero but v h not zero Comparing equations (1) and (2) we get

(Iid 1005,Iodof1*! iOOS,Mir l*ur 2009. Mtn 2009,Grw 200% 2009 2010)*F v. - v

Ans. (1) Parallel;At-ft m

and acceleration > ar*ll the Velocity ol the car h Increasing along a straight path then velocity v

pat allot.-•

m v, * mvrF- AtAnti parallel(ID -»

then velocity

v and a» < deration a are anti parallel, lor example when the brakes a

a straight path where m v, represents the initial momentum while represents final momentum.m v 1

> Thus we can saySecond law in terms of momentumThe time rate of change of momentum of body is equal to the appliedapplied to a moving car.

(Ill) perpendicular to each other

are mutuallyII the tar is moving In a circular oath then the velocity v andperpendicular . Am. Impulse ^ ^

_

When a large force acts on a moving body for a short Intc

's called Impulse.delation

rval of time then the product of force and*ft

(Iv) v Is zero but a Is not zero

If moving far Is stooped bv applying sudden hard brakes then at that Instant the velocity of tfit 4becomes zero hut the acceleration is not zero.

- ft I * F xtAs f°rce is the time rate of change of momentum So

-4 «4

m vf -mv,

a is zero but v is not zero

*> ^ .When car moves with uniform velocity thru v is not zero but a Is zero.

aa.7 Motion with constant velocity is a special case of motion with constant acceleration.statement true? Discuss.

(v)

IS tbbF- t

(Kwp 2005. f ** 20°*' Lhr X0°9 )•*°r F x t * m v f *mv,

PHYSICS- XI (Subjective)

117of flightTime

is called .time Q^n.ghr " ''om ,he place °f Proiection to the place where it Just to

S = v,t +pt’O = (VjSinO) t -^gt1

^ gt’ * (v,sin 0)t

2v, sin0 %

1=m v, - m v (/ = Fx /)So the impulse is equal to the instantaneous change in momentum of the body.

Q-3-iO State the law of conservation of linear momentum, pointing out the importance of isolated

i

Explain, why under certain conditions, the law is useful even though the system is not compuisolated?(Mtn 2003,D.G.Khan 2005, G

= O. If v,sin 0 is the

*Ans: Law of conservation of momentum

Total linear momentum of an isolated system always remains constant.Importance of Isolated systemLaw of conservation of momentum holds only for isolated systems otherwise it is not validApplication for not completely Isolated systems

t *When the effect of external force (like frictional and gravitational forces) is negligiblycompared to the forces between the interacting objects, then this law become applicableQ.3.11 Explain the difference between elastic and inelastic collisions. Explain how would a bouncing btfbehave in each case? Give plausible reasons for the fact that K.E is not conserved in most cases?(Bwp 2005-2008. Grw 200S. thr 2009!

gmu * Range of the projectileMaximum distan e which & pro * life covers in the horizontal distance is called the range of projectile.If v. cosO is horizontal opmponent of initial velocity then range of projectile R for the total time of flighttcan be expressed as.

R = vu * tputting values, we get#

Ans. Difference between elastic and inelastic collision:In case of elastic collision the K.E of the system is conserved while n case of nelastic collision the K [ $not conserved. 2v sinG

X

gBut the total linear momentum and the total energy of th ? system remains constant m both typescfcollision.Behoviour of bouncing ball:When a hard ball is dropped onto a marble floor, it rebounds to very nearly the initial height. It loses' •negligible amount of energy in the collision with the floor. Then such collision is appro* ziety r Jelastic collision.But if the ball is not able to reach the initial height then there is a loss of kinetic energy and sue?collision is called inelastic collision.Solid reasons for loss of kinetic energy:In case of inelastic collision, the loss of kinetic energy is due to• friction of ball with floor• friction of ball and air• sound

,' (2 sin0cos0) ox * Rg

_ v sin20R = — (2 sin 0 cos 0 = sin 2 0)) gMaximum RangeThe range of the projectile is maximum when the value of sin20 has maximum value. The maximumvalue of Sine function is one. Thus

sin 20 =120 = sin ’(1)20 = 90°

<90°e = —2

0 = 45°v,2 sin20

Q.3.12 Explain what is meant by projectile motion? Derive the expression forthe time of flighta.Sothe range of projectile.

Show that the range of projectile is maximum when projectile is thrown at an angle of 45°^b.

Rg

v,2 sin2(45°)horizontal.Rma* =2006-20«(rw JJ 2003,D.G.Khan 2005,Mir Puf g

call**’Ans. The two dimensional motion under the constPit acceleration due to gravity and inertia is v,* sin90°projectile motion.A foot ball kicked by a picExamplesA ball thrown b , a cric ;cer.

m n

PHYSICS- XI (Subjective)Chapter 3 [Moti *, School118

119fear from B to C:Q 3-13 At what point or points in its path does a projectile have its minimum speed, its maximum \foti°n 0

' The g«Ph furthcr le,ls us that ,he ^l°city of the carthe value of acceleration is zero, that isfear from C to D:

(Gr* mains constant from 5th to 15*h second, it meansAns:

a = 0.Minimum speed:The speed of the ba 1 is minimum at its maximum height because at this point the verticalof velocity become zeroMaximum speed:'He speed of ba.i is maximum at>) its oo nt & projection

) the point just to hit the around (point of landing).( a) Afhat is meant by ballistic trajectory?

< *o. owed sr a" :jn-po //e^ed and urgjided project ec '- e pa*' fo< owed o / *.re powered and dec project^c. 32!" fo oweP 0/ -n-po //ered ded projectile3- pat- fowowed by powered and gu ded projectile.

Am: a s the correct ar.swer

Molioo ocomThe velocity decreases uniformly i0 zero from 15“ to 19“ gL^The acceleration of the car during last 4 second is - " '

Av = v 2 - v, = 0 -2 0t 2 - t , Aa =

1 9 - 1 5A t

9-5p* negative sign indicates thz

Toul Bis^ncc Covered by Car:Distance covered * Area of AABF +

velocit car decreases during these four seconds.

of rectangle BCEF + Area of ACDE

0^14 '"* ) What happens when a system of two bodies undergoes an elastic collision? Identify the correcianswer.5 i"or-end." o* the *> /st changesp *'emor rtturr of the s/stem does rot changec ~'e 906«cor e to rest after a> rgond T>*cr«rg/ co*fervafctf aw s rotated

Am. fs rS the correct answer

3535553A 1500 kg < argretarding force.

Given Data:Mm of the car - m - 1500 kgInitial velocity * v( * 20 ms* 1

' nal velocity * vr 15 ms'1

Time- t •3.0 s

to 15 ms 1 in 3.0 s. How large was the averagevelocity reduced 20 ms

'•/ r >-r 2004'240t'2009. 0 ftJO-Wl 200* 200« Mtn 2004 200) *O'* 200) 2008, 8» 2 '-7 2008,Kd

ASolved Examples f * fii»d:

Average retarding force F * ?I^•oilation;r.%»The vcfaAtlysate graph of a car mo.mg on a straight road is

*• 3,7. Describe the motion of the tar and find theAltaic ewvrred.i

A , / ,' / (Jmg to Newton' s sound law in term of momentum.4

t i Lc£!iTo

tM*t 0 ' of rar fror A to BI 'hrig valuer we gC|

rig v;

i.L500 * I* - 1500 / 20. . v in 5 iecondt.Vapf V 4f* % r>M <>- f Hart /ou. rev *.-4 .fi vcFiCily intrsswrimmvmipaudwkmm&mkt 3

f - 22300 - 30Q00AvAf

ty - 750a -of

h )

^"GTsowl^gaiivr fejjy, 5how* that the force ii retarding force.

tto Chapltf > (MOUQB 4 y- 53 PHY8IC8 XI fSubjtttin ,scb“}!^> simple 3,3

I wo spherical India of 2.0 kg unci .VO lift masfCi arc: moving towafd* each other with vilotilltl ^and 4 iiis» 1 respectively . What must he the velocity of the ftnallci lull aftci colliik^^^velocity of the bigger ball is 3.0 nitGiven Data:

Mass of smaller ballMass of bigger ballInitial velocity of smaller ball = V| - 6.0 msInitial velocity of bigger ball “ v2 * 4msFinal velocity of bigger ball = v' = 3.0 ms

v > * 1 007 - 0.141nu0.07 * 0.140 07

vf ° -3ms

A 90 , 2 1=* mi 2.0 kg

* m2 = 3.0 kg ii

Now-i 2 m ,' =v;i n > , ni 2

Putting values, we get2 x 0.07

0.07 4- 0.14

To Find:Final velocity of smaller ball = v[ ?

Calculation:According to law of conservation of momentum.Momentum before collision = Momentum after collision

mi V|+ m2 v2 = mi v[ + m2 vjAs both the balls are moving toward one another, so their velocities arc of opposite signLet the direction of smaller ball is positive and bigger ball is negative, thus

niivi + m2 (-v2) = nij vj + m2 (- v' )

*

2HEH5E A .

A lOOg golf ball is moving to the right with a velocity of 20 ms"1. It makes a head on collision withan 8kg steel hall , initially at rest. Compute velocities of the balls after collision.Given Data:

Mass of the golf ball = mi = 1OOg = 0.1 kgVelocity of golf ball before collision = vj = 20 ms"1

Mass of the steel mass = m2 = 8 kgVelocity of the steel ball before collision = v2 = 0

putting values, we get2 x 6 + 3 x (-4) = 2 x v J + 3 x (-3)

1 2 - 1 2 * 2 v J- 9

2 v[- 9 = 02 v;= 9

v!= 4.5 ms"1

To Find:orVelocity of golf ball after collision v, ?\ docity of steel ball after collision v* ?

Calculation:

or

i

IMJUMUtot* 2 m 2A 70g ball collides w ith another ball of mass 140g. The initial velocity of (fie First ball is 9

perfectly elastic, what would be tkm i - m 2Asms /v . -

putting values, we get0 . 1 - 80 . 1 +8

V ) + v2right while the second ball is at rest. If the collision werevelocity of the two balls after the collision? ^ m| *. m 2 m|+ m 2 J

Given Data:Mass of the first ball = ni|* 70g = 0.07 kgMass of the second ball = m2 = 140g = 0.14 kgVelocity of the first ball before collision = Vi = 9 msVelocity of the second ball before collision = v2 = 0

vj -vf - (As v2 0)x 20 + 0

-7.9 x 20-i I

v'-- 19.5 ms'1NowTo Find:

Velocity of first ball after collision * v[Velocity of second ball after collision = v':= ?

= ? ro 2 "mi V:km|

2m|iv - V|+i r i|4 m 2 JPuttin8 values, we getCalculation:

As the collision is perfectly elastic, so( 2m

fa » 0.0 (A« vj- 0)I x 20 * 0v: «Jni | - m 2 1v i + v;v . = 0.2i

\m l + 2 Jm ! + m 2Puitvng values, we get

V|" Fi * 20vaf* 0.3 ms'1

> [.jr* NT ''i f *-&-V'

PHYSICS- XI (Subjective)122 Chapter 3 [Motion Ar Schol^5a 123iWTBWByin ,1 = O0)^(sin30^ )2~~

2*9 jTA hose pipe ejects water at a speed of 0.3 ms 1 through a hole of area 50 cm 1 . If the water str Lwall normally, calculates the force on the wall, assuming the velocity of the water normal to th * *is zero after striking.Given Data:

h = 900 x (Q,$)2*a||

19.6EZ]T5 m a;-iSpeed of water = V|= 0.3 ms

Area of the hole of water = A = 50 cm 2 = 0.005 m2

Speed of water normal to the wall = V2 = 0Density of water = p = 1000 kg m 3

Time of flight:(ii)2v,sinQ

t =As

Pulling values, we getTo Find:Force exerted by water on the wall = F -?

Calculation:Volume of water striking the wall per second = rate of flow

= (Arca)(Vclocity)= 0.005 x 0.3= 0.0015 m3 Is

NowMass volumeMass of the water striking the wall per second = ~~- =density x — : Pulling values, we gettime .time

R = (3Q)2sin(2x30° )9.8

-n _ 900*sin60°9.8

900 x 0.866

= 1000 x 0.0015r*

j= 1 . 5 kg/sForce exerted by water on the wall is

F.K R =9.8

R = 79.5 mlV tl -"»l)le.VHIF = (1.5) x (0.3)

F = 0.45 N In example 3.7 calculate the maximum range and the height reached by the ball if the angles ofprojection arc (i) 45° (ii) 60°.[HHHEHE

in a direction 30° above the horizon. Determine the heigh' Gi \ Cn Data:-lA ball is thrown with a speed of 30 msto which it rises, the time of flight and the horizontal range. Speed of the ball = v, = 30 ms

Angle = 0 = 45° and 60°Acceleration due to gravity = g = 9.8 ms

-iGiven Data:

/ -iSpeed of the ball = v, = 30 msAngle = 0 = 30° with horizontalAcceleration due to gravity = g = 9.8 ms"2

r2T° find:

(i) Maxmimum range = RHeight attained = h = ?

To Find: For G = 45°max(i) Height attained by the ball = h = ?(ii) Time of flight = t = ?(iii) Horizontal range = R = ?

Calculation:

( H ) Maxmimum range = RHeight attained = h = ?

‘1 For 0 = 60°max

C,,Cul«<ion:For height: 0)(1) Maximum Range (For 0 = 45°)

v‘>sin2G ’^v, sin 20

As 2g R =Putting values, wc get g

Putting \ alucs, we get Exercise ProblemsR9.8

V h » lu *M,Uf u •Guiding verlitully Ml ih« i t,l,r ground. i< ‘tone droppid. Ho» |»

del*:jft,ufll velocity ol the helicopter Jg

vertical <Jistance covered by lfie

M 900uin%R

v< * 1 n u tutgitf *A !‘X. >> /'• *# •' . tt* rttfei Ilf gruei3.1

9.8900 x I GivenR

9 8jk * 91.8 m|

(There is negative sign, being site to initial velocity)NowAcceleration due loHeight (For 0 45°)

I ,, find:11me i2g

Cikulation:Putting values, we get nation of motion

(30)W 45"h *

2*9.8

h = 900 » (0.707 )2

19 6 - 1 5 6 8 = 19.61 -v1 (-9.8) I 7

F-8 22.9 nil(ii) Maximum Range (For 0 - 60°) - 156.8 - 19.6t - 4.9 / J

„_ v*sin20K — — i— 4 9 /1-19.6 /3-156.8 -0

4.9( / - 4I - 32)-08/ J- 4t - 32-0Putting values, we get

R _ (30)>»in(2-60‘) making factors , we get41- 32 -09.8

l(t - 8) 4(t-8) -0„_ 900*»inl 20#

R — _ _ (t - 8Xt 4)-09.8t - 8 - 0, t - 8 see900 x 0.866R - t + 4 - 0, t - -4 secOR9.8

|R - 79.3 ml But time cannot be negative, thus( i .c time taken by stone)Now rscs

Height (For 0 - 60*)t ln't cruph OP » rtfluht wd of motorbike.'

io 20 | 20 1 20 0v, iln'8 the following data, drew » velocity-h - 2gf 100Velocity ( m,*

'1 ) 18015012090Putting values, wo get 60300Time (a). (30r (f /n60’ )h " 2 x 9.8

l the gruph to calculatethe initial acceleration

| ^ 'he final acceleration andfc ) ‘he total distance traveled by the motorcycle

900 x 0 866h - - 4- 34.4 n

Pv-1HChapter J «n0ft ^ (

||» HY5ICH \ l (Suh ) ritlvr )I,sM*Li akulatum:

127A03f tv 'm the given data, \ve have to draw giaph ami then find,(t'^ Ihr initial acceleration < ), ?O') l lu - rtnnl acceleration <i, M

I he total distance covered S ?I tom graph

(a ) - Initial acceleration a ,slope ot the graph OA

K proton moving with ipoed < » r i « , )^ nlMwith a *|»« cd of 2 0-10*«|

Ci|» UMt «|. Aaiititokiig ,,„j “ Ugh “ 0020 i in tliiik *lu fl of jH»p*r md

‘OuhtalloM, hod i « tilt () i* ttoo -Oil lion

t "I 10tiiutntc*

( 4 kni h > through ( In papri

( ,iv » o l >‘H «

r9

w0 & M tf toft u!1 V Init ial * pvrd Of pftJUin ~ v

DliUnu c covered S| inal speed of cla

Mii

AllIn find:OH

Kiiard.dioiI inic taken

v . - vl , t

( rtli iihillHii i

10 0Ulu;g on of motion‘i.

(»0 02060

o, 0.3.1ms ‘ jI 0, J -M 014

2 - 0 02 ' 101(b) Final uccclcnition ar

I / I slope of the graph *CE’ -96* I 012a -CP0.04 ^ 101 m

DEa *-2400M014 mi*2

0- 20U s

IK 0 ISOm -20

11)

a-- 2.4*10' W;

Now using the equation .for calculating time Vv.•v. +ot

Or

rV, -V,Q ,--0.67/m '

Negative sign shows that the velocity decreases during last 30 seconds.The Total Distance-Covered

I he total distance covered is numerically equal to the area under VCKH ity- time graphS (Area of AOA13) 4 (Area of rectangle AllDC) 4- (Area of A CDE)s ~ (f MX An ) 4 ( DD)( AB ) 4 ~(D£)(C 77)

ta

Siting value, we get

2» 10ft -l*107-2.4*10'1>8*10*

«2.4*10^|t 3.33* lOnseeJ

(c)-t-t “

1- vo() x 20 4 9 0 x 3 0 4 - x 3 0 x 2 02 2S 600 4 1800 + 300S » 2700m

Or S » 2.1kni

S =tlirni . what is the

rwl with u ,pri*» K comprcMcd bchvwi.after the spriug !»•* been released .masses m ( and mj arc initially at

1 »tio 0f t )lc magnitude of their velocities

Fust mass - n* iSecond mass mj

Initial velocity of mi" vi *• 0lniiiiii velocity of m: *" v2 ~ ^

* A*-*

PHYSICS - XI (Subjective)128 Chapter 3 SchoJPn.v ,= 0 F = Mal o find VJ = 0

FI :n.il velocity of mr = \ = ?Final velocity of mj = vj = ?

v'Ratio of their velocities after release = — = °

Or n = —M%

putting values, we get

l .Ox 10-I 7V,V, a = l .Ox 10" 12v:

m ,C alculations: -17 *120 = 1.0 x 10

According to law of conserv ation of momentumInitial momentum = final momentumnijVj+injYj - m, V; *‘(-m2Vj)

Where negative sign shows that V| and v2 are opposite in directionml (0)+m, (0) = m, vj +(-m2v' )

0 = m, v; + (-m:v 2 )

m , v; = m2 v'

a = 1 .Ox 10"5 ms"2

A hoy places a fire cracker of negligible mass in an empty can of 40 g mass. He plugs the end witha wooden block of mass 200 g. After igniting the firecracker, he throws the can straight up. Itexplodes at the top of its path. If the block shoots out with a speed of 3.0 ms'1 , how fast will the canhe going?

3.6

v

& /Given Data:

Mass o f can — m i = 40 g- 40'x 10 kgi .

Mass of wooden block = - 200g = 200 x 10 ' kgFinal speed of block = v 2 = 3 ms'1

* V

which is the ratio of their velocities after the spring has been released . To find:Final speed of can = vj - ?

As both can and block arc initially at rest, so their initial velocities arc v, = 0 and v:= 0According to law of conservation of momentum,Total initial momentum = Total final momentum

n^v,+111^2=111,vJ+mjVj

mlx0+m2*0=m1 v]-*-in2v'20=m,v|+m2v'

2

3.5 An amoeba of mass l .Ox 10 kg propels itself through water by blowing a jet of water throughtiny orifice. The amoeba ejects water with a speed of l .Ox lO'4 Calculations :and at a rate of 1 .Ox 10' IJ *gfAssume that the water is being continuously replenished so that the mass of the amoeba remainthe same.

ms

(a ) if there w ere no force on amoeba other than the reaction force caused by the emerging jet,what would be the acceleration of the amoeba?

( b ) If amoeba moves with constant velocity' through water, what is force of surrounding water( exclusively of jet) on the amoeba?

< • i \ cn Data : -m:v2vi =Mass of amoeba =\f = l .Ox 10‘i:A*gSpeed of ejected water = v = l .0* l 0Jms‘!

.

Mass of water ejected per second = — = 1.0 xl 0'1 ' kgs

mi

-200 x 10'3 x 3/V 1 -340 x 10

/ vj = -15m / secThe negative sign shows that the can will shoot in the opposite direction to .hat 11 bio

1 » » find:(a ) Acceleration of amoeba = a=?(b) l urcc of water on amoeba - F*?

head on collision with a3.7. - iAn electron ( m = 9.1xl0~3' kg ) travelling at 2.0x 10 mshydrogen atom ( m = 1.67x10-77 kg ) which is initially at rclastic and a motion to be along a straight line, find the 'cloeih of h > d

cn data:

undergoes a

c.,t. Assuming the collision lo be perfcc.lvatom?

< ikufifion:1 = mass per second x speed of ejected water

IQ - _| < v f»ht

Mass of electron = /n, = 9.1x10 31kgVelocity of electron — Vi = 2-Ox 10 msMass of hydrogen atom - )7h = 1 *07 x 10 kgVelocity of Hydrogen atom before collision

Futung values, we actF* I -0* 10 ’* 1.0* 10* 1- i

F- I 0* 10 \= VJ*0Acceleration of amoeba

Sew ton* a second iaw ol lior.

m. .I

i i - :

*r4I Tilm r •^»>.“HMMMM ’••V - v*~m -1i y,« ' *1|n r .v

i* V *• 1%. . ' '*v130

f PHYSICS- XI (Subjective)Sebo^ 131To find:

Two blocks of masses 2.0 kg and 0.50 kg arc attachedclastic potential energy stored in the spring is 10Jdelivers its energy to the blocks when released.data:

Velocity of Hydrogen atom after collision = v 2 ?(J at the two ends of a compressed spring. Theind the velocities of the blocks if the spring

3.9.Calculation:As the collision is perfectly clastic, so

[nym,] v[m, +m2 ]

Given Mass of first block = m, =2.0 kgMass of second block = m } = 0.5Elastic potential energy of spring = P.E =Initial velocity of mass m,=Initial velocity of mass

2m,v' = I V.+K+m,] I

putting values, we get2 x 9. l x l0"" x 2 x l 0 7/

F+°V,“i ‘9.1xlO'11 +1.67x10"

36.4 x 10'”1.67 x 10-”

Vj - 2.178 x 104

:=0

To find:/ __V2 Final velocity ofmi - vj - ?

Final velocity of m2 - vj « ?Or2

Or v'2 - 2.18 x |0W Calculation: According to law of conservation of momentumInitial momentum final momentum

m,vl+m,v3 = m,vj + m2 VjPutting values, we get

2.G*<HO.5*0“ 2.0*vj+ 0.5*v'0- 2v; + 0.5 v;v2 ‘=

According to law of conservation of energyLoss of P.E. - gain in K..E.10 - im.vf +inijvJ

3.8. A truck weighing 2500 kg and moving with a velocity of 21 ms 1 collides with itatioun e»weighing 1000 kg. The truck and the car move together after the impact. Calculate theirIvelocity. come*Given data:

Mass of the truck * m, - 2500 kgInitial velocity of the truck v,-21 ms *Mass of the car ml 1000 kgInitial velocity of the car v2 =0

- 2Or — vi0.5(1)

To find:common (i-c combined Jvclocity after collisions ?

vj-v',- v-7l-C

:2Calculation:1According to law of conservation of momentum

Initial momentum final momentumm,Vin2 v,- m( V; m,vjni|V| 4 m2vj - iri|V 4 rn2vm,v,im^ Vj-Cffi, 4 m2 )v

-*2»v.: 4-*0 . 5*v.J

2 1 2 11 0-putting values , we gel

lO-v^v?220“ 2v1'1+0.5v1"20“2v,/J+(0.5)(4vl );

2O-2v1,1+(0.5)(16vl'>)

20-2v't,+8v;*iov;’-20v,'1"2v;-1.41m«:r

Pu«ing v»lu« of v { In equ. (1), we B«>v *l -4 « 1.41v i--5.64mi

Ornr

As Vj/-- 4v^v miV'mivior

111, 4111,putting values, we get

Or2500x2141000-0v250041000

2500^212500+100052500

V" 3500

Before CollisionOrVOrm,+lul

After Collisionv - 15m» 1

Or •I

J? .J'\r > &...VJ x»Ul

_1 ftS»|

132 - - ChaPter 3 tMotio^A Foot* ball is thrown upward w ith an angle of 30° with respect to the horizontal. To thm pass w hat must be the initial speed of the ball?

^ Angle of Projection with horizontal = 0 = 30°. jRange of football = R = 40 m

Acceleration due to gravity =g = 9.8ms

•s PHYSICS- XI (Subjective)

5|3.10 iro*» «» 1 0 = 0x1-*- — x9 8xi

l 0 =0+4.9r'

, I 0

Given Data:

i49-l

i = 2.04To find:Or [ t = l .42sec.|Now horizontal distance i

X = V„X l

Putting values, we gelx = 21 x l .42x = 29 82/,I

Initial speed of football = v/ = ?Calculation;

Using the equationD

_ vi 2 sin20K

Vg2 Ryg

V = —sin2 0

Putting values, we gel

v>

or x = .Himf

For Calculating v (magnitude of resultant velocity)We needv , * and vfy W

R x

40*9.8 * ( As horizontal velocity remains same)Sin2(30°)40*9.8Or v.2 =sin60°392

V ( v =0+9.8x 1.42

vf ) =!4ms 1

Now resultant velocity =v=^( vfx ):+( vf >):

putting values, we get

v 2 =Or' 0.866

v? =452.65Or-iOr v =21.27ms

•iv = 21.3ms v = >/< 21 )2 +( l 4 )2

v = V441 + 196v = >/637

IA ball is throw n horizontally from a height of 10 m w ith velocity of 21 ms\How for off it hit nground and with what velocity?

3.11

Given Data:

A W*., M «•-— «*300A:/;I /I-1.(a) How long was it in air? . , fhe bomber at the instant the bomb was(b) At what distance from the point vertically below the bombe

^Jdropped, did it strike the ground?b'ven Data:

- iInitial horizontal velocity =vix =21msInitial vertical velocity =viy =0 -Height from the ground - y -10mAcceleration due to gravity = g = 9.8ms" 2

3.12 was

To find:Horizontal distance = x = ?Velocity to hit the ground =v=?

Height of bomber = y = h = 490m

Horizontal velocity of bombcr =vtxInitial vertical velocity of the bomb =v,y -0

jy.

Acceleration due to gravity = £ -9.8/iw

-1v. = 21 ms 300 x 1000 = g3 3/f|v60 x 60

- i

=300mh 1 =ICalculations:The first thing is to find the lime to reach the ground from verticalfall,Using the following equation

y^t+^gt2

Putting the values, we get

y

i T° find:=r = ?( a ) Time taken by the bomber in air

( b ) Horizontal distance covered by bomb=.r = ?

M 134' , pHYSICS — XI (Subjective)

S 135( altulahon*:( A ) f or calculating time, using the equation,

r%t*T*i*Cutting values, we get

490 = 0 xf +i x 9.8 / / 2

2490 = 4 9/ 5

,* „490

T° flD<)To pf°vc ,hat ra/lge is iame for ^Ble of projecti

^ is same for angle of projection G2 * 45°.$45° + 4

°ICUUWe have to prove that range is same forv? sin 2b

' 4 and 0 - 45'- 4 . As, formula for range isv* R - gor Cate (i):- »5-

y4.9 vf stn 2(45W) »n 2 f 4S’- -9 jt:-100 l R i -o t1- iOsec gv

u? Sin(9Qf -2f )g

Since *in(90o*8j* cos 9

= ^ cos( 2<pj

X tf b ) Horizontal distance coveredx = 83.3x 10* = %33m

Yes the bomb will hit the ground

= X = V B X f

R2 g313 Find the angle of a projectile for which its maximum height and horizontalGiven Data:

rar.ge remains jnaffec:ed when the angle of projection exceed or fail from 45’ byrange are equal.Ma> i iT. - j.rr: r:cigh* horizontal range

Angie of projection = 6 a ?To find: 315 A SLt. Ifsubmarine Launched ballistic missile) is fired from a distance of 3000 km. If the Earth b

considered flat and the angle of launch is 45* with horizontal, find the velocity with which the: is fired and the time taken by SLBM to hit the target

Range of ballistic missile = R = 3000 km = 3 * 10* mAngle of launch = 0 = 45*

CalculiUoits:Cota Data:vWoAs maximum height

2** tfmd:v JUR20fUngrofprojcctile

initial velocity of missile =v -?Time taken by SLBM to hit the target=t=?

iAccord;sg to g,v«n ctrxfc:. XL

H - R > ^dculaiioo:v :sin20As R =29 g g

%mb : Hv = a—sm29OrOr 2stffe0*62values, we getOr tic#*4aa8

3MQ*y9.g _ 29 4Q0f

1~

iiua*-*?) «nW

vJ =2$U*l<fv =5.42*10W

I

v g5.42kms'|Tl®« b> SLBM L*.(tottJ rim* of night)

2v.»in0

v3*O = 4tttoh

b*4OrOr9=1361 (4)Or

Or

or fail »hnrt of 45* by equal amount*, th«^114 exart 1*

iV4JF< uet, we get

i.

138 1

PHYSICS- XT (Subjective)SchoJiH.s 137

2x5.42 xl 03 xSin45°t =

9.8Chapter2 x5.42* l 03 x0.707

t =9.8

7.66x 103 WORK & ENERGYt =9.8

31 =0.78x 10t = 780sec.

780 .t = min

g Objectives^earnor

60Y =13min

Understand the concept of work in terms of product of force and displacement in the direction of the

force.

2. Understand and derive the formula Work = mgh for work done i

surface.

Understand that work can be calculated from

Relate power to work done.

Define power as the product of force and velocity.6. Quote examples of power from everyday life. .

7 Explain the two types of mechanical energy.

Understand the work-energy principle.9' Derive an expression for absolute potential energy.

1

* * * * * * * * * * ** * * *gravitational field near Earth'sin a

ATP FOR Part - 1 under the force-displacement graph.area3-4.

N

Alternative To Practical 5.

OnPHYSICS, CHEMISTRY,

BIOLOGY8.

10* Define escape velocity.&tial energy of a body is equal to gain In kinetic energy

Understand that in resistive medium loss of potenof the body plus work done by the body against friction.COMPUTER SCIENCESGive examples of conservation of energies from everyday life

[NewPaper Pattern For Intermediate Examination!13.Edition for 2014 is available in Market

1S! Chipur 4 f\vs PHYSICS » XI (Subjective)orV 4

Loduct|on(i(?rgy of Object changes if an exchange of energy occurj

^ transfer can occur due to aforce or due to an e4ianto of /L.

. tra^fer of energy via force is a process called doing work. Doing work Is the act of transferring the energy,

unrk then, •* transferred energy. Energy transferred to the object is positive work. Energy transferred from

I )r,e object is negative work. -JI (S 0tten thought in terms of physical or mental effort. In Physics, however, the term involves two things

farce III) displacement

What do you understand

139

Chapter No. 4etween the object and its environment.

v*uI I J i l lnr « r* ii g

f [ i ie term work? Explain.Su at

Do You Know ?42? Work has the same dimension asthat of torque.I

s f a l3 7 s* <X *v.< < £

5 Work Done by Constan

Work donemagnitude odisplacement.Mathematical form

nr ii c- 3i£, I . I& j yV 2

ce£. '

2 ^s nstant force Is defined as the product ofthe component of force in the direction of

on a body by a co

f displacement and t-3 -J

a.

fit£ —<flu

*-$,**£ > -va

let> >

S O) E F = constant force applied on a body.d = d> $place • • of the bodyy = angle between force and displacementThen the v. rk done on the body is,

W = (F cos 9 ) dW = Fd cos0

s muS*= |

1“ l $I* § 33 - jTw

H * IUJ Fig. 4.1!XON Du C

1«u J3 WOJ JO JUIfl i sV0*' dtuop jo aicgz o

5o u0 OR3x -*u c nl|i*i l l-•a b i

=8 ORo uac- —c tI* W = F . dAnother definitionWork can also be defined as the dot product offorce and displacement.Unit of Work

_^0rk 15 a scalar quantity SI unit of work is Nm called joule.jUnitlon of joule

one&Wer i

tE -2 5

UHn BEECEEEDD1 joule =l<r ergs

* X

Is•S II

t/i

tt , g. ,| 1=73-, r;k nsi i c

P yo co o 5i><£ -03c .2; v o it = o _scs .5| 1* 1* 1 1 1

t

newton force acts on the body and the body covers a distance of one

the direction offeree, the work done is said to be one joule.ensi°n of work

X

Ulm

\Theditension of work is [MLJTJ]casessPecial

I 1) If 0 <;90°, work done is positive.2) If 90°<0 180°, work done is negative.3) If 0 =90°,no work is done.4) If 0 = 0°, work done is maximum.

*»

m &S.

140HJl ElfLit'VorU

Scholar’* PHYSICS- Xj

So work done for the first interval is

I MV,=F,.Ad, =FjcosO,Ad,

Graphical representation of WorkGraphically, the area under the force-displacement curve represents the work doneby force.If we plot graph between force and displacement then,

Area under the graph = (OP)(OR)

14

Simila^Y/BAW,= F;.Ad:=F,COS02 Ad3

and up to nth interval

AVVn = Fn .Adn = FucosGrNow the total work done in

W = (AW,+ AW^OR W =(F,cosO,OR

1= FdSo = W

If force Fand d.

makes an angle 0 with horizontal. Then the graph is plotted between F cosO ng the body fi point a to b is.Q.2 . +How much work Is being done by upward force whenthe pail moving forward?

a person holding >2 Ad/4^...+FncosOnAdn)ycosO;

How to calculate work graphicallyTo calculate the work done plot F cosO versus d. Area under the graph is divided|into n rectar , les from a to b. Area of each rectangle represents the work done £-luring that interval. The total work done is equal to sum of areas of all therectangles.For more accurate calculation of workThe vork done can be calculated more accurately, If we subdivide thedistance Into a large number of intervals so that each Ad becomes very J**a .i.e. Ad-» o • LI

In this case no work is being doneReason

-*because the angle between F and d is 90°.W = Fd cos 90°W = o

i-L

SoI

W/W//////r* «««Q.3 How much work is being done on the wall (shown in figure)?

A<0 work is being doneReasonbecause the displacement of wall is zero.

So W = F (o) c o s OW = 0

W- lim tF.cose.Ad,In this case the total area of rectangles H equal to area under PcosGd graph.

Dtipi«c«m*nl d —Pig. 4.1

verses

—Q.4 HOW can we calculate the work done by variable force?J *5 ^gravitational »!•>«« 11 *con.erv.tlv.field.0«S h o w

0,pathfollow.d.Wwork done In gravitational 1|u,lona|liitfgiWork dona along a clotart PBt^ *n1—

on a body 1*Gravitational field

gravit*#011*1forci *Tht ipace around the $arth IffCq,/fd the gravttatlonaljleld- ,.vltatlon«lfi,d

kupositive

Vl*1convention for wok done in rtl0nii

wQfkl»11MMI...»»*'>"•* „«««•"*”’• displacement It perpar‘lcul|f\ ***iaitro.

vWork Done by a Variable Forcein many cases, the force is not constant, but it varies In magnitude o?directk>n orIn Doth.e g.

Force of gravity acting on a rocket moving away from earth.• Fo'ce exerted by spring increases by the amount of stretch.

consider the path of particle In xy-plane from oo.nt a to b as shown In-e:usfigure ,

D v.de the oath- * nto n

~^e forces actmg Curing tnt ;e Inttrva s T p

^'2 fo'ce .J cons de'td to

5 tz ace~«nt.

» .<snort intervals of displacements Ad .AJ;

,F, respect* vtfy.

* projumatF constant 4or each interval of

...Ad, .A particle actedyartablf forca. $tne pet* ano*nto point oa

V'V

- A HOT9HH U ; *4 - 'i;i*

r -4Krjfr'

42 Chapter 4 fw.PHYSICS- XI (Subjective)

Sch^i-Vork Done in Gravitational Field.et oS consider a body of mass m .The body is displaced from A to B alongj **e'ent oaths with constant velocity in gravitational field as shown in figure,

tat'ona - ‘orce acting on the body F=W= mgPath - 1 (Work done alon& path ADB)A O'K cone by gravitational force along path ADB is

W ~ VVA-»D + Wo_

8

\VA^D = mguVDWX)0

143Wv gAy.cosISO’+ mg^= -mgAy, - mgAy,- mgA^

__= -mg(Ay, +Ay2

Wv = -mghw

Thus, the work done along curved path IsWAB = W„+ Wv

= 0 + (- mgh)

WAB = -mgh

+. + mgAy#cos18U°

...- mgAyn [v 80° = - 1)OB+ Ay„)OB

I v (Ay, + Ay;patn2C OB8ri'

CMJC I 4j* '

\ 0 )mg* patnlA , From equations (ij, (2) and (3) ;t js deapath followed, i.e.

WAB = -mgh

t workaD w* is independent of the- mg i, AD ) : o 'S'*////;/ ////// /7777777777T77

= 0 Fig. 4.8 WAQ8 = W/.CB =Work done along a closedAnd \V = me DB cos!SC

Cor der a closed path ADBA. The body!moved from A to D D to B and rh.n

from B to A. the total work ' t0 8 and then= mg (D8) (-1)

= _-ngU[v cos*8cr = -1][DB = h ] IKis equal to sum of work done along these paths.

So.Sc WA^D = mg(AD)cos90°

= mg (AD ) (0)

z ~ -

= O +• l-rr &n)For Your InformationW*c* = -mgr.

Path -41 (Work cor.e a ong path ACB)- =: £-= .:r. rms a z ~ z path AC3 S

= O(1) 1 hp = 746 wall = 550 ft - lbv:* W:~ B = mg(DB)cosl 80o

• 2 ~ < = mg (DB) H)= -mgh

'

•v _ - = m e AC co-si SC3 |‘" r- :ne curved pathV>.= mg(Ay. )cos0°*mg(Ay2)cosO0 -mg(Ay,,)cos0°

+ Ays) fvcosQ°=lJ[v (Ay, Ay2 Ay„) = h)

"6 - *

“ “ s'rn. C3 :P$90°

= -5 -2 ~

= mg(Ay, + Ay2 +- = mgh

-WD^B + WB-A

= 04 (-mgh ) «• (mgh )= WA-D

s —A = 0

2r — “ r or*done along a closedP*lh **rvatlve Field For Your Information'= pat*

- t; ca s- at* •r< dor.e r/ (; s-'. e t a r . 7 / ce t'e :atr -i':.em ca ress as r 7 f ?r?.

A'/ F m.'» a mg repi

•m »-* 'f : ?«i rep* c«».se ~ gt :: >»- *V '7' i z“ i ntJtys Sc

W,*5/ rr r r t c C'g c a i t t p i

pV 7 V' t i 7' ; .< ~ 7i 2

? r» jn - »TT.rflffiigis called

******* **0ne U lndependent of the path followed Gr»/tta\<r<* ’oc*E>o*t »c to' rg'eye*&90tflC 10TC0Non ConBir / tkY* Fpr£ g §

frK.tono' fo'enA* r«r»<«nc*Tart#**1 -n a itong

tor&ajP'WV’*=0O ot B 'kwlMB fewest of •rnensy

ORis called conservative*hich work done along a dosed path it iero,

T Conservative force1. Work done by conservative

independent of path followed.2. Work done by conservative force along aclosed path isExamples

force is-•V' Utsonil fieldV t

zero.of conservative force are

gravitational force, elastic spring force andelectric force etc.

3.

* f7' '/

J

PHYSICS - XI (Subjective)145

that Instantaneous power,P=F - vQ.6 Define power and instantaneous power. Give its unit.

'proof

ut FIs the constant for«acting on a movine t

the body

s

I Then the power

Power

Power is defined as work (lone per unit time. ya s constant velocity of

ant is given byOR

Power is defined as the rate of doing work.

Average PowerIt is defined as total work done divided the total time taken.Mathematically

delivered to the body at any inAW

p = limAl-*0 At

F - AdAi-Q AtP = lim AW = F.Ad )

ORAW

P.v =At P = F | IOR

Where At-»0

At = time takenAW = the work done andAd -*[ v lim = v ]P=F • vOR A«-0 At

Instantaneous PowerAnother Definition of powerPower car also bo defined as the scalar product offorce and velocity.AW

Instantaneous power is defined as the limiting value of —following the time t approaches zero.

as time At,Do You Know

Q -9 Define energy. Give the two types of mechanical energy. It lakes about 10° J to

make a car and the car thenuses about I * I01' J ofenergy.from petrol in its lifetime.

AWP = Lim,ni At-»0 AtSo,

EnergyThe ability of a body to do work is called energyTvpes of Mechanical1 Kinetic EnergyKinetic Energy .mlledkinetic energy.I'Juirgy possessed by a body due to its motion is <thematically __

Unit of PowerPower Is a scalar quantity. 51 unit of power IsJoule/sccond called watt.Dimensions of power are [MLaTJ]

Definition of wattThe power Is said to be one watt If one joule of work Is done In one second.

Define commercial unit of electrical energy. j ^

energyFor Your Information

2. Potential energyApproximate Energy Values

rncrci (J >Source

Q.7jn. ui5 « 10

Bumino 1 Ion Ct’.lfBurning 1 III/ -* peirolK E. of C.W a!

1 t 10*

1 Running Person *1K.E = — mvJ 3. tG'lOknih

Commercial Unit of electrical energy

The commercial unit of electric*! energy Is kilow'ott-hour.Kilowatt-hourKilowatt-hour Is (he work done in one hour by an agency whose power Is one

kilowatt

-ST*ner&Y Y be translational rotatioAPPfj2<Ktent,a'Energy

, osition„n cncrgy possessed hy a body because °f&v^' ’1E^condition 15 calledpotent

c'o"'° P o t e n t i a l Energyne3r theenergY due t0 6rav

Fivtirm of on.’ '»tomof uf /inuirn

* t . o'air -ynuji* of .vrHW 10 '

C » 10 *

in aforcefield or because

i ikWh IOOO W * B6oo *'GC -1000 J/sec * 3600 sec

•3600000 J

» 3.6 x 10 J

So surface of the earth at a

for VnarInformulionAn body tr.a> c«cm>without having momentum

P-F. = mghIkWh 3.6 M J

PHYSICS -XI (Subjective)Chapter 4 [\Vt)ru & r 147

146

E is the energy of mau^ I

^ T^Tabsolute potential energy. Derive relation for absolute P.E.oT;£ nr***-*- /of body of mass m at distance r

Elastic Potential EnergyThe energy stored in a compressed / stretched spring is called elastic po

energy.then

E-JL2m

/> = 'JlmEl

Elastic potential energy = -kx 2

Where k spring constant and x is the extension.

The units of energy are the same as those of work.

potential EnergyAbsolute

^ ln t£Zof Absolute P.E ^Q.10 State and prove work-energy principle. Igw

Calculationbody of mass m is displaced in space from point 1 to N in the gravitational field. The gravitational force

LCl anQl remain constant during this displacement.elation P.E = mgh is true only near the surface of the earth where gravitational force is constant. As

itational force varies inversely to the square of the distance from the surface of earth

^,tnver^|Bl9||B law j,

TidbitsAH the food you eat inor*day ha* about the tan*energy as 1/3 litre ot octroi

Work-Energy PrincipleStatement «

Work done on a body is equal to the change in its kinetic energy.

Proof

grav, so it can not be applied in this case.

i.c. F«.order to calculate the work done by gravitational force, the distance between, to N is divided into equal

small steps of lengths Ar,so that the value of force remains constant for each step.

Work done duringi'! step (1 to 2)Suppose I^

Letm = mass of bodyv, = initial velocity of the body

F = force applied on the body

d = distance covered by the body

V/ = final velocity of the body

The work done on the body is

Work done = F dNow, according to equation of motion

2ad = Vf2 -vi

2 (as

m = mass of the bodyM = mass of earthr,= distance of point 1 from the center of the earthr,- distance of point i from the center of the earth

Calcuh on of r

-•stance between the center of this step and center of the earth isNo

(1)r =2

l Also displacement of body from point1to 2 is

Ar = r:-r, _r2 = Ar + r, _

S equation (3) in (1)fjjf Ar +r,

- v,!)

And according to Newton's second law of motion

F = maUsing equations (2) and (3) in (1), we get

Work done= ma

(2)2a

d= (2) 4lOR (3) 3 H

(3) Usin21

r =0

,- 2r. + Ar1

r. r,OR Work done = 2

2 2r Ar- — 1r = — L + —Work done = final K.E. — initial K.E. = change in <.E.

OR 2 2Ar (4)

faringboth sides, we have

4 VO

Noteearth

If a body is raised from the surface o ’ arth, the work done changes its

gravitational P.E.

If a spring is a compressed the wor. one on it is equal to the increase in

elastic potential energv

I\ **

• : * “ ’V

1«43

'4 9

«r -— _

* ;lr« ^G V- 7'-»-c* k,

4

v '-ce — "" v, r s >"• s-sr se ?-e^erred "•'r* ca— *d * '

'' = c: - * .irT<-Ms— ^ '•'c*5 cca -as -sed

sros * s^5 :-s* --as -sed I' re *~ae r -say>:e*'i'e -at

f % is£ 5'-i j

= n' -r *

j G*r. ra• *c*w -re grirTIi*TcrT~rr*'^e r a* r e center q* thisrec s

Do You Know? - _GMm: * *•- r.L ‘-CMx

ss *U = r.GMrr.r = — :— . acso; jte .- '5 /Stationr:i cote'tiai of body at distance r from the center o* the earth, s(6ir

-GMm (9)r, r 2 " - *e -avet GMmr =

= rAisoiute P.E on the surface of Earth.vr - a bod / lies at the surface of the earth then,r = R. So, equation (9) becomes

U _ -GMmr ~

R.'<‘ere R U: rg -j us of the earth and the negative sign represents that gravitational field of eartn for; tractive.

’ f ,v-- r .ne body moves away from earth's surface,r increases,U increases, (i.e. it becomes less negative.)' nen the b°dy falls towards the earth's surface,r decreases,U decreases, (i.e. it becomes more negative.)

calculate P.E. choice of zero reference point, JcCho,ce of zero point is arbitrary., T^Ca 1 tal<e ^e surface of the Earth or the point at infinity as zero P.E reference.

Thedrfference°f P-E from one point to another is significant

in P.E as we move a body above the Earth's surface will always positive

.(7)

.s the do-e d.' - gia step :

WM2 =FAr= F Ar cosi8o'

(v F isopposite to A r )

=-FAr

ho)

mass mThere more energy reachr:Earth in 10 days of sunlightthan in all the fossil fuels «3

the Earth Hcte :

=-G“^-0TJ

[using equations (2) & (7)]

3LTih Vi.

Ww =-GMm ---_ ri r2.Work done during 2nd step (2 to 3)

W2_, =-GMm —

fl *3 _Work done during last step (N-i to N)For last step work done is,

Wu2 = -GMm

Tj meescape velocity and derive the mathematical expression for escape veioaty?

%PeVelocity

. .eM. That

finWaieaCWnBacerta‘I!!it ja

'"'smorC ,'ethe ground-

WJ^ .I*»» i***££.»**»”'l 'Wiensbody is thrown upward. it’ lheipitia' ° ,he bod> ,l*'

'0rr-*acting downward, if we '°crat cert3'n ve 0basing the initiai veiocitVrthetoC,fy.w to^\ar velocity is caiied escape

^0nlhefor escape velocity

F 0f abody°*^^know that the absolute P-v

0ut of the earth '5 gravitational field is called escape velocity*

velocity of a body with which it goes

%-,_N = -GMm|— J

Total work done from point 1 to N

Ww +W2^+ 7

r* J^-1

- W rface of earth isN-I-N .

M l tXI (Swhjiitiw)PHYSICS 151< (watk 4 r

1S0

5* intertonvertlon ol potential energy and kinetic energy'GV. ro 0)

| jr-‘l,** s«•* Q-11R

As t-v>cv c.'r < v'.- t o - s • 'st - o \* * e-G. > tL> L . becomes:oro.GMm'i GMm 4UP

version of Potential Energy andSc = ointer -conR

ic Energya body is

>-$ the -mta \ S. neeced by the body to reach infinity (i.e. out ofgrs\ ts:: 3 field' s

Init 3 K.E =

KineticConsiderresist'surface o

P.E-m f r | «K.R -0 L, r

fallinE freely under the action of gravity thr

Suppose a body of mass m i* at rest at heighta 5on-*/e tr e

A

medium,f earth.

ition A

GMmR For Your Informationl GMm At P05'potential ener

Kinetic energyThe total energy at A is

E^/nglt + 0

EA - mgh _ —[

(1

. & K.E. when body falls

gy of the body = mgh

of body = o0« -® v«a — —

RWheren = the mass of the body; M = mass of the earth and, R = radius of earth

2GM

MMiie Escape speeds (KniM RE = sag (h - * )

K-E =* rr.£*h

MoonMercuryMarsVenusEarthNepyuneUranusSaturnJupiter

32 4t43

OR 5 0 Gcv *

Cb - x >5K. R 10.4 P.E = 0K_E =:righ *1122GM

Change in P.EAt position B

224OR ( 2)v =~ V R 254370This is express on for the escape velocity of a body.

Another formula for escape velocityAs the gravitational force for a mass m placedmass is

61let

covered by the body = xdownward distanceheight of body from the surface or eart

Potential energy of the body = mg (h-x) = mgh -mgx= (h-x)

the surface of the earth ofon

1 (2 )Kinetic energy of body - — m v„Where vfa is the velocity of body at point B.

Calculation of vBy equation of motion

2ad=vJ -vjv 2

r =2a d+v*

v 2 =2gx +0

v2n= 2gx

Putting this value of v2 in equation (2), we have

GMmr — :—

R -But F = mg

BGMmSo mg =

R 2

OR.GM= vB ,d=xanda-g]OR g =

(v v,=0,vrR: OR

ORGM=gR2OR

Thus equation ( 2) becomes.l-m (2gx) = m g x2

2gR2 K.E =Vm = s°. totalR energy at position B

EG = mg (h - x) + mg xER = mgh -mgx + mgxEB = mgh

= V2gROR

(3)Value of escape velocity onEarth

R=6.4*iOl,mg = 9.8m/sec; andAs ^Position C [Just before hitting the ground]distance covered by the body = h

^ote * kocty from the surface of earthn*,a’ energy of the body = o

mvj

= N/2 X 9 . 8 x 6 . 4 x 1 0*So v«= 0

= 11.2 x ioJ m/sec

=11.2 x 103 m/sec KinetjOR (4)c energy 0f body =|v>tc=n.2 km/secOR

I KChapter 4 [Wo1'> 2

PHYSICS XI (Subjective) 153Where v< K the velocity ol body at point. C.Calculation of v,liy pqu.'ilion of motion

2a d vj - v/v; - 2nd i v'vf- - 2gl» » 0

vj»2ghPutting the. value of r, In equation ( /1), we have

Energy SourcesI ^enutlcally .total energy * P.g.+ K. £.,COnstant

ThisI* the speCial C3Se°f conservatlon of energy

Need for new sourcesI ln dally life we ob5erve manV energy changes from

||energy transfers heat the environment anhlch Is useless. So, useful enero,/ a**. . -

served. That Is why we ne

SiKirmewnItUKum.- wablu*4

CoalNatural Gai

HydfOoftCftcWindTsd*»Cfeoincffi.al"Hiomac-jttuultghtCthrjnol/Mritf liif ‘of *

0.1i one form to another,id energy is lost, in the

Uran urnOr! nhaiATil* i.jftd*.OK

OR 17 v, 0 V , V, , d hand a g | At l*5t 3' Individual fields rn./ run off*’f'er.i;//niyii: wfior,made fromDio rr.ann

of hrol wis con

OR form•fif 'd'./eri6fgies *

>

Discuss the different non-conventional energy sources.K I , m (zgh) in g h Q.15

So, tol.il i-m-iKy .»t position Bi t o mg hl r rngh Non-Convct i ional Energy Sources

These sources art- not very Cbmmoq these

r ..•I...OR ('>) days. Some of them are givenConclusionII .t body falls hnm .i height h„then .» t height h, above the surfac < > ol earth

loss In IM . (.aln In K I

i • * •

below;

1.Energy from Tides| Gravitational force ol the moon produces

water in the sea roughly twice a day.

ExplanationWater at high tide can be trapped in a basin by constructing a dam. Dam Is

filled it high tide. Then, water Is released in a controlled way at low tide to

derive the turbines. The dam Is filled again for next high tide and the fall of

water also derive the turbines. This process Is used to generate the current.

• • i « »v vS?I '/"» Ufi>*//•|pf I•..— rfW I •*. * •-' ** * »tides in the sea. The tides raise thei

1 I 1mv,2 2

mv/mgh, rngh,>i

* I* *«-*•»WU— . «*M-.»>11»— *

2m N/ ' v/) 111mg(bi h,)

Where v, and v; are the velocItlt - s at height h, and h, respectively.P I . and K I . are Interconvertible but the totalenergy remains unchanged.

• In the absence of friction,loss in V E gain in K.E.

• In the presence of friction,loss in I1 I - gain in K E * work done against friction

mgh = — m v 2 + f h

ORvf

i',T w»r« — nli.w.1H> A»»w *******mio *•••'••. •a.* */'*’® *"**"••

PI® . A.< S

* — — *—Tha anergy of thesa waves can generate electricity.

“=MSU,.a...«Wo — -Saltir's duck.

produce strong water wives.

. i.p.t or Yo»f InformationQ.14. State law of conservation of energy. Why new sources of energy has to

be developed if energy is conserved? II* (TI|*.J it.tMif) C'l'lumm fnpuUlionlw|>ti anifuuunnif iwfutil IUUUCVJ*AM*BJii#lc»iciciy mimniiAS thobervitt

f«)uw Mngm/ingil)«u*> af

«U§> fe an ulwmatodrfomovaibfiVMBSMi«uiciemy by,

tfiitjltunl eUiruil<33>Conservation of EnergyStatementEnergy cannot be destroyed. It can be transformed from one form intoanother, but the total amount of energy remains constant.

Conservation of mechanical energy

K.E and P.E are the different forms of mechanical energy,

total mechanical energy of the body is equal to the

K . E and. P. E.

P.E may change into K.E. Similarly, the K.E may also be

Chang into P.E, but total energy remains constant.

(I)rtr /»WJ£ PIHJflftru*vtmml me

~~ —toftI lntr>d</

nifnknvrt t>i /bp)MMfr tmlmH h <i[h> «, t xat

c*ur,,ni TOI#MJ* ng1 (41 JMAg »kn*aihnmn_ If " ~ ]

I

For Your informatloj'Orliilni*1Source

Sourco ofenergy

l!.1'" Parts (1) Duck float (2) Balance float.

!W,V|*n.rgy products th.mov.m.nt In duck float r.latlv.to balanc.float.

Th* f*'«,v.motion of duck float Is us.d to run al.ctrldty

SunSunSunSunSunSunMoonHorth

.SourctaBio massFossil fuelsWindWavesHydro electricTidesGijolhormni

Thei :m ofThe

w

iJ-*

154 _Chapter 4 [\\u - XI (Subjective)PHYSICS 1553 - Solar EnergyS<jn ts the major source of energy- on earth. ‘Solar energy at normal incidenceoutside the earth's atmosphere per second per unit area is about 1.4kW/m:

,wkick is referred as solar constantWhile passing through atmosphere, thetotal energy is reduced due to reflection, scattering and absorption by dustparticles, water vapours and other gases. On a clear day at noon, the solarenergy at the surface of the earth is lkYV/m2.Solar energy is used to heat the water by solar reflection and thermal absorberscan also use to generate electrical energy. For this purpose, the surface of thecollector is blackened to absorb the heat energy. It can heat up the water up to70' C.To get higher temperature; we use the reflectors or lens of large size-.Photo voltaic cellsSunlight can be converted into electrical energy by photo voltaic cells . Theymade up of silicon. Electrons in the silicon gain energy from sunlight to createvoltage. Solar panels are expensive but of long life time and have lowcost.For cloudy days or nights, electric energy can be stored during the Sun light inNickel cadmium batteries by connecting them to solar panels.They are used to power satellites. They are also usedweather stations and rain forests communication and also in solar calculators.4- Energy from BiomassIt includes the organic materials such as crop residue, natural vegetation, trees,animal dung and sewage. Biomass energy refers to the use of materials as fuel.Two common methods of conversion of biomass into fuels are direct combustionand fermentation.Direct Combustion MethodDirect combustion method Is applied to get energy from waste productcommonly known as solid waste and confined it into chamber and ignite it Theheat produces is then use in a boiler to run the turbine of generator.Fermentation MethodBio fuel such as ethanol is a replacement of gasoline, which is obtained byfermentation of biomass using enzymes and by decomposition through bacterialaction in the absence of air. The rotting of biomass in a closed tank called adigester produces biogas which can be piped out to use for cooking and heating.5. Energy from Waste ProductsIt is probably the most commonly used conversion process, in which wastematerial like wood waste, crop residue and municipal solid waste is burnt in aconfined container. Heat produced in this way can be used in a boiler to producesteam that can operate turbine generator.6. Geothermal EnergyThe heat extracted from inside the earth is in the form of hot water or steam. It

be generated by following methods.

*ive Decaydio3 of radioactive elements, the energy heating the

*11'1'16 , Heat of the Earth|ResWua rocks within lOKm of the earth are present

// i Hot >8ne°“ { gnergy from interior part of the earth. The^^IlMOO-Cormore.

I nression of Mater.al

' C inside the earth, the compr,

R3 ocks is constantly

c *«*UUUQ£n form. They

temperature of these

y dull bUckd materials cause for the generation ofin deep

heat energy,

m some places,

raised to high temperature and pr-vents. The steam produced in

purpose.Geo thermal energy may causecolumn into the air. They usually occur in volcanic region and erupt with irregular,„.erv3ls. This extraction of geo thermal seriously disturbs geyser system bvreducing heat flow and aquifer pressure

V by

****— *»**.? ground is in contact with hot rocks and isssure in form of hot springs, geysers or steams way can run the turbines or for heating

water bene

are

ce hot geyser releasing with an explosiverunning

Do You Know ?in remote ground based Populationcan te rexiunfif I

(I JPccpk iiscnuss tranquil(2)UcgccUvnral.Solar.h\ dcricctricaJanJ windpwg}*alternative fern's ofera#

FORMULAE

W = F . dWork done by constant force1

W = XFj Ad,cos0Work done by variable force2i=l

gas

irp = —P = F . V3 I Power /

AWp,v AtAverage Poweroa

AWp = L im,nl AJ-*0 At

e -m*nur* and Instantaneous Powerrotting plants:

P1 K.E = TK.E = - 2mKinetic Energy 2p.E = mghGravitational Potential E 11 :2W=^mv,- --vWork Energy Principle

surface of earth)r

-GMmU =Absolute P.E r

GMV = s/2gR V = REscape Velocity

1— mv 1 fh1 -O mg * -m|v,mg(h( *hj) amConservation of Energy

can

- >\ • , v

IMill

156

Multiple Choice Questions Absolute P.E of a body at infinity is13' (a) Zero (b ) Negative

(c) infinity

(c) pull of earthvbj mass of satellite

l5. When speed of a body is double then. ^ N°ne

(a) K.E is doubled(c) P.E Is doubled

,6. Consumption of energy by a 6

(a) 120J(c) 30J

17. Dot product of force and vclo<(a) Pressure(c) Momentum

(d) None

(d) Both a & bhour possible answers to each statement are given below, lick < S) //„- < , ,, i r f ff||

Which of the following types of forces cannot do the work on the particle ?(b) Gravitational force(d) Centripetal force

1.(.1) frictional force

acceleration is doubled(d) Its momentum is doubled

ond is:(b) 60J(d) 0.02J

(c) Elastic force4Relation between horse power and watt Is:

(a) ihp - 766 watt(c) ihp = 736 watt

2.bulb in

(b) 1hp » 746 watt(d) ihp * 796 watt

Proton,Neutron, electron and a particle have same momentum, which particle hav(a) a (alpha) particle (b) Proton(d) electron

3.ty is:c Rroater Kl) (b) Work

(d) Power(c ) NeutronSlope of work-time graph Is(a) Work(c) PowerWhen two protons arc brought together, then(a) Their K .E. Increases(c) P.E . Increases

4 - 18.(b) Displacement(d) None

(a) 180 W(c) 1.8 W

19. If we go .hove the surface of earth, the gravitational constant5-(b) Becomes zero(d) Varies directly to the square of the distance

(a ) Remains constant(c ) Varies inversely to the square of the distanceThe typical source of wave energy Is(a) sun(c) earth

(b) P.E. decreases(d) P.E. remains same

6. A porson holds a bucket of weigh 60N.Ha walks 7m along the horizontal and then climb up5m. The work done by man Is:(a ) 720J(c) 300J

vr 20(b) moon(d) mars(b) 420J

(d) NoneANSWERSThe SI unit of power (watt) Is equivalent to

(a) Kgm*s**»(c) kgm1!1

7.10. u9. b8. b(b) kgmi“J

(d) None7. «6. c5. cl . d 4. e2. h 3. d

20. a19. II18. cl17. cl16. itU.b 15. clU. b12. c 13. u8. Two manes of igm and 4fm are moving with same K.E. The ratio of their linear moment srai(b) 1 ; 2 Ik.(a) 1 : 1 6

(c) Va : 1 Short Questions of Exercis(d) 4 : 19. Pick out tha conventional source of energy:

(a) Wind anergy(c) Tidal energyEscape velocity of•body depends upon the(a) Mass of tha planet(c) Velocity of the plant

11. KWh Is tha unit oft(a) Mas!(c) Power

11. Which onellnon-renev. >•energy tot *7(a) Wind(c) Uranium

1 friend. A car Is stationary with Its0 situations similar?

talking to aQ.41 J--ho,d* *> « of groceries while standing stilln# runnlng.from the standpoint of work,how are these tw

n t>°th casts work done Is ura.

Reason;p9rson and the car both are at rest.So the

Work done f dcosOW •P(o) co*8

SoW •0

f work don* becomes zero.

(b) Coal energy(d) Bio mass energy

10. Am.(b) N iss oft u body(d) None displacement is ggrft. ( > •«•* d •0)

(b) Work(d) Time

(b) Tides(d) All of these

r

v .\

V *1-1 |S

Chapter 4 [u;n:158

PHYSICS- XI (Subjective)gchola^iCalculate the work done In kilo joules in lifting a mass of 10 kg (at avertical height of 10 m?

159Ste3dV Vel°city) thf|(Mtn 2006-2009, Lhr 2006, Fsd 2006, D.G.Khan 2006,Bwp 2007-2008

Q.4 2A ball of mass m is held at a height h, above a table The * hi

0ne student says that the ball has potential enerev me h ** T* 3* 3 he‘Bht ^ ab0Ve *he fl°0r'° correct?

B¥ mgb'' bUt another saVs that it is mg (h,"* h,) Who

goth of them are correct.ReasonSince P.E ** always measured with respect to some referencepoint. Therefore, we can say that first student has measured P.Ewith respect to table top (mgh,) while the second student hasmeasured P.E with respect tofloor = mg (h, + h2)

Q.4-6LhrJo°9,GrwAns. Data: m = 10kg

h = iom

W (kJ) = ?Calculation: As the work done is equal to increase in its P.E. So,

W = mghW = 10 x 9.8 x 10

W = 980i

Ans-h, + h,

h,

/

n,\ /

980 When a rocket re-enters the atmosphere, its nose cone becomes very hot. Where does this heatenergy come from?

kJW Q.4.71000Result: W = 0.980 kJ {Grw 200 j 2009,Mtn 2004-2009, Lhr 2005-2006, D.G.Khan 2005, Bwp 2004, Lhr 2010-2011)

Source of heat energyWhen the socket re-enters the atmosphere, then some of its kinetic energy is used in doing workagainst friction with dust particles and air, which is converted into heat. Due to this heat energy, itsnose cone becomes very hot.

Ans.Q.4.3 A force F acts through a distance L. the force is then increased to 3F, and then acts through a fur%

distance of 2 L draw the work diagram to scale.Ans. Force displacement graph:

As 2*53 jnce' the force displacement graphwork Pone by the body. So

W = (F)(1) * (3F)(2L)W = FI+ 6FLW = 7FI

3F.is equal to the t

F 2F- Q.4.8 What sort of energy is in the following:(a) compressed spring.(b) water in a high dam.(c) A moving car.

4ns. : A compressed spring has elastic potential energy.b) Water in the high dam has gravitational potential energy.

F -(Sgd 2003, Mtn 2004, Rwp 2005, Bwp 2007, Lhr 2009)

O L 2L 3Ld

Q.4.4 in which c a s e i s more work done? When a 50 kg bag of books is lifted through 50 cm, or when *?kg crate is pushed through 2m across the floor with a force of 50 N? I(Lhr 2010*2" c) A moving car has kinetic energy.

^•4.9 A girl drops a cup from a certain height, which breaks Into pieces. What energy changes are

involved? (6 200^2006)

Ans. Energy changesA cup thrown from certain height losses its gravitational potential energy and gam its K.Estrikes the ground then a part of this kinetic energy is used to break the cup and rest of .he gy

converts into;(') Sound energy( •* ) K.E of moving pieces(iii) Heat energy.

10 A boy uses a catapult to thrown a stone whichPossible energy changes?

iAns.Case !:Dato:

Case II:Data: . When itma = 50 kg

d = 2mF = 50 NW2 = ?

m. = 50 kgh = 50cm = 0.5 mW,= ?

Calculation:W,- m.gh

= 5 0 x 9 . 8 x 0 . 5= 245 J

Calculation:W2 = FG

= (50) (2 )= 100 J

accidentally smashes a green house window. Ust the

(GrwWi)Result: More wcr < s done in COSK

Am, p<Mlble Energy changes:

,Ihe f0"0win8 enerev chan*eS 0CCUf' fha.ll5tic potential energy change* IntoWhen the boy throws the stone th s#d t0 freak the window into pieces.When stone hits the window, » P*of Kln^c ene

(UwWo.jRest of the energy converted Into txaLSBUW*

Q 4 5 An object has 1 J of potential energy. Explain what does It mean?(Lhr 2004, Sgd 2005, Mtn 2005, Fader;! 005, Mli * 2006, Grw 2005-2009, Bwp 2008, Grw 20H

Ans. We know that energy is defined as the ability of a body to do work. So an object haspotential energy means that tr body has r paclty to do a work of one joule.

one \°(111)

*

k

Chapter ^[w0160 PHYSICS- XI (Subjective)li<v 161

Solved Examples Initial velocity of brick = V|= 0

Initial height

Final height

To Velocity at height 3m above ihc ground = v - >

Calculation:2

= h| = 5m= h2 = 3mm

A force F acting on an object varies with distance x as shown in Fig. 4.7. Calculate thby the force as the object moves from x = 0 to x = 6 m. C|v*rkJ

Given Data:A force acting on an object varies with distance x as shown in Fig. Loss of P.E = Gain in K.E.

To Find: lmg(h|-h2) =Putting values, we gel

v; - v;Work done W ?Calculation:

Total work done total area under the curve= area of rectangle + area of triangle

= (4m x 5N) + ^ (2m x 5N )

= 20 Nm + 5 Nm= 25 Nm= 25J

Exercise ProblemsfflLilJFkfrtA 70 kg man runs up a long flight of stairs in 4 s. The vertical height of the stairs is 4.5 m. calculiits power output in watts.

Given Data:4.1 A man pushes a lawn mower with a 40 N force directed at an angle of 20" downward from the

horizontal. Find the work done by the man as he cuts a strip of grass 20 m long.

Given Data :Mass of the manTime takenHeight of the stairs K h

= m =» 70 kg* t = 4 see.

4.5 mApplied force = F = 40 N

Angle = 0 = 20°Length of strip of grass = d = 20mTo Find:

Power output « P ?Calculation:

To find:Work donc= W= ?

W Calculation:As power ** P = —twork = W — F.d -FdcosO

W = FdcosOP.msHort

Putting values, we get Putting the values, we getp - 70 x 9.8 x 4.5 W=40*20*cos20°

W=40*20 >< 0.93

W=751.68J

W^T5* K)^3087P *

4P « 771.5 wattp - 7.7 x 10? watt

A ra»n drop(/H = 3 35*10'5 kg ) falb vertically at a constant speed under the influence of HiM

!ric,r° fgnvi ty and friC,i0n- ln faUing ,br0U8b 10#m’ mUCh"“ j ^Or

rest position 5.0 m above the ground. What i*A brick of mass 2.0 kg * Iroj ped fronat a height of 3.0 m above . ground?

Given Data:Mass of brick

(,lvcn Dj aL ata:*3.35x 10 ^Mass of rain drop = m

= m « 2 kg

m

k

Chapter 4 [w„162 PHYSICS-XI (Subjective)

SeboJ^ 163Height h 100m

W2-mg.h cos 0°=mghwork done lor 3 brick when place on lo the two hri

W,=mg.(2h) cos 0"=2mgh

To find:Work done by gravity *» Wg ?(a) work done by friction Wf ?

( nlculution: 'imjjh

Bmqli

7m/ jhflmgh

5mgh

4mqh

3iorjh

2mtjh

1mgh

* •Work done by gravity - Wg F. h

Wg - FhcosOAs gravitational force is along the displacement, therefore angle between mg and h is 0ft

So Wg * mghCosO°Wg - mghWg-3.35x 10‘5 X 9.8* 100

Wg = 3283x 10‘*

(a ):- SimilarlyW4 - 3mghW3 = 4mghW6 = 5mghW7 = 6mghW8 = 7mghWo*8mgh

Wio' B 9mgl

o

Wg = 0.0328J

Work done by friction =Wf = F. hW,= f hcosG

(b):- Toial work done isW = W, + W2 + W3

• W4 + W5 + W6 + W7 + Wt + W9 + W10rAs friction is opposite to displacement, therefore angle between frictional force and h is 180°.

Wf =fhcosl80°

Wf =-fhAs the drop is falling at constant speed so f=W=mg

Wf =-mgh

W = 0 + mgh + 2mgh +3mgh + 4mgh + 5mgh +6mgh + 7mgh +8mgh +9mgh

So W = 45mgh

Putting values, we getW = 45 x 1.5 x 9.8x 0.06W = 39.69W = 4Qj (approximately)

bl A car of mass 800 kg traveling at 54 kmh is brought to

retarding force on the car. What has happened to original kinetic energy.

( c o s180° = -1 )

Putting values, we getrest in 60 metres. Find the average

Wf =-3.35x 10'5 x9.8x 100

Wf = -0.0328J Given Data:Mass of car = m = 800 kg

Initial velocity = V , =54 km/h =

Final velocity = v, =0Distance covered = d = 60 m

Average retarding force — F — .What happened to original KkE.= ?

As according to work-energy principle.

Ten bricks, each 6.0 cm thick and mass 1.5 kg, lie flat on a table. How much work is required to

stack them one on the top of another?54 x 10004.3 -15 m/s

60 x 60

Given Data:

Mass of each brick - m -1.5k gHeight of each brick = h = 6cm - 0.06/rcNumber of brick = n =10

find ;

0)( ii)Ca,c»l»tion:To find:

Work required to stack them one on the to; of another= W=? 0)

iFd = — mvkulation:

korx wi.: be done on (he fiut bn. 'yr ± fiat on table ( i.e W, =0) . When we place other bricks <>njone. work is done in the form of V . accordin. o their height. Hence we must add all these

k ’o gel the total work done in terms or P.E. 7 us work done for 204 bnck when place on to

22

Or Fd =Puttimg values, we get

k

1S4 £!S!E!2L1 * K pHVSlCS XI (Subjecti « • )

1MlHeight ol the lank /7 ] 0/77l imc taken - t 20 minDensity ol water p 1000

FfbO) ; V 800(0 * (l >V )Or

l2(X)*cc

ypump^f ?

CH.U -too ^ v ::M400 x 225F s -

foHo60U ) Increase in P . E( b ) Power ddi^r©

F = -1500NO:Negative sign shows that it is retarding force.As the velocity of the car is decreasing* so its kinetic energy also decreases and becontdue to frictional force

Calculation#:(ii> e* zero massdensit

4.5 A 1000 kg automobile at the top of an incline lOmetre high and 100 m long is released and rolhdo« n the hill W hat is its speed at the bottom of the incline if the average retarding force dfriction is 4S0 N?

Orue to

OOx 1000Gisen Data:100000 kgMass of automobile = m = 1000 kg

Height of incline = h= 10mLength of incline = s = 100mAverage retarding force = f= 480 N

ncrease in P.E .= mghP.E .= 100000 ^ 9.8 ^ 10

[Increase in P.E .= 9.8x 1 O^JIncrease in

To find:Power delivered by pump-

p= worktime

(C)Final speed of automobile at bottom of incline = v -?Calculations:

.AsUsing the relation loss of P . E = gain in K.E - work done against friction

mgh = mv2 + fd

mgh - fd = — mv2

t

p _ 9.8 *10*12001

P=8166.6P=8.16^ 103wattPutting val ues, we get

1000 x 9.8 x 10 - 480 * 100 = -xlOOOv2 P = 8.2kWOrroad friction and air resistance in propelling an

inc develop?2 A force (thrust)of 400 N is required to overcome

automobile at 8« kmh'' . What power (KW) must the engineGiven data:

98000- 48000 = 500 v2

50000 - 500 v2

50000 Force ( thrust) aF = 400 N

Velocity of engine* v = 80 km/h

v2 -or80* 10°° X 22.22m’ s60 x 60

500/ - i 00

0 m*IWm’ vf water* patmptd Irum a rmrvoir nil« a lank* 10 m higher than the rtMTVCtf t ,n

/ f #f water « )M# k%m findthe tfttrove in V V0M pvtre? 'J ‘m e4 by t uurn*

Wind;

Mutation:

>:or

Pfrwcir P ?4 A

kI c r<» IPM±

m « a

I »A % p F . vOr P (• v to* 0

,r i *f i d veloc ity are in the u,m*p i v c o s <r

dirccUOOA *, f ,* ' ' ,/n/ v an** ' wa •&/

•*

I k f

•«Ch*pt»r_4 [\v „

s pHVSICS XI (Subjective )

>' j. ,.Ov • * V ^Tl

? -m' 2i"

& 167

in I (i! diver ai h » vSfK*(^ulu« i0,,s:? SSSO watt

? * $ > > ' K>’ watt Cium ol K I .. = loss of p. E.Or r,g air friction)Or P = P|VS jmv-smg(h,-h , j

v-=2g(h, -l^Pulling vulucs. we gel

v 2 =2x9.8(10-5)

v 2 =19.6x5vliar

4.1 How large a force us required to accelerate2.9\ iO’ms through a distance of 5.0 cm?

an electron ( m » 9.M0"kg ) from OrW to « sPccd 0|Given data:Mass of an dcctroa =Initial velocity * v =0Final \ velocity *\ ,«2» lQ

,m/s

Distance = d = Scm -Q 05m

m = 9 i MO"" kg

v=9.9m/sTo find:

4,10 A child starts from rest at the top of a slide of height 4.0 m.(a ) What is his speed at the bottom if the slide is frictionless?(b ) If he reaches the bottom, with a speed of 6ms'1 , what percentage of his total energy at the top

• of the slide is lost as a result of friction?Given Data:

Force required *= F = ?Calculation:

As work energy principle isWork done = change in the K.E. of the object

1 2— mvInitial velocity = v, = 0Height = h = 4m

1 2 1Fd = — mv

To find:1Fd = — mv 2 As v, = 0 speed of child at the bottom if slide is frictionlcss = v= ?

°k age of total energy lost = ?

if velocity at bottom is\ - 6ms

(a )i2( b)Putting values, we gel

- iF x 0.05=^ x 9.1 x lO'31 x (2 x I 07)2

*

— x 9.1 x 10“ JI x 4 x 10M2

Caliuluti,^sPced of child = v = ?

^0r frictionlcss systemGain of K .E = loss of P. E.

ms:

1- ' 0.05-I- ^ 0.05-18.2*10 17

F.IHXIO 1

0.05it - 3.6x )0 IS N[

Ijjjjj ihrA diver weighing 750 N dives from a bo.* rd 10m above the surface of » pool ofconservation of mechanical energy to find bin peed at H point 5.0 m above the waterneglecting air friction.

I-mv:=mgh

v* =2ghV = yfiPh .4.9 r>utti n8 values , wc gel

V = -/2x9.8 x 4

v - V7MGiven Data:

Weight of ihc diver mr 750 NHeight of a point above ti. m I hi- HimI ( eight of .i point above the

v * 8,8m/s«8*|0 f,h Ol Energy « ?•urfae' t • h 5 m

168Chapter 4

PHYSICS - XI (Subjective)"S! ScboJiLf.In the presence of friction

Speed = v/= 6m/s

P. E. at the top

1

= mgh= m x 9.8 x 4= 39.2 m

Chapter

iK. E. at the bottom = — mv' ' CIRCULAR MOTIOI2I

= 2 x m * (6)2

|Learning Objectives^i= 2 x m x 36

= 18= 39.2m - 18ms 21.2m

Lossof energyTotalEnergy

21.2m” 39.2m= 54%

Loss of energy Describe a

2. Define angular displacement, angular velocity and angular acceleration.3. Define radian and convert an angle from radian measure to degree and vice versa.4. Use the equation S = r 0 and v = rw.5. Descri: • qualitatively motion in curved path due to a perpendicular force and under stand tf

centripetal acceleration in case of uniform motion in a circle.- e. ive the equation ac =ro2 = vVr and Fc = mco2r = mvJ/r.Understand and describe moment of Inertia of a body.Understand the concept of angular momentum.Describe examples of conservation of angular momentum.

10 - Understand and express rotational kinetic energy of a disc and a hoop on an Inclined plane.Describe the motion of artificial satellites.

12* Understand that how and why artificial gravity Is produced.15, Understand that the objects In satellite appear to be weightless.1^* Calculate the radius of geo*statlonary orbits and orbital velocity of satellites.

Dwcrlbe th* Ntwton'j»nd Elnst«in's views about gravity.

r motion1.

% loss of energy x |(X>

% loss of energy x 100

‘9.9.Get Scholar Series books

PhysicsChemistry

BiologMathematics

(Subjective .& Jt ’ective)

11.

lu

— I— —170Chapta 8 tn pHVSICS XI fSubjrf i » v»^Mlt&L\

Chapter No. 5 * - Motion0,lon a)an object in u urtulur path y« 0*11*4(A

ths (lira Unn of Ibe body umtihuoutly change* \

0f(U^r

I IK( I motion

temple*1) A stone whirled around *.T/,r^2) A satellite In orbits around tin- ,. 0f3) Motion o1electron in atoms f[4) Motion of CD's. /5) Motion of circular saw blod#

6) Motion of ceiling fan7) Motion of merry go round etc^

1It3 t ! If8 8 VIIs £ > iSI '|& °I a

i *-»inoi

Q.1 Define and explain angular displacement. Show that 1 radian =57.3°I«3 * l<2 </ic2 co

^ <— .2 «»Angular Displaceme ntTltc angle subtended at the center of a circle by a body moving along thecircumference in a given time is called angular displacement. It is denoted by6.Explanation

a.2 0 § fk *

% O 2C

Iz5 1(c,

£plane of circle due to motion of the pa ,

tQ the p|ane of circlethrough the center of the circle, which is pe pwhile OP is rotating.

^,e Q wjth x.axis. AfterSuppose at any instant t the position of OP, ma' - S

x.axis. Thus thesome time t+At its position be 0P2 making an \.&(JfQp during time intervalangular displacement can be defined as t e ang e^ vector quantity.4. For very small values ofAO, the angular displaccmcn

%hConventionAun\acement AO Is positive whileFor anticlockwise rotation of OP, the angu ar

negative.for clock-wise rotation the angular displacemenDirection of Angular Displacement , ment/ we use the ‘rightorder to determine the direction of angular displacefond rule'

Uni^

O cdu n

HUMIItlEIf ^ill 1u “ aM 5K &o 25 2< Q3 a

•r? Oo 5O Ee E cEu Ew •OLi•f 2 aF.

V LU

—oLx 3

=1«IS 3

— ES| 8 c§| tf tis s s a s S .s

o n = c u #

*oU Degree

In one complete rotation, arotating object subtends anangle of 360 degree. If thecircular path 1$ divided into360 equal parts then theangle subtended by eachpart at the center of thecircle is equal to onedegree.RevolutionA complete round trip o'the body elon| thecircumference of the circleIs called one revolutionRadianIt Is the angle subtended byan ere at the center of thecircle whose length 11 equalto the redlus of circle.

I1*.2 3 cv w> ou 5 w«o oi£ Ei1

— . Ti 1 1 rr gBo t3 .J'S a -g t-

gs;i!" '3.2 g, S IIg& 3

c d

J jrhe S|unit ofnl^e cente

a,°gular displacement Is radian. It Is the angle subtended by

dr unitj^ ° drC,e Whose,ength 15 eQUdl t0 the r*dluS of Clrde‘,rnenS|0nsS are degrees and revolution. Angular displacement has no

*1an arc

172 Chapter 5 [Q\n^w pHYSiCS- XI (Subjective)Relation between linear displacement and angular dispiacemcntSuppose S is the length of the circle of radius r whichmakes an angle '9’ at the center of the circle. Then interms of radian, it can be expressed as.

Length of arcRadiusof circle

A S, .9 =- frad)

173explain the angular acceleration?

Singular AccelerationrateofchangeofanKU,ar „. .

Average Angular Acceleration- we switch on dm electric fan ,hp

thtimtlal angular velocity and c„u ’

actively. Then the average an -~~Wlnal a

the ratio of ( .jJ

0called(rad)

uIar mccleration.city goes 'creasing, if o>,

velocity at time t, and t,iring tl

rOr

accelerationS = r 0trti Is can be

ulur velocity to the total timeige infined asThis is the relation between linear displacement and angular displacementRelation between radian and degreeIn one

Do You Kno* FOR YOURINFORMATIONcomplete revolution the linear distance covered by a particle is equal tothe circumference of the circle i.e., S « 2nr and the angle in radian would be

a s 2*r0 = — =

When angular velocity ofthe body Is Increasing thenlingular acceleration k alongthe direction of angularvelocity and if angularvelocity k decreasing then'd' k opposite to thedirection of angularvelocity.

= 2n radianInstantaneous angul.i

jthi Instantjneous angu cderation can be defined as the limiting value of“ at (he lino* jn /ervo 1 ,\J upproai he\ to zero, Is called Instantaneous angular

(deration

r rSo. 27t radian a 360° 1 revolution

360°Or 1 rad * 2 nAs re fftttftumitafiangle e. r; lays out a anpedctenoeS **

360°Or 1 rad • Au>2 x 3 . 1 4 a c l i m —Al

S oO r iradian = 57.30 rr,:„ *— -— — — i i w * “ 01

MUtbn

UMI ri,n.ulM acceleration is ud/tec'. 1» 18 MTORYOU* |04 Derive the relation between enguler and 'ln**r ve|ocl,lc,INFORMIX*tr »’iM ,u " *same «Ml#r><# * pf,»***r;,* ^covers thecJnpl*i«,r nT >

Axis of RotationAll portlcN". of .1 rollingbody move*. In circles, flicline joining the ront/c*. offhoie Circles k called axis ofrotation.

Q.2 Define and explain the term angular velocity.

Angular VelocityTime rate of change of angular displacement Is called as angular velocitytSuppose A9 Is the angular displacement during the time At. So the averageangular velocity can be expressed as,

AOw„m —AtThe Instantaneous angular velocity can be defined as the limn • value ofAO/ At as the time Interval At, following the time t, approachr to lets.

AOSo, •llm —* »-*•At

ORShow that v •ru)

ta«lonbetween Angul»r *n<»U n,,r§p #ngul»r

withCon,lc,*r » r'|idbody routing »boul * ** <jl»t»r'c*' ,r°vl direction of ar'8uU'rr** ln » »i«W body »t P,rp,nd ln|» flrtd. th*Svelocity v. A, the «1» of «*•«“ "'tyujalwaysremains the sarne.West

ci5 “ ’

oDirectionAngular velocity is a vector quantity. Its direction Is along the Mg of rotation andcan be determine by right hand rule.UnitThe SI unit of angular velocity Is red/sec. It is giso measured in revoiutlon/minand degrf?e/sec. The dimension of mgular velocity ls[T'' J,

Point p moTh

lf f#fer§n *** thf0u*h 9 ^Stance RiRj«AS In time At during Its circular

** c#n,,ne^ 9n angular displacement AO during time Interval

Fla 3.4(«)

AS •rAO""'•bo* s,d«* by At

*174

, PHYSICS - XI (Subjective)So0oUf

d a have been replaced with 0, to and urespectively. Thus‘ntions for linear motion

Vf rVj + at

s B v,t +J4 at2

175AS AO-— * = r - -At At

Applying limit \t v o - *AOASIim lim r

*» .0 At *• At.. As AOlun r InnA> >° At

K/ » rt»J

o.

2aS = vf2* v,2

tion for angular motion

(l)f=0>|+Ut

• 0= o)|t +Yi ut2

2(i0 = 0)f2 - (0,'

Equation of angular motio^tho angular vectors have the same direction. So they can be treated as scalars.Q.7 Define and explain he centripetal force and derive the relation for It?

* i

A« ... faOr Equa

In vector formI or limit At > 0 the length of the arc P,P2 becomes very small and its directionrepresents the direction of the circle at point P,. Thus the direction of velocity vat any point P Is always tangent to the circular path. So the linear velocity atpoint P Is called tangential velocity.

Derive the relation between linear and angular acceleration.

v to x r

only if axis of rotation is fixed. In this case all’ Q 5

Relation between Linear and Angular AccelerationAs the reference line OP Is rotating with angular accelerations. The point P also has alinear or tangential acceleration. As we know,

Av r A (u

Centripetal ForceIhe fore which bends the normally straight path of a particle into circularpath is culled centripetalforce.

ORA force which compels a body lo move In a circular path Is called centripetal

Dividing both sides by AtoAV AO)

forceAt At

ExplanationOn both sides, we haveConsider a body attached with a string moving In a circular path. If the string istapped (broken) then It would not continue to move In circle. Observation[howi* If the string snaps, and when ball Is at A then It will follow the straight

7P«h AB. So In the absence of some force which pull* It toward the center of|rclMht ball will not continue to move along circular path. It will move along“ >•ten

Applying limit as At -> 0

lim - - - lim r —A t * 0 *1 " *0 A t

AVSo,

Avlim — * r limAI *0 At •i .« At 8*nt at that point.

*lmp,il of centripetal force:j; Forc« acting on electrons In fixed orbits around the nucleus,

kjPorCQ acting on artificial and natural satellites.Forc* acting on earth around the sun.

I^,or c*ntrlp«tal acceleration end centripetal force

ejjgOrIn vector formWhere a Is the tangential accelerationNote:-Advantage of rotational motion over translational motionOn a rotating rigid body, the points at different distances have the same angulardisplacement, angular speed and angular acceleration. While tf cy have differentlinear displacement, velocity and acceleration.

0 “ « x r

Q.6 How can we write the equations of rm n In case «. ngular motion? AV.(1)a •—

feqUlred bV par!‘CletlT.°V.*vff°s7ncA.tth! »pe«d of the

At

Equation of Angular Motiontquetion of angular motion are slmllai thos* » r> linear motion except that S, v

t. * if Jt,»-v ' ;-v > • m _ _ . *•

aW;rm i

* JJ pHYSICS \I (Subjective)si I'"111

177JA * - sJ|0n for centripetal forte

W F, * ma,V

As

& (%jMk. Tw^‘ Hi .-****WfZlSZJZj&L. AiS

ter' J-- ••; •<- 5f5 ; %r v. 'j..v. , • .v

f''frrV / v 'JtrKQtfa' hr **

~v>-3 (;r recsr-es &v»Att ac —And

i= r> v

.mvFc =A *Or soU)i= v r$

m angular measurements, this equation b\o ,\t we : 3 «. 3 triofrg e A?QR such that is parallel and equal to v and PR ispa.-ai 3, and equal to v, As radius of the circle is perpendicular to its tangent, sov , and v, areequa ; to the angle <QPR.

Moreover | v , | = | v. | = v and OA = OB*

(radii of same circle)5o rrorn isoscsies triangles GAB and PQR, we can write,

QR _ ABPR

~~

OBAv AB

So, TfJIVft7ri2..2mr w7 cifo ir to# 'ji DHftripaiJ

force is provifcrf fort* offriction, rt'oea vptaid of i caris .ncrasod a’, a *. jrn. ;t*n

(v = ro)r LFc = mrb)1perpendicular to OA and CB respectively. So the angle ZAOB is

Or /Unit of centripetal force is Newton and dimensions are [MLT2].

Define and explain the moment of inertia?

frictional forcesufTiciem to provideoenir.petal fore -.. Tncrefort.the cer will ino . -: awa> fromthe trad. To avoid tie side-slip,the outer edge of the roadis raised through some angle,called the angle of me:motion.

The speed of 'he car al the

turn is v= ^ rgtan^where <p is angle ofinclination or banking.

is nett

Point to ponder

Q.8

mI

Or Moment of Inertia (Rotational Inertia)Moment of inertia of a particle is defined as the product of mass of particle andsquare of its perpendicular distance from (pivot) the axis of rotation.

It is denoted by I and is given by

(3)v r

when At— »0, point ‘B’ is very close to 'A', then length of arc AB nearly equal lineAB. i.e. AB= SEquation (3 ) becomes,

Av S

nttmtaitlaYou may feel scared ai the loptfroller coaster ride In the :3fcbutyou never fall down even^ I = mr2

ExplanationConsider a mass attached to a massless rod which can rotate about a

frictionless pivot 0. Let the system be in horizontal plane. A force F acts onthe mass perpendicular to the rod. So,

F='maThis force rotates the mass m about 0. As the angular acceleration a can beexpressed as

You clo not fall out of yourseat when you go ups'(3:

down on a roller coasterbecause 0; many forces ontlu in going loop on a rollercoaster the centrifugal *»centripetal forces, the force

of drag, gravity. »friction. Due to cenirifiig

force, your body is prcSSwith the outer rim.

v r

A V = S —r

Putting this value in equation (2), we get

Or (4)

Ji)Sv

a = v —rS2v a =ra

(5)a = —equation (1) becomes,r

Where a is the Instantaneous acceleration, as this acceleration is caused bycentripetal force, so it is called centripetal acceleration ac.Direction of acceleration

(2)F= mraMultiplying both sides by r

rF = mr2aSince rFEquation (3) becomes

T = mr2(x . u , mrJ The quantity mr:

r-*JSSSi«« — “-'„,tar mkram./«WI = mr2

Two cyhnoof*one twin iw ithe orea!«ruSince PQ is perpendicular to OA and PR Is pi pendicular OB. So QR is parallel

to the perpendicular bisector of AB. As acceleration Is parallel to Av when AB -y

0, So direction of centripetal acceleration is along the radius, towards the centerof the circle . So,

represents the torque.

DO \ Ol KNOWis known as

pl3y$ in linearMoment at uwrtl.* rrpiv

ihe icndeucy to maintato .»1«state of rot nr ttiitc ofuniform angular motion

Here F isThe instantaneous acceleration of an object traveling with uniform speed in acircle is directed towards the center nf the circle and Is called centripetal

acceleration.

The direction of centripetal force is in the direction of centripetal acceleration.

-1m J

17$

Dependence of moment of inertiaNs shows that moment of inertia depends upon the mass m and the square of

nerpe^o,cu ar distance from axis of rotation r. Practically, it also depends upond stributton of mass and position of axis of rotation.Moment of inertia of a rigid bodyVoshy the bod es have non-uniform mass distribution. Consider a rigid body^ade up of n' small pieces of masses m^ maof rotation O

PHYSICS-XI (Subjective)lar’s 179

^Ifine angular momentum? How can we relatP jt with moment of inertia?

Momentum (Moment of linear momentum)r

cross-pr°duCt of position sector r with rcsrfg to axis of rotation and linear momentum p of a rotatingUed angular momentum.at distance rvr2 - from axis

prt\cU '* caOR

Article Is sa*dt0 h°ve anSu^ar momentum about a reference axis if it moves in such a way that Us angular4

n changes relative to that reference axis.MP**

• m. ^or HOOQ

!•*mia body of mass m moving with velocity v and linear momentum P

to origin 0 Then its angular momentum can be expressed as

| L- r 'P

r 9 rl Considerrestivem4 ( c) '•

m2— Vm4 O

r is the position vector at that instant with respect to origin.-•

The magnitude of L is given by_ ^ L = r p sin0

= r(mv) sinG

^ L = mrvsinG

Were 6 s the angJe between r andP.Direction

The Sre-von of angular momentum is perpendicular to the plane containing r

r 'i ,t can be determined by right hand ru e.

Where rSoi-U disc or cjfeidr(a) (b)

5.10S<.rccse the ood\ be rotating angu ar acceleration a So.vagr tude o'torque acting o- rrv

= r%r?aa.:

M(since p = mv)!

2r<d)secorc — ass

= nva'a.=or Ttf -ass

ie total 3 g.ver byl unit r .

* S vr*of angular momentum is kgm’/sec or J-sec and dimensions are[ML‘1 J -= ' ~ t.

= - r - -VaV Show that kgmVsec JsjR-H.S = j sec

= Nms, m

*re ccrTr s 'i' >s ? r-e — asses -state ^ argL,ar acceeratioc e.= au = a fsayj (v J = 1 Km)

(. 1 Nm = 1 kg ^ou = a> *

'V*1)aX -Y

5*IVA .ms

M

Z-nnt x V — t* Ir Or t- f a r

d *err.ajr -i - car »/yesMdK. « LM.S.**r momentum of a particle moving in a circle

rvnrsi :I *.w*- *»•»*-»«*”«T«— h •sp**^ g ****e*n r and tangential velocity v I* **

L * mrvs4n90* 1 0 * 9^ JL •mrv(1)

»- W angular retooiyco

fTatpift -he c * s s*>d

- fl/ 5f JDtMn**l (A$ v * f»1if*' r -r(fto)

l - rr*J «> f$Uf 4

C3.

180

Another definition of angular momentumrefine and explain the law of conservation of angularThe product of moment of inertia and angular velocity of a rotating body Is called angular momentum?

Angular momentum of a rigid bodyf Conservation of Angular MomentumConsider a symmetrical rigid body rotating about a fixed axis through center of mass as shown in f °,eternal torque acts on a system then the total angular momentum of theparticle of the rigid body rotates about the same axis in a circle with same angular velocity 'oj' ^n° Remains constant.momentum of a particle of mass m, particle,

^ totaj = L,+ L2+ = constantL’ m'r ’ (° nUop°nd»( gxplalJatl^ jaW 0f conservation of angular momentum is apparent if a singleThe e 6C .

|nning body changes its moment of inertia. For exampleisolate !f the board with a small angular velocity, Upon lifting off from thepUShf the diver’s legs and arms are fully extended. So the diver has a largeb°ard' of inertia '1' about this axis. But when the diver' s legs and arms are inclosedTuck position,Mmoment of inertia reduces toI,.

According to law of conservation of angular momentum.Li -8

For second massL2 = m2r22G) a diver

For nth massU = mnrnJca

Rg.5.13| Aman diving from a diving board.

The total angular momentum can be expressed as,L = L, + L2 + L3 +L = nv^o) + m2r22G)

L = (nv^co + m2r22 .

+ Ln..+ mnrn2 co+ mnrn2)co Wliy does the coast***>,

syslom slow down asinto the beaker?

f " Or hci), = l2w2

So diver must spin faster for small value of moment of inertia to conserve[ •••L = Xm.r2‘ “V 1=1 i=l

EXPLANATION: angular momentum.

This enables the diver to take extra somersaults. The direction of angular

momentum is along the axis of rotation which remains fixed. The axis of rotation

of an object will not change its orientation unless an external torque causes it to

do so. Earth rotates about the sun experiences no sizeable external torque act on

it, so Earth's axis of rotation remains fixed.

rotational Kinetic Energy and show that (K.E)rot =

So, L = I coWhere I is moment of inertia of the rigid body about axis of rotation.There are two types of angular momentum

Spin angular momentum

Angular momentum of a spinning body (i.e., rotating about its own axis) is calledspin angular momentum (Ls)

Orbital angular momentum

Angular momentum of a body orbiting in a circular path is called orbital angular momentum (U)

So the total angular momeptum of a body is equal to the sum of its spin and orbital angular momentum

When water drips in;:-1beaker, the mass ohcontents in (he boltincreases which inert*the moment of biertiifcto increase in mow"inertia, the angular w*decreases according to >law of conservation 1

momentum.

% ICO2. AlsoQ.11 Define

write down its practical use.

Rotational Kinetic EnergyThe energy possessed by the body duerotational kinetic energy.Expression for rotational K.E.Suppose the body is spinning or W t a W J w eto • In order to find the total K.composed of small pieces of masses 2

lf mass W has distance Vi* frorn 3X15 0 r0 3

K*E. of mass m, is given by

(K.E,)ng v, = r,ro, we

to its rotation about an axis is called

axis with constant velocity

consider that it Is

1=“ mw,

get rotational K.E

(K.E,)/trt - ” m,(r,c«>)2

BY Putti «1.11*

rrt*Point object :When the orbital radius is larger as cor. p

object

.red to the size of the body, the body may br* conside

?

182

1= “ m#iVK-E. -*= X 'r»ar220‘S-miiarty (CE^ = X m*r,2»

putting in e' Intrusting Informationp £ = Xmv2 + X mv2Hence nxatona < E of the w~>oie body ismgh = (X + X ) 2- Xrrv2

®2)< E -* = CX r -r/o3 - X m2r32<D3

2 + 1 :mg*1 = I 4

*4gb^=T l

mv= X (m/-2 + m/j2 + *m»r^) o2 - Z H

*1L1^Kl«k.= X to2

A ~ cr s the rotatiooa CE. o* the body.Use of rotational ICE.Pract>ca ,• -otato'- a K_E. is used by wheels, wr ch are essential parts of manyeng ies. A fy -»heel itorci energy between the pcn*er strokes of the pistons sothat energy is uniformly distributed over the full revolution of the crank shaft andhence the rotation remains smooth.

K-E«^ « KL.

4gh (b)( 2 )v, Rotatonal ccH' sion - the ciutcn

Speed of a rAs rotational Y .

:."*ng in • q.h)

for disc is ^ EXPLANATION:The two discs with momentof inertia are spinning withinitial angular velocities asshown in (Fig a). Rotationalcollision takes placebetween the discs and bothof them combine as shownin (Fig b). As no torque is

the discs.

oypi = X m fFor Your krfornuDor

P.E = X m / + X mv3

mgh = (X + X) mv3

*h = v2

v = \ ghcsuaton (2) and (3) shows that the velocity of the disc on reaching the bottom

me inclined plane is greater than hoop. —=77 1 0.13 What are artificial satellites? Find the expression for minimum velocityT*V -- f»r**r* &** I __ and period to put a satellite into the orbit?53SSCS3

Q.12 Find the rotational ICE. of the <£sc and hoop. Also derive the relationsfor the velocities of disc and hoop moving down an inclined plane. 2(3 ) acting on

therefore, by law ofof angular

the combinedJPRotational ICE. of a Disk

AS ! consenaiionmomentum,

discs rotate with angularvelocity ©t having totalmoment of inertia (Ij +• It).

(ICE).* = X la1

Therefore,+ fees = (li + h)°r

As "orent of nertia of a d*sc is i = X mr3

(K-E)*,= >4 P4 mr1) to*(K.EU. = X mr* e.1 JESSES i 1WMal Satellite

Ihal orbU around dearth.^^are kept MB theirIfcWUte can be launched from earth's surface to c,;^^jre near the earth have

***»00. aw;tvvclosea satellite of mass ’m’ is moving wit

p-*ntripetal force acting on the satellite is

mv2

F = ”F

Sc

*e know / = ro

Rotationai K- E. of a Hoop:As *e enow

rW cto* to the earth is«fW critical velocity.(K-EU =As ~o^r-e-* of '.ferta of a hoop s

I x mr7 to the earth in a circle of radius R-|1LEW = » I'nT1) ®J

IK.EU rXmr' a1So.

. = rm](Dfck-EW = Xrrr/

te being provided by it weight Hencemv2

mg = —O e., R = 6.4 *10* m )

Rolbngt 20^ -neves downward on an rc' ned place *rom height h. Then it hasMAtht roUtionsi and translabonal mortons.If no energy is lost in friction theru0rT'

Qj tyi ejijc c * JOp on rea n ' ng the bottom of the inclineat the top

[v F ' ^)aodtoca energy*r vst Cr-e ep< ja to the potenoa eoerfc

P.E = ( K£) mmm+\ , "I (D

184

PHYSICS- XI (Subjective)v2

Sch^i 185Thus (2)g = RWhere v is the orbital velocity and R is the radius of earthThus from equation (2 )

v2 = gRv = VgR

t Weight

r3lly weight of an object is measured by a spring balance.

Tht rcadi"S of d" *P * ba,anCe Wh‘n the object is acceleraM up or <lown isHedapP '" WC‘8CC

vent weight is equal and opposite to the force required to stopAPPin the frame of reference.

»*0^ [ ADparent weight of an object in a lift f^ insider the apparent weight of an object of mass m suspended by a string and

balance in a lift, as shown in figure. The tension T in the string can bed with the help of spring balance.

. When the lift is at rest or moving with uniform

APParCnGene

As it fromR= 6.4X106 mg = 9- 8m/sec2,

Putting values,we getfalling

v = V9.8x 6.4xl06v = 7900 m/secv = 7.9 Km/sec

This is the minimum velocitycalled critical velocity.Calculation of time periodThe time period can be calculated as,

circumTerenceof thccirclevelocity of satellite

0-H springmeasureL A MWl,ie

°f <bomilk** A wiellnc vsuh U-JU , CASE 1of *boui Mtna ' of^ l velocitymovts mCirn,i*i«tn 1 ijft js at rest, Newton's second law tells us that the acceleration ofMi. A Mtcllitc Mb i v/i 'On Uic

vdocK) of 3 0 . 0 0 0 j object is zero. So the net force becomes zero. If 'W' is the gravitational force

K j acting on the object and T is the tension in the string,

ihc F nh ai a J Then, T-W = ma

'• 24 ckwc oftnajJform ihc GloW hmmSyticm.

ornecessary to move the satellite into01 orbit. This is

at resta = 0T = w

FI* 5.17(a)

km As a =0T -W = OT = W

or T = mg2/TR

(Ast =~)T=v

Result:Hence the apparent weight of an object is equal to the real weight for observer

inside the lift.CASE :. When the lift is moving upward with acceleration 'a'.' .n the lift is moving upwards with an acceleration a. So the upward force oftension T is greater than downward force of weight W thenThen the net firce acting on the body is

T - W =maT = W +maT = mg + maT = m(g+a)

As 7T = 3.14, R = 6.4 x 106 m, V = 7900 m/secPutting values,we get

2(3.14)(6.4 x 10*)T =7900

T = 5060 sec5060T = = 84min(approx)

60 OR w - r « /rwT * w - m*

Fi»5.17(b)

ResultIf the satellite moves at height ’h' from the surface of earth. Then he gravitational acceleration d#

inversely as the square of distance from center of earth. Thus, higher the satellite, the slower will^ 11

speed and longer it will take to complete one revolution around the earth.NoteOoses: oro:t» ng satellites oroit the Earth at a height of about 400 km.

Result;shows that the apparent weight of object Is increased by

^ln^ actual weight.?**When the lift I* moving downward with accelerationV.

the object is accelerating downward along the weight. So the we g

^ ls greater than the tension T.W - T = ma-T = -w maT = W -maT s mg- maT * m (g - a)

s^°ws that the apparent weight is loss .0vjfH equal to ma.

an amount of

Do you know /

Your apparent weightdiffer* from your trueweight when thevelocity of the elevatorchanges at the startand end of a ndt. notduring the re*t o' thende when that velocityis constant

for #What are real and apparent weight? Find ne apparent weight in different casessuspended by a string and spring balance in an elevator? .

<

Q.14

Or

Real and Apparent WeightRea! WeightJt is the gravitational pull oj the arth on the object

Similarly, the weight of an object on the su face of moon is the gravitational pull

of the moon on the object.

the actual weight by a

J

,Fj-mwmmmmm

— KgJr .m

Mr*%ilon for Orbital Vtloclty~~~

•*****°' m,momg Mlt— M "“ •'•••‘•irSss;'—‘‘""L".' e.— to

Chopin,,j jf1*7T-NCAM IV When a lift Is falling frealyunder gravity;

Now wc* iunsidar that the lift Is fftlllhK und*1 KMV,,Y * *‘WM

I •W MMI mg inft

I mg n>KT * o

Mifl <y

i

froMi. «rtdtr»» *»*•nif«*•Result:

So. thr appeiant wftlght of objec t shown by the %prinn baUnc e iti** ro lht obj* tseems to lit* weightless lhut it It siot ** of weightlessnessI Iplftln the phenomenon of weightier In satellites and grevtty t « ««ttyttem.

0M"Vr

IqvitingeQUfttlon (I) end (y), |

m.vV K

V -QU

oOrWeightlessness In Satellite and Gravity Free System

When .1 satellite Is fulling freely In spate under the «n tlon of fou»- , * attract n *earth# the sun or some distance star then the every thing w <th.n tn» w-bo In state of weightlessness.(4)

Ffcy «1»Will WhtrfCius of the orbit

W Iyt » Radius of the earthH «Height of the orbit above from the equator

?*ui the A \ of satellite is not important in describing the satellite orbit. So Ift*t speed of satellite Is less than the orbital speed then It will not be able to

^tyohi around the earth and fall back to the earth.7 What It "Artificial Gravity"? Derive expression for frequency of spaceship required to provide the

artificial gravity? __

lo show that the earth's satellite Is .» freely falling object. *e cons.dor thebehavior of projectile thrown parallel to horizontal surface of the earthabsence of air friction. in the

If th« projectile Is thrown successively at larger speeds, then dur.ng < ts free fall tothe earth, the curvature of the path decreases with Increasing horizontal speedsIf the projectile Is thrown fast enough parallel to the earth. The curvature of thepath will match the curvature of the earth as shown In figure (I)In this case projectile will stert orbiting around the earth. The spaceship .*accelerating towards the center of the earth at all times because it orbits aroundthe earth. Its radial acceleration is simply g‘the free fall acceleration. In fact thespace-ship Is falling towards the center of eerth at all times, but the curvature ofthe earth prevents the spaceship from hitting the surface of earth. As spaceshipIs like a free fall object so all the objects inside It appears to be weightless, thusNo force is required to hold an object falling In the frame of reference of the

» space satellites. Such a system Is called gravityfree systemQ.16 What Is orbital velocity and derive an expression for orbital

Artificial Gravity . ,^Artificial gravity b the gravity like effect pro

the orblt for a longerUpliWlon:

ltu....,pict. If the *P»c«hlP vhl To overcom.this

1« no fore* on tht space craft In1*r'v for MtroniuU Pr*sent ln 1 * normi||y by astronauts.,Jm*. this weightlessness craates * lot o Pr° Wp t0 perform the *xp*[ ,oor 0|,p»c«shlp m much the"‘"'cully,martificial grevity 1» createdIn the P

jfldMirU,force on thiltroniut then Is pressed towards the ouJJ"'1W|Y as on the earth,tension for f

biting satellite by spinning it around Us own axis.an or

3T:" lb own central axis with anguler speed to .le,,tlon a, is

Orbital VelocityIn IMU. *< *Orbital velocity Is the tangential velocity to ju/ sateUlu • orbit around the100 km ebove

ipc^earth.island with •OR i

A2 000 kmhMcCandlas* tWV00The velocity of satellite with which It revolves round # earth is called orbital

•pact from i n

buttle and becam* O*Th^r'th end some other planets revolve round the sun In nearly circular orbits.

of motion If called orbltel totlon. Artificial satellites also revolve first human»hc l-'artA.This type

around the sun.

<r* *:

188\ PHYSICS- XI (Subjective)

189a* = Ro>2

5ut o = — , where T $ the period of revolution of spaceship.

=R! T

OrS For Your Information!t A geostationary satellite orbits

the Earth once per day overthe equator so it appears to bestationary. It is used now forinternational communications

_ 2TH (2)So r*c Or

T -p is period of revolution of satellite that is equal ' • So^satellite also complete one rotation in exac

Equating equations (i) and_(2), we get4*'7' surface

RI7 1 GM2nr3P4IJ X [As - = f I Satei'-rtetr>*

T rEXPLANATION;To Create artificialon the

both sides, we get

4xV GM= 4 x: \ Rf: Squaring,ac

space ship. s.space ship is routed **about its own axis Th,

centripetal force is ^on the astronaut. He tven3 force of

a. if: =Or aL r4rR_

f = — ^2x\ RAs *'orcc of gravity oro . oes the necessary centripetal acceleration So,

S c = g

GMT2 Eartnr 4r

GMT*naans(centrifugal force) oa isouter rim of the space #:and presses its floor iotasame manner os wtdftaEarth.

Do You know?r =Or 4K21 GHz = 105 Hz

This equation gives the orbital radius of the geostationary satellite.

Substituting the values,

G = 6.67 x 10'" Nm2/kg2# M = 6 x 102,1 Kg

T = 1 day = 24 hours = 24 x 60 x 60 s = 86400 s

6.67 x 10"" x 6 x l0:i x (86400)* ] ’

12x \ R

Hence f = —

/. er trie scaces^ o rotates w th this frequency, then artificial gravity heips the astro nauts perform theeasily. .

What are geo-stationary orbits and geo-stationary satellites? Find the orbital radius of >

stationary satellites?SoQ.18 r = 24 x (3.14)

s= 0.423 x 10 m= 4.23 x 107 m

£‘0rh1si£i*e °f0r r0b^°n°f 5alellUe* equal 10 thepeTi0d 0fr0,ati°n fear,h^ t r7 7 1! h e^h a!7U a v e s d s t a T d7e c t f y 3bove surface of earths he.ght

^equator comes out to be 36000 km. _The satellite which completes its one revolution around earth in 24 hours is called geo-stationary sate 0^9 Write a short note on communication satellites

is IS^ \ %ORA satellite whose orbital motion is synchronized with the rotation of the earth about it own axis

geostationary satellite. H-" Allllfl’'his type of s a t e l te s the one whose orbital motion becomes equal to the period of rotation OT e

cs axis.

9M'PulCOCMP

, 5oi« Communication Satellitel Acom Tht who* Earth can M covt^a

by just thret gao-atatoruuy

iataiiitM

-S-SS5-of;r;^beam, in straight line and can pass eas.ly through

**l,s Provlde the ener«y t0 ampHfV 8ndhr*^MIS"'viaVawllite from*arth stations transmit and receive the

Countries.

i

sate te re^ a ns a ways over the same point on the equator a: :ne ei th spins

Applications mCommoncauom sa'.fl" 1 *®*INTELSAT VI

^ sed in communication system, weather obser . ation ano other military uses.S-cn satellites a^eExpression of orbital radius of geo-sta onar

satellite:I in

ed necessary for circular oroit is g *n byAstheoroita

GM (DV = v rspeeo must be equal to »verses speed of stellite in one day. So,

Sut This

rH

Ch«Pt»f s tI K . PHYSICS- XI (Subjective) 191

Sate’ tt Offaniiat > on (IWTOSAT)

T*e argest sate&t* system., managed by t:6 country is ca ed NTELSAT V

news ntornabona: tebscomrr **n ca? on sate *e O'g*^ ;atior

T worts at t*e m<ro>%*ve frequences o* a. 4. ti and 14 GNJ and :*“ <t capacity

cy 30 000 two *ay te epN>ne c»rcw t p:u$ t*rf T V c* ar<- < s

C *c WXat a t t tSe Newton's and finstem v ews ab<KJt gray- tat Ion>

ORMU

0 = *Relation between S, r and 9r

Average angular velocity

instantaneous angular velocity Lr?.;tNewton s Theory about Gravitation

u x t oM-*0

/4afACCOr^ng to NrWIO' k >t *vn porrade <v Acoa «irtrcf/>rft / Average angular acceleration a,v =

*> AtIf

<y r4r duiwr Nr-r-*<**in WCVTIS

£ rtste n $ theory about Grav tat onAocQrdacig 10 i ftstmitr*eor> Grvv.f> a

AIDinstantar-ccus ^naflftarJoArjbrn V>«nu a =.u-*o —

Atacceleration

v - rcaRotation betweenNr. r and w v = cox r

”0 observe tNs *y mxrt !-ve s ace is a f ^ n 'atfcec sited * a *#**> we g^t *-.-g : t cx-r\e> as sKow^ n 4|yf- £ -ste - theory *e 00^ t spaa* o* ‘ore* o* gravgy act -* 0^ boo es but»* say that tcc -es and g*: -avs ?^ov« a <rg geocevo .iem to tf>estra ght "ei - pi^go<m«cr>l * curved space t ->t

&**tft*K*s between s tws

a , - ra a , = ax r

1 . 2a9= coj -co420 cj,t +-at

vJ»«‘7Centripetal acceleration1mvHtwtO*a F,“ "TCtntr.peta force: *•0- grtvty W g>t no fir ntor o' w*y

:%J 84 *° bWd'ltheory tt1 y$ t*( ease- of teaceet *t 9tro*g gra> ta* o-a *e si|> but ? *w S •» a -s me reason *orK0> "g t4# *« e'te W3.a‘t *»

So tf«K» sfw> 0m.1 matIr»* ty a^ ecceeraton art n•*•f"'- lg-t by a doAruto a^ ownt met cou

j*»* xtwton 1 meo'v based on 00a of lg*: is •imtf~ of.MP*W t*Ut «*•6«- wOs d M MfKtM >»r .tv'

»-• *• t»wn» tN«> tf« 5iW V sr« t twvc.M f M i j t iictyfi “|to Hawtpn 1

M-d ng of fterbgnt by t*a1 :-a of strart sta-#tCMiiby tr>a grMiy fll-aas^ ad durt^ a soar tc ose - ^9*9 rx,~ ,t i o s :o-ate- £ *I:I- 1pradete- 'atfr t*an hm+mn

^tYc't £ -tta^itr*o-y «ia congee « uicav

1F,•mnoCtntripatal forca1

Moment of Inertie ofaTorque on a pe/tlderoteting t•mr'a

I i1••— ml-Moment of Inartii of a tWn rod

'4 1part;as a sol

V" Moment of Inertia of a imrJI-v^fn 2Moment of Inertia of a d »c

I-Moment of Inertia of a spheres •la

law of motion forL•mvTiinGL» rpiinOnTuiniNc mioimtiov

«/ ty Wi ugr aeneciee n a•0 *>• 0« »4 «ty tiM we see tne star ^ tf»e

O « ' ?4S MKC Jo» tg«i ••tjr«e.'vrg tne

er >

192

, PHYSICS- XI (Subjective)l 193Rotational K.E.21

Multiple Choice Questionsl1 2 2— mr coK£- =4Rotational K.E. of disc22

lFnur possible answers to each statement are given below. Tick (S) tl,c correct answer:K.Ew =^mrVRotational K.E. of hoopv/ 23

iVelocity of hoop falling froman inclined plane of height hVelocity of disc falling from aninclined plane of height h

Velocity of sphere falling froman inclined plane of height h

Right hand rule for rotating bodies Is used to find direction of

(a) Angular velocity

( C) Angular accelerationThe mud flies off the tyre of a fast moving car In the direction of:

(a) Parallel to moving tyre (b) Antlparall to tyre

(c) Tangent to moving ty r e^^

(d^Jon ^hes^

3 The correct SI unit of angular momentum is

(a ) kgsnr*(c) kgm*sj1

4 A 30kg fly wheel Is moving with uniform angular acceleration. Ifl

moment of[?. 120 - - v

(c) 2kgmJ

5. Moment of inertia of 100kg sphere and having radius 5cm is:

(a ) o.ikgm2

(c) sookgm2

6- The hoop and disc have same mass(a ) K -Ehoop = K.E|W K 'Eh„op = iK - E

Critical velocity of an artificial satellite is.(a) 7-9miles s~1

(c) 7.9kmlr1

In angular motion, Newton's 2nd law of motion is.

(a) F = ma

(c) T = la9' A man of weight W is standing In an elevator w c

aPparent weight Is(a) mgW mg + ma

10‘ W a body of mass 10kg Is allowed to fall(a) Zero(c) 9.8N

1.(b) Torque(d) All

^ 26

\/ 27 V-Mv = v/gRCritical orbital velocity (b ) kgms-1

(d) kgmJs-JR

Time period of close orbitingsatellite

2TTR28 If radius of flywheel is 2m, then ItsT =\

Apparent weight of an objectat rest or moving up withuniform velocity

(b) 30kgmJ

(d) 6okgm229 '= mg

Apparent weight of an objectmoving up with uniformacceleration a

(b) 5kgm2

(d) 2.5kgm 2

and radius their rotational K.E. are related by equation:

(b) K.Eh00P = 2K.Edlsc

30 T = mg +ma

Apparent weight of an objectmoving down with uniformacceleration a

31 T = mg-madisc

(d) NoneApparent weight of an objectfalling freely32 disc

</ 33 -1(b) 7- 9»<ms(d) 790ms

Orbital velocity of a satellite- 1

Spinning frequency of asatellite about Its own axis34

AP(b> F = At

(d) All of above

h Is ascending with uniform acceleration a than It.(b) mg- ma

(d) ma - mg

, Its apparent weight becomes.(bl 89N(d) 10N

Orbital radius of a geo-stationary satellite35

pUVSjCS M (Subjective)194195

If external torque is itro. then wtuch of these quantity is constant(a) Angular rhomentum(c) Linear momentum

1The acceleration due to gravity on moon is ^ th of that on earth, what wth be ttijon moon. If its mass on earth is m?

m<*' 7

Short Questionsof Exerciseti.

(b) Force(d) None of these

the difference between tangential rf.heel of known radius, how will you f,nd t

angular velocity, if one of these is given0>i W*‘n° for * *U.

( Lhr 2010-2011)

*is OiffWK*between tangential^• Angular velocity w is djff?latuuavi the circl * "

;ular 1(b) 6m

g«is o/ rvtotloq Jhile tangential velocity v is directed along

m• unit o< w is rad/iwtiiIc) m w, 7

The value of angular momentum of a body 1« ma» ,mum if 0 i,ecjuel to(») 0*(«) 90”

termination13by using the following equation.

petal force and why It must be furnished to an object if the object is

(Rwp 2006,Grw 2006,Lhr 2009)

JfAds the normaUy straight path into circular path is called centripetal force.Or

corr.pe \ a t>c-j * to ^o.t* along a circular path is called centripetal force.

mv:F<

(b) 45*(d) 1S0*

The diver spins faster when moment of Inertia becomes(a) Smaller(c) Remains the same

a circular(b) greaterId) None of these

When a body moves In a circle, then its linear and angular ve*octy are(a ) Parallel

15

(b) Perpen4«cv a#

(d) None(c) AntiparallelTorque per unit moment of Inertia Is equivalent to(a) Angular velocity(c) inertiaThe moment of linear momentum is called(a) Torque(c) ImpulseThe ratio of angular velocities of the hour hand and minute hand ' a watch

(b) 7^:(d)

If the earth shrinks to half of Its radius without change In » ss, the duration of

V.4thematicailY.16.1

(b) Angular acceleration(d) Radius of gyration

Significance: of the circular path.and directed towards the centerit is perpendicular to the tangential velocity

Without centripetal force body will move oiong the tmtUl-17.(b) Couple

(d) Angular rpomentum <Ul What is muniby moment of InortU? E*PlainK* »l*n,fici,nccLhr 2010- 2011)

and the square of its(Dwp 2006-2008, Mtn 2005

massthe product of it*18.

*"» Th,moment of m«rtia of * 0*1*1« is defined as

perpendicular distance from axis of rotationfi>f* (for a point mass)

(a) i :i(c ) 43200 : t

IMathematically.Physical Significance:Tht moment of inertia plays samemotion.

hich is played by mass during linear

is the measure of rotational inertia of a

bod* iIZ 7while mass determine the linear

Moment of inertia determine the flfflWtff «i»

•ccalerttion.$4 *>>« Is meant byMimomwtum?MMMW«» cons.rv.tion of snguler mom.ntum?

T 1 (Mir Pur 2004, Bwp 2006-200*.Mtn 200$)

locity of a rotating body is called angular momentum*

19.motion w

rotf during angular(a) 6hrs(c) 24hrs

(b) Uhrs(d) Shrs nt of inertia

nr acceiersiis.it!

The ratio of angular frequency am linear fr « ncy is(a ) 2n

20.

> ) *

d angular v*&roduct of moment of inirt’a an

** d e n o t e d by L .Mathamatfca'hr\l. d 2. c 4. d3. cn. * 12. c 13. c 14. k

r

, PHYSICS- XI (Subjective)197

L = r x pthe direction of the following vectors in simple situations; angular momentum and angularStateL = rp sin 0 velocity-

L = mvr sin 0Law of conservation of angular momentumThe total angular momentum of the system remains constant, when no external torque actsMathematically,

Uotai = Li La L3 +

Q.5 - 5 Show that orbital angular momentum Lo(Federal 2005,Mtn 2006,Fsd 2008, Grw 2008,Mir Pur 2009,Grw 2010)

Let us consider particle of mass m moving in a circle of radius r as show infigure

(Grw 2009,Grw 2010)

AflS' Direction of angular momentum and angular ve|or't

The direction of angular velocity and angular m

determined by right hand rule.8 "omentum is along the axis of

Right hand ruleGrasp the axis of rotation in your right hand u W Merect thumb will represent the direction ofangXtm **** of then the

Q.S.8 Explain why an object, orbiting the earth I, u ,

out why objects appearunder ^ explanation to point

, which can be

°n it= leu = constant

= mvr

Ans. Proof: * (Fsd 200s)

P Explanation:When the cuject is thrown horizontally fast enough from a certain height, so that the curvature of its

path will n atch with the curvature of the Earth then the object simply revolve round the Earth. Now,

the motion of the ob act is under the constant acceleration due to gravity (equal to centripetal

acce ’eration). I-e ice we can say the orbiting body is freely falling body.Weightlessness of the body:A freely falling body moves only under the action of gravitational force so that the object is said to be

in state of weightlessness.

Ans.As Lf = r x p

Lo = rpsin 0(where 0 is the angle between r and p)

So Lo = r(mv)sin8Since p = mv

-4or

Lo = mvrsinBAs the angle between r and v is 90°

l_o = mvr singo0

= mvr (1)Lo = mvr

' Q-5 9 When mud files off the tyre of a moving bicycle, in what direction does it fly? Explain it

(Lhr 2005,Rwp 2005,Mtn 2005,Mir Pur 2006-2009,Bwp 2008, Fsd 2008,Grw20ii)

Hence| Ans. The mud flies off along the tangent to the tyre.

Reason:When speed of bicycle increases then adhesive force (sticking force) between the mud and the tyre is

not sufficient to provide the necessary centripetal force so the mud leaves the tyre and moves along

tangent to tyre.A disc and a hoop start moving down from the top of an inclined plane at the same time. Which

W'H be moving faster on reaching the bottom?

(hence proved)

Q-5.6 Describe what should be the minimum velocity, for a satellite,to orbit close to the earth around it

(D.G.Khan 2005 Mtn 2009, Lhr 2010-2011)one

Critical velocity:The minimum velocity needed to orbit a satellite close to the earth is called critical velocity.Calculation:

Ans. ( Federal 2005-2006, lhr 2006, lhr 2009)

Ans, O/sc will be moving faster on reaching the groundProof:Speed of hoop at the bottom of inclined plane is,

Vhoop = >[s^sPeed of the disc moving down the inclined p a"e

VdlK =

Consider a satellite of mass m is moving with velocity v in a circle of radios (i.e radius of the &Since gravitational force provides the

mv:So mg =

necessary centripetal force

ror vs «= gR

or v = y/gRR (rad' J5 of the Ean»r s ifrh3

V 9-8 x 6 4 K io‘v “ 7900 m/set

or v •7.9 kmr1

v «=or

3

VdiK « 1.15Vh00P

VdiK > Wool*H*nce

%f *uT '

'''"•crM' ,,198

^^yfllCR * 1 (TiubjwD**)

Phytic *l rNton

Slnct the moment of Intrtia of disc I* inwHor !* KnM * t,r moment of iiunua < >f ult? (

vrlot » tv of disc IN greater (h.m hoopQ.S.11 Why does J diver change hit body positions before and after diving In the p0o|}

(fedenlioosAns. The diver changes his body position to nuke fxtffl iom^iQuIti

Solved ExercisesVr

ikJJ,i

An fkttrit foil routing « j w k ,

dtecIrraOunlnb* uniform,fMtit vcki*v*'3! ***** ** '«* «* . tA , i„,„ ,,

|)>U; — “ ^ 1*’" l” 1*1^* •*** Ww*UIM^^ng*

Initial annular velocity - <u. - 3.0 fey g

Final angular velocity - at,- 018.0 sec. <

7-^vExplanationWhen a diver lifts off from the diving board, his legs and arms are full extendedmoment of Inertia is large (lt) but angular velocity ( w,) Is small.When the legs and arms of the diver are drawn into the closed tuck position, Its momentreduced considerably so that Its angular velocity will Increase to conserve the angular mom® ' ^

llW,* l2U> 2

(,»vrM

thlic,l ime - 1 -

To Find: 9»(i) Angular accele(ii) Number of rcvolttl^F'Angular displacement

ron*e «?

Q.5.12 A student holds two dumb-bells with stretched arms while sitting on a turn table. He is glveniJ Ci|cu|»tion:until he Is rotating at certain angular velocity. The student then pulls the dumb-bells towiidJ * C

^ Aschest, what will be the effect on rate of rotation?ar

(VO),a = — l

Putting values, we get0 - 3

a- — -1 8

3a =-18

|a = - 0.167 rev(ii) As angular displacement is

9 = cot t + - a t2* n

Ans. Rate of rotation increases when student pulls the dumb-bells towards his chest.Reason:According to law of conservation of angular momentum.

lu) = constant .When student holds two dumb-bells by stretching his arms, then value of angular inertiaiis j"So the angular velocity is decreased. When the student pulls the.dumb-bells towards L

angular inertia decreases. So the angular velocity is increased. But during this whole Proceangular momentum remains constant.

2Putting values, we get

1(-0.167) x (18)2

20 = 54 + (-0.083) (324)0 = 54- 26.40 = 27 rev (Approx)

0 = 3 x 18 +

ss the

round a cornerquired for global A 1000 kg car is turning

radius of the circular path is 10 m, how large

to hold the car in the circular path?Q.S-13 Explain how many minimum number of geo-stationary satellites are ri

T.V transmission. mi."**(Fid 200s,D.G.Khan 2005-2006,Mir Pur 2009,Mtn 2009)

Minimum three correctly positioned geo stationary satellites are requiredfor the global coverage of T.V transmission.Explanation:As each satellite In geostationary orbit covers jUJinaWt so thewhole populated Earths surface can be covered by three correctlypositioned geo-stationary satellites

correctly

Giv«n Data:Mass of car = m = 1000 kg

Velocity of car= v * 10 ms"1

Radius of circular path = r = 10 m

Ans.

PlClftC oc«fl

Find:Force required = F ?

in circuit path will be centripetal fow,^•aleulatlon:byMtillRH* The force required to keep the car in c

-s\,w/ if - ' £ -/I.>V

Iif *pHVSlCS M ( Subjective )

[ r#K*U^n:1 ( jilfll

201

SvV f* - the lorwularU * mvj-

S- orbiUl speed md vw - - (as s 2nr)

Fusing V A.--CX we *ct ( 1 )!0W ^lOpF, here v,

When S 2*r •distance travelled in one year10

1000 % too l ime period T.F« • 2>rr10Vo = —thusFg - 1POOPS

r, * i » i d iia values, we getPutting

0 n2 * 3.14 *V o *3.16 *A ball ried to the end of •tiring. ts swung in a % trtictl circle of radius r under the*j ihowa n Mg. 5.7. What will be the tenaion in the string when the ball Upath and its speed is v at this point?

9.42 .4•dlon ofgn^« the point Astiv« * ;

: -ImsVo *Given Data;putting that inAs the boll is moving in a circle, thus the force acung on the ballthe required centripetal force.

At point A two forces arc acting on the ball.(i) Tension in the string Tfii) Weight of the ball W’

4^2 981 * 10% 1.5 * 10"* 1.5 * 10*

must provide

10”r* lo^kWQT

To Find: ~n*.

A dhc « iihout flipping rolb do* n a hill of height 10.0 m. If the disc starts from rest at the top of thehill , whit iv its speed at the bottom 11

Tension in the string - T - ?Calculation:

These forces (Le. T and W) act along the radius at Point A, so their vectorsum must furnish the required centripetal force.

T + W -Gnen 1' ita:

Hcife, of hill - h = 10 m2 T« Findmv

'peed of disc at bottom " V s ?Calculation:

Using the formula

r2

T - _ wor But W* mgr2 -i?.r mvI « — mg

MiPutting values, we get

vT - m 8r

' i4 *9.8 * 10If

^g, then tension T will be zero and the centripetal force is just equal to the weight. 3

v - >/130.67F- 11.4 msj

The mass of Earth is 6.00 * I 0U kg. The distance r from Earth to the Sun is 1.50 * 10" m . A»**Irom the direction of the (North Star, the Ear » revolyes counter-clockwise aroundDetermine the orbital angular momentum of the Earth about the Sun, assuming that It travelcircular orbit about the Sun once a year (3.16 > 10%)

Given Data:Mass of earth * m s 6.00 xl 024Kg

r * 1.50 x l 0Mm1 year 85 3 16 x l o' s

•— •»-*—"•P«riod of one revolution in dars° Take mar< of the tartn rvi -y*R “ ^00km.CLl''" Djjtt:

^c'ght of satellite from earth* h •384,000 km‘k*e a r t h - M - 6.0 * 1024 kg^lUs of earth - R -6400 km

DistanceTime - T -

To Find:Orbital angular momentum ft trth •Lo “

1

is"*•& *- :vr -Ju 4

*'V . AA 1 »

202

's PHYSICS- XI (Subjective)To Find: 203Period of one revolution in days = T = ?

Calculation:Total distance of satellite from earth center = r = R + h

r 6.67 x l0“n x 6 x 1024r =

4

cn simplification, we get

[JT23 X 107m|r = 6400 + 384000 = 390400 kmr = 3.904 x 10*m

por orbital speed we know thatAs v =

hPutting values, we get

66.67 xl0“n x 6.0 x 1024v =83.904 x 10 Earth

v = V10.25 x 10 s

v = V102.5 x 104v = 10.1 x 102 ms"1

2;rr

864 x l O 2Orv = 0.03074 x ]

Or v - 3.074 * 10'Now T = Or v = 3.074 kmsv -1Or v = 3.1 kmsPutting value, we get

j m 2 x 3.14 x 3.904 x !Qg10.1 x 102 Exercise Problems

j m 2 4 . 5 1 x 106 2.427 x 10ft seconds10 .1 5.1 A tiny laser beam is directed from the Earth to the Moon.If the beam is to have a diameter of 2.50 mat the Moon, how small must divergence angle be for the beam? The distance of Moon from the Earthli 3.8 x 10Mm.

Given Data:

T.2.427 K 10660 x 60 x 24

days

imirirrDiameter of beam = length of n.-c = 3 = 2.50 mDistance of moon from the earth s rs 3.8 x 10 m

Divergence angle e 0=7

As S«rO

Or 0e-rPutting values,we get

0--^3.8x 10’

To flndiKiullo nml TV NIKIIUIM bounce from M »ynchronou» aalclllte. This untcllllc circle, t h e ! irth once In«hour*. No If (he aatelllte circle* enMwnrd above the equator, It stay* over (be name »pnt onbecauie (he Earth I* rotating «1 the *ame rale, (a) What I* the orbital radio- for a «ynchre»*'aatelllte? (b) What I* It* tpced?

Given Data:

Calculmlonai

Mass of earth - M•6 x 1024kgTime period - T - 24 hours - 24 x 60 x 60 sGravitational constant - G - 6.67 x 10"nNm2 kg“2

To Find:(a) Orbital radius of satellite - r - 7(b) Speed of satellite- v - ?

Calculation:

0-6.6*10 °rnd

5,2 * Kmmophonc record turntable accelerate,from ret to aa angular velocity of 4S.0 rcv/mln In

G|Vcn What I* It* average angular acceleration?fOMT1^A* r -( ) [

filial angular velocity - a>i * 0Putting values, wet got

'v4 I204 •s PHYSICS - XI (Subject ive )Chapters [Q -Sj.ru, Si-hjili

T« r,otl:205

45*27t 45x 2 x3.14Final angular velocity = cor =45.0 rev/ min = - 4.71rad /sec (a ) Torque acting on cylinder = r ~ ' >( h ) Angular acceleration of cylinder ~ a ~"U

6060Time = t = 1.60sec F

To find: latiuns:OiM1t= rFsinO

As r and F are perpendicular to each other So0 =r = (( ).2( )( ( ).6( ) )s intXr

Average angular acceleration = a =? AsICalculations:

- 0)rAs a=t t = ( ). 12Nm

putting values, we get Now for CX4.7-0

T = I ( XAsa = 1.6inder. moment of inertia = I =-mr:T = — mr ua = 2.95rad / sec2 s

A body of moment of inertia I = 0.80 kg m2 about a fixed axis, rotates with a constant anvelocity of 100 rad/see. Calculate its angular momentum L and the torque to sustain this mofGiven Data: 0001

Moment of inertia = / = 0.80kgm2

Angular velocity = co =100 rad/ sec

5.3

Putting seines, we gelW

:

2 x0.125x0.20x0.20

ju = 1.2 rad/seelTo find:

Angular momentum = L =?Torque = r = ? Calculate the angular momentum of a star of mass 2.0 x 10 "Cg and radius 7.Ox 10 km . II i t5.5

makes one complete rotat ion about i ts axis once in 20 days, what is i ts kinet ic energy ?Calculations: ( liven Data:

As L = Icoputting values, we get

L = 0.80x100L = 80 kgm 2 /secL = 80 Jsl

Now torque = T = I aAs angular velocity is constant, So a = 0Thus

Mass of star = m = 2 x 10 '" kg

Radius of star = r = 7.Ox 10' km = 7.Ox 10 in

Time for one rotat ion = T=2() days = 20 x 24 x 60 x 60 = 1728x 10 see1 » lind:

Angular momentum =L= ?Kinetic energy = K.E =?

Simulations:As L = Id)

For sphere ( s tar ) = I = — mr" and o)=—•Thus

r = (0.80)(0) 2 K•*>

IV = ol T5

5.4: Consider the rotating cylinder shown in fig. 5.26Suppose that m = 5.0kg,F = 0.60 N and r = 0.20 m.Calculate (a) the torque acting on the cylinder, (b) the angular acceleration of the cylin cr*

(Moment of inertia of cylinder = — mrGiven Data: „

•> 2 71I - = — m i x —5 TPulling v alues, we gel

2 x 3.14— x 2 x 1 ( ),,‘x t 7 x 10' )' x 1728 x 10 '

543U0.1424x10

F= 1.4x I ( )J:JSForce acting on cylinder « F 'ONRadius = r = 0.20 i

Ws.w£!l2£tor 5 ...’06

p^ySlCS XI (Subjcc( ive)^ Set^ 207l Kin .isv 2\ . • \ \

g-Ihusk I r

v2 =rgOr

v/rgK i =i h s - mi: si i ; vallies, we eel

v=r Orputting values, we get

v = Vl 000x9.82 x 114 r172Sx U )

;v - O ' 10 'k I I 0‘ »M ? - Iv = 99ms; 5'I'lie Moon orbits the Earth so that the same side always faces the Earth . Determine the ratio of itsspiii angular momentum (about its own axis) and its orbital angular momentum.( In this ease, treat

the Moon as a particle orbiting the Earth). Distance between the Earth and the Moon

is 3.85 x 10K/?/ .Radius of the Moon is 1.74 xJ 0^ /?2

\ I-

Distance between Earth and Moon = r0 = 3.85 x IOxwRadius of moon = rs = l .74 x 1 O^ m

k I =:5.4SxlO .1

K . l -. =:.5xl (Pj]5.H

A 1000 kg car travelling with a speed of 144 km h 1 rounds a curve of radius 100nccessan centripetal force.

.''.ti (liven Data:m. Kind the( .in n I lata:

Mass o|ear = m = 1000 kg

Speed ol ear = \ = 144 kmh | To find:i 144 x 1000 L,=40in see

Ratio of spin and orbital angular momentum = -?3000hRadius ol *. in veil path = i = 100 m

1 u I m 11 : Calculat ion:The spin angular momentum of the Moon about its own axis is

Ls =Isweeii inpet.il loiee = \\ = ’

( ah illat ions:i— 2

As for Moon (sphere) Is = — tTirtm — mrs2to ( 1 )\s

The orbital angular momentum is given by

Lc,=I„a)

Lo=mro2“Diving equation ( 1 ) by equation ( 2). we get

L _ ^ mr>mr>

Dulling \ al l ies, we gel

1000( 4t ) i

As for Moon (point mass) I„- mr*

(2)100I = I 0000\

1 = 1 0, MEN\\= l .O .s IQS- 1

^ lv.it is the least speed at which an aero plant;can execute a vertical loop < > l 1.0 km rudiu'

then, will he no tendenev lor the pilot to fall down at the highest point?

R uhris of loop = 1 - 1 0 kill = loot ) MlUeelciaiion due u» gravilx =*j = ,x m/see *

Angular speed to is same for bothPLo>.7

Lo 5 r02putting values, we get_k = 2 ( E74 x_l 0V

‘ k 5 (3.85 x 10* ):

L. _ 6.05 xIQ12

7.37 x I 017

( l i v e n Data:

t n I iiul :Speed ol aeroplane = \

(. .denial ions;

\\s

k1 r1- ~ 8.2* i (TkSN .ei aeioplane exeeule i eneul ii loop : ! n eeni pela I loiee is supplied h\ gravity.

Sasj^jci, | m,|, niliiU's on ils.vis one,a «l«> - Suppose,by son,,prot-ov. ,b,Kllrll,1 . . .. i w .1 iirtsi'iil Mow Iasi ^ill H lie rotating then?radius is onl\ hull as large as ul prtsini.I < M sphuc l =2/5MR )

200

PHYSICS XI (Subjective) 209

Radius oTearth r - 6400 km|leight of circular orbit h 900 km

( iivfn Diilii:fnno*1'-I mu- pci'IIHI 11 = 2-i hums

Moment nl iivm.i « » 1 .splicic -I - ~ MRs( )n!_* in.il itidiiis R , R

K.KIILIS «illei conn.icimn = R =\R

Orbital speed = v = ?->)’\1R tut ions:s C*RU GM

As v=r (

of c irthWhere r is the distance fror = R + hr = 6400#$

r = mOkni

centre

2 Mi RMnmvni nl mciti.i altci cnniiaClimi = I - _ MK )00)

11cneeIn lind:

/* = 7300x 10’/;/.||§]*

Pulling values in cqu. (I), we getjviiud nl ml.iimn alici contraction =

( .ih nl.i l inns:6.67XI0 W102J

Vcmding in law nl cnnscr \ aimn nl angulai momentumV 7300M0’• "'i -i 8^ 106(•)

( 1 ), ' 1 4 * 10?m/sccto.v-7.4km/sec2 71 2 K

and m.=\ s ( , ) — —T

Pulling values, m ci|u.t I ) wc gelIn

M« >s2 n ' MRs

T.4I4 :

I =24O!

I =6 hoursI lencc earth would eomplcic us one rmaiinn in 0 hmns

9 ~.>.10 \ \ hat should he the orbiting speed to launch a satellite in a circular orbit 900km iilxn,tli,Mirim,»1 Hi,larlli.* ( Ink,mass nl' Hi,Imill as 6.(1 X lO^K

/ T-and its radius is 6400 km ). *'sC vi %i’ll Data:\

Mass nl eaith = M = (u( 1 - 1 ( 1

- XI (Subjective)PHVSICS 211pl/s

Chapter 6

FLUID DYNAMICS

Leanil

Understand that viscous forces in a fluid cause a retarding force on an object moving through it.

Use Stoke's law to derive ar express-on for terminal velocity of a spherical body falling through a

viscous fluid under laminar conditions.

Understand the term steady flow, incompressible flow, non viscous flow as applied to the motion of

an ideal fluid.

Appreciate that at a sufficiently high velocity, the flow of viscous fluid undergoes a transition from

laminar to turbulence conditions.predate the equation of continuity Av = Constant for the flow of an ideal and incompressible fluid.

ppreciate that the equation of continuity is a form f principle of conservation of mass.

J' ^erstand ihat the pressure difference can arise from different rates of flow of a fluid (Bernowi.Effect).Duive Bernoulli's equation in form P + 'A pv2 + pgh = constant

Explain how Bernoulli effect is applied in the filter pump, atomizers, in the flow of air over an aerofo

Venturimeter and in blood physics.

^ve qualitative explanations for the swing of a spinning ball.

wmrnm' K * i V >. ,

pHYSICS - \l (Subjective)r ' s

213Chapter No. 6fWid^ suMW«>«that can HQW from one place to another is called a fluid.

andggfiare classified as fluids.

fluid dynamics

=«—~—u

lX.

is called fluid dynamics.JZZ The Dra°cl

Conservation law in fluid dynamics

,w study of dynamic fluid is relatively complicated, but analysis can be simplified by makinggumptions. For this purpose we also use the following conservation laws.. Law of conservation of mgs% which gives basis of the equation of continuity

• Law of conservation of f nerqy fives the basis of Bernoulli's theorem.

t*.r

a fewG» .O-KM — * ]£

coWhat is meant by the term viscosity ? Explain, and fluid friction or drag force and state Stock's Law?o Q »VE 11TJ0

I ~ © •*> O32C z. 3 Viscosity (n )

Viscosity:of iry of

<© 3£ cr *o *3

>. 9o O’lbi s >5. O <-LW <

5ftV ffect between different layers of a flowing fluid is describediViscosity measures, how much force is required to slide oneliquid over another layer. It is denoted by q. The SI unit of viscosity is Ns/m2.

0. .*5- ,n terms,ayerofthe

< - c he*3 C5 >C E1 ~ *5 *H Samples

• Substar ike honey and thick tor have large coefficient of viscosity. So they can not flow easily.Substance like water and milk has small coefficient of viscosity q. So they can flow easily.

c h*a. 2:—QW --ru 3*TD

I TZ *a Unitc o• The SI unit of coefficient of viscosity is karri 's ' or Nm *s• Its dimension is [ML Y1).

Viscosity of liquids and gases• Liquids and gases have non-zero viscosity.1 Viscosity of gases increases with increase in temperature while for solids and liquids it deceases.

u.O -o

3 *o oE£rt

JBEE- E>> u -c:.-ara sc toon Q.2c. jr What is meant by drag force?r 1©v O'> (N <15»s *>EZJ

Dra8 ForceAn object moving through afluid experiences a retardingforcc called dragforce.

ORtion of a body moving through the fluidr rilled rirag- forco or fluid-faction.

h*

u -i t YgTgF ch~7f3TmThFor examplepWhe" we stick out our hand out of the yyindow of a fast moving car,we feel a force opposite to our motion.

ac^ors on which depends uponCord|ng to Stoke's law,drag force is given by,

a reo c* c mu

* 1?V } Ucu cu.

^ •reQ

F = 6 n q r v

r5/7..r* '

^•A1

214

\s PHYSICS - XI (Subjective)215So drag force is a velocity dependant force. This equation shows that drag force

depends upon 4m =i) speed of sphere (v)

size of sphere (r )viscosity of medium (q)

3ii) orin-°n 0r;:> equation (1) becomes.

Henceiii) (Vr < )v. =Q- 3 State Stoke's Law. What are its limitations?

67iqr

( 4Trr ’)pgStoke's LawThis law states that the drag force F acting on a sphere of radius r movingslowly with velocity v in a fluid of viscosity q is given by

F = 6 jtq r v

vOR 3 x 6rcq r

ForVoVmconiti.*

LimitationThis law is valid only for spherical bodies moving slowly. For high speedsdrag force is not simply proportional to velocity.

What is terminal velocity? Show that terminal velocity of fogdroplet is directly proportional to the square its radius?

v. = MMaUrialonstant]OR bnAw

fanAt

IVftlrw

Ofyc*»fWi

fl /v,OMO ORQ-4 Thus terminal velodty is direr! !y proportional to the square of the radius of droplet.

Q 5 What is the difference between steady and turbulent flow?

o«oi

i*

mTerminal VelocityThe maximumcalled terminal velocity.ExplanationConsider a fog droplet falling vertically downward. The drag force of air increasesas the velocity droplet increases. The net force on the droplet is.

Net force = weight - drag forceF = mg- 6nqrv

ma= mg- 67iqrvWhen drag force becomes equal to the weight of droplet then it will start tomove with uniform velocity, called terminal velocity (v,). So its accelerationbecomes zero. (i.e. a = o)Thus the above equation becomes

m(O) = mg- rvt0= mg- 6TIT) rv,

67rr|r vt= mg

Can You Do That? Fluid FlowLet us consider the flow of the fluid through the pipes. It may be either streamline or turbulent.Laminar Flow

and constant velocity of an objectfailing vertically downward is©%

line moves exactly along the same path asIn laminar or stream line flow each particle alongfollowed by the particle which passes through that point earlier.Theflow of afluid in which every particle of thefluid moves along a smooth path is called laminar flow.

a stream

Turbulent FlowThe irregular or unsteady flow of thefluid is called turbulent flowWhen the speed of flowing fluid exceeds a certain cnbca/ value, the flowbecomes extremely irregular and complex and it changes continuously with time.Thus,

A table 'eims ball can oemade suspended in n?

$!feam of a > r coming fre* (a) Streamline* (laminar flow)

According to figure, ’ , ^T,.rhulent FlowDifference between Laminar Flow asmoothjeoth wh,ch. In laminar flow, each particle of fluid moves along

does not change with time.smooth but continuouslyIn turbulent flow the

changes with time.1 In laminar flow, stream lines 42-On-flow it does not happen.

rhaneesin turbulent flow, the velocity of flu.dchangesBSD^°te (Steady flow condition)

F.XPIWATIOVftoifl i**ii)c.a/r is jjccicii

noss.Ic with hiph \clout}-therefore, according «'I Bernoulli's equation. lJl<

pressure at the nirnlc «

decreased TJic tennis hull

is suspended duearound it .

OR Turbulent flow

HQ 6.1(b)

mgOR v each other while in turbulent0) I0 ere*6rcr)r

pressureWhere v, = the terminal velocity andT) = coefficient of viscosity.

Relation between terminal velocity and radius of droplet iniersect each other. Thispor steady f|0w different streamlines can neveimassdensity =Since c°nditi0n is called steady flow condition.volumemOR p =V

OR

,s pHYSICS XI (Subjective)216 £^Pter 6 [,.|„.

'%i * * *. ». , o,„„H „Q.6 What is an ideal fluid? Ifc0nserV/rn, that flows into the bottom of the pipe through A, in time Atof the liquid that flows out through A2 in the same

mass5t be equal to

Therefore,Ideal Fluid massA fluid which satisfies the following condition is called an ideal fluid

1 The fluid is non- yiscous i.e., there is no internal frictional forcebetween adjacent layers of fluid

2 The fluid is incompressible, i e its density is constant3 The fluid motion is steady

4 Irrotationol flow.

Am, = Am2

iA,v,At = pjAjVpAt

PiA,Vi = p,ApvJThis equation is called the equation of continuity

Since the fluid incompressible, so the density is constant1J

Thus, the equation of continuity becomes Ap' ~ P2 = P <say) -

pA, v, = pA2 v2

As the water falls, its speedincreases and so its cross sectionalarea decreases as mandated hy thecontinuity equation

EXPLANATION:Conlimuty equalinn is A|v, A 2 \ J

OR Av - constant

constQ.7 State and explain equation of continuity.

Ax —A =A, v, = A,v2

Av = constantis called flow rate.

vv

WThe product Av

State and explain Bernoulli's Equation.Q.8

Bernoulli's EquationBernoulli's equation is the fundamental equation in fluid dynamics which relates

the pressure to fluid speed and height.

Aij •

AX2 •Am P2

ii V 2

Amh

?r‘

StatementThe sum of pressure, K.£. P^r un‘]t u\ lil* ..incompressible, non-viscous fluid JlW'WPoint (tloii }> a stream Hue.Mathematically

and P.K. per unit volume of an

is constant at eachExplanation in steady stateConsider a fluid is flowing through the non-uniform pipe The particles in thefluid move along the streamlines in a steady state flow as shown in figureAt lower end of pipe

— pv i pgli =constant1* iLetproof

^ consider the steady flow of an incomprcssP,pe time At. as shown in figure below.

^0rk done at upper endat upper end

area of cross- section of pipe = AI fluid through a, non viscousdistance moved by fluid in time \ t = \ x j

velocity of fluid = v

density of fluid = | » _

volume of the flu- d contained at lower end = Vmass of the fluid contained at lower end = \mThen

\mi = p j V jOR Am -. =|i: A \x .OR \mi = p i A, v _ \t ( Y Ax = v \! i

At upper end of pipeLetarea of cross-section of pipe = A ^distance moved by fluid in timeAt = \ xvelocity of fluid = v -

density of fluid - p,volume of the fluid contained at upper end = Vmass of the fluid contained at lower end - \m.Then

\rn = p A < j , \tj

C1B Chapter ? [» luil*“lI) PHYSICS- XI (Subject ive)liir’Na y|K>Area of cross section of pipe = A,The pressure of fluid = P,Force exerted on the fluid = F,Distance moved by fluid in time At = Ax,Velocity of fluid = v,Density of fluid = pVolume of the fluid contained at upper end = VThe work done on the fluid by the fluid behind it is given by

W, = F,Axl

219

Do You Know?in K.E. = A (K.E) =2

mv2?

* ~ mv’2

Changing in PE-= A (P.E) = mgh,- mgh,e h2 and hiare the heights of the upper andI

law of conservation of energy8V work done = change in K.E. + change inV r

I P,- P i ) — = ” rnv;2 - r

both sides by — , we get,

Change

y >.>Smoke

AirV

Interesting Inform; 2Bum

f lultiplYinS m A chimney works besl when it is tall3nd exposed to air currents, whichreduces the pressure at the top and

OR [ P, =» F'= P'A'IA,_ (l) [ v A x, = v,At]

W, = P, A, Ax,Pt - Pa =- (V - - pv,2 + 'Jh - pgh,

1A— IIrun . J rr r r rsU .r-OR W, = P, A, v,At

iu the upper region aunusphericpressure is small am] the smokem the tall chimn.c) rises upward

0 jrk done at lower endat lower end

Area of cross section of pipe = A2The pressure of fluid = P2

Force exerted on the fluid = F 2

Distance moved by fluid in time At = Ax2Velocity of fluid = v2

Density of fluid = pVolume of the fluid contained at lower end = V

larly, the work done on the fluid by the fluid ahead of it is given byI W2 =-Fj AX2

P, * - PV,1+ pgh, = P; + -- PW + pghj

This is the Bernoulli's equation and it can also be expressed asI ,— pv + pgh = constant

A stream of nir (6 )passinq ovfir a ^'dipped m a liquid will cause HIP I,W" o rise m the lube ns shown Thrs oKrciis used in perfume bottles andsprayers.

P +

Q 9 state and explain Torricelli's Theorem.

EXPLANATION^ Application of Bernoulli's EquationTorricelli's TheoremTorricelli's Theorem is Bernoulli 's equation with certain assumption made.Statement‘he speed of efflux is equal to the velocity gained by the fluid infalling through the distance ( h;

union of gravity.ProofConsider a large tank of fluid having small orifice (hole) A on it, as shown in figure.

Due lo greater speed ofair innarrow part of Ihe lube.

decreasedpressureTherefore, the liquid in llietube rises up in the tube atshown in the figure.

is

// j> under theW2 = -P2 A 2 AX 2 [ ••• => F 2= P2A2]

A 2w2 = -P2 A2 v2At

negative sign shows that this work is done against the fluid force.work done

(2 ) [ vAx2 = v2At]let

1v,area of cross-section of upper end of tank = A,area °f cross-section of lower small hole = A2sPeed of the fluid at upper surface of tank = v,sPeed of the efflux = v2pressure at the upper end of tank = P,ressure at the lower small hole of tank = P2

^eiSht of fluid at the upper endei^t of fluid at the lower endf e o w.p

lng to Bernoulli's equation

Pi pv,2 + pgh,= P2 + ~ pv2a + PSh>

^uation of continuity

Pinet work done both at upper and lower end isW = W, + W 2

W = P,A, Ax,At- P2 A 2 AX2 AtW = P,A,v,At - P2 A2 V2 At _

>rding to equation of continuityA,v, = A2v 2

A,v,At = A2v2At = V•:e, equation (3 ) becomes

w:= P,V - P2 VW = (P,- P2) V

W=(P,-Pj) —

h,-h:V VA'

r * TO= h n= h2i

(1)I (4)I “Ve

I OR

Alsc

(5) [ v (>=-=> VV - ]

A,v, = A 2v2P P( A,>> A 1A ,nge in K.E and change in P.E

act of this workdone changes K of fluid ar d a part changes its.e, / ’itational P .E. So,

(2)v,= — — Vj * 0

Pressure will be the same because they are op the atmosphere So

k.

P, = P,= P (atmospheric pressure)

Hence. equation (l) becomesl

~ P(0)J + I'gh, = P + - pv/ + pgh;P *

Pgh, = - pv/ 4- pgh?OR UsingtheseDividing ootn sides by p, we get pA +i- (1000) (0.20)2 = PB + ~ (1000) (2)2

Igh, = - v/ 4- gh>

,•7:r::::PA + 20 = PB 2000

PA - PB = 2000 - 20

PA - PB = 1980 N/m2

that the pressure in narrow pipe where stream lines are closer

is smaller than wider pipe. Thus

eed is high, the pressure will be low.

/I

OR - v,' = gh, - gh2 ORORA Ii'I•:I pump

cui'SHu.ii iu m mur.cmu. *tlu»\ .1)I-M ill WrtHtt llUlltllt,.,Hows '•' -.let note U\isi.v.titop tnpiitSSUCettoafAf,^iiti tluueUm: I'ws m l«u,the side IIIIK: Ihe a« *«w.'iei lOiU-l'uu muiut(»,viIP1 < itt'i"U't; luwHll».’wlitlti.pitU'P

nv/ = g(h,-hj)

y/~ = 2g (h, - h,)

OR This showstogetherWhere sj

^7l How a dynamic lift is produced in an aeroplane?OR

OR v . - y j Z g i h - h ) (3)

This is Torricelli' s theorem. iAns-12Note Lift on an aeroplane

The lift on an aeroplane is due to the effect, where speed of fluid

is high, its pressure will be low.

ExplanationThe design of wing deflects the air in such a way that

• Stream lines are closer together above the wing than lower.

Air is moves faster at the upper side of the wing than the lower side.

Pressure is lower at the top of the wing.Hence, the wing experiences a net upward force.

Q.12 ' How swingproduced in a tennis ball?

The speed of efflux is independent of the direction of flow whether theopening is directed upward, downward, or horizontally.Speed of efflux of liquid is the some as the speed of the ball that fallsfreely under the force of gravity through a height (h, - h:).

• If the hole is pointed upward as at 'B' shown in figure, this K.E. wouldallow the liquid to rise to the level of the water tank.

a In practice, viscous energy losses would change the result to some extent.

Q.10 What is relation between pressure and speed of fluid in a horizontalpipe system?

KMH.ANATION:AI the coiuirjciKm

IS Iircilllf injv. dualspressuretherelore, ilk- .ur iWmthe lube.

tdtimrMil,ill

Deflection forceFaster air.lower pressure

Swinging of a ballright (sh0wn in figure) inI ^-*r-->«» — “*• ol

due to friction between the ball and the airsic)e.The speed of air at one side of the ball *

ei smaiier thaIn this case, the pressure at one side of ba —I; other side. . cuutntt which deceives an

This gives an fytro curvature to the ball nownopponent player. —[ 5 13 What li venturi relation? Explain

Relation between Speed and Pressure of the FluidIf here the speed is high, the pressure * */// he low.Proof

Suppose that water flows through a horizontal pipe system as shown in figure.

The area of cross- section of the pipe at A is greater than the area of cross-section at B

According to equation of continuity. The speed of fluid at B is greater than at A.

Applying Bernoulli's equation at points A and B, we have

Slower air.higher praaaura

Spinning balln that at the 0

1Pn 4- |U , * pglf , In * IHII P£l» i

WhereP.v = pressure of water at A

Pft = pressure of water at B

vA = speed of water at A

vQ - speed of water at B

P = density of water = 1000 kg/mAs the pipe system is horizontal, therefore h = hQ. Hence average P.E. is same at

both ends. So equation ( i) becomes

1 Vanturl Relation ,th lnere0» of w/otWUnt of door.au of pr.ssur.^hodiontal pipe Is called venturi efft^ # ^oriionial piPe *yS

* *uPPosa that an Ideal fluid flows t

of the fluid In

i as shown In

Chapter222 SiS-sD, v PHYSICS- XI (Subjective)cboUrJ

d pressure (BP)area of cross- section of the pipe at At is greater than the area of cross-section at A;The speed of fluid at A > is greater than at Aj. (By equation of continuity ]

Applying Bernoulli' s equation at points A and B, we have

1 ,p» * 2pv '*

223

Bloo r:r “»a- “»-« *c-Thevessel, Blood vessels are not rigid.. under normal conditions, the volume of the blood is sufficient

the vessels inflated at all times.

. so the pressure of the blood inside the vessels isatmospheric pressure.. During each heart beat, BP varies between a maximum (systolic) and a

minimum (diastolic) pressure." A person's blood pressure is usually expressed in terms of the systolic

pressure over diastolic pressure (mmHg), for example 120/80.

Systolic Pressure.• it represent -- the maximum pressure exerted when the heart contracts.

• The value of nigh blood pressure (systolic pressure) is 120 torr

Diastolic Pressur• It repres

I

T

(1)•M>Sht = P2 +~Pvi + P«h: o keep

WhereP: = pressure of fluid at wide part of the pipe

P ? = pressure of fluid at narrow part (called throat) of the pipe

Vi = speed of fluid at wide part

v - = speed of fluid at narrow part

p = density of fluidAs th£ pipe system is horizontal, therefore h, = h.Hence average P.E is same at both ends So equation ( 1) becomes

greater Do vou know?thanBlood pressure is ameasurement of the forceapplied to the wails of thearteries as the heartpumps blood through thebody. The pressure isdetermined by the forceand amount of bloodpumped, and the size andflexibility of the arteries.

»1

^7 *•*%!' 1 )*

1 1p. +^p v r = p

1 2'

1 2- p V j - 2l>vi

ftSftr

* S m pressure in the arteries when the heart is atthe mP l - P2 = For \our information

rest.• The value of low bli

Blood pressure is usuallymeasured while you areseated with your arm restingon a table. Your arm shouldbe slightly bent so that it isat the same level as yourheart. Your upper armshould be bare, with yoursleeve comfortably rolled

iressure (diastolic pressure) is about 75 -80 torr.

OR P i - P2 = - P( v;- v; )

This is called venturi relation which used in venturi meter to find t ^ e speed of fluidSpecial caseIf A!»A 2 therefore, vj« v ?.Then, according to equation of continuity

AjVi = A 2v 2

J2 ) NoteThe value of blood pressure Increases with age due to decrease In the

flexibility of vessels.Unit of blood pressureThe blood pressure is measured in torr or mm of Hq.

Relation between torr and pascal1 torr = 133 3Pa = 133-3 N/mj

up.

A ,OR — - \ . = 0v. = Measurement of Blood Pressure

Sphygmomanometer is used to measure the pressure of blood dynamically in the vessels.

Steps to measure blood pressure

: .• When the external pressure becomes Ijinm than systolic pressure, the vessels fall down.• The flow of blood is cut off .

• Head of a stethoscope is placed over the artery.Systolic pressure . ..' Open the release valve to decrease the external pressure gra ua y

• \yhTn externa, pressure becomes equal to sj«{o«^ressure, the vessel

opens a little bit1 A first surge of blood flows out of the narrow

Sgeed.B As the flow speed is high, so the flow is turbuleiL-

1 t A I /

Hence equation (2) becomes,

Pi- P2 =\ p(vj2 - 0)

Pi- Pj* - pv22 ( 3)

Venturi-meterThe device which measures the fluid speed is called venturi meter

venturi relationQ.14 What do you know about blood pressure? How it can be measured dynamically?

rkmg principle of ven u

opening of vessel with highBlood Flow

5 120 -Biood

| 80• Blood is an incompressible fluid >Density of blood is nearly equal to that of ater.

u£ 40

• Viscosity of blood increases three to five tin that of water due 0.” One beat

to high concentration of .1 bloc cells i. 30%) 1Time (•)

£ Pter 6 [pi PHYSICS - XI (Subjective)224 225

Diastolic pressure Multiple Choice QuestionsNow decrease the external pressure further till it

becomes equal to diastolic pressure.• The vessel gets normal. ^jrnur possible answers to each statement are f>iven below. Tick ( S) the correct answer:

If the radius of falling body is doubled, then what will be effect on its terminal velocity?(b) Decreases by 2 times

'r

(d) Decreases by 4 times

The blood flow changes from turbulent to laminarThe gurgle in the stethoscope disappears.This is time to record diastolic pressure. I.

Increase by 2 times

Increases by 4 times(a)

(c) \

A ten meter high tank is full of water. A hole appears at its middle.The speed of efflux be:

(3) 5 ms ' (b) 10 ms ’2.FORMULAE]

(c) loo ms ’ -1(d) . 5.11 ms

The maximum drag force on a falling sphere is 9.8 N. Its real weight is:

(a) IN W 9'8 NF- 67trqvDrag force 3.l

vt«— r*V =J1 6THT|

(d) 0.0 NTerminal velocity of fog droplet2 (C) 4.The effect of decrease in pressure with increase in speed of the fluid in a horizontal tube is known as:

Bernoulli's effect (b) Torricelli's theorem

(C) Viscosity effect (d) Equation of continuity

The SI unit of flow rate is:

(a) m; s

(c) m 3 s ’When the temperature increases, the viscosity of the gases

(a) Decreases

(c) Remains constant

"Dynamic lift" is related to:

(a) Bernoulli's theorem(c) Equation of continuity

A gale [ i.e.,very strong wind] blows over a house

(a) In the downward direction(c) ZeroWith increase in temperature,the viscosity of.(a) A gas decreases and a liquid increases

(c) Both gas's and liquid's decrease

The viscous drag on a small sphere moving with

9il4.V A|V1=A 2VJ— constant Av =constantEquation of continuity3 (a)

t

P+ pv2 +pgh=constantP,+JPVJ +Pgb,“Pj+^Pv:+P8h:Bernoulli's equation4 5.(b) ms-1

(d) m3 s~ J-1

Torricelli's Theorem v2 ~>/2g(h,*h2)5 (speed of efflux)

P. -P^PC^-vf)Venturi's relation6 (b) Increases

(d) None of these%(b) Archimedes' principle

(d) Pascal's law

.The force due to the gale on the roof is:

(b) In the upward direction

(d) horizontal

(b) A gas increases and a liquid decreases

(d) Both gas's and liquid's increase

a speed v is proportional to:

(b) Vv(a) v

Id) v’their terminal velocities are:

(b) A : 6

1(C> ^Two fog droplets have radius of 2 •3.(a) 4: 9

IV

PHYSICS-XI (Subjective)lar'sScho 227(d)(c) 2 : 9 4 : 3

12 . Which one Is the venturi relation?A Al v:= \'2g(h, -h:)(a) (b) Explain what do you understand by the term viscosity?

Q.6.1(Lhr 2004, Fsd 2005, Mir Pur 2004-2006,Ihr 2010-2011,GrwJon)

(C) A,V, = AjV, (0 ) None of these Viscosity• viscosity of a fluid is the measure of its resIsSancejoJlow. It is the frictional effect between different

layers of a flowing fluid. It measures that how much force is required to dide one layer of the liquidover another layer.

Art*1

IX Laminar flow usually occurs at:(a) High speed

(cl Very high speed

The SJ. unit of co-efficient of viscosity Is

(a) Kgm*VJ

(b) Low speed

What Is meant by drag force /hat are the factors upon which drag force acting upon a small sphereof radius r moving down through a liquid, depend?(d) None of these Q.6.2

Drag ForceAn object moving through a fluid experiences a retarding force called drag force.

(Federal 2004,Mtn 2009)Ans.

Kgm’V(b)FactorsAccording to Stoke's lav d ag force is given by,

F = 6 TI qr v

This eq jation shows that drag force depends upon

i) speed of sphere (v)

ii) radius of sphere (r)

iii) coefficient of viscosity of medium(ri)

(c) Nsm'J

The relation between Nm‘J and torr is. a 1 torn = 13.33 Nm'2

(c - ton- = 1333.0 Nm’1

(d) Both b & c15 -

lb) 1 torr = 133.3 Nm'J

(d) 1 torr = 1.333 Nm'J

now the *‘u d speed changes if the diameter of a pipe is increased to double?16.s Re^s n samec Reduces to ha

(b) Increases to double(d) None of these

Q.6.3 Why fog droplets appear to be suspended in air?(Federal 2003,Mir Pur 2003-2009.Sgd 2005,Mtn 2004.Rwp 2006,

Fsd 2005-2008,Bwp 2007 Grw 2009. Lhr 2010-2011)h"gh concentration of red blood cells increases the viscosity of blood from(a) 2-3times that of water(c) 2-4 bmes that o( water

T7-(b) 3-5 times that of water(d) 5-7 times that of water

For which position, will the maximum blood pressure in the body have the smal est value

Ans. ReasonTerminal velocity of a fog droplet is

mgvt =

6xr)T

18.= Stand ng up rightc Stand’.-gon one's head

The value os piooo pressure _a 'creases: Renan sameSphygmomanometer b used to measurea r JA soeec

ic B*tood density

(b) Lying horizontal(d) Sitting relaxed

AS the weight of aTgdroplet Is V^m!L *1««• ^^Thus, terminal velocity is very small and hence the droplet appears to be suspende

Q-64 Explain the difference between laminar flow and turbulent flow?

LUr 2008, Bwp 2008, Grw 2003-2008. Lhr 2009)

with age.19(b) Decreases(d) “ .c' e of these

(Federal 2004, Mir Pur 2004,Mtn 200520.

fb 3!ood pressured) B o t h b & c

/- o c?5and describe some of its applications’J M ^ State the Bernoulirs relation for a liquid In motion9- b2.b 3.b 4. a 8. bi.c 7. a5. c 6. b ( lhr 2009.Grw 200*1

20. b19. a12.a I3.b I4. dI I- a 18. bJ5. 17.h16.d

*ns. e and the potential energy per un4Bernoulli's relationpor an idea! fluid, the sum of pressurevolume at any point along a stream

' ins constantlint ilwayi'ema

1

V

228

,»s pHYSICS- XI (Subjective)i 229

P f — pv -f pgh= constantExplain the working of a carburetor of

Bernoulli's principle. Interesting Informationa motorcar usingQ.6.10Applications Air(ii) Lift on an aeroplane(iv) Blood flow

0) The swing of ballWorking of carburetor

Q.6.6 A person is standing near a fast moving train. Is there any danger that he will fa||

(Sgd 2003-2004, Lhr 2005-2008, D.G,Khan 2005, Bwp 2004,Grw 2005, Fsd 2004 )Ans. • Yes. there is danger that he will fall towards the train.

(Rwp 2005,Grw20ii)(iii) Working of carburetor

The carburetor of car engine uses a venturi duct to feed thecorrect mixture of air and petrol to the cylinders.

Air passes through the duct and along a pipe to the cylinders,

petrol is mixed with air by a small valve at the side of duct.The air through the duct moves very fast which produces lowpressure in the duct It draws petrol vapours into the air stream.

t / u~Ans- Almonolinnoerfwnuf* Ier«l*«Kjrn

“'"'ards j,? 1/

Gas

ReasonWhen fast moving train passes near the person, speed of air between train and the persAccording to Bernoulli's relation

where the speed of fluid is high, pressure will be low.So pressure between train and the person decreases Hence large pressure behind thhim towards the tram.

<X6.7 Identify the correct answer. What do you infer from Bernoulli's theorem?(i) Where the speed of the fluid is high the pressure will be low ?(ii) Where the speed of the fluid is high the pressure is also high?(iii) This theorem is valid only for turbulent flow of the liquid?

Ihfi carburetor of .i rjtt ringing u».o-.n Vonlun dud 10 thn oorw.lmi/of air and pulfol to »*> r.ylu/Wr. A.«rift drawn dirouqfi IM duo tint) ai*»nqa pip** In ft -n cytiOdom A tiny trtlwt litIt .** of durj ift M with petrolfhc air thrr/ jqh the dud movoft varyfaV C/Mtimj Iw prtwcure In t>*lduct, whitch dram patrol vapour intotbr» -Uroarn

wun.For which position will be maximum blood pressure (systolic

pressure) 1« the body having the smallest value, (a) standing up

right, (b) sitting (c) lying horizontally (d) Standing on one's

0,6.11e Person pi«U,

(Grw 2010)

The correct answer is (c)^(; t ) standing upright, systolic pressure has maximum value in the neckAns.

(b) Same as above(d) In thi . c - » se, the systolic pressure has maximum value in the legs.

orbiting space station, would the blood pressure In major arteries in the leg ever be greater

than the blood pressure In major arteries in the neck ?

(Mtn 2004, Mir Pur W;Ans. '/tatement ( j) is the correct answerOAJ . H fwo row boats moving parallel in the samr* direction nre puller! towards ^ach other. Explain?Ans. Reason

0,6.12 In an

(O.G.Khan 2006)

Ans. blood pressure would be same.

IrTa orbiting -.pace Nation, everything Is In state ol wlahilSiiDSii So, pressure will be same in major

arteries of both in neck and legs.

Ac epend of water bet ween the t wo boats Increases 15o/ According to Bernoulli's relationwhere the speed of fluid h high, Its pressure will he low.

be,prev.ur * - b* -i //‘-eo die two boats derreoses , Hew «* procure difference is produce d which pullsir *

boats towards each other.(Bwp 7003, federal 2005, Mir Pur 2006 2009, Grw 7009 2010, Lhr 2010 2011,On*

UcfioMlon forcJSolved Exercises

lilftmnle ft.If ,i .Ir air.love urrt'i*4 fif «;O f>.0 f /plain, how the swing Is produced In a fast moving cricket hall?

( f"darral 7W>, f ul )<* />, f j f , Kf ,.,n 2005, Bwp 200/, Ihr 7009)Ans. bwmg in a Cfickfrt ball

When tip* ball movoc forwards ac well a *; spins, tin- speed of airit ', d I l f Mitt broorne - greater / Ofr.p.ui-d to ||„- other / rr o-lifig ^to Bernoulli' relation,where the speed of fluid It high, Its pressure will he low .bo af dial Mde, tin po v.UH - of air < . wra A- ner fm < . . . 4-Tfjis gives^n

Mjr /., ’ un to the balldeceives the batsman.

A tiny water droplet of radius 0.010c* d«c«nd« through -ir * high

terminal velocity. (liven that „for air - 19 a ...‘hK m and dcn,ty of water p 1000 kgm4-blvin Data;

r 0.010 cm - I .O x i o ’mon RudillN Oi water droplcl1

13criuily of water p 1000 kgm

19 * It) " ksm ' s...ul—Splnn'rtjVisLonity of air r \

Find;I ertninal velocity v, V' * 'd swing, ( his which ( filiation:Iermiiuil velocity IN Riven by

V '

:«*

230 Chapter gPHYSICS- XI (Subjective)

231Vl « ish im9T1 v * — ^~~rt A x p

values, wet getPutting values, we get

2 x 9 . 8 x C l .0 x l 0"4 )2 x 1000V( — 9 x 19 x I 0'6

2 x 9.8 x I 0 ~8 x 1019 x 19 x 10'6

19.6 x 10 ~8

171 x l O'6

vt = 0.11 x 10ms

putting1

v = 0.5 x3.14 xlO"4 xlOOO

v,= 0.5v * 0.314

- i|v » 1.6 msvt--1

IBBBIHEBVVater flows down hill through a closed vertical funnel. The flow speed at the top is 12.0 ems' . The

flow speed at the bottom is twice the speed at the top. If the funnel is 40 cm long and the pressure at

the top is 1.013 x 105 Nm'1 w hat is the pressure at the bottom?

Or v, =1.1m/sec

Given Data:Flow speed of water at top = vj

Flow speed of water at bottom = v2 = 2vt = 2 x 0.12 = 0.24 ms

Length of the funnel = h = hi -h2 = 40 cm = 0.40 m

Pressure at the top = Pi =1.013x10sNmDensity of water = p = 1000 kgm"3

A water hose w ith an internal diameter of 20 mm at the outlet discharges 30 kg of water in 60iCalculate the water speed at the outlet. Assume the density of water is 1000 kgm 3 and its (ionsteady. . , affg

Given Data:Internal diameter of water hose =d = 20mm = 0.02m

d = 0022 - 2

= 12 ems" 1 = 0.12ms"1

-i

Internal radius of water hose = r =0.0 lm To Find:Pressure at the bottom - P2 - ?

Mass of water = m = 30 kgTime taken = t = 60sDensity of water = p = 1000 kgm-3

Calculation:According to Bernoulli's equation

Pi + I p v2

+ pg/? i = P2 + ^ P v2+ Pgfo

2 i 2

or P2 = Pi + pg(hi - h2) +|P ( v -v2 )

or P2 = P, + pgh + p ( v 2 -vj)

Putting values, we get

P2 = (1.013 x 10?) + (1000 x 9.8 x 0.40) + -P2 = 1.013 x 10s + 3920 + 500 (0.0144-0.076)

P2 = 1.013 x 10s + 3920 + 500 (-0.0432)

P2 = 1.013 x 10s +3920- 21.6Pj- 1.01.3 x 105 + 3898.4P2 = 1.013 x 105 + 0.0389 x 10!

P2 = (1.013 + 0.0389) x 10*

1.05 X 10’ NrrT3

To Find:Speed of water = v = ?

Calculation:

Mass flow per second = —t

30* 1000 x ((0.12)2 - (0.24)2)-1— = 0.5 kgs

60Cross sectional area = A = K r2

= 3.14 * (0.01)* = 3.14 * 10*4 m 2

From equation of continuityAv = Rate of flow

volumeOr Av =sec

fasimass As volume•Thus,Av “ see'densityMass

DC. ry

1Or v - Sec area * dm.

Exercise Problems26.1 Certain globular protein particle has density of 1246 kgm-3. Or ri v11 falls Ithrowater(7 j = 8.0x 1CP Nm 2s ) with a terminal speed of 3.0 cm h\Find the ugh d,2h^ radius ofth — — V.=eHitGiven Data:

Density of protein particle =p=T 246kgm'3

Co- efficient of viscosity = r|=8.0x 10^Nm’2s3Terminal velocity = v( =3.Ocmh'1 = =8.33xl 0^m/s putting values, we g! 100x60x60

. 2 _ ( 1 X 10 X )2 ( l )21

d 22 =4.76xl 0'6

d 2 = 2.18xl0'3md2 =0.2cm

To find:Radius of particle = r =?

Calculations:2gr2pAs v9T|

r2-9T1V,

I

OrOr

2pg The pipe near the lower end of a large water storage tank develops a small leak and a stream ofwater shoots from it. The top of water in tank is 15 m above the point of leak.a) with what speed does the water rush from the hole?b) ‘ If the hole has an area of 0.060 cm2 , how much water flows out in one second?

6.3Putting values, we get

2 _ 9x8xl (r4 x 8.33xl (r6

' 2 x 1246 x9.8r2=2.46xl 0'12

r=V2.46x10‘12

r=1.567xl (V*

Given Data:Height of water above leak point = h = 15mArea of the hole = A = 0.06cm2 = 0.06 x10"4 m2 h.

To find:Speed of water =v = ?

VVolume of water flow per second = ? (i.e. flow rate)r=1.56x 10'6 m rt h( =o h=h,*h,

Water flows through a hose, whose internal diameter is 1 cm at a speed of lm/s. What shouldthe diameter of the nozzle if the water is to emerge at 21m/s?

6.2 Calculations:(a) According to Torricelli’s theoremGiven Data: W2g(hrM

v=>/2ghInternal diameter of hose = d , = l cm=l x 10'2 mSpeed of water in the hose =v, = 1 msSpeed of water emergence =v2 =21ms

-lPutting values , we get

- lV">/2*9.8x 15v**>/294v^lTjUm/s

Or Ivl7m/s|

To find:Diameter of the nozzle=d2 =?

Calculations:According to equation f ntinuit)

of w»ter flow pet itc-.AvV0» Rate of flow — •volume

Rate of flow•Avd

IVI=AJV2 ai td2

1; 'w

• - • • • ' . . <'1 SIS. X! V ' V>., w

BUTh,

’ , - rnrt*.yWbS;-1 •

A\9R -S kW234

PHYSICS ~ XI (Subjective)235= 0.06x10^ x17

= 102 X10“ 6/MV *

Hence volume of water flowing oui per second i .e.Rate ot flow =102cm ’’ /sec

h, * h:= h(say)

To find:pressure difference = P2 - P| = AP = ?

ra|culations:According io Bernoulli ’s equation6.4 Water is flowing smoothly through a closed pipe system. At one point the speed of v

while at another point 3.0m higher , the speed is 4.0 ms 1 . If the pressure is 80 kPWhat is pressure at the upper point?

er noa at lower

fru1

N<Given Data:H- ISpeed of water at one point = v , =3ms

Speed of water at second point =v 2 =4msDifference of height between two points ho - hi =3Pressure at lower point = P, =80kPa = 80,000 Pa

•i; 3m

6(1

= x'

xl .29 ( ( 450)2 — ( 410) 2 )= O. t .45(202500 — 168100)

= 0.645(34400)

= 22188Pa

= 22.188x103Pa

p-V

To find:Pressure at upper point = P2 =?

Calculation:According to Bernoulli ’s theorem

Pi +^PV +Pghi =P:+^Pv::+Pgh;

^PV +PghrPghj

P; -Pl = 22kPa

p_’ =p. +2 PV|

2'

P;=P|+~ p(v|:-V::)-pg(h, -h|)

Or The radius of the aorta is about 1.0 cm and the blood flowing through it has a speed of about

30cm"1 . Calculate the average speed of the blood in the capillaries using the fact that although

each capillary has a diameter of about 8xl0"4 cm, there are literally millions of them so that

their total cross section is about 2000cm2 .

6.6

Putting values, we get

P: = 80000 +|xl 000(3:- 4 2 )-1000x 9.8 x 3

= 80000 + 500(9-16)- 29400= 80000 + 4500-8000- 29400=84500-37400

= 47100

Given Data:Radius of aorta = ri = 1 cm = 1 X 1CT2 mSpeed of inflowing blood = V| = 30 cm/sTotal cross sectional area = A:= 2000 cm"

= 30 x 11) 2 m/s= 2000 x 10~W= 0.2 m2

Average diameter of capillary = d: = 8 x 10"4 cm = 8 x 10 m

= 47.1 x 10' Pa To find:P,= 47kPa Average speed of blood =v 2=?

Calculations:Applying equation of continuity

A , v, =A 2 v 2

the top of the wing is

differenceLAn airplane wing is designed so that when the speed of the air across450 ms 1 , the speed of air below the w ing is 410nts ' . W hat is the pressurethe top and bottom of the wings ’ ( Density of air = ( 1 . 2 9kgm )

6 . 5 bet'veen

A , v,or v2=Given Data: A 2Speed of air on the upper surface = v , = i 50ms nr,'xv iiVn= A:“ '""g values, we get

V"V

1-.< <»

236PHYSICS-XI (Subjective)

3.14x( lxlQ-Yx30xl (r:V'

_0.2

3.14x30 x10"Speed of air past the lower surface =v, =160m/s

Density of air = p = 1.29kgm'!

Thickness of wing = h 2-h, = lm0.2942 xlO = 471x 10 °

f o f»Dtl:0.2Speed of air over the upper surface jying =v2=9= 4.71x10 4 =5 x|( ) 4 ms ' ( Approx ) w

Cg)cul®t»ons:According to Bernoulli’s equation

P,+Ipv,J +pgh =Pl+ipv3J+pgh2L m

Or P, - P2 =^P(vl -v|) + pg(hI - h 1 )\ kk ' K - '

Putting values, wc get

6.7 How large must a heating duct he if air moving 3.0 m/s along it can replenish the300 m' volume every 15 min? Assume the air’s density remains constant. n a

Given Data:- iSpeed of air = v = 3ms

Volume of air = V = 300/ / ? ^Time = t = 15 min = 15x60 = 900 sec

1000 = -X 1,29(vj-(160)’) + (1.29X9.8X1)ITo find: 2Size of ihc duct = r = ? ( i .e. radius of duct) 11000 = -X 1.29(Vj - 25600) +12.64

2

1000- 12.46 = ix 1.29( v\-25600)

987.54 = (0.645Xv 22 - 25600)

987.540.645

1531.07 + 25600 = vjy\ = 2.7 xl 04Vj = 164.71

v2 =165ms'1

What gauge pressure is required in the city mains for a stream from a Ere hose connected to the

mains to reach a vertical height of 15.0m?

Calculation:volumeAs Rate of flow = = (area of cross - section) (speed of the fluid )

limeV

Rate of How = — = Av= v3

a -25600tV

Rate of flow = — = TIT" vi

tVr =Or

IX7TXVOrPutting values, we get

6.9300r 2 = 900x3.14x3300 Given Data:r2 = 8478 Vertical height = h = h,-h2-Ah 15 m

Density of water = p=1000kgmr2 =0.0354r =0.188m =0.19m To find.r » 19cm

Pressure difference = P2-Pj AP ?I*|r = 19 cml ofabo*An airplane design calls for a lift due to the net force of the moving air on the wM floW, PNm'2of wing area. Assume that air flows past the wing of an aircraft with strea overtb* uPjJspeed of flow past the lower wing surface is 160 ms'1, what is the required spsurface to give a lift of 1000 Nm‘2? The density of air Is 1.29kgm and asthickness of wing to om netre.

pfadtti6.8 ons:Using Bernoulli’s equation

Pi +~ Pvl1+P8hi=Pj+'^PlVjJ +p8h3

that the pipe is of uniform cross sectional area so0^ SoVl « v2= V)(say)

ins same throughout iuthe speed of stream remains

Given Data:

Pressure difference on the w » % P, -r =1000N/m:

A

Thu*. IV ipfcll

l > . 1 » , fij?h , w,h

=f^l> . I>, PK<h, 1«, )o»

&9WAT 7Putting value**, wr jtrl

IV \\ IOUOXV.S( IS)

IV !47OO0Nrn 1

(AP*l~

47x 1O^Pa

OrOSCILLATIONS

Or

Scholar’s YjtCUrtl

PHYSICS Investigate the motipn of cj> oscil 'ator using experimental, anal/tical and graphical methods

Understand and describe that when an object moves in circle the motion of its projection or tne

diameter o' the circle is simple harmonic.Show that the motio/i o; mass attached to a spring is simple harmonic.

Jnders’.and that the motion of simple pendulum is simple harmonic and to calculate its time period.

(Objective)Are also available

t

1.

3-4-

Understand and use the terms amplitude, time period, frequency, angular frequency ad phase

difference.Understand and use the terms amplitude, time period, frequency, angular frequency ana pnase

difference.Describe the inter change between kinetic and potential energies during SHM.Describe practical examples of free and forced oscillations.

to the effects of theDescribe practical examples of damped oscillations with particular referencedegree of damping and the importance of critical damping in cases such as car suspension system.

240PHYSICS XI (Subjective)

oscillatory Motionrid fr0 moiton °f a body ab(,ut «

library motion.periodic Motion

The oscillatory motion that repeats itself after equal intervals of time Is

periodic motionExamples

Chapter No. 7me“ n P0si,hn « catted oscillatory or

O <u

8 a rDC <*-O Oic/a 2J2 §

called

GO

S

Q.C of vibrating bodiesThe motion of mass suspended from a spring

2. The motion of bob of a simple pendulum

3. A steel ruler clamped at one end to a bench oscillates when the freeend is displaced sideways

1*<uor.C

G-.-oC

I2O O CL/>if!u ^ iTrrrl TO

totsi m * « n *111U E *<L>

a A steel ball rolling in a curved dish4'-a .2Restoring force

The force which brings the System back to its stable equilibrium position is

called elastic restoring force.Mathematically

t/i

* 4l 4l*(Vibrating objects)

Fig. 7.1B-3.§ *2OON &>1

GO S *!“ 3'5i-i fcCuc

O£ 5*•n<u F =-kx

It is equal and opposite to the applied force

How to produce oscillationIn order to get oscillation;

A body is pulled away on one side from its equilibrium position and then

released.The body begins to oscillate (vibrate) due to restoring force.Under the action of this restoring force, the body accelerates and it

passes over the rest position due to inertia.The restoring force pulls it back.Since restoring force is always directed towards the mean position, so

the acceleration is also directed towards the mean position.Requirements for oscillationTw° requirements for oscillations are

1- Oscillating system has restoring force-2- Oscillating system has inertia

aO 'LlH 2»-a u-CL, £u03 00V) tnCu o cr. a ccS3 o oou CG ooC/3 E IS) Coz 2 8t/3 «r-Jb> 2 s £ 5:2 -5 ^c > + =13O ui|x

^EST3u £

<m

o o1 3 EP3 Uif <uaa o UJ

O -q \Da Hi1ll«.B

OQ 1 i i io ui•a 0-o 3 c

Iliao 1 -s8 ie s3 C -u "S

s .9 <

coCG

i .sI

2‘" GOc NoteS

vibrating bodies produce

^ere are many phenomena in nature, which are explained onancj

ration and waves. There are many large structures sue^ these

. 1 ees, which appears to be rigid. They actually « rat* vibrations, whileso the architects and engineers take into account these

gning and building of certain structure.

A violin string produces waves.\ ri f waves, e.g.the concept of

e 1l|i 1Ifa £ *

6 8cj c*2 0

I 1E

II•M

£>s facts,H desiQ

242 ,s PHYSICS- XI (Subjective) 243

Define Hook's law and simple harmonic motion? What Is restoringforce, derive the relation for acceleration of mass attached with aspring?

Q.iElastic Restoring Force

deforce which brings the body bock towards its mean position is coiled eiostic

' ' storing force is represented a F, is

KThe pegative sign shows that Fr is directed opposite to x.~

yVhen the mass Is released, it begins to oscillate about the equilibrium position

as shown In figure, such type of oscillations are due to restoring force and

inertia. This type of oscillatory motion is called simple harmonic motion.expression for acceleration

The acceleration a produced in the massm due to restoring force can be calculated usingsecond law of motion

f — e

F - m a

Comparing equations <i) and (i|, we getk j

m a = -k x

OS There h0»Hook's LawAccording to Hook's law, within elastic limit, the applied force is directlyproportional to the displacement.'Mathematically

-+ -»F cc x

.0)OR F = k xWhere k is constant of proportionality, known as spring constant.

Spring ConstantIt spring constant is defined as the force per unit extension. Its SI unit is Nm 1 anddimension is (MT2]..

. Simple Harmonic MotionThe oscillatory motion, in which acceleration of the body at any.Jnptant isdirectly proportional to displacement from the mean position andfflFrectcdtowards the mean position, is called simple harmonic motion

Examples

(2)

k -a x

m

a = -constant r

Hence proved)a«- x

Define the ollowing terms related to SHM.Motion of simple pendulumMotion of mass attached to a spring.Motion of a swing.

1-(b)lnstantaneous displacement(d) Vibration(f)Frequency

(a) Wave form of S.H.M.(c) Amplitude(e) Tune period\g) Angular Frequency

2- Movwnent of Paper

3- UH .frrfiConditions for SHM Tt-S

The system must have inertiaThe system must obey Hook's law

The system should have elastic restoring forceThe system should be frictionless

1. <#.p ; . H-2. - h -

3 Wave form of SHMThe curve representing the variation displacement with time is called wave from

4~of SHM,explanation

Q.2 Show that motion of mass attached with a spring Is SHM.Consider a mass spring system with vertical arrangement in such a way that pen

Cached with mass m form the trace on the strip of paper mov.ng at constant

is obtained which shows the*oeed from right to left.

^ Provides a time scale on the strip. The sine curve

^nation of displacement with time.

' ' 1 cahed wave form of SHM.he Point A, C and E show its mean position w

Motion of Mass attached to a springConsider a mass m attached with on* end of the spring. The mass n can

move freeiy on a frictionless horizontal surface as shown in figure

When mass m is displaced through a distance x from mean po ’ion by a

force F then,

Accord- ng to Hook's law.

bile B and D represent the extreme

Position.

i&i s r s s-— “to elasticity, spring oppos *he applied f e Th opposing force is

ca ed restoring force

7

neous displacement

244

it >$ usually denoted by x. The value of instantaneous displacement imean position while it has maximum value at the extreme positions.Amplitude (xQ)The maximum value of displacement of thUs mean position is called amplitude.It is denoted by x<>- In figure, the amplitude is theVibrationOne complete roundvibration.

12^ ^r- s PHYSICS - XI (Subjective

I TA , B, A , C, A

•s 2ero at

e vibrating body on either sides ft

measure of line Bb or Dd.

rom

trip of a body about its mean posaion i\ called oneThe motion of body from mean position to upper extreme position, fromextreme position to lower extreme position and back to itscalled one vibration. So according to the figure ABODE showsthe body.Time Period

uppermean position isone vibration of

The lime required to complete one vibration is called time period.It is represented by T .Its unit is second.FrequencyThe number of vibrations

c ;(•)

completed in one second by the body is calledfrequency. It is the reciprocal of the time periodIt is represented by f. The unit of frequency is heru or vib/secl nAt t = 0 pointer is at position A then at position B, A, C and back to A at instantT/4 T/2, 3T/C ind T respectively.

A circular motion point ' P is moving in a circle of radius x* with uniform angularfrequency o» Now consider the motion of point N, the projection of ond.ar- ter DE The levels of D and T are similar to points B and C.With the motion of P on the circle, the point N moves to and fro on DE. LetP0«nt p is at 0,at t = O, the projection N at instants 0, T/4, T/2, 3T/4 and T will be*r 0,0,E and O respectively.

ResultHence the comparison of motion of N and P, shows that it is a copy of pointer's

^on. Hence the motion of projection of particle P moving in a circle is SHM.

^ 5^ Derive the expressions for instantaneous displacement, instantaneousvelocity and acceleration of the projection of a particle moving in a •circle of radius Xp. ‘

‘et N be the projection of a particle P moving in a circle.Aguiar frequency of P

e angle subtended by OP at any time t = 0 = cotad,us circle = Xo

stantaneous Displacementr°m figure (in right angled triangle OPN)

ONO P = s l n 0

°N * OP sine

or cycles /sec.1= —T

OR f xT = t(i.e. product of frequency and time period equals one)Angular FrequencyIf T is the time period of a body executing SHM, its angular frequency (©) is givenas

24CO =

Tlco = 2 n( — )T

CO = 2 TtfNoteBasically, angular frequency is the property of circular motion In SHM itprovides an easy method to determine the instantaneous displacement andinstantaneous velocity of booy executing SHM.

= co

4 Q-4 Show that the motionpath is SHM

of projection of body moving along a circularI'lln

SHM and Uniform Circular MotionConsider a mass m attached with the end of a vertivibrates simple harmonically with period T, frequency^ S“ Spended sPring- It

The motion of the mass is displayed by a pointer p,.the amp,itude x<,. C:

0 (alternate angles)][v <OPN = <0,0P =

99

-. 4- ->' .W'»- . •* :* F, *...S'* £ **if * •

' - fei - .L -V-CT/ a**246 * >. %! .iM

PHYSICS — XI (Subjective)But ON = x and OP = = x0 247

values of PN and OP in equation (3), wegetSo x = x0 sin AV *;- x 2

As 0 = <ot cosOX o

|x = x0 sin oq 0)values of cos 0 in equation (2), we get

puttingThis equation shows the displacement of pointer N at instant t.Phase angle (0)

The angle 9 which gives the states of the system during one complete cycle is colled phaseValues of 0

The wave form of SHM is shown in fig1(c). In whichAt the start of cycle,0 = 0.When1* quarter of the cycle is completed,0 -n/ 2.When half of the cycle is completed,0 =rc.When three fourth of the cycle is completed cyde,0 =371/2.For the complete cycle, 0 = 2 x

v = x 0coXo

oh/x 03 - x2V =OR

f Velocity of NDirection o

direction of velocity depends upon the value of phase angle

When iwaries from 0? to 909 then the direction of v is 0 to D.The1.

• When it varies from 90« to 2709 then the direction of v is D to E.• When varies from 2709 to 360 -' then the direction of v is E to 0.

Special cases J |• At mean position fi.e. x =0) the velocity is maximum [i.e. vm„= wxo)

• At extreme position (f.e. x = *<,), the velocity is zero (i.e. = 0]

(3) instantaneous Acceleration (a)r.e acce er ihon at any point P moving along the circle can be expressed as,

ap = Xo to2

It is always directed towards the center 0.The acceleration of point N will be component of

*cc* oration ap along the diameter DE as shown in

fyure.

2.

1

4-

5-Note

por each quarter of the cycle, the phase of vibration is changed by x / 2 radian.(2 ) Instantaneous Velocity (v)The linear velocity of po nt P at any instant t = vpThenS nee the motion of * 4 or diameter DE is due to the motion of P on thec»fde

vp - X* co

FOR Y01KINMlHMAXilil a = ap sinOThe /e oc V of U *. actual / the vertical component of velocity / p in the

<krect>on parallel to DEtsin 2

''e component of velocity para el to DE is/P > n(90°-0)v *

v * vP COS0OPv * /</ > COS0OP

OP a /,//> COVJftFrom r grt .5 ' jr *d tf «ar g «r OPS

PS00*0 OP

App ymg Pyt' agO<*a' ’heofem, for calculating va ue of Pf 4(Off - {w<;'»(ONyIPHf (OS')'' - (ON)'

* 5 PHYSICS- XI (Subjective)a = -to2

\ §cb0,ar

249_ eqUation gives the disp..cement of SHM, but in,his^ rtingits m0t'°n f°rm extreme position instead of thein fi6u1__a =-constant x ' [v (n2 = constant]

a a -x

This equation shows that the acceleration is directJy proportional todisplacement and is directed towards the mean position which is the property ofSHM.

So we can say that point N is performing the SHM with the same amplitude, timeperiod and instantaneous displacement of pointer Px,

case the point N ismean position,as shownOR

Discuss the motion of a horizontal mass spring system and find thevalues of acceleration, angular velocity, timereplacement and instantaneous velocity.

Q.7period, instantaneous

|pA Horizontal Mass Spring System

Consider a mass m is attached with a spring, while the other end of the spring isfjXpH with a rigid support and it Is capable to oscillate on a friction less horizontaltable.Whenabout mean position is SHM. pInstantaneous AccelerationLet at displacement x the restoring force F produces the acceleration a. Then

-* -4F - ma _

According to Hook,s law,

Q.6 Define the phase angle.

the mass m is displaced from mean position and released then its motionPhaseThe angle (0 = cot) which gives the displacement as well as the direction ofmotion of point executing SHM is known as phase.OR

0)The angle (6= aX ) which determines the state of motion of the vibrating point iscolled phase. I-Note (2)-k xThis angle is obtained when SHM is related with circular motion. Comparing above two equations. We get,Displacement in terms of phase FlB. 7i(D

F - maSpecial caseLet at t=0, the point P is at Oi and N is at mean position. Then the initial phase of m a =-k xIrotating radius OP is zero hence the displacement at t = 0 is zero. It is considered k -as a special case. x

mGeneral case(concept of initial phase) Angular FrequencyLet at t=0, the angle made by rotating radius OP with the reference line OOi= 4> As we know that

from SHMAfter a time t, the radius rotate through angle =cot. The angle made by rotating The wavea = -(o2 xcosineradius OP with the reference line OOi at time t is (tiJt+0) is sine or

Comparing equations (3) and (4), we getSo, the displacement at time t is given by curve.x = x0 sin (o>t + <))) k V

Initial PhaseNow the phase angle is

0 = tut + 0

When t = 0,0 =0, So p is called initial phase.

Now taking initial phase as 90° or - ,\ 2 ) then displacement >

x = x0 sin (OJU90°)

OR X - X0 COSOif

« “ l . . “ > ••

,

J

HEK^r !i v- C _

V i, ACiS). 5 *3TS. Ws- *' V V -• v250

PHYSICS XI (Subjective)Time periodAs the time period of mass m having SUM can be expressed as,

T - 271251I

( 10)Cl)

• |rnum Velocity

Joclty of vibrating masslsminlmum at

"J K - Kv = G)(0)Vmin s 0

between maximum velocity (v0) and

Putting values of < >, we get2nT «

V C l )/: So,

m inT « 271 J—OR Relation

velocity (v)Using equation (10) in equation ($);we get

instantaneous(6)k

FrequencyAs the reciprocal of the time period is called frequency. So,

f -I = v„ ( 11)I

This is the n lation between maximum velocity and instantaneous velocity.Q 8 What is simple pendulum? Show that the motion of pendulum is SHM.

Also find i elations for its time period and frequency.

Putting value of T, we get

J_ k_2n Vm

Instantaneous DisplacementThe instantaneous displacement x of mass m is

x = x0 sin cot, Simple PendulumAn ideal simple pendulum consists of a small heavy mass suspended by a weightless, flexible and inextensibleit' r (j fixed with africtionless support and medium.iddically, the above mentioned conditions are incompatible and we use the light weight and less extensible

string.Motion of simple Pendulum is SHMConsider an object of mass m attached with the end of a light weight string,l-fingth of the pendulumThe length of the pendulum t is the distance between the point of^Pension and the center of the bob.WorkingWhen the pendulum is displaced from Its mean position through a smallUfl* 0 and released then it starts to oscillate to and fro about meanPosition.

^Ponents of weightthe Welght m8 into two components mg cosO and mg sln0.

°ther force In this case is the tension T In the string.ST!^ T are equal and opposite to each other. So, they cancel the

0 ®ach other.!.e.,

N"«forc.m8 C°,0 BT

,lck towards Its mean position. So, the re

Putting values of rn we get

x » x0 sin t

Instantaneous VelocityThe Instantaneous velocity v of a point performing SHM Is given by

Putting value of <o, we get

\Cv

O \Wl- x3OR v »

TX

V

mg cos 0ET3V" x„!

1/

mg sin HMaximum VelocityVelocity of the mass Is ma-'mum a* he mean po ion 0 where x o, so aboveequation becomes.

mg

»r?J: J

-mmmW 'H . - f

It,ir•1k. V=.252i

PHYSICS- XI (Subjective)Scbol^Dependence — - <

_»-

F = -mg sin0Negative sign shows that force is directed towards mean position.Also we know that

d) 253of Time period

F = maComparing above two equations, we get

ma =-mg sin0

(2)penNoteTjme period of the pendulum is independent of mass.FrequencyAs the reciprocal of the time period is called frequency.

f = -

Putting value of T, we get

OR a =-g sin 0For small value of angle 0, sin 0 = 0

So, a = — g0 (3)TAB

[v5 =r0 =>0 =- ]0=From figurei

\r(8 )0= * 2n \ t —

[ v 0 is small so arc AB = x ]i Q.9 What is second :enduium? Calculate its frequency and length.Second PendulumThe pendulum whose time period is 2 seconds is called second pendulum.

So T = 2 secFrequency of second Pendulum

So equation (3) becomes,a = ~g £

t

f i )xa =- (4)/

la = -constant x x [vi = constant] Ti l/ m

2/ = 0.5Hz

len£th of second Pendulum

As T = 2n l —

OR a cc - xThis proves that the motion of pendulum is SHM.Angular FrequencyWe know that for a body having SHM,

a = -to2 xComparing equation (4) and (5), we have

— co2x =

(5 )g

2 ( t'1 Or T2 =4TII u;

t-$4TTAs T = 2 sec

4 x (3.14):f = 99,2 c m

few that the ( mechanical) energy 1» conterved In SHM.

OR

gOR (6)CO =

0.992mTime periodAs the time period for SHM can be expressed as,

T = ^LOr

3*Putting value of co,we get

T =

* V\ot_ ^oe_ r>A-\— -< Q At-

*<> A ocofi.JCJL ^Cfc\

4*C_xrT\ rv* r g, -'tjyj

C rV *•“ >* "I vibrating man tpnng Vrri.n^rnriWn^(jb|f

:• i*••»--Ult> change in velocity will be very small.is icro,

254PHYSICS- XI (Subjective)

Instantaneous P.E.Let for any instant t the mass m is at a distance x from mean position,So according to Hook's law

255

maximum K.E.« E is maximum at mean position where

1n*' / . vF = kxWhen displacement = o

Then F = oWhen displacement = x

Then F = KxSo average force is

1K.E 2kx I2

1(K.E)max ~ — k*02Minimum K.E.The K.E is minimum at extreme position where x = Xo0+kx total «n«ryy2 >IF = K.E

= 5^0-0K .12 A/iftO /

1F =“(kx)

Hence the work done in displacing the mass through displacement x isW = Fd

j1

l/ \ P.E.// \l>(K - E)min = 0O

Total EnergyAt any position, the total energy is sum of partly P.E. partly and K.E.1

= - (kx) (x)

= ” kx2

So,*. E = P.E + K.E

E =Work done appears as elastic P.E. So,

1E =“kx2 +“kxo2 -“kx2(P-E) ,nJ = ^kx2

rMaximum P.E.At extreme position, the P.E is maximum, as the displacement at extremeposition x = Xo So,

i

’ Jithe total energy of vibrating mass spring system always remains constant.Noteduring the oscillatory motion where the K.E Is maximum, and the P.E Is lero andwHsn the P.E is maximum Is K.E Is zero. The change of P.E and K.E withdl*pl»cement Is required for maintaining the oscillation.

Aperiodic exchange of energy Is the property of all oscillatory systems.What are fret and forced oscillations? Also define driven harmonicoscillator.

T(P.E)

2

Minimum P.E.P.E. is zero If the displacement x o, l.e., the mass Is at mean position, thus

P.E. A(O)12

(P.E.)_ = 0Instantaneous K.E.We know that Jr,» Oscillation*

/ H Is said to be executing free vibrations if It oscillates tvlih Its

»M>out the Interference of an external force.Mndulum vlbratM fr.ely with It,n.tur.l »*»•"***d,p,"d‘

•••ngth of the pendulum.' wo*d",u»ni

** taU t0 ** executing force4 vibrations /tr*ce of an external force*

(K.E)**“ mv2

5lnce v » x

(K.EU- -So,

with the

r!lSo,

/

4 - r'

' 266

2^! ^r^PHVSICS X! (Subjective)For example• If the mass of vibrating pendulum is struck repeatedly, then forced vibrations are pf0(j(• The vibrations of factory floor caused by the running of heavy machinery is another e

Driven harmonic oscillatorThe physical system undergoingforced vibrations is known a\ driven harmonicoscillator.

257

tuning a radio (E,ectrical resonance)0f radio is a good example of electrical

a We turn the knob of a radio.. it changes the natural frequency of electrical circuit ofrecelverof transmitter. _ — .. NOW the resonance Is produced and energy absorption is maximum

• Hence a station is tuned. r«r w ^I cooking by microwave oven

• . Resonance plays an important role in h

Igignonce. To tune a radio,HTuning

/v

until it becomes equal to the frequencyA

Q.12 What is resonance phenomenon? Explain it with examples?

0ResonanceResonance is the specific response of a vibrating system to a periodic forceacting with natural vibrating period of system.

ORThe phenomenon in which the amplitude of a vibrating body increases whenthe frequency of an applied force is equal to the natural frequency of theharmonic oscillator.

eating and cooking food byb microwave oven.. The microwavwater and fats molecules in the food.

• This incri ases the- internal energy of the molecules.• They get heat up a

vWave

Stirrer guideproduced by are absorbed due to resonance by Magnetronc trAy

** Ti /

Cookingcavity ~id is cooked Power

t NoteThe wave length

I and frequency is

Experiment to demonstrate resonance

Consider a horizontal rod AB is supported by strings S, and S:.Three pairs of pendulum aa', bb' andcc' nsuspended to rod 'AB'.

• The length of each pair is same but different for different Da.rs.• Displace pendulum c is in a direction perpendicular to the plane of the paper

A small force acts on all the pendulums through the rod AB.

• All the pendulums will oscillate with the pendulum c but with a slightperiodic motion.

• The pendulum d,whose length and hence period is exactly the same asthat of c oscillates with larger amplitude equal to c.

• The amplitude of other pendulums remains small because their natureperiods are not same as that of the disturbing force due to rod A 9.

of the microwaves produced in this type of oven ist2 cm2450 MHz

TuiDoor and Choke

Cross- sectionof oil damperQ 3 Wh3i are damped and undamped oscillations? What is damping? Chassis weigh!

v Vajve ^

<£> OildamporSuspension

spring "y]r amped OscillationsOscillations in which amplitude decreases with time due to energy dissipationare called damped oscillations.Explanation

amplitude of the oscillating body gradually becomes smaller and smaller

^use of friction and air resistance. As the energy of the oscillator is used up intong work against the resistive forces, that is why the amplitude decreases withtoe till it becomes zero.

^plication

Wcation of damped oscillation is the sMck_gbsgrbiL °< a car which

“'P'hdes a damping force to stop the excessive oscillations.0aihplng

J"1**is the process by which energy is lost by the oscillating system

^damped oscillationsosc,|| l0ns in Mhlch the omplitude remains some wi

>"i- 1.... - -TS'"1'dtl0n* of an ideal simple pendulum is the examp

o Platon

iitrz

,0mnv. «. “ PT*Tf#****!

to r»»» __jzxsxs&mExamples

Motion of swing (mechanical resonance)

• A swing is a good example of mechnlcal resonance.

" WP aoolv a periodic force on swing

• When the frequency of periodic force 6»cents e \ual *o the naturalfrequency of the swing, resonance 1$ producedSo energy absorption is maximam.

u Hence, the amplitude of vlbrat. n Is I creasevCollapse of suspended bridge

On a big span bridge the soldiers crossing the -idge are ordered to break theirsteps. If the frequency of steps .*nc»u« vith r jral frequency of the bridge.Then there is a chance to c *o e the briot due to resonance.

rtf

/SAIW3ssisi-r,- UMn.iu..i

viulcfl* . *YU**®,#brflW “p Z*«*> «*> ‘

Graph between arnplltuda and tlma.

S ' /IrtMUvnc)

II* !ofrvtjucnc<c»

develop

main 'P,inibe brldpcinto lhe wJ,cr

ta.

ith time ore called undamped E<

fb) Damped•Vo

Graph between amplitude and Uroe

l le of undamped oscillation.

PHYSICS — XI (Subjective)259

Q *4 What Is the effect of damping on the sharpness of resonance?

for mass spring systemof ©ValueSharpness of Resonance

• The amplitude of vibration of a body increases when the damping issmall.

• Thus, the presence of damping prevents the amplitude from becomingsufficiently large.The amplitude decreases rapidly at a frequency slightly different fromresonance frequency.The amplitude as well as sharpness depends upon damping.A heavily damped system has fairly flat resonance curve.

Example to see the effect of dampingAttach a pendulum having very light mass such a pith ball and another ofsame length with a heavy mass of equal size such as lead ball.Set them into vibrations by third pendulum of equal length and attachedto the same rod.It is observed that tbe^mplitude of the heavy ball is much greater thanthe light ball.So the sharpness of the resonance curve of resonating system dependson energy loss due to friction.

iod of mass spring systemexecuting SHM w?Time Per T=2*J£L«

1 ks ncy of mass spring systemexecuting SHM

[^JTntaneous displacement of

mass spring system executingSHM

a

27tVmE frequeS<

skx=x0 sinJ— t

Dr,vi"0 m

Vm \ x„Instantaneous velocity of mass /

spring system executing SHMVk 2 - x! JV = xoV m

Maximum velocity of mass springsystem executing SHM

Instantaneous velocity of massspring system in terms of

maximum velocity executing SHM

kv„= xnO 0 m

v = v.

Restoring force for simplependulum V4 • F = -mg sinG

m«ui bob Pith bob MM

Acceleration of simple pendulum a=- \ i

' fiValue of a) for simple pendulumFORMULAE]

Time period of simple pendulum

/ frequency of simple pendulumHook's law F =kx

Restoring force instantaneous P.E. of mass spring^ system

Maximum P.E. of mass spring^ system v/

instantaneous K.E. of mass spring

^system

Ma*lmum K.E. of masssystem

F =©t A Tv/^k -a = x (P.E)m

In /G>=Y \/ 0>=27tf' \ jAngular frequency

sin(o)t+*'Instantaneous displacement ofbody executing SHM y/ X = Xox = x<, sinG x = Xo sinwt

Instantaneous velocity of body /executing SHM * V coN/xo: - X

2

Total energy of mass springacceleration ofInstantaneous

body executing SHM

Time period and frequencybody executing SHM

a = -to* x

271s/ T = —

(0

\ PHYSICS XI_(Sul>]rcHvr)

261Multiple Choice Questions ^he dkUnca cov«f«d by .. body Inbody:

(a) 1° cm

(c) 15 cm

rs!r* “““““r***p«— »/-— «*«-(b) 8 1W (d) 4 j

If the length of second pendulum Is L, then length of pendulum having a periodi sec will be:(b) 2L(d) 1/ 4

on** tomP^Hi Vibration It 20 cm. What it the amplitude of tbe

(b) 'ftrnW 7-Scn-i

r. / .- msw*ws J wA >artfarw aft rim . • «. 1 u-\ ( * ) ths fflwci a/i.YHVf

1. 'N? a^veforr*' o' 5 " V s|j) Standing wsve(e) Square wave

(b) Sine wave(d) None of these

Force need to produce and extantion of one metre in a spring is called:(b) Strain(d) None of these

When the amplitude of oscillation is doubled then energy of mass spring system becomes-(b) Four times(d) Six times

A simple pendulum suspended from the ceiling of a lift has time period T when the lift Is atWhen lift falls freely, the time period is:

Infinite

2 J( to(c) 1 J

2. , Vla) Stresslc) Spring constant

L / 2(to(C) *L

14, A spring of spring constant K IIs cut Into two halves. Then the(b) K /

3. spring constant of each part will be:K(a) Double

(c) One halfM s (d) 2Kto 4K

15. The time period of the hour hand of a watch is:

(a) 24 hr(c) 1 hr

16. Which of the following quantities is doubled on doubling the amplitude of a harmonic oscillator’(a) Total energy(c) Maximum velocity

17. Which of the following characteristics must remain constant for undamped oscillations of the

particle?

4. rest. (b) 12 hr(d) 1 mmT(a) (b)

g(b) Kinetic energy(d) PE.(d) 8(c) Zero

T5. In SHM, at extreme position

(a) Velocity(c) Kinetic energyThe working of(a) T.V.

is maximum(b) Accerelation(d) All of these

is not based upon the principle of resonance:(b ) Radio(d) Bulb

(b) Phase(d) Velocity

Time period of the simple pendulum at Karachi and at Murree are related as.(b) TK < TM(d) 2TK = 3TM

l!l' A simple pendulum is oscillating in a lift. If the lift starts moving upwards with uniform acceleration,

the period will:

(a) Acceleration(c ) Amplitude

6.(a) TK > TMW TK = TM(c) Microwave oven

Natural frequency of simple pendulum varies inversely:(a) Its mass(c) Square of length

7.(b) Length(d) Square root of length

Total distance travelled by bob of simple pendulum in one vibration is:(b) Square of amplitude(d) 4 times of amplitude

(b) Be shorter(d) Can't say anything

(a) Remain same(c ) Be longer,n order to double the period of a simple pendulum:8.

(a) Amplitude(c) 2 times of amplitude

When K.E. of SHM is maximum, its:P E. Is zero

(C) Restoring force Is zero

In damped harmonic oscillation, which on* f them decreases?Amplitude of vibration

(C) Both amplitude an(J energy

(b) Its length should be quadrupled

(d) The mass of its bob should be quadrupled(a ) Its length should be doubledto The mass of Its bob should be doubled

9. ANSWERS(b) Acceleration is zero(d) All are zero

(to 10. c9. d8. du 7. d6. d5. b2. c 4. »3.1) 20. b19. b18. b[1.1, I7.i-16. C15. b12. d 14. d13. <110.(b) Energy of vibration(d) Neither amplitude nor energy

i( to

J

262PHYSICS XI (Subjective^SchoolShort Questions of Exercise 263

£T = 2n /—gQ.7.1Name two characteristics of simple harmonic motion (i) Effect of doubling the length

When length becomes double, the tiExplanationIf £ =2 £

(D.G.Khan 2005,Fsd 2005,Mtn 2006,Bwp 2007-2008,Lhr 2010. me Period Increases V2 times .2011,6Ans. Characteristics of SHM

• Restoring force is directly proportional to displacement from mean position.• Acceleration Is directly proportional to displacement from mean

mean position.• Total energy of system is conserved In SHM.

O.7.2 Does frequency depends on amplitude for harmonic oscillators?

ifposition and Is directed toward T = 2nsthe

T'= V2

(Mtn 2005-2009,D.G.Khan 2005,lh T'= -Jlj(ii) Effect of doubling the mass *

When mass become doubled the time period remains sameExplanationAs t .me period of simple pendulum is independent of mass So it does not change with mass.

r zoo,)Ans. No, it does not depend upon amplitude of harmonic oscillator.

ReasonIn case of simple pendulum is

I g27i V £

Q.7.6 Does the acceleration of a simple harmonic oscillator remain constant during Its motion? Is theacceleration ever zero? Explain?In case of mass spring system

f - JLjE27t \ m

(Federal 2004, Grw 2005, Lhr 2010-2011)Ans. Np it doesnot remain constant.

ReasonThe acceleration of the body executing SHM is

-* -»a--CD

1 x

These equations show that frequency of simple harmonic oscillator is independent of amplitude.

Q.7.3 Can we realize an ideal simple pendulum?(Rwp 2005, Mir Pur 2006, Bwp 2006, Lhr 2008, Grw 2OO9-z010 *c' ' (1)

No, we can not realize an ideal simple pendulum.Ans. acc- xThis shows that acceleration varies directly withdisfitocsmealZero Acceleration .„«/Won where the value of displacement isAbove equation shows that acceleration is zero at rneanj^zero (i.e.x = o).

*- -.---— — ”*”a **driving force?

(a) Phase Angle , as motion of point executing SHM. Itit is the angle which gives the jj(sci2£S2J£2 -determines the stateofmotjoaof vibrat-ngbody.

Reason I .An ideal simple pendulum consists of point mass suspended by massless and [nextensi!>l£ string

practice, it is not possible.Q.7.4 What is the total distance traveled by an object moving with SHM In a

time equal to its period,if its amplitude is A?(Fsd 2006, Bwp 2007, Rwp 2008, Mir Pur 2009)

The total distance covered by the body is 4A.Ans. (Sgd 2005,Bwp 2008,Grw 2008-2009,D.G.Khan 2006)

ExplanationTime period is the time during which the vibrating body completes one

round trip. In one round trip

Ans,

A * *Total c stance covered = A+A+A+A = 4Ael* (•>)t hapP

What happens to the time period of the simple pendulum if its length is doubled? ^is doubled? ** it is actually the angle which the rotating vector makes with reference line.

Q.75the suspended mass 2010)

(Lhrat 8,Bwp 2.oo7, Rwp 2008, Mir Pur 2009, Lhr 2010,Grw

eriod of a simple pendulum is,The time pAns.

?'d.a

264

pHYSICS- XI (Subjective)Q.7.8 Under what conditions does the addition of two simple harmonic motions

which is also simple harmonic?The addition of two simple harmonic motions produce a resultant, which iwhen,

ConditionsThe two simple harmonic motions have;

1) same frequency

2) Same nature (i.e., mechanical waves cannot be super posed with electrom3) constant phase difference

265Explain the relation between total energy,with SHM.

. The total energy for a body oscillating with SHMAH5* explanation, At extreme position, the whole energy is in form of P.E., At mean position, the whole energy is in form of K.E., At any point between mean and extreme position, total energy of si

p.E. and K.E.nj.UDescribe some common phenomena in which the resonance plays an important

Potential energy and kinetic energy for a body oscillating(Federal 2005)

Q.7.11Ans. 's alsoalways remains nstant .

>le harmonic oscillator in sum ofagnetic waves)

role?Q.7.9 Show that in SHM the acceleration is zero when the velocity is greatest and the velocity |sthe acceleration is greatest? (Lhr 2008, Mir PUAns. (a) Acceleration is zero when the velocity is greatestIn simple harmonic motion, the instantaneous velocity and acceleration can be expressed as

zer° whenr 7009)

(Grw 2005-2009)Ans.Tuning of radio (Electrical resonance)Tuning of radio is a good example of electrical resonance. We turn the knob of a radio. It changes thenatural frequency of electrical circuit of receiver until it becomes equal to the frequency of transmitter.So resonance is produced and energy absorption is maximum. Hence a station is tuned.

2V = a = tuxMotion of the swing (Mechanical resonance)A swing is a good example of mechanical resonance. We apply a periodic force on swing. When thefrequency of periodic force becomes equal to the natural frequency of the swing, resonance is produced.So energy absorption is maximum. Hence, the amplitude of vibration is increased

Q.7.13 If a mass spring system hung vertically and set into oscillation, why does motion eventually stop?(Mir Pur 2004-2005,Lhr 2006,Mtn 2009,Lhr 2009)

At mean position (i.e. at x = 0)a = io2(o)

a = 0 (min)

V = «A/.V;- A-

V = (ox0 (max)Thus at mean position acceleration is zero when velocity in is greatest.

At extreme position (i.e. at x = x<,) Ans. It eventually stops due to damping.ReasonWhen a mass spring system vibrates, it gradually loses its energy in doing work goainst frictionalforces So amplitude of vibration becomes smaller and smaller and hence the motion eventually stops.

a = toV ( max)V = U\/A: - V;

V = 10 (0)V = 0 (min)

Thus at extreme position, velocity is zero when acceleration is greatest.

Solved ExamplesSo in SHM when a = 0, v is maximumAnd when v = 0, a is maximum

A block weighing 4.0 kg extends a spring by 0.16m from its unstretched position, i he block isremoved and a 0.50 kg body is hung from the same spring. If the spring is now stretched and thenreleased, what is its period of vibration?

Q.7.10 In relation to SHM, explain the equations;( ii ) a = -to 2 x(i) y = A sin ( c t + 4>)

Ans. Civ'n Data:y = A Sin (cot + 4>)

• Wave form of SHM is sinusoidal

• y = instantaneous displacement

" A = amplitude- 4> = initial phase angle

• cot + 4> = phase angle

Acceleration of SHM depends upon directly proportional displacement and directed to

0)Mass of the block = mi = 4kgLength of the stretched spring = x = 0.16 mMass of the body = m2 = 0.5 kgT» Find:Period of vibration = T - ?

'“ Ution:(ID The formula for time period is

T = 2nTo find value of k, we use the Hook's law

if (1)position. a = instantaneous acceleration. x = instantaneous displacement. angular frequency

266pHVSICS - XI (Subjective)

F = k xor k = —Xor

267

" ring, whose sPr*nS constant is 80.0 Nm 1 vertically supports a mass of 1.0 kg in the rest^ S'tion. Find the distance by which the mass must be pulled down, so that on being released, it

P°S, ass the mean position with a velocity of 1.0 ms'1.

Data:

i

As F= mtgx

4 x 9.8k = Giv*11Spring constant = k = 80.0 Nm-1

Mass = m = 1.0 kgVelocity of mass = v = 1.0 ms-1

0.16 -ik = 245 NmPutting values in equ. (1)

0.5 To Find :

Calculation:T = 2 x 3.14 Distance by which mass is pulled =245T = 6.28 x 0.045fT = 0 28 s

V = XoCOAs

ThusWhat should be the length of a simple pendulum whose period is 1.0 second at a place h9.8 ms"1? What is the frequency of such a pendulum?

W m Is x0 = vorGiven Data:

Putting values, we get

x0 = 1.0Time period = T = 1.0 secondAcceleration due to gravity = g = 9.8 ms"2 [To

UoTo Find:

Xo = 1.0 X Vo.0125

Xo = 1.0 x 0.11Length of simple pendulum = l = ?Frequency of the pendulum = f = ?

Calculation: p<0 = 0.11 rriUsing the formula

fi Exercise ProblemsT - 2n

Squaring both sides 7.1 A lOO.Og body hung on spring elongates the spring by 4.0cm.when a certainobject is hung on the spring and set vibrating, its period is 0.568s. what is thema8.i.of the object pulling the spring?

T2 oCooGiven data:- Hi o<3tor

An1 Mass of the body = m = l 00g =-l^L = o. lkg

Extension produced in the spring

Time period = T- 0.568sec (when mass in is suspended)

Putting values, we get 1000_9.8 x (1.0)2

4 x (3. l 4)2

9.8 x 1— = 0.04m TT«w= x = 4.0cm = 100

/ - 4 x 9.85 To find.9.R

Mass of the object = m / ?39.4

^•culation:f - 0.25mNow frequency

According to Hook’s lawF ~ k x

Thus kx * mg

lf - - also F - W - m gT1f - —1.0

k\

^ -. f

f .• ' •A- BH

£hapter 7 , j a r's PHYSICS M ( Subjective )

EZHEN^I(i ) when m is attached to th

268 269

Oror

.v e spnng .ihen

putting values, we get

, 0.1 *9.8k=0.04

k - 24.5 NmAs the time period of mass attached spring is

Putting values, we get

•i T=2x3.14

&T 2*Squaring on both sides Maximum s

T l - 4 n* AOr

4iJlotting x alucs. wc get

m. 24 5 v (0 ShftV4 x ( V 14)

• -=50 crVs5

m* •0 200 kg Z J \n 8.0kg bod* execute* SUM with amplitude 30cm. The restoring force is 60N. When thedUplacQBKfil Is 30cm. Findi (it period

m \ccclcrution . vpeed , kinetic energy and potential energy when the displacement is 12

In 200 cm{Or

V K'ad of 15.0g elongates a spnng bx 2.00cm. If body of mass 2*4 g is attached lo the »pnai*nset uito xibraiton w i t h an amplitude of 10.0tm ‘maximum « p ed of its vibration .

\2cm.

w h a t will be its (itpenod (»0 springtin n data:

of body = m =$ 0k £Amplitude = \ = 30cm = 0 30mResu>nng fcrce=F= 60SDisplacement = x =30cm = 0 30m

Given data:I v\ad - m * 15 Ogm - 0 015kgExtern.c« pcwiuccd -\- 2 Ocm -0.02mMa*> attached to the spnng •m - 2*4gm -0 2^vg

Amplmak •x*- 10.0cm -0 10mtil Period = T = '0 » ) Acceleration = a = ?(in iSpced = v = 'U > ) K E = ?(v) P.E = ?

When the displacement. x = 12cm = 0 12

To find:(i) Ume penod •T * '(u) spnng constant - k ?

tin) maximum sreed •ve - *Period

nlmg to Hooks law . F =kxc(u) Aocorcmg to \ \ »v s law

F « k\ ikx -mg

k.-l Fa k=—n w s

IO r 6 0k_

0JOfc*200Nm

x »Or•\ xahstSk we gc*

k - - 0 1 ** 3

ym

270 * PHYSICS XI (Subjective)

sc^ 271Now using the formula f * » r time period ol mass prinj: system

K . l- = 7.56 J( ,r2

kPutting values, we get formula lor P t. is

I beXI 5 2*.1.14

I 6.2 K - ->/OTO4I 6.28 - 0.2I l .25(isec

I )second

2p.n = - k \2no *)

:vP.E -"200«PT^P.E = 100^) 1 ) 144

Or

( i » ) \ nclcr4lion: X .4.0kg U dropped frid the maximum did

k A block of max*k •I960 Nni Hi

rom a height of 0.K0 m on to a spring of spring constantancc through which the spring will he compressed.> "

(.nrnd*'*:

\ s I Ia c » \ m

byk yt the block m= 4kg!h OJfflm

Spring constant - k ~ !960 Nm

^ Maximum distance B X ?

Or assva 1 (eightm- i

Mulling \ allies, we gel200 |o find:- 0 1 2a .»

H1 Oin sec ( ilculation :i i P.E =mgh

P. E M.O * 9.8 * 0.80 1

P E * 3l .36joulcsW hen the block is dropped on spring, the spnng will be compressed through Maximum distance

that gravitational P. H will be converted into elastic I* h .gravitational P.E•clastic P.E

mgh = ikx;

AsNegative sign shows that acceleration m directed n » \ \ ,od > the me ui p- - H. n

( iii ) Speed:I he speed of the bod \ executing SI l \1 is given b \

r~V l'» v \ -N

So

bI luts.

Putting values, we get Putting values, we gel

31.36 2- x i960 * x.J*>A.

31 36= 980 * xj3 I 36

Or

Or 980x.1 -0.32Or

Or R = 0.l8mj15 A simple pendulum is 50.0 cm long- Wh.l vs ill be its frc.,«encv

*= 9.8 m> J?

wb <„f \ ihrat ion at a pile*

- ( - 50.0cm = O.50m

= 9.8m/ s2l ength of simple pendulum

Acceleration due to gravils - gI ofind:

sfS f

frequency of simple pendulumA

272

PHYSICS - XI (Subjective)Calculation: 273For simple pendulum, formula for frequency isTotal mass = m = mj + m2m== 1300 + 160 = 1460%Spring constant of each spring = k =20,OOONm'

Combined spring constant =k=4k =4*20.000=80

Frequency of vibration = f = ?

putting values, we get

2x3.I 4/o.509.81 m 1 fbecause they are connected in parallel )f=

To fmd:

Calculation:1f= — % 19.6

6.28k11 f =—f= x4.427

2TC V m6.28Putting values, we get!f=0.70Hz!

1f =2x3.14 V 1460" 6 A block of mass 1.6 kg is attached to a spring with spring

constant 1000 Nm"1, as shown in Fig. 7.14 The spring iscompressed through a distance of 2.0cm and the block isreleased from rest. Calculate the velocity of the block as itpasses through the equilibrium position, x=0 if thesurface is frictionless.

f =— V54.794NL/1fc=—6.28HzfGiven data:

Mass of block = m= 1.6 kgSpring constanr=k= 1000 NmMaximum displacement = xc = 2cm =0.02m

Find the mplitud frequency and period of an object vibration at the end of a spring, if theequation for its position, as a function of time, is

x=0.25cosf — It“ uwhat is the displacement of the object after 2.0 s?

7.8Fig. 7.14-1

To find;Velocity at mean position = vo =?

As the velocity is maximum at mean position, soCalculation: Given data:

x=0.25cos — tk 8v =xn0 0Time = t =2secAmplitude = Xo = ?

mPutting the values, we get

To Find:1000vo= 0.02 Frequency = f = ?Period = T = ?Displacement = x = ?

1.6

vo = 0.02 x>/625v

0 = 0.02 x25 Calculation:Amplitude:

As given displacement is •

x = 0.25cos ( — )t

vo = 0.50m/s

(1)A car of mass 1300kg is constructed using a frame supported by four springs. Each spring h^spring constant 20,000 Nm . If tw o people riding in the c? r have a combined mass of

the frequency of vibration of the car , when it is driven over a pot hole in the road .Assumeweight is evenly distributed.

Mass of the car = n?, = 1300 kgMass of two people = m2 = 160 kg

7.7 8find And general equation for displacement is(2)x = xocos cot

Comparing equation ( l )and(2),we get

1 - = 0 -5m|From above equation , angular lrcqu.ncv is

Given data:

T*' *I-*

74

rcqucno :

Chapter 8Or

I 11/ WAVES16imc period:

Now T = -/ Learning Objectives jir=17

/ 161' = 16 see] Recall the generation and propagation of waves.

Describe the nature of motions in transverse and longitudinal waves.c.11 ulation of displacement when t 2sccon

Understand and use the terms wavelength, frequency and speed of wave.uumj due in equation ( I )

Understand and use the equation v = f>..v=0.25 eo>( — ) x 2

8 Understand and describe Newton's formula of speed of sound.Derive Laplace correction in Newton's formula of speed of sound for air.

.v=0.25 cos(— )Or4 Derive the formula v = v0 + o.6lt.

Recognise and describe the factors on which speed of sound in air depends.x 0.25 cos 45"Or

Explain and use the Principle of superposition.Or .v = 0.25 * 0.707Undc- rstand the terms interference and beats.Or A =0.1768Describe the phenomena of interference and beats giving examples of sound waves.

licncc |.v = 0.1S'/ /]Understand and describe reflection of waves.

Describe experiments, which demonstrate stationary waves for stretched stings and vibrating air

columns.Explain the formation of a stationary wave using graphical method.

Understand the terms node and anti-node.Understand and describe mode of vibration of string.Understand and describe Doppler's effect and its causes.

satellite and radar speedRecognize the applications of Doppler's effect in radar, sonar astronomy

traps.

.> t

pMVRlCn xi (Nuiij* . iivt- )/ ( !

Chapter No. 8i/JjpIV*V * /> O'* ""W fa truHijerrtd from

(!,<•! . If ! niiU ^( }nli ill urn. > luuhpo,t vnergv without u**‘

t natur* of wave may be different, but the mechanismenergyis sameTypes of WavesWflves are of three types.,| Mechanical waves

lk Mnvs which " material medium far theft htonagatlonmjhliniMl waves. OR

rk „.uves which propagate by the oscillation ofhanical waves

for example

Water waves, sound wa

2) Electromagnetic 1

[he waves which arc produced duej to oscillating electric and magnetic fields andthev require no medium for their propagation are called electromagnetic waves.Thesewaves propagate due to oscillations of electric and magnetic field.For exampleRadio waves, light waves,micro waves, x-rays etc.3) Matter wavesThe waves, in which the energy transfers in the form of kinetic energy of veryast moving particles, are called mater waves or de Broglie waves. These wavesre associated with particles in motion.

I v 'lace tov * • - .\ - > oihpoitiny m

I'/ wl*kh itw 3TZ1 titporu

« 3-, - S >.i JiiDT 111|

in i|Pm utP!T’

H1•— • Ai

r-,1141 are culled1— ^Uz5 r uterial purtides are called/ ma%* o a

2IPa

You K/tOW 3|l me Uit/asonK waves areparticularly useful forundersea communication anddetection systems. Highfrequency radio waves, usedin radars travel just a fewcentimeters in water, whereashighly directional beams ofultrasonic waves can be madeto travel many kilometers

isIsO “ I?3

53

aS—i. 2 f.s *3

£35^ - «

< V u.5 5>- o.— =»2

” a *— JMia>* >as

£ -sJ -rz -

> § §c Radioelectromagnetic waves andultrason.c waves are mechanicalwaves. The mechanical waves(ultrasonic waves) can producedisturbance in matter andtherefore, can travel for longdistance through matter (water )

without any loss in energy - Theelectromagnetic waves travelthrough matter for very shortdistance and lose their energybecause they are absorbed.

waves areu 2 -E -« =~ S-J — onG _E <V•J

for exampleWave associated with the motion of electron._L±S 2. =a >

r 3 O g o >*U — £> = > N What are progressive waves? Give its types.|I -5 « m ar m— ~

Progressive WavesThe waves which transfer energy' by moving away from the source ofdisturbance are called progressive or traveling waves. i»|r— ^Sampleonsider two persons holding the opposite ends of the rope. Suddenly one

^s°n gives a -erk tQ the rope The distur5ance in the rope produces a pulse^

f

.,ch moves toward other person.When this reaches the other person it pushes - if7

Jhand upward. So the energy and momentum transferred from one person^

r‘Vards other person.This is an example of progressive wave.

Jerking of hand is its source and rope is the medium for the propagation of- wsves.

iI-— -

"•nds of waves- two kinds of pro6 assive waves

278

|ar*s PHYSICS- XI (Subjective)L°ngitud(ii ) Longitudinal waves(i) Transverse wavesTransverse WavesThe waves, in w hich particles of the medium ore displaced perpendicular to thedirection of propagation of waves are called as transverseTake a loose spring coil (slinky spring) for illustration of motion of source in

generating waves in a medium. Slinky is the soft .spring which has small initiallength hot relatively large extended figure. Consider a horizontal spring systemwith its one end fixed. When the free end is moved from side to side, a pulse ofwave having a displacement pattern as shown in figure, which will move alongthe spring. This shows that displacement of particles is perpendicular to thedirection of propagation of wave, hence transverse waves arc produced.Longitudinal Waves (Compressional waves)The w aves, in which particles of the medium are displaced along the directiontf propagation of the waves, are called as longitudinal waves.If one end of the spring is moved back and forth, along the direction of thespring. Then the waves are produced in which displacement of the spring is alongto the direction of propagation of wave and such waves are called longitudinalwaves, as shown in figure.Note

279

Crestm "I»K the mean level /, „,Trough

called<u

The portion oj the mm- hchw its mean level /r mil /

Amplitudetrough.

mplitude is the maximum dispiacemen, of poin,|

Creslcrest. CAwaves.

Througha crest or a trough ofThe a

the string.Wave LengthThe distance between two consecutive crests or tw„

,,,/re length, it is denoted byX. ° trough is k,Time Period

^„cal,ed time

Which produces it. * *° be the t,me Period ‘he oscillator

Transverse waves Amplitude -fi*edend

town

i - o

n nM 1 r# -Longitudinal waves

*Speed of WaveWhen a wave progr sses, each particle in the medium performs SHM. The timethat the crest require: to moves a distance, of one wave length is equal to thetime required for a point in the medium to go trough one complete oscillation. IfV be the speed of wave, then

Distance covered

wavesFX* -M

W-V = Interval of time(Why, sound waves in air are longitudinal in nature?)

Both types of waves can be set up in solids. In fluids, however, transverse wavesdie out very quickly and usually cannot be produced at all . That is why; soundwaves in air are longitudinal in nature.

v = X/T

i - =nQ.2 What are periodic waves? Also discuss its different types?Where f ir. the frequency of the wave which is same as the frequency of oscillator(crest or trough) which produce it.Wave ProfileRelation between path difference and phase differenceConsider the snapshot of the periodic waves moving through the medium. AsanY distance x from the reference point then phase difference can be described

<J25>,A IPeriodic Waves

The waves which are produced by continuous and rhythmic disturbances in amedium are called per italic waves.These may be transverse or longitudinal in nature. A good example of periodicwaves is an oscillating mass- spring system which executes SHM.Transverse Periodic WavesThe periodic waves in which the displacement of particles of medium isperpendicular to the direction of motion of waves are called transverse periodicwaves.

A

as

P

Points C and C, as they move up and down are always in the same state ofOrations, (i.e. they have identical displacement and velocities). There are manyD0|nts or particles along the medium which are vibrating in phase. The pointsSitouted from one another through distanced, 2X, 3A. ... ore oil in phase w ithi h other.Some

Experimentlet us consider a mass spring system which can vibrate vertically as shown infigure . A long string of uniform thickness is stretched hor,?ontally and its one endis attached with the oscillating mass m. Due to oscillation of mass spring systema transverse wave is produced in the string.

AI

Points are exactly out of step, for example, when point C reaches itsMinium up wall displacement, at the same time D reaches its down wall®*plac

l>obu separatedfrom one another through a distance ofV2.»J2.M2.... <*v"Pixtsilc in phase.

V

ement.

The wave appears to be traveling on the spring, from its one end to the other. Inthis case each part of siring vibrates at right angle to the length of stretched

The crest and troughs are being repla - . d 1.7 one another periodically andstringwaves appear to be traveling.

r09

? ••r

280

vb0„,

2*'Longitudinal Periodic WavesThe periodic w.iviu In whk h p.utk le *. of the medium vibrate* along the cflrei tjr>„of motion of waves are called longitudinal periodic waves or compression^periodic waves.ExperimentTo explain longitudinal periodic waves, we take an example of a springwhich is suspended by the help of threads. Longitudinal wave is producedin this spring by applying horizontally varying force at one end of thespring. This force produces compressions and rarefaction in the spring asshown in the figure.In this case the various parts of the spring vibrate along the length of the spring(or along the direction of motion of wave). When spring was undisturbed then allthe suspension threads were vertical. But when the longitudinal wave isproduced in the spring then these suspension threads are displaced. Theirdisplacement is same as the displacement of corresponding parts of the spring.The graph of displacement of various parts of spring and corresponding values ofthe distances of these various parts of spring, from its one end is shown in thefigure.

ulcjUUo'i "< inoiiuhJi </» 1

^ to a****® rvxjuiw w'Lperalu/tt «j ihe air during a comprmiun',.todutimai change)

,< on,.- ->i» HnKS ,

fltlt.vioOii.ii

$0 PV = constant

the pressure increases from P toSts&3a, jtJ&ZMilt'-r'jpfxrMiurnrf-.iiJirIfpriQlau

rfOUt1-rVt

r>5Ut

Vi/henyto V -AV. According to Boyle's Uw,

PV = (P + AP) ( V - AViv

•s 'rr/r

PViPV - PAV + AP V - APAV

Since changesneglected. Hence above

PV = PV -

-vacAP and AV r th< So their product APAV can oe

ition 25*GIG332Ar072Helium0 = -PAV +

PAV = VAPVAPAVAP

OR ’-26C*far

ORFor Your Information

AP volumetric stress values DI constant[whereP =-0R = H]A V /V A V / V volumetric strain Types of Gas ! T

P = ESo equation (V becomes

(3)ORWhat are the factors on which the speed of sound depends upon? Whatwas Newton's formula for the speed of sound? What was drawback init,how it was corrected by Laplace?

Q.3 Monoatomrc 1 e~

Dramatic 1 40

Polyaiomic ' AS‘

Atr T.P, for air P = 0.76 mHg =i.oixi05 N/m2

p =1.29 kg/m3andSpeed of Sound in AirSound waves are longitudinal waves and their speed depends upon

compressibility (i.e. elasticity ) of the medium• inertia (i.e. density) of the medium.

If E be the modules of elasticity (reciprocal of compressibility) anddensity of the medium, then the speed v can be expressed as,

1.01x 15sv =

1.29v = 28om/sec

experimental value of speed of sound is 332m/sec. The theoretical value is“ o0ut •ess than the experiment value.drawback in Newton's FormulaD ing a compression the temperature of air does not remain constant but

"creases .i.e. it is an adiabatic change.LaPlace Correction[apla«assumed that compressions and rarefactions in air take place so rapidly

r *! heat °f compression does not able to transfer to the neighbonng cooler

ieB,°ns- Therefore the temperature of the medium does not remain constant.,s ar» adiabatic change.s case,Boyle's law can be takes the form

H r-A A:r: --— — -v

A^), so

be the

Speed of sound in solids is much greater than in gasesReason

Since molecules are closer in solids than in the gases, so hey respond more

- quickly to a disturbance.

In other words, so the speed of sound in gases is smaller hah in solids because

the gases are more compressible and thus have smaller modulus of elasticity.Newton's formula for the speed of sound i< airIf E be the modules of elasticity ar d p be the density of the medium, then the

speed v is

Hi

PVr = (P+AP) (V/-AV)rE (1)v =

\ P

f 282

C Molar heat capacity at constant pressureMolar heat capacity at constant volume

PWhere r

SirlCCYv)' y

/

V c <,n0•X (l > 4-Mut

n) Efl/ iH<i. >

COAtAVP ! P AP) Ic >' . ti lociV . )< .

how-$By Binom e expansioni ; «

( n IIX1

(1+- x) s 1 -mx +n — f^ J

\AP~~ ( P-AP) ! - i i . A \

Hjiuic auumgliei powers ol —A.-ASi) ~h neelecl iney v V

i A VP - P-yp- APAY+ AP - 7 -

V /

\A P A Y\P and AV both are smallNiru e so neglecting the term y -\

NoteThe spdensity(3) Eff.Whendensity

AVP P -VP t APso,

I r

V* \4 l \

OR AP TP JTenVAP \P S’IC'•N71* [ whore - - L

V\ \\i /\ shamAsyP - t

So equation (4) becomes

Ti’So the s1 letV - I —

y

f At S T PP = 0.76 -mHg = 1.01 * 10s N^n2 p - 1.29 kg/m3,

Value of y is dif ferent for different gases. For 7n?rY=l41( for diatomic gas)SI-r,M -- i —" \ :

AVI H x 2S0v - 333 ni/sec

which is close to the experimental value of 332m/scc .

v =Di'viejjlng t

How the variation of pressure, density and temperature effect thespeed of sound in a gas?

Q- 4V

Effect on Speed of Sound in Air(1) Effect of Pressure

vl*\ -Av \ •'

\wMibjccti

proportionalf' ,d density of gas also increaseVof Density

ve)sadensity is directly

to thes> so the

‘P^ssure.— **nen pressure ofspeed ol sound remains same.i)

c0nstant temperature and pressure ofeloeity is inversely proportional to thesmaller the density, greater the

::*4 gases havsquare

speed,i

! mg sameroot of their

value of y[flOVVS, / theI >

nerisiti!

i - e. les whichy P /

(b)

V P To Uyr* i9**r9tt>r

iv = x/ yi’ r e candipVP

As themechanicaltransfer energytherefore cause fltckcrimi ofcandle (lame.

1v = constant sound/Pwaves arcwaves ihc\ c«m

amil i

V oc -

VP -.if — •V

AV -t:V Cut'0- *4

- AT

lote t-.-0'

'

if speed of sound in hydrois sixteen times

j)Effect ofiremperatureVhen a gas is heated at constanttensity is decreased

-- , ... . -"A..yen is four limesas that ofoxyye/?

to its speed ihisity of oxyye/ j i in ox\ycn hemuse:

pi

pressure then its volume is increased and

yPi ftv = ft

Pihe speed of sound is increased with the rise in temperature.

v0 = speed of sound at 0°C

vt - speed of sound at t°C

Po = density of gas at o°C

Pt = density of gas at t°C

So d)V < »

yP (2 )And V'f

P,%g equation (2) by (1)

yP / pvI

yl’/ PoVo

(3)P„v f

P.V0

284 . pHYSICS - XI(Subjective)Settop 285If V0 is the volume of gas at 0° C and Vt is the volume at t C. Then

V, * V0 [1+ pt]Where p is the coefficient of volume expansion. For all gases, its value is

about

state and explain on principle of super position. For Your InformationQ.5WbVM 1

Superposition Principle

Ifa particle of medium Is stmulta„eousi , ' kthe resultant displacement of , Upon h.Y a n u m f, r

Then by super position principle thPresultant dispiaceme'n't h

'

v = y>+ ya+v + Vnmembe'

a' shown in figure. When tflft ^^°f Spri"&

time of overlapping, the displacements of wavesZ 2 then durin«»«(c). After having crossed each other they again aHn “P 35 shown in figure

continue their motion along the spring in theirrZZ °ri8'nal shaP« a"d

figure (d). We can study three Important cases nfP V<? directlons as shown in

Cases of Superposition principlesu -position Principle.

StQCXTXn'r ,nd

f

Ce the phenomenon of interference

zzzszzsr * ~~ j

l Y*vm» 2A.

273Stipor POA »»<<

lV, = V0 | 1+ — -•lovv.u th Y°rt tho Mnmiltant wnv»I

V - Y, * Y,-1VV. -* Suporpoilllon of two wavnn of lh« anm*frtciuancy which nr*» oxacily in phaawv„=- [ v p =andAs

PPc \AMvn» 1JXJ.

1+—P, Po L 273

B L =I+J_*So

r«afor than tho »h«od of xn-ind(4)OR Wh.it happens when .1 |PI plan? |n,Concoicie tiles l.ister than the speed olsound?273P.

\Ahvo 1 and wav* 7 super posedRaaullant waveA conical surface ol

energy sweeps over the groundsupersonic plane passes overhead, l| nknown as sonic boomEXPLANATION:

concentrated sound Y a 0Suporposiuon of two v/avan of tho somofrequency wnich am oxnclly out of phnnn

Using equation (4) in equation (3), we have tt 1

(i)t°C + 273vtv t(5) Or1+ (a)

zzz:h-,h~-:r|'LnCh, C*SCi ‘he *^pe of a cone with the movingbodytit,ape*. Jel airplanes, missilesmovmg with speed greater than (he jperi

ol sound, propagate Shockwava ih:trad

behind them, producing a so called stmt

boom. Shock waves are longitudinal w. i

with larger aro,-.!/fudej that

Possess much pressure Both c-x/iifiswu

and supersonic objects produce this typv

shock v/aves. The sound waves producedby airplanes travel out in all directions but

may become crowded together at ihcMas the airplane increases its speed. Flg(B|

The supersonic airplanes leave a cone

shaped trail of sound waves behind it. A*cone-shaped wavefront produces a thw*effect on reaching the ground. They causjan increase of pressure ol aboul 2 P**1

per square foot for about one-hal »

second/ followed by a decrease In PreJJ“Jbelow atmospheric pressure. when .edge of that cone intercepts the S,oU

below, we hear a " sonic boom"*hlc 1

ociated with an. aircraft "bre3kWJfl(t,sound barrier" . The sonic boom

,Jthe total effect Of the concentration6

^surface of the aircraft's radiate ^ j(

energy, which would fad*,,edirections at subsonic speeds. 11 (fh

possible to hear two sonic booms ^same aircraft , one from the lea

and another from the trailing .^j

cone never intercepts the pt0’e . jti1'1thus the aircraft passengers do n

sonic boom.

0°C + 273273 VoVo (ii)Tv (c)(6)1Or (iii)T0Vo

Where 273 + t°C = T = Absolute temperature corresponding to t°C

AndThus, the speed of sound varies directly as the square root of absolutetemperature.Speed of sound in air at t°C

Q.6 What is interference? Describe its types. Also write down theconditions for constructive and destructive interference.273 + o°C = T0 = Absolute temperature corresponding to o°C

Audio generatorlike sound but

T s,InterferenceWhen two identical waves meet each other in a mediumthen at some points they reinforce the effect of each ot urand at some points they cancel the effect of each otter. x ’•This phenomenon is called interference.

S. . --=3 RA 1 =Power amplifiertV

1 +As 273Vot

Or *;273

^=5ine the Binomial expansion and neglecting square and higherBy using _

powers, we have :CROL

1 t^= 1 + +2 A 273Vo asso

1[ ••• o«rV = v + —° 546

332

=1+ nx +V

So

(as v0 = 332 m/sec]V, = v0 + 546 F|0- 8.8 (b)

°f Bound*»v«*Point Z" P* ar* P°,nt* of constructive Inferforonce

p» end P.are points of destructive IntorfensnceVt= V0 +0.6U

This equation shows that with one degree Celsius rise in^speedofsound is increased by appr 7mutely 6Trn/.scc (orTl "^^Or

Piera.286 5'z'

ts pHYSICS \l (Subjective)Types of interferenceThey are two types L

o Constructive interference• Destructive interference

Explanation , jn fjmjre. It consists pf:C°rZZlXXZZor production o.horntonic sound wo.os

of fixed frequency.An audio-generator.

A microphoneA cathode ray oscilloscope (CRO) is

input signal wave-form.Constructive InterferenceThe micro-phone is placed at various points in front of loud speakers,

n f- gure. At P„P,and P5, a large signal is seen on CRO; as shown in fig. At these

points compression meets with compression and rarefaction meets with

rarefaction. So the displacements of two waves are added up at these points anda large resultant displacement is produced. We.can find the path difference at

point P, between two waves is,AS = S2Pi -SiP,

c 287of interference f' 1 >

for destructive interferencethe f ^“ h difference is mid integral mulling (>/ , . .-i c "on

i» W"»

,WIA

'Vtveitteids of

mv interferenceAS = (2n 1) -

AS = (n + ^ )Xdm

What are beats? Explain It with the help of examplegraphically ? Also mention some uses of beats?

i

where n=0

wFig. 8.8(c)Constructive

OrQ

attached to microphone to see the as well asQ- 7

440$ < > AAMAMAAAAMAI I !

as shown

ll lic' ii two nijves aJ slightly different frequencies, traveling in the samedirection overlap each oilier then there is a periodic variation of sound betweenmaximum and minimum loudness called beats.Consider two tuning forks A and B of same frequency say 32Hz are sounded (C)

separately then they will produce pure notes. But when they are soundedsimultaneously then it is difficult to differentiate the notes. The sound waves oftwo will be superposed on each other and will be heard by the human ear assingle pure notes. If the frequency of tuning fork B is lowered slightly by loadingit with some wax, say it becomes 30Hz.No -, if A and B sounded together, a sound of alternately increasing and

ceasing intensity will be heard. Such a not is called beat, which is due tointerference between the sound waves from A and B as shown in figure below.At some instant X is the displacement of the two waves is in the same direction.The resultant displacement is large and a loud sound is heard.After time !4 sec, the displacement of wave due to one tuning fork is opposite to

displacement of waves due to the other tuning fork. As a result, a minimum

^placement is produced at Y. So a low or no sound is heard. After next V* sec,lhe ^placements are again in the same direction and a loud sound is heard

^ain X It represents a loud sound is heard two times in one second because

^ frequency difference is 2Hz.C°nclusibn

totingf °{ bt a,S per second “ e<lual 10 t,ic dlfference between frequencies of

e atically,

f! B !diimi

Coseisin

X 1 Sy 1 z

tw< — S4 4!I Resultant

= 4-X-3-Xn inanit v Formation of Beats

= K No«Condition for constructive interference dec' • nenever path di fference is an integral multiple of wave length, displacements of

w aves are added up. This effect is called constructive interference.where n=o, ±1, ±2,±3,...

tote/ 11a

AS = r\ Xf h eDestructive Interference:

At P; and P4, no signal is obtained on CRO, as shown in fig. At these pointscompression meets with rarefaction and they cancel the effect of each other soresultant displacement becomes zero. We can calculate the path differencebetween the waves

Aft<t h edispthe*teiiAt PthefAS = S2P2 - S1P2

= 4X - 3- *

Ihi2

%= -A MathSimilarly at P4, f«- fB = n = number of beats per second

Cm* frequ?n'V difference is greater than 10H1. then it is difficult to recogmzeAS = S2P4- S,P„= 3U - 4 X

= -U% 1,565°f beatss.\tfM re used to;

1

PHYSICS - XI v-abjecthe<-szr.~er~ suer as :arc y *: ' -' -^5* * « = ": = =§= T

3 -cce r * t -cv* .e~c-

I*3SrEC “*KuCO : ' i~ ~ tm ~ z J* CC "t “ -""1 "C zeszz 5r» “ *r*rC

i :,C -r ro r *<K *rc» r •brae « toe .i ric.cs ar«*r. r -*sc

239T*e ?r - g car re- :e *:.r.ec :: of Stationary As.es

r e s ;:enx»t'or of two'-= c r.-e of such r*o wa*es

*" s r opposite

— B~ 4 2*0 ~ =3- f g-'e , a arc b).5TO--

&! l«:i r* t*e eerier of

5 ejection of Aaves

*rw ft

¥• “ = - - H9B MOfBOH BWI WUBmm ms9m ID *.~ t *: - * :

~cC.' f«r i;K* r «c z*+'s 'f'>ne::J- -as «re-«-exir arc wa» *rgty ^r« "c4rl “ a*

:.* • r r * * -: r r j: :* : • •: • * : r ;£ - : : : * * - *. :*

*s '•e^Veaac fw n« b<x*>5Arv

: ;• ~ e_»-

ror- bOo?v3ary o* denser

JZ' -s :-s* e a o-g > r* t sc i ••^cse or-e s f » *d to a rig-d seeped or. a

ro— s_ -sce of a t i r e other en: o* r s sp-rg s h-ee to:>c a:e -Stare s to wtueft a crest or thresh « produced oil ft ftbints on the

sz rg ts e- c - :;* a cs rv e-c 3 C« -eac- ~g at z~ z 3 spr -g e*e<ts a

•:rtc or r-e *

* c support to :'c:-ce s- F -C:QT - t_ Sot the ng r >oc * “ as

e < e r*r-s t. sc c e»ets ec ~ a ar - op-po*te -ear. o- or the spring :.e to *- ch

rest i zsar - rted tTtxjg- *~ c ? ra.e oat* Viend 8 to A.

- e*ec: a- or waves bo-u oao o* a rare — ec»rc-c- e- r 8 of ergs' * w spr ~ g » r a §“ !sr- f 3*:^ eeo ts e^ z «

vc:rre " £" i i s zzzr ert to c~o - a rest Is o-oc-cec r t a-.ekjor re sc ~ c *r0“ - ts e ~ t - to ex 3 «\ -er re rest p-odjeed reac“ es :* e

:;r:r. of str “ t " ~ c * sr - g te 'c '3'e '•e: _ ^ to ~ot g .e feact o- to r e

s r * £ Sc a rest s nr'erted rac t as a rest c-~ re so^“ g

re er or ©r »;

DO You Know? /s^oe-pose to each other, we want to find out the

ssc aretnents of the points\2 B-4.5.6 and 7 shown in fig. it dear that points

*.i3 etc are at a a.stance >74 apart from each other. The resultant

- ace^ents at these are ca c- a:ed b> applying the super position principle.i cl shows that the resultant d^sc-acement of the po > nt 1 3- 5.7 at the instants

:=0- T/4.T/Z, 3T/4 and T. t can be seen that the resultant displacement is a/ ways

at a 1 the instants.‘i ci shows the resultant c splacerent of points 2 4 and 6 at instant t = or T / z .

- 3? 4 and T. These points move ^ -th max-rum d'sp’acerr.ent from mean

»it>ons.Perties of stationary

A s t a n f l i n g-w j v a c s t i^ r ni s fcrrr.-ac w t's n t w -5 a n g t h

>A.2 \.p<of the string s an integral. multiple of half wave-fenglh:

f1- There a^e points of m.ed um in stauonary wa-.es which permanenoy snow otherwise no atandingH

^ ero displacement are called nodes.

r* oomti rwo s -:ce>sive nodes are in phase with each other.

3- each point along the stationary waves vibrates with different amplitudes.L There are points of medium in stationary waves wn-ch have maximum

amplitude are called antinodes.I ff* distance between two consecutive nodes is /J 2.

• distance between one node and next antinode »s 74.• energy remains standing in the medium D tw fn

nodes r^ain at rest, so energy cannot fto* through these po-nu,

h-l-y stationary waves are a-so called standing waves

^£hsrgy osollates between P.E. and ICE.Between nodes.

e - - 5 e' d

| V W- 2 r-2” s.e-se * c - a -a-e* — ed s - e-: z ~ 3 - ~ se'

-e: jtr : s rejectee s :~ tra: : - ^ce^ goes a ph-ase ::na“ ge D* ' 30 '

I3t“ :“ c c'Cr a* I• a transverse treeing - s Denser — ec _ — s rodent cn a rarer-e: *. s ^ectec wth out s » c~3-ge « . * - se J TB: ' z ^e- e“ te

Define states =ry waves ano *x?w tn: _ -e o<ocu: i? G ve proper es of

stationary waves. Also de-~ne -ode anc :Mi -ode.

I\PL \TM>N:If L is the \ensit-. of the smer^A*>d X;, X; BO i-; 2« C*wavelengths in ihc ftgiises a. cand c respecnvrh thatmvnFig a,L = X»2 =1 >« X^o?r. Fig bv L = X-T - \:-2 =

X:=2 » Xy2from Fig c,L = X72* *x,:= 3 . x,:Thus the leagtfi of ibe mr > Lis an wtegpA (1,2 md 3>multiple ofhalf ' c

wavelength.

nodes because the. That is

C 58

<EPStationary Wave When antinodes are at their ertreme P04'00^j^eneW -^-E.

at passing through equ-l'hnum pos-bon.the who« ene gy

^ -rnr produced by e sup*r nkiom of /w identical ntfvvrsrfcein opposite iirecnom is CJJ. *wr Mn

290

y.cl,niar^ PHYSICS XI (Subjective)Commonly the standing waves are produced due to superposition of awave traveling down a string with its reflection traveling in opposite

direction.291

d Mode of Vibrationceconwhen the string is plucked from one quarter of its length

t0loops as shown in figure. If be the

frequency of vibration in this mode, thenf = Ai+ *u.

Nodes the string vibrates;th and f2 be the K3>£DIThe points of zero displacement on stationary waves are called node.

. *

The points of maximum displacement on stationary waves are called antinodes.

Q.io Show that frequencies of stationary waves in a stretched string arequantized.

Prove that for stationary waves in a stretched string fn= n f,

waveAntinodes

2

ORX2 = LOR

Thus, speed of wave vis\j = i2 X 2

Stationary Waves in Stretched StringLet us consider a string of length 0. It is stretched and is clamped at its two ends.The tension in the string is denoted by F.When the string is plucked and then released, two waves are generated whichmoves in opposite directions along the string. Both of these are reflected backfrom the clamped ends of string with opposite phase to generate stationarywaves on the string.As the two ends are clamped with rigid supports, so these do not vibrate and weget nodes at these ends.Speed of waves on stringThe speed of wave depends upon tension F in the string and mass per unit lengthm (i.e. thickness and nature of wire).

Putting value of k 2, we get

f,=

{since — = f, ]f2 = 2f,2C

F Thus when the string vibrates in two loops, its frequency is double than when itvibrates in one loop. f2 is called second harmonic.Third mode of vibrationWhen the string is plucked from one sixth (1/6) of its length then the string

vibrates into three loops as shown in figure-. If X3 be the wave length and f3 bethe frequency of vibration in this mode, then

, X3 X,c = — + — + —2 2 2

(D- v =m

First mode of vibrationWhen the string is plucked at the middle of its length then the string vibrates in asingle loop as shown in figure. Such a mode is called fundamental mode ofvibration.Distance between two consecutive nodes = —

2If X,be the wave length and f, be the frequency of vibration in this mode, then

Ni*2L N

2

u3( =2

2 2Cor x-7X, = 2C (2)So the speed becomes,Thus, speed of wave v is

v - f3 X3v = f1 X1

f,= 7O R

3vputting value of X,, we get

“5* putting value of v from equation (1) in equation (3 ] /e get

V. =— m

f3 = T72 C(3 ) v

f a= 3 —

M

»292

PHYSICS XI (Suhjcclivt)

Modes of vibrations in organ pipe open at both ends,ll5 consider an organ pipe of length ( which is ooen h. ..^Jplete freedom of motion so it acts as antinode.

h er,ds' 5 at the open ends air molecul«have

fundamental Mode of Vibration

this case there is only one node at the middle of th. 'pjn,jntlnodes at both the ends. If X is the wavelength of soSSf,

[J --L+-L 41

The frequency f 3 is called third harmonic,

nth mode of vibrationIf string vibrates in n loops then,

fn = n

293

— Unf,\V. )

curves which represent theAnd wavelength is disp

V. where n = 1,2,3,4,S

So the stationary wave have a discrete set of frequencies f,, 2 f,, 3 f,,which is known as harmonic series. The frequency f, \\ known as fundamentalfrequency, and the other are called over tones.Note vThe stationary waves can be set up in the string only with the frequencies ofharmonic series determined by the tension, length and mass per unit length ofthe string.Waves not in harmonic series are quickly damped out.

A-n =n e. As both ends of pipe are open, so there are two

nfi,

\ :4 4for Your Informant X .

2

t Or

Iff,is the frequent y of souncv = f,;, Vfi = r

th.n velocity of sound is

Q.11 How can we change the frequency of string on a musical Instrument? ORi

'iifUwThe frequency of a string on a musical instrument can be changed either byvarying the tension in string and length of string.For exampleThe tension in guitar and violin strings is varied by tightening the pegs on theneck of the instrument. Once the instrument is tuned, the musicians vary thefrequency by moving their fingers along the neck. By doing so the change thelength of the vibrating portion of the string.

putting value of A.,, we get

f,«—2CTht frequency is called fundamental frequency or first harmonicSecond mode of vibration:In this case, there are three antinodes and two nodes.If /.< is the wavelength of sound then

ft ^v2 ^* 2T = — + —4 2 4l+2 +0,

4 rx 2 = e

is the frequency of sound, then speed becomes,v = f2 X ,

nR value of A.a,we get

f, =-l

Air+*«.

iSSSiSSS^maintain» "toady ofcllldlon

Q..12 Find the frequencies produced in organ pipe when it Is(I) Open at both ends

Closed at one end

t -

OR00

provided With< idled n rcsoitnior. T,icbe in the form ol « •^periodic movement b

^by «»ou,h p|cc,e '1cr8|or «i'dpiece nets ni n lsupplies the^ ,il)0, w10 mmnlm" n,c * * ".S

Stationary Waves in Air ColumnStationary waves can be set up In air column inside a pipe or tube. A commonexample of vibrating air column Is an organ pipe.Organ pipeAn organ pipe is a wind Instrument In which sound Is produced, due to setting upof stationary waves in air column.It consists of a hollow long tube with both ends < vpen u with one end open andthe other closed. There are two types of organ pipes

Open Organ pipe: It is that organ pipe wi. e both ends are open.Closed organ pipe: It i - that organ pi; whov ne end is closed.

OR

f'utti

2vair columnmoulh-piccclip and ids uprcsonuting pip*

Or fj = —Itor

V.0) V

(ii)

mm MB

-.r— T 1

..' rT i 3K2i. .> • 4 >fa >P . * 5> t B ‘4

294PHYSICS - XI (Subject^Third mode of vibration

For three loops, there are four antinodes and three nodes. If / _, iswavelength of sound, then length of the pipe is

X, X, X,£ = - + — + — + —

4 2 2 41+ 2+ 2+ 1 V

295

50 the speed becomes.V = f, X,

Or f, = v/>Fi9. A As X,=Stati. fi = v/4(waves inboth Pnr' The frequency f, is called fundamental frequency

Second Mode of vibration

Second mode of vibration contains two nodes andwavelength, then length of the pipe is

f = A,4

I _ XX >anti-i s If X3 is the

2V 1= 3(>74) /, = 3../4/)

'M = (b)

3So the speed becomes,

V = fj >.j

U -T-

3v If fj is the frequency of sound, then speed becomesv = f3 X3

u =iPutting value of X}/ we get

*3 =^7

f3 = TT2f

vu= 3 2? OR

v[since — = 1', )2C

The frequency f 3 is called third harmonic.nth mode of vibrationIf air column vibrates in n loops then,

3v

f3= 3 (^- 1

or

or= n f AC1

So [since — = f,]AC

?!S ls called second harmonic or first overtone.|h,rd Mode of vibration:w 'r rT,°de of vibration contains there nodes and three anti-nodes. If ?^ is the

length, the length of the pipe is.CM2*

And wavelength is2f

Xn = where n = 1,2,3,4,5 ...So the longitudinal stationary waves have a discrete set of frequencies f,, 2 f„3f »,

nf„which is known as harmonic series. The frequency f, is known asfundamental frequency, and the other are called over tones.Modes of vibration in organ pipe closed at one endLet us consider an organ pipe of length ( which is closed at one end Then at theclosed end we get node while at the open end wo get anti node.Fundamental mode of vibration:

Fundamental mode of vibration has one node and one antinode. If X, is thewavelength of fundamental mode, then leng h of the pipe is.

/ = 5(i. /4) /, = 5.V4/)n (C )

Stationary longitudinal waves in apipe cloud at one end. Only oddharmonics are present.1 x5 X,

4 2 2

c =4

4OK ACi’-flffs 'sthe ^squency of sound, then speed becomes,

296 hlj«r^PHYSICS - XI (Subjective!

0mber of waves received by observerf -I

297

v = fs Xs

Putting value of Xv we get

f* ~J7

in one second isthenOR w

Casel(When observer moves towards station

| let observer A moves towards the sou “ "ary source)

rhe relative velocity of the waves and the oh! en

N0w, thenumber of waves received bv ot» ' °v + u»-

'y°bSCrverinon,»cond is

5vf' = M

U* 5 )

orV + u 1 '^r

or 4 * V’

|f5= SfJ (since -V- f, ] (using equation 1)So \ >

Fio »-1«An otwarvw moving wWt velocity u,toward* a (tstkvwy aouroo hoar* afrequency f. that n greiitar ttwjr th«•ourw frequency.

Which is the frequency of third harmonic or second overtone

nth mode of vibrationIf air column vibrates In n loops then,

fn ® n(— 1 > 1 so

Af )

fn n f, ResultThus the apparent frequency of sound heard by the observer will increase.Case II(Observer moves away from the stationary source]let observer B movers away from the source with velocity u0, thenT*- relative velocity of the sound and observer = v - u0. Thus, the number of

^ves received by observer in one second is

And wavelength isOoserver

AtK * - where n = 1,3.5 - 7, for this longques. see onazeemguides— •

nConclusionBy studying the both these cases, we conclude that the pipe which is open at

both ends is richer in harmonics.Q.13 What is Doppler Effect? Discuss its different cases.

Fifl a.io

Un observer moving with vetooty u,j rwjy bom atnoonary scum hears a(frequency /, that <s smaler than the

Jscores frequencyDoppler EffectThe apparent change in the frequency (or pitch) of waves due to the relative

motion between the source and observer is called Doppler’s Effect.Note

v-u. (Using equation 1)

^ rV

V-f,,=This effect was first observed by John Doppler while he was observing rhe

frequency of light emitted from a star . In some cases the frequency of emitted

light was found to be slightly different from that emitted from a simi ar source

the Earth. He found that the change of frequency of light depends upon

motion of star relative to Earth.

Example

v - uSince - < 1vso ffi < f

on Result

^se in6 apparent frequency of sound heard by the observer will decrease

HWhen

The pitch of whistle of an engine coming towards the platformappears to become high, to an observer standing on theplatformThe pitch of whistle of an engine joing away from the platformappears to e . ome lower to on obsr er standing on the platform

D en source moves towards the stationary observer]observer C with velocity us thensource moves towards the s t3 ' t h ,s decreased- ln1 ,s"*«* are compressed and their waveleng*^^second: * compresse(j in a ^stance equal to v2)

Different casesConsider a source of sound S at rc emits so 1, d waves having wavelength X .Let

Meed of the sound for a stationary observe is v. Then,

(\v rh<VSlCS-XI (Subjective) •

pnber of waves received by observer D in one second is

fo = Y”

""D

value of >-o from equation (6),we get^

299298 hir’s

For Your InformationWhen source and observer move towardseach other with velocities u, and u.,respectively, then waves are compressed m

a distance equal to v — u,in one second andthe restive velocity of the sound andobserver becomes v t uf In this case boththe relative velocity and the wavelengtn ofwave changes. So, the apparent frequency

Thuswavelength of sound waves for observer is ,, the

flit)*-the Hi A\ -11 ( 3)A-c =

putting thet/.At = ?-T

Ac = A ' A/w

OR Fm -t ?(,M 1» f. V

. • v ,r -1v inr„rStic,,’ is*vWhere t

n- r,«.« ?r:"* '"•iv

«,i' . V + u„v -u,=AA = decrease in wavelength in one second called Doppler shift .So V -t- u V * u'“.-J * 4 rr = - >/ V - 11,I r

US When source and observer move away eachother with velocities u, and u5 respectively,

then waves expand In a distance equal to v

u, in one second and the relative velocity

of the sound and observer becomes v - u0.in this casethe wavelength of wave changes. So, the

apparent frequency is

(4)A/v = —1'

Thus, the number of waves received by observer C in one second is

k' Tputting the value of >.c from equation (3 ) ,we get

Since

So both the relative velocity and

Jj** apparent frequency of sound heard by the observer will

Jfcrcuw v -u„ fV + ut

V_

Uo U v -u„ =the uses of Doppler's Effect. r =Write down -v + uQ.14 Yt\

f c = - m( v - UK )I

(',] Radar SystemJAOAR IS an acronym, it > sRanging.

*adar is a devicesystem uses to determine ,«rnnl;»ne

radio waves which are reflected from aeroplane

fte system.I1 reflected waves

p:;;;^ct -^ng away from the radar system

11«) Speed of SatelliteI speed of satellite can also be determiFe reflected back after colliding with the

hg) 1^PPler's shift in wave length of these radiation

> system use*1 \ Sonarl ndNavigation and Ranging)objects under water by an

depends upon the111 1^ •* a technique for detecting the Lances in water,

observed, which is"llratonic waves because they can r apparent change ' ons and mines etc Akn

** speed of the target and the ^^^e^ubmannev antisubmarine weaponsWw's shift. In this way we can locate an

^Pth of sea can be measured.

Plane approachingV 1'Pats navigate and find toodt*echo location.

f t = i (5)

t l v - u derived from Rsdio Detection AndEXeUNAllON!

during flight, theyuluavnind sound in user •

of pulses and ’he rellciictlwavesobjects provide the nai

information and iocatico t'lthe food

ceding\ince >1 . The radar

which transmits and receives radio waves

the height and speed of aeroplane. This systemand received by

\ -ufrom the i ' L 'nn? w

E (b>

f c > fSo dir

ResultThus the apparent frequency of sound heard by the observer will increase.Case IV[Source moves away from the stationary observer]When source moves away the stationary observer D with velocity u$ thenare expanded and their wavelength is increased. In this case the waves expand ina distance equal to v -u* in one secondThusThe wavelength of sound waves for observer D is

v +u*

have shorter wavelengths, then the aeroplane is

. If reflected waves have longerFig. AA frequency shift is used in a

radar to detect the motion of

a aeroplanen :i)usln;

j couplingk»' . T.R * radiations from Earth. When these

the earth. The value ofwaves

ined by sending electromagnetic

satellite, then these are received on

estimation of speed of satellite.BxMT**\mTr.'invilH'fl .

*ii)'ia*vU- 1

Ol'.vjtl <

T' IO dopph i effect CM J

I'lionifyi wotxf i!w through

l* rtene$ Ultrasound JIreqmr' i«s SMHz to “ Jdirected towards tt>e angry

rarpiver deeds the ttmthsiorml The at parent "**“JUttpeniis un the volooty ,)f’l,n /

’ti'CHidJ

/-D = w(G)f

AV U-A0 = — 4- —OR KSiM!i

= A + AA

velocity Of blood^detected by

effect by noting ncyfshift (change in

Whereii

=AA = increase in wavelength in one second c, lied Doppler shift. So

4k = -L

1

(7 )I

300PHYSICS ~ XI (Subjective)

(tv) Speed of StarBy comparing the line spectrum of light coming ( ron a distant star an<j

the light emitted from laboratory source, Doppler's shift can be

measured to calculate the speed of star with respec o Earth

JT .3 r •> moving toward', tiir «Mrth show blue r.hih]

301

between speed of

V ^\oond and temperature

^^^Tfor constructivev,= v0 *0 6lt

; away from sourcewhere

/ Condition for destructiveIn sound wavesBlue shift

The frequency of light emitted by the star inm . ( » .e wa «ri<ng

decreases) if it is moving towards the earth, a pared to the li*emitted by stationary star Thus spectrum is I towards

udvelcnglh i.e to the Blue end of spectrum,wh»ct * i * l B ue 'tRed ShiftThe frequency of light emitted by the star dec * . (i r * i.a * 4

increases) If it Is moving «» way from earth Th t\ „r-;trum i

towards the lonyrr wa\ i lrnyth \ « tow.u ' . ' < •

which is called red shift

AS = (n + - ) X

where n=o,± 112,± 3...Interferencet r

Beat frequency

transverse waves on'1 IV

rTundamentjl fraqu* 'itatlonary waves oil firing

on strlngf nth

f */ A*•!* fn = n U

where n = 1, 2, 3, 4, —4 vV

/ Wave lefifth of Stationary

' wivfS on str .nj | nth mod*}—Note n

As astronomers have alto discovered that all th< - 1, > 'r where n 1,2,3, .«na away from ut and fcfrequen|yof longitudinal

v/itjro iry W* n organ pipe. ..

their red shifts, they have estimated their speed fn = n f,f, « n

(v) Speed of Car where n = 1, 2, 3, 4,

Microwaves are emitted from a source »n form of ' rt bunts [Kh b . 7avt It t’h of longitudinal

^/ stationary waves . n organ pipeOver at oolh ends (nth mode)

KIn their way The reflected bursts uc drtc tm tht detector If xin nDoppler s '.hill i 1 »bsei veil Hy wh ch Speed of <HitUlOjimd.MtoOWtW where n 1.3,3, -car by computer program. frequency of longitudinal fn » n f,

f,attonary waves In organ pipe » n where n•t, 3/ 5, 7, ****

doted at one end (nth mode)4 /Wave length of longitudinal

Relation between velocity, *- —stationary waves In organ pipe nfrequency and w.ivelongth of where n •1,3,5-

wavesRelation between frequency

and time periodRelation between phase

dlffeteiue and path difference

Speed of sound

Boyle v law for Isothermal P\ jnstanlprocess

IPNewton's formula for speed v

1#!Boyle' s law for adiabatic

process

Speed of sound in air

302

SMI%\ 4 t -4

Multiple Choice Questions tK. • r , 4<-

Four possible answers to each statement are given below. Tick ( ) the correct answer;

Longitudinal waves can not pass through:\t -Solids(b)(a) Liquid

Gases(d)(c) Vacuumperimposed, the velocity of the resultant wave:When two identical waves su

\i -increases(b)(a) Decreasesis zero(d)Remains unchanged(c)

Waves produced in organ pipes are: 18.Longitudinal waves(b)(a) Transverse wavesAll of above(d)(c) E.M. wave

than in winter.Speed of sound in summer 19.(b) Increases(a) Decreases

Can not tell(d)(c) Remains sameNewton calculated speed of sound using 20.process.

Isothermal(a) (b)AdiabaticNone of these(c) ' (d)Both (a) and (b)

times its speed in oxygen.Speed of sound in hydrogen is(a) %4

1. <11.

(c) (d) 342Error in calculation of Newton's formula for speed of sound is about:

(b)(a) 35%10%(d) 27%(C) 16% IThe frequency range of human ear is:(b) 20 Hz - 20 MHz20 Hz - 20 kHz(a)

(d) 200 Hz - 2 KHz20 Hz - 200 Hz(C) Q.Beats are the results of:

and destructive interfered(b)diffraction of sound waves constructive(a) Ai(d)destructive interference None of these(c ill travel 'If the time required for the tuning fork to make one complete vibration, the wave in air wdistance equal to:

XX (b)(a ) Q24

(d) 2X.(c) X called:The term which can tell us the stage of vibration of the particles of the medium is(a) (b) PhaseTime period(c) Wavelength . (d) Amplitudeit is common characteristics of all types of wave motion without the transport of partld«s:

(b) Drown(a) ParticlesMass decreased(d)(c) Energy transferred , produces:

When a string, which is tied atboth ends is plucked from the centre of wavesLongitudinal wavesElectromagnetic waves(b)(a) Transverse waves

(d)(c) Standing waves

z '/

in a stretched string,ifspeed of the wave is doubled,. |

rl"* Propo,tiona|, <d)

2(a) the tc^slon will be:(c) 8 (b) 4frequency of a stretched strlTensionReciprocal of length

A string fixed at two ends vibrates ifirst overtonefundamental

Two sounds of the same frequency \(a) amplitude(C) loudnessWhich of the following phenomenon can(a) reflection(c) diffraction

- - ». * . ” - - v :

Presence of moisture in air:(a) Increases speed of sound(c) May increase/decrease thDoppler's shift measure the change inobserver.(a) Intensity

Velocity

6(a) othe:(c) (b) 1 inear density

n one se«mp.n»Square of tensioneSment.The standing wave pattern Is:second over tonesecond overtone

w»(a)

(b)(c)Id)m air must Have same:(b) Intensity(d) wavelength

not take place with sound waves?interferencepolarization

(b)M)

(b) Decreases speed of soundDoes not have effect

of the wave due to relative motion of source and

e sound velocity (d)

(b) FrequencyWavelength(C) ?

(d)ANSWERS

4.b 6, a2. c 7.c 8, a 9.b5. fe3 b 10. c

19. a17. 18. 20. d16. c14. b 15. c12. c 13. c; •• .

Short Questions of Exercise ’ v *r‘+y - r *

. > •

\ 1 •"

mon with transverse waves?What features do longitudtnal vaves have in com

(Federal 2005,Mir Pur 2006-2009,Bwp 2008,Lbr 2010-2011,Grwzor? )

Common FeaturesDOlf1 types of waves traaSBSS. e‘,CI* 7'

^u^rtinn. refraction and reflection.

^types of waves can produce In^mSS^- '' b|e f“

both these wavesF* determination of the speed of wave, from.microphone Is fed Into*.***«<*N*.^bode ray oscilloscope,^ * * thode ray oscilloscope controls.

**the 5ame adjustment of thenete7

a) Which trace represen ** freqb) Which trace represents the

Iuency?

c . rV • r

v-* » • - V*-Jo Xi

304Ipr^ PHYSICS- XI (Subjective

Why d°es sound travel faster in solids than in(Mtn 2006,Sgd 2003-2005, Fsd 2006-2008,Rwp 2006Sound travels faster in solids than in gasesReason:Speed of sound is given by:

2 apteLitv, Scbo 305

Q.8-6 §ases?Grw 2003-2004, D.G,Khan 2005,Mir Pur 2004, Bwp 2006)

Fifl.823 C 0BA E

(Lhr 2004,Bwp 2009, Fsd 2008,Gr* 2008 Wh6re 'uomDared to sa<- '"MiT P "^densitV °f+«"• Although the density of solid isgreater as compared to gases but modulus of elasticity for solids is much areefer as compared togases. So

•*°09)Ans. Loudest NoteIn fig. (d), the amplitude is maximum, so loudness is maximum for this case.

Maximum Frequencyin fig (b), the number of waves is maximum, so frequency is maximum in this case

Hence. sound travel faster in solids than in gases.

Q.8.3 Is it possible for two identical waves travelling in the same direction along a string to give rise to 3

stationary waves?(Mtn 2003-2004,D.G.Khan 2005,lhr 2006,Fsd 2005*2008. Rwp 2006,Grw 2008-2009-2010)

No. it is not possible

ReasonStationary waves are produced only when two identical waves traveling in opposite direction along thesame string superpose.

How are beats useful In tuning musical instruments?Q.8.7(D.G.Khan 2005-2006,Rwp 2005, Bwp 2006-2008, Lhr 2006- 2010-2011,Grw2011)

Ans. Tuning of musical instrumentsIn ord^r to tune a musical instrument;

Sound the ir strument against a note of known frequency. If the two frequencies do not match, beats

will be produced. Adjust the frequency of the untuned instrument by tightening or loosening the

string. vVhen no beats are heard, the instrument is said to be tuned.Q.8 8 When two notes of frequencies hand f2 are sounded together, beats are formed. If h> fz

what will be the frequency of beats?

f»+ fifi- f2

Ans.

Q.8.4 A wave is produced along a stretched string but some of Its particles permanently show zero

displacement. What type of wave is It?Ans. These a^e stationary waves.

Reason*

On i n stationary waves some points of the medium permanently show zero disp 'em'

nodes and some points show maximum displacement called ontinodei

% (fr f2)(Federal 2005,Mir Pur 2005,Bwp 2008,Rwp 2008)

Ans- Since beat frequency is the equal to the difference of individual interfering frequencies

Beat frequency is fi - fj. r . , , ,M.9 AS a result of distant explosion, an observer senses a ground tremor and then hears the explosion.

Explain the time difference?

(b)(a)(d)(c)

• -0,8.5 Explain the terms crest and trough, node and antinode7

(Mtri 2003-2004-2009, fid 2005, Rwp 2006, Bwp 2009;Grw 2010, Lhr loto-tf’1! (Lhr 2009)

ExplanationThe speed of sound is given byAns. Crest >.Craft

£Tre port on of transverse wave obove the equilibrium position scalled trougn

TroughThe portion of transverse wave below the equilibrium position Is

ca’ied trough.NodeThe points o' iero displacement on star, onary waves are callednodes.

iThrow/ than the speed of sound in gases (air ) due to

the ground tremor first andsensesAmplitude

then hears the explosion. jn cold air?8 to Explain why sound travels faster In warm air t

^^S|d 2005> Bwp 2003, Federal 2004Nod® ,Mtn 2009, Grw 2009)Antinode

A|' *«ason '

The speed of sound is given byAntinodeTr e pO'nts of maximum :placement n stationary yvaves arecalled antmodes

306 , PHYSICS - XI (Subjective)Scboi^i 307

gssnn®® , . . ,B5> s^c waxP, the beat ^ "f°Und 'ha' by l"a,'inKHz, determine the frequency of B when loaded Per SCC ' ‘hC frcquency of A “ 320

Data:

Since gases expand or) heating. So the density of warm air decreasp* Henceequation, the speed of sound will be greater in warm air than in the cold air.

Q.8.11 How should a sound source move with respect to an observer so that the frequendoes not change? ,

0

If the relative velocity between the source and the observer is zero , there will be no chfrequency of sound.ExamplesWhen the observer is at origin and sound source moves along the circumference of the cidistance remain the same and the frequency of sound does not change.

e hentheir

Source and observer are moving in same direction with same velocity.

^cording to aSGiven Frequency of tuning fork A = fA = 320 Hz

Number of beats per second before loading = n = 4Number of beats per second after loading *= n'= 6

Frequency of tuning fork B after loading = f„=

sound(Bwpzooj,**ange jn

p 2008)Ans. To Find: w..

?Calculation:Case 1: Before loading:

fA - fD = ± nAsfB = fA ± n0f

Putting values, we getf„= 320 ± 4

Solved Examples fn — 324 Hz or 316 HzBy loading B, its frequency will decrease. If 324 Hz is the original frequency, the beat frequency will

reduce. On the other hand, if it is 316 Hz, the beat frequency will increase which is the case.

So, the original frequency of B is 316 Hz.Case 2: After Loading:

f A - f B= n'or fB = fA-n'

fB = 320 - 6

fB = 314 Hz

r.xatuple 8.1Find the temperature at which the velocity of sound in air is two times its velocity at 10°C.

Given Data:Temperature = T0 = 10°C = 10 + 273 = 283KVelocity of sound at 283K. = v2»3 = v0

Velocity of sound at T= v,As velocity of sound is two times its velocity at 283 KI *hcrcforc

vt = 2v0 = 2V283

To Find: Thus f0 when loaded is 314 HzTemperature = T = ?Calculation:

• Using the formula1 *amok- 8.3

A steel wire hangs vertically from a fixed point, supporting a weight of SON at its loner end. The

diameter of the wire is 0.50 mm and its length from the fixed point to the weight is 1.5 m.( alculaic

the fundamental frequency emitted by the wire when it is plucked?

(Density of steel wire = 7.8 x 10Jkgm J)

Glvcn Data:o, 3-. _ LLV3» 283 Weight = W = F = 80 N

Diameter of steel wire = D = 0.50 mm = 0.50 x 10 1ni

D 0.5x 10'’ .. 1 <v jr = — * 0.25 x 10 m

2v TOr 283 ’v 2„ Radius of steel wire

Length of the wire = = 1.5 mDensity of steel wire * p * 7.8 x 10 kg m

T 22 - 2283

Squaring on both sides

4 = —283T- 283 x 4T- 1132KT « 1132 - 273 * 859C°

Find:Fundamental frequency r fi ?

308

Calculation:As formula for fundamental frequency is

Second harmonic \\* ?£1Um -rz 0) ( aU’u *at'wn:2f When Pipe »» open at both end*.(a)

•' rvere m is the mass per unit length of the string.

So first we bave to calculate the value of m.

.As volume of the wire = Length * Area of cross section of wire,

vo(lime of the wire = i x rcr2

Using the formula.

fn

For fundamental frequency

AndMass of the wire = volume x density

M= l x rcr2 x p

t * KT 1X pMMass per unit length of the string = m = — =

m = Ttr2 pt 340

2 x 1170 HzPutting values, we get

m = 3.14 x (0.25 x 10~3)2 x 7.8 x 103m = 1.53 x 10~3 kg m

fz = 2f,- i

f2 = 2 x 170Putting values in equ. (1), we get f2 = 340 Hz

For second harmonic n = 31 80f. = f3 = 3f,f3 = 3 x 170f3 = 510 Hz

(b) When Pipe is open at one endUsing the formula

1.53 x 10*12 x 1.5

hi 80f, =0.00153

fi - jV52287 .5

f , = 1(228.6)

fi = 76.2 Hz

A pipe ha* a length of Im. Determine the frequencies of the fundamental and the first tv

harmonics (a) if the pipe is open a both ends and (b) if the pipe is closed at one end.(Speed of sound in air * 340 ms"1)

Given Data:Length of Pipe = t * 1 mSpeed of sound = v = 340 m

f.=-4fFor fundamental frequency n = 1

lxvThus '•- IT

Putting values, we get340f|= ~~~T4 x 1

f, = 85 HzIn this case only odd harmonics are present.So, for first harmonic n = 3

ft == 3f,f3 = 3 x 85f3 = 255 Hz

To Find:fa) If the Pipe is open at both ends.

Fundamental frequency = f|» ?First harmonic = f2 = ?Second harmonic = f3 = ?

fb) If the Pipe is open at one cod.Fundamental frequency - fj - ?First harmonic- fj-?

for second harmonic* n = 5

f5 - 5 x 85f5= 425 Hz

j

310

pHYSiCS - XI (Subjective)311

A train is approaching a station at 90lunb soonding a whistle of frequency logo ..the apparent frequency of the whistle as heard by a listener sitting on the pUtfo *the apparent frequency heard by the same listener if the tram move av*av fromsame speed? (Speed of sound 341) ms ') * ', 4 lK w

Given Data.

Speed of train * U, = 90 Kmh 1 *

i ^ v.aThus

V=f,4orPutting values, we get

90 * 1000 - 23 m*'* v-200 * I 0‘« 150060 * 60

v *3 0 - 10‘m iec•lSpeed of sound » v * 340 msFrequency of source f 1000 Hz A* ihc transnuttcr emitsK

To Find:(a) Apparent frequency (when train is reaching a *uik*vcr> oKerve.') - f -(b) Apparent frequency (when train is moving away from the ob .erven - f - >

Calculation:(a) When train is approaching toward listener then uting the formula

vI • fV Us

12 Two speakers *re arranged u shown in Fig. 8.24. The distancebetween them is 3 m and they emit a constant tone of 344 Hz.A microphone P is owned along a line parallel to and 4.00 m(ram the line conacctiag the two speakers. It U found that

is beard and displayed on the CROWhen microphone is on the center of the line and directlyopposite each speakers. Calculate the speed of sound.

Cisco data:

Putting values, we gel

r -pi* )l 340 23 ) * 1000

f ' - — X 1000315

f ' - 1079.4 11/.(b) When the train is moving .sway from the listener ’.‘ter . • e ^

l nr of maximum loud

v fr *-Distance between speakers 3mTone frequency * f * 344 .HzDistance between speakers and line of motion of P « 4m

lv UsPutting values, we get

140 Tt find:r * IOOO340 * 25 Speed of sound 3 v ?

For tone of maximum loudness or the condition for constructiveinterference, the path difference must be 0,± M,±2A, ±3A,At middle point *0* the path difference between two sound waves is*ero (as ,S|0*SjO) thus at that point construction interference takesplace.

t ilculaium:340f - x 1000105

r - 931.5 Hz

Exercise Problems0 200

The wave length of the signals from a radio transmitter is l 5f'0m and the theWhat is the wave length for a transmitter operating at 1000 kHz and w i t h * •waves travel?

8.11or the next point Pof construction interference the path difference between waves should be X.

X * path difference * S2 p] — 5,/?,Now we calculate values of S1p]

From right angle triangle SiS1 pi

__“ V(3 16*^25*5m

Sc-Given data:

Wavelength of the signal- =* 1500mFrequency of the signal * f, =C )kHz *20(3 \ 103 HzFrequency or transit uc. *U\ *lr= lOOOx10 Hz

To find:Wavelength the transit ;er =X;='Speed of radio ves = v ?

i l• 1*J m

312

HKTCtbre path difference — .V, / '

X- 5- 4X* l m

This is the path difference for constructive interferenceNow v = fX

putting the values, we getv « 344 x l

PHYSICS (lubfrrti r jSchoW*Or

312' f u Uf » d:

(^*cn length ofih vre<^dbyone birdwidwMi(b) - ( when “ inc^|^y*r4«*0«ekmvngfcka^ )When length of the wmk reduced by oonhird without cfcuswf the lot*.:When tension in the string is constant, then speto remains constant hencev = f 'X' '

original wavelength, then the reduced wavelength is.— Xft, V3.

Olculalion:(»)

- tv = 344m5

and8J A stationary wave is established in a string which is 120 cm long and fixed ivibrates in four segments; at a frequency of 120 Hr. Determine itsfundamental frequency 7

WThus fX = fIf X is the

both ends. ThJ wavricngth an(1 Hu r -x -Given data:

Length of string = f = 120 cm = 1^0 = t 2 m

Number of loops = n = 4Frequency of vibration in four segments = f4 = 120 Hz

100thus equation (1 )becomes

300)1 = r*-xor ':^bo=|r

r=450HzWhen tension is increased by one-third without changing the length:-As the relation of fundamental frequency for

To find:3

Fundamental frequency = fx = ?wavelenght = X =?

orCalculation:N (b)As the string vibrates in four segments and the distance between two

consecutive nodes is ~ > so the length of the string is £ stretched string isa.ill2/ V mI / (2)

1when tension is increased by -F, then/Or X.I 32 !F'= F+-F

F'= jFr-±JE21 V m

1.2 3Or X = —2 t*X = 0.6mIf string vibrates in n loops, then frequency of stationery waves will be

f„ =nf,f4 =4f, As n=4

Or f =

, 4 P1 I£f =-21 V m

JMOri 4 f =>

120

Af. = —l Or4 as f=300Hz|fl =30Hz

The frequency of the note emitted by a stretched string is 300 Hz . What will be thethis note when:(a) the length of the wave is reduced by one - third without changing the tension.(b) The tension is ncrease 1 oy a le-third without changing the length of the wire.

*Frequency of th^ etched sir g = f = 300Hz

frequent Thus f -1.15*300

*...» -----— -“•"- “ “ “harmonic when it is(»)- open at both ends( peed of sound =350mss )

8.48 . 5

(b)- closed at one endGiven data:

- t ii .4

i thr1 -VI I

IT ' * r:• s>314

r*s PHYSICS- XI (Subjective)Sch°iai- -Maximum length = /Speed of sound = v

Frequency range =? i.e./i = ? & =?For organ pipe open at one end only ,

f_ nv

Given data:31Length of organ pipe = / =50cm =0.5m

Speed of sound =v =350 nis max =-l

~ 340msTo find:To find:(a)- (when pipe is open at both end)

fundamental frequency = f, =?)

Next harmonic frequency = f2 =?)(b)- (when pipe is closed at one end)

fundamental frequency = f, =?)

Next harmonic frequency =f3 =?)

-

Calculation:

" 41Minimum length

amental frequency, put n = 1£lm“ 4x30x10‘3

= 2833Hz

For fundCalculation:

When pipe is open at both end:-The frequency for nth harmonic in open organ pipe is

fB = —So the fundamental frequency is

f_ 1x350

1 2x0.5

min(»)

. 1 x340when n= l ,2,3,.•••••••••••

Maximum lengthFor fundamental frequency, put n = 1

put n = 1nvf\.mn 4j

f =350HzimaxNext harmonic frequency i. e. n=2 is

2 21- v 350f‘“ T“ oTf2 =700Hz

(b) When pipe is closed at one end:-When the pipe is closed at one end , then frequency for nth harmonic is

n 41So fundamental frequency is

1 x340f =4x4

f,,„=21Hz:Result

So the fundamental frequency ranges from 21 Hz to 2833 Hz.8.7 . Two tuning forks exhibit beats at a beat frequency of 3 Hz .The frequency of one fork is 256 Hz.) Its frequency is then lowered slightly by adding a bit of wax to one of its prong. The two forks thenexhibit a beat frequency of 1Hz. Determine the frequency of the second tuning fork.Given data:when n=l,3,5,7,..Frequency of first tuning fork = = 256HzBeat frequency before loading = 3 HzBeat frequency after loading = 1 Hz

Frequency of second tuning fork =6= ?

To find:C*lculati

1x3501 4x0.5

(Put n = 1)

on:- f,=175HzNext harmonic frequency i.e. n=3 is

r 3vfJ = TT

As f, -f2= ±nOr f2 =f,± n

Putting values, we get41fj= 256± 3

To| f2 = 259Hz or 253Hz"7^*OX 259 Hx as correct answer }&*£*££********!££&'oaded with wax, the frequency of fliX fork must "J Wow 256^ re.25^254

number of beats produced per second will mere.se and will be greater then 3 beats

3*350*r 4x0.5' Ik[f,= 5_2$Hzj The ml»lBlttl11

A church organ consists of p*nes open at one end, of different lengths.n0(fi

is 30 mm and t^e ’ongr . is 4u Calculate the frequency range of the fundamen-i(speed of sound 0 IS )

Given data:Minimum Ic l min * 30mm -30 x 10 m

>* * iV * A1*1 a %.‘.A fV* ‘ft's V r>. mfc\. .w . &>-*1

316 • V.’Ti. J *

Since the number of beats per second decreases on loading first fork is one, therefifrequency of second tuning fork.Thus

Scholar'8 PHYSICS-XI (Subjective)ore 259 i317Speed of sound =

Time = t = 50 secv ~ 340 /fry'1Correct frequency = f2 =253Hz As (254-253=1 Hz)

Two can P and Q travelling along a motorway in the same direction. The |caa steady speed of 12 ms the other car Q, travelling at a steady speed of 20

"V*’ ?^to emit a steady note which P’s driver estimates, has a frequency of 830 H/ YVhQ’s own driver hear ? "

8.8 To find:,v*«,0un‘‘its h0ni***** K

Speed of source (i.e.train ) = Us=7Distance covered by the train i= S ?= ?Calculation:Using Doppler, s fonnula, when

f'=( 7T)fVfUsPutling values, we get

1140 — (

- I(speed of sound - 340 ms ) source is moving away from listenerBotrai (JOm» ')Given data:

- 1Speed o f c a r P = U p =12 msSpeed of car Q = UQ =20 ms

Speed of sound *= v « 340 msFrequency heurd by P’s driver - fp =830Hz

340- i)1200340+U,

340x12002T

• •

P-1 I Or 340+U,

340frU,-357.89(Js = 357.89-340

1140OrTo find:OrFrequency heard by Q’s driver - fp- ?

U,- 17.89msDistance covered by train after 50s

-iC a l c u l a t i o n:Speed of Qrelative to P UI"UQ-Up- 20-12 -8 ms 1

Using Doppler’s fonnula, when source is moving toward listenerAs the acceleration of the train is uniform,

distance covered -S*vBVx tsince v V

0+17.89v fr- ^ - 8.95ms'1

S = v„xt2Vv -U•/ so

Putting values, wc getF QOrS -8.95x50• /

P u t t i n g values, wo get S - 448m340 The absorption spectrum of faint galaxy Is ifoTh'” !Jwstslsngth of 397identified as the Calcium a line Is found to be 478 nm. Tbwhen measured In a laboratory.•) l. the galaxy moving toward, or away from the Earthb) Calculated the.peed of the galaxy relative to Earth?

(Speed of light-3.0 « 10'ms )lr> data:

fo830-340-8 nm340830-(~ )fo

8 3 0 x 3 3 2fqOr 3 4 0wavelength ( laboratory measured )-X-397 nm - 397* I 0Jm

Apparent wavelength - 5i'-478nm-478*l04mr^r 0f '‘ t"v-o-3.0*lO’ini'1

*)1« the glaxy moving towards or away from earth*?h) Speed of galaxy relative lo earth U, 7

f -1104711/

A train sounda Ita horn bafo». It sets off from the station anti an obaerver ,ui0

platform aatlmataa Ita fraquancy • i 20011/ The train then move, off and afcelartlFifty aeronda after daparture, thr driver aotinda the horn again and the pW«- „tV*sstl,nates the frequency at 1140 !|,. t afr„|,te ,r.in ,,,rc)|50 a after departure.

the station Is the tralo aft. ?

(ip««d of wound -340 mi 1j

n.9

M

^Maa,a what*o0 ipiU of lightA.v•fX.

original frequency,or c*

(ih.n dates - 170011/Original frequency oiApparent frequr ricy•f I I 401 (

i tn

f

\ > .Vl

=p • ' -

318 £!liPter 8 ru PHYSICS XJ (Subjective)

55!

f =‘\3 x 10“

397x 10’f= 7.56xlOMHz

For apparent frequency,r=-

Chapter 9f=

PHYSICAL OPTICSV

3 x 10sr=

478xlOvf'=6,28* 10, 4Hz

AsA'> A or f'<f f so galaxy is moving away from earth.Using Doppler effect. When source is moving away from observer, then

(a)- Understand th wavefront.I 1.(b)- State Huygen's principle.

Use Huygen' . principle to explain linear superposition of light.

Understand interference of light.Describe Young's double stilt experiment and the evidence it provided to support the wave theory of

light.Recognize and express colour patterns in thin films.

2.v fr« 3-

v+Uk As v = c4-

c ff = 5.

^c+U

Putting values, wc get» /

6.3x 10»

3 x lo'+U,

6.28(3 X 10B +US)= 3 X 10* X7.56

6.28X3X 108+6.28U,=3X 10* X7.56

6.28U,=22.68 x 10" - 18.84 x 10*6.28U,=384x io6

x7.56xl0“6.28x 10'*= Describe the formation of Newton's rings.Understand the working of Michelson's interferometer and its uses.

Explain the meaning of the term diffraction.Describe diffraction at single slit.

7-8.9.10.

Derive the equation for angular position of first minimum.11.Derive the equation d Sin 6 * mX12.Carry out calculations using the diffraction grating formula.13.

_ 384x10°‘ 6.28

Describe the phenomenon of diffraction of X-rays by crystals.14.Appreciate the use of diffraction of X-rays by crystals.15.

u. .^xioW associated with transverse waves.Or 16. Understand polarization as a phenomenon

Recognize and express that polarization Is produced by a Polaroid.17.

Understand the effect of rotation of Polaroid on polarization.

Understand how plane polarlz.d light Is produced and detected.18.19.

r

m Chapter 9

Scholar * s PHYSICS- XI (Subjective) 321

Chapter No. 9 The branch of physics which deals with the nature of light and its differentphenomenon is called physical optics.

cLight is form of energy which produce thesensation of vision.in 1678. Huygens's, an eminent Dutph scientist, proposed $hat light is theform of energy which travels in foflfn of Waves. yWhat Is a wave front? '^ ^

Xft oil the points of waves have same phase of vibration is

V

c

i *

Q.l _V M

Wave FrontsThe iurfocrjn whicknows os

Explana

Supposedirection fcith speed

re with

*ront

*3^> the light

V\i 1tted from a point source propagates outward in all

After time t, the waves reaches the surface of an

enter as S and radius as ct.

(b)V'&vo frontSpherical wave fronts (a)and piano wavofronUs (b)spaced a wavelength apart.Tho arrows represent rays.il nai

I these points from the source is same so all the points on

of the sphere have the same phase of vibration. Such as surface isAs the distanthe surknown as wave front.

Note

T e wave front from a point source are spherical

>

*>

Thus wave propagates in space by the motion of wavefronts.

The distance between two consecutive wave fronts Is one wave length.rw

Ray of Light

The line normal to the wave front which shows the direction of propagation of

light is called a ray of light

Do You Know /

3/r*a **0m4»r*ierf l»<y» *(>r>0f>C*wvwftonl

(1) Spherical wave front

The wave front in which the light waves are propagated in spherical form with

the source is called spherical wave front.for a point source of light in a homogenous medium, the wave fronts are the

concentric sphere of increasing radii.

U) Plane wave front

EXPLANATION:

The wavefronts far away

from the source are very

large spheres,

portion of the sphere will be

the plane wavefront.

A small

At very large distance (i.e. at infinity) from the source, a small portion of

spherical wave front will become very nearly plane. Such a wave front is known

plane wave front as shown in figure.

322 Chapter 9 [ph<~rfrnlar*s PHYSICS- XI (Subjective) 323

Q.2 State and explain the Huygen's principle?Conditions for detectable interference pattern

The following condition must be met, in order to observe the interferencephenomenon;Huygen's Principle

If the location of the wave front at any instant t is known then Huygen's principleenables us to determine shape and location of the new wave front at a later timet + A t. This principle has two parts;

Every point of a wave front may be consider as a source of secondarywavelets which spread out in forward direction with a speed equal to the speedof propagation of the wave.

The new position of the wave front after a certain interval of time can befound by constructing a surface that touches all the secondary waveletsExplanation:Let AB is the wave front at time t.To determine the wave front at time t+ A t, draw secondary wavelets with centerat various points on the wave front A'B' and radius as c A t, where c is thespeed of propagation of wave. The new wave front at time t + A t is AT3 ’ whichis a tangent envelope to all the secondary wavelets.

The interfering beams must be monochromatic.The interfering beams of light must be coherent.The sources should be narrow and very close to each other.The intensity of the two sources be comparable.

1.2.3-4-

I. Monochromatic Sources

r..r:“ ria^ ^« —Coherent SourcesThe monochromatic sources of light which emit waves, having a constant phase

difference, are called coherent sources.

ii. 1Kc c

* How to obtain coherent sources>A common method ‘o obtain the coherent light beam is to use a monochromatic

to illuminate a screen containing two small closely spaced holes, usuallyCAI A sourcein the shape of slits. The light emerging from the two slits is coherent because asingle source produces the original beam and two slits serve only to split it intotwo parts. The points on a Huygen's wave front which sent out secondary learn the diagrams and

mathematical derivationDescribe the Young's double slit experiment for demonstration of and read the theory onlyinterference of light. Derive an expression for fringe spacing. “

wavelength are also coherent sources of light.Discuss the interference of light, discuss its different types andconditions for detectable interference.

Q.3 A0 Q-4

FIQ-* 7

construction f0f

Interference of Light WavesWhen two identical light wave travelling in the some direction are superimposedto each other in such a way that they reinforce each other at some points(constructive interference) While at some points they cancel the effect of eachother (destructive interference). Such phenomenon is called interference of light .

Types of interferenceThere are two types of interference(i) Constructive interferenceIf crest of one wave falls on the crest wave, then they support each other Such a

interference in known as constructive interference.For constructive interference

Path difference = mX where m = 0.1,23(ii) Destructive interferenceIf crest of one wave falls on the trough of the other wave, then they cancel eachother. Such a interference in known as destructive nterference.

For destructive interference

k-vtarv* J A ST urn Ctfm*** Young's Double Slit ExperimentIn «01. Thomas Young performed the interference experiment to provethe wave nature of light. A screen having two narrow slits ,s dlummatedby a beam of monochromatic light.The portion of wave front incident on the slit behaves like the sourcesecondary wavelets. The wavelets leaving the slits are c0 .Superposition of these wavelets results into the series o rig anbands which are observed on the second screen placed at some distparallel to the first screen.

i

Conditions for Maxima and Minima,n order to derive the equations for maxima and minima, an arbitrary

Point P is taken on the screen on one side of the central point O as shownln figure. AP and BP are the parts of the rays reaching P. The line AD is

drawn such that AP=DP. The separation between the centers of the twoslits is AB=d.

distance of the second screendetween CP and CO is 0, It can be proved that the angle BAD=0 byassuming that AD is nearly normal BP. The path difference between thewavelets, leaving slits and arriving at P. is BD. From right angled triangleADB,

:1

from the slits is CO=L. The angleflgfnr ^ |V.i

Sort...m1

1 TPath difference = m + — X , where m = 0,1,23/ we can write

324 £^apter 9 [phy»ica|crholar's PHYSICS- XI (Subjective)J 325BD

=sin0ABBD = AB sin 0(AB = d)

Path difference = BD = d sinQConstructive InterferenceIf the point P is to have bright fringe, the path difference BD must be an integral 6fmultiple of wavelength. So,

p

T* Fringe Spacing:

jhe distance between the centers of two consecutive bright or dark fringes iscalled fringe spacing .

f o r Two Bright Fringes:

fn order to find the distance between two adjacent bright fringes on thescreen mth and (m+i)th fringes are considered

Position of the mth fringe ym = (m) —

mor y

iB 0(DL

BD = mXd sin 9 = mXThus

Where m = (X,1,2,3,.Where m is called the order of the fringe. At central point 0, the path differenceBP-AP=o, so there will be a bright fringe at 0 corresponding to m=o. This is calledzeroth order fringe.

%Destructive Interference:If a dark fringe is formed at P, the path difference BD must contain half integralmultiple of wavelengths. According to the condition for dark fringe.

(2) d )XO•••••*

fringe = ym+i=(m+i) Im-position o'

Then.

An interference pattern by monochromaticlight in Young,s double slits experiment.Ay = 0Y+1-rf ) —a

AXL

Ay = —a

For Two Dark Fringes:

In order to find the distance between two adjacent dark fringes on the screen

lBD =[m + — ]X

)d SinQ = [m+ — ] X

m =0,1,2,3,Position of Dark and Bright Fringe:Let y is the distance of point P from the central point O and a bright fringe is

formed at P. Now from figure, we can write from trianglePOC_ OP _ yOC L

(3)ThusFor Your fnfoftnntion

Where#

(m+-) th and [ m +~) th fringes are considered.1 ,

Position of ( m +-) th fringe = ym = (m+-) J3 . , 3.(XlA

Position of (m+ — ) th fringe = Ynwi- (w + ^ ^ JThen,

ooi04362O K-0 04 0.1060104 0140y.i*»0.(71am

tanQ

y= LtanQ(for small 0 sin0 * tan 0 )

y= LsinQ _ Ay

3 fXL'l , 1J XL= (m+ 2

)ur(m+ 2)lTAyPosition of Bright Fringe:

From equation (2) l

-mmX Aysin0* —dXL. Thus equation (4) becomes Ay

Hence the bright and dark fringes are equal width and equally spaced. The

2; T Z e — foMhe^elpac.ng

can be used for the determination of the wavelength.

XL m = 0,1,23... .y = m —dPosition of Dark Fringe:

From equation (3) \

i-H;sinO

iThus equation (4) become.

y = m+ — —l 2 J dm = 0,1,2,3....

326 Chapter 9 h>|,Scholar’* PHYSICS XI (Subjective)Q.5 Explain the phenomena of interference of light in a thin film? 327

Interference in a Thin FilmExperimental arrangement:

The thickness of the air film between plano-convex Ion . and glass

slit is almost zero at the point of contact /0' and it increasesgradually as we proceed towards the periphery of the lens, Thusthe point where the thickness of the ajr film Is constant will lie on

the circle with O as center.

inA transparent medium w hose thickness is very small (Comparable withthe wavelength of light ), is called thin film.Examples: s

G "o) oil film on the surface of water,surface of soap bubble,cracks in glass plate.

—Light beam from a monochromatic source 'S' becomesafter passing through the convex lens 'L\ This beam of light falls

on the glass plate G. Some rays are partly reflected normally

towards the air film and partly refracted through G. When lightrays falls normally on the lens, these rays are reflected by the top

and bottom surfaand interfere each

When the light reflected upward is observed through a

microscope. "M ’ focused at the glass plate G, a series of dark and

bright circular rim • are observed, as shown in figure. Theseconcentric rings are Called Newton's rings. •

Dark Central Spot:

At the point of contact of the lens and the glass plate, thethickness of the film is effectively zero but due to reflection at the

lower surface of air film from denser medium, an additional path

difference of k / 2 is (or phase change of 1800) introduced.

Consequently the center of Newton rings is dark due to

destructive interference.

Ilel (a)(ii)

Interferons 0,' ;°";;o' n— h(Hi)Explanation:

Consider a thin film of a refracting medium. A beam AB ofmonochromatic light of wavelength X. is incident on its upper surface. It ispartly reflected along BC and partly refracted into the medium along BD.At D it is again partly reflected inside the medium along DE and thenalong EF as shown in fig. The distance between the beams BC and EF willbe very small, and they will superpose and the result of their interferencewill be detected by the eye.It can be seen from fig. that the original beam splits into two parts BC andEF due to thin film which enters the eye after covering different lengthsof path. The path difference depends upon;

1. Thickness of the film2. Nature of the film3. Angle of incidence

If the two reflected waves reinforce each other, then the film will look

bright. However, if the thickness of the film and the angle of incidenceare such that the two reflected waves cancel each other, the film will lookdark.

Interference of White Light:If white light is incident on a film of irregular thickness at all possible

angles, we should consider the interference pattern due to each spectral

colour separatelyBut if the thickness of the film and the angle of incidence are suc^ fhar

the destructive interference takes place from one colour Then theremaining colour of the white light will make appearance on the film. .

Discuss the formation of Newton's rings. Why does the centralspot of Newton's ring look dark? ‘ - fl

Fig. 9.6 (a)

Experimental arrangement forobserving Newton’s rings.Do You Know? s of. the air film. As these rays are coherent

(her constructively or destructively.V’«

'Vv‘

L\

The vivid iridescence of peacockfeathers due lo interference of,thelight refler. tpd from its complexlayered surface.

Interesting InformationFig 9.6 (b)

A pattern of Newtons rings due tointerference of monochromatic light

Describe the principle, construction and working of Michelson's

interferometer. How can you find the wave length of light used?Q.7

Michelson’s InterferometerMichelson's interferometer is an instrument that can he used to measure

distance with extremely high precision. Albert A. Michelson devised this

instrument in 1881, using the idea of interference of light rays.

Principle:,ts forking is based on interference. When light from a single source is

spitted into two parts and then interfere it forms an interference Pattern

Construction and working:

The essential features of a Michelson's interferometer are shown

schematically in Figure

Coloursare dui to inferencewhite tight. For Your Information

QQ-6 /

A photograph of NfictveVInterferometer.

Newton's RingWhen a plano-convex lens o_ long Inca! ten th is placed in contact with

a plane glass plate, a thin m 'll * is enclosed between them to formcircular dark and brightfringes * u wn as Sen ton 's rings.

c.hoUr’> PHYSIC? \1 (Subjectc) 329Ray Diagramm

(2) M.cheHon measured the length of the ^tedard met* in terms ofwavelength of red cadmium light and proved^atStandard meter = 1553^63.5 wavelength of light

{3) If light of wavelength >„= 400 nm fs used, then it can.thickness upto 10 4 mm (or t

(4) It Is used to observe the in

Moveable Mirror

\ m

sure the- f

wf !*•<!Mirror

Q,8 What Is meant by diffraction

Also discuss the di ough a narrow slit ?% O t J I X*

Diffraction of LightThe property 0/ hendmi; of light around obstacles and spreading of lightinto the geometrical shadow of an obstacle is called diffraction.

Schemaic diagram d a Mcftdeon’t inaurfmvnjtrcScreenWorking

Monochromatic light from an extended source fails on a half srivered glass plateG, that partially reflects it and partially transmits it Th* refected per - or Mx cdas I In the figure travels a distance t, to mirror wh.ch reflects the beam p*.towards G,

ShadowObjectEipUn-Jtions:

Consider a small andlilluminated by a point source of l-ght ‘S' The shadow of the

object is receded on a screen as shown in figure. The shadow ofthe ipbencal object is not completely dark but has a bright spot at- ts center it happens only if the light bends around the obstacle.

This phenomenon becomes prominent when the wavelength of light Is large asco pared with the s ze a* the obstacle or aperture of the slit.

« d.ffraction of light occurs, In effect, oue to the interference between rayscoming from different pans of the some wavefront.Diffraction of light through a narrow silt:The e*oerimentai arrangement for stub> ng diffraction of light due to narrow siltIs shown in figure.kit us consider a slit AB of width d is Illuminated by a parallel beam ofmonochromatic light of wavelength X. A small portion of the incident wavefrontP*sses through the narrow silt. Each point of this section of wavefront sends outsecondary wavelets to the screen. These wavelets then interfere to produce thediffraction pattern. It becomes simple to deal with rays Instead of wave fronts.0n>y nine rays have been drawn where as actually there are a large number of

Sh steel ba; i of about 3 mm in diameterShadow

^ Bright spot

Bending of light caused by itspassago past a spherical object

The half silvered plate G,partially transmits this portion that f naJfy rnves at theobserver's eye. The transmitted portion of the ongmal beam labeled »s < navelsa distance L, to mirror M, which reflects the beam back toward G;same piece of glass as G, Is introduced in the path of beam >1 as a compensatorplate. G, therefore equalizes the path length of the be.im l and m g’ass Th.two beams having their different paths are coherent They prod uce mterferersceeffects when they arrive at observer's eyesThe observer then sees a series of a parallel Interference fringes. In a predic t

interferometer, the mirror M, can be moved along the direction perpersd eu sr to.Aits surface by means of a precision screw As the length L,is changed, thof Interference fringes is observed to shift

0Point lo I’oiuk JHold two fingers closetogether to Form t slit Lookat a light bulb Through thesilt Observe the Pattern oflight being seen and Thinkwhy It is so.EXPLANATION:The lending of light rays aroundthe small openings or obstaclesIs called diffraction. We see thediffraction fringes through thetwo flngeri close together like

iatt<

If M, is displaced through a distance equal >72. a path difference of.F \ ~ x xthis displacement (i.c — — * A.) Is produced, i.e. equal to k Thus'* fringe is seenshifted forward across the line of reference of cross wire in the eye piece of thetelescope used to see the fringes. A fringe is shifted each :,ne the m.rror itdisplaced through X/2. Hence, by counting the number m of th; nges whichare shifted by the displacement L of the mirror. We can write th equc= : n.

L = m —

site

IA r»Y*.ut consider ray 1 and 5 which are In phase when in the wavefront AB. After

tht*« reach the wave front AC, ray 5 would have a path difference 'ab * say equalto Va. Thus, when these two rays reach point P on the screen, they will Interferedestructively. Similarly, each pair 2 and 6, 3 and 7, 4 and B differ In path by >72#nd wlt| do the same.iri order to find the value of pith difference ab, we consider the rignt anglelr‘ingie aAb, as shown In figure.

Si«R »

21K =

m

Very precise length measurements can be made . s an interferometer.Uses:Michelson's interferometer is u;ed lor the following purposes:

(1) it is used for the detc .. nation of w elength of light.

orl*lg19.0

SR<*

T + . «

•9'

3JO Mitipttr j,HiU' A'*1 * XI (SulijttHvi'; 331AH,.b hot Aft dinOnr t *-» rrijft7 For your Information

•'» *1 fi./n r„jr;,||(pl M /'v ,f ,‘ ' ’ / ''OK, y,f 'V < Of,MXU0V« * Ml’ . W^J)f^4p poth t)y

/ vvh«-n ‘fv- y *irriv< - ./ > >' lh* - / // /II therefore,i/ iN-rh-n • , < i t - jVH /(,rnUny.I rl,J'

,1|f,rl ' ,

path d«f fvr "fix for > on^nif.rijcor(«,/( f ullvr * nr/c chould lx - inli-gr - ) l in jlt »|

Path /lift /ibfrom figure, -»'- b " r,gM '

CiriOill) •!'. MnO

d ltMO* )J

f or (JctUru' live tot* rferenc « ( fir *.) minimum) , ;»b2

M* f ,' » • nqtniMjn for fir r minimum ' »n b < - / /rI!t <*rf .n,

thuc ah if'J ion.>. f.7/0fit" «• r /i”i 1 -,r* ),* /«), >! '(I.)

• 1 .^ i/ -» , - jMX vj» ,<| 0 10 (*,'Mlf) .

y..)»»•» • { Amir -J|nJ,|,,A„„7,,,.^^

h'I / ngU-.1117 J. fnfereefmg Application

Of «Hu. f » /

In K* n* / * l, » he r /mdlfronc for different ordmc -,f minim,. V(ffj

mth<-r vdi of f of ,/ 'M (i if » f f jy t- n I , /tHmO in /

/yfif - M ffi M, ' /, ) {,Il-o region b« r **ef> *» • / * //o ' on;*- 'oh /•» minima both above- ..f,)J below

‘be ) * fill ) !) <• l / f l y ) >\ I

d*rk ,»od bright fring* . // / ( it * r- ntnil bright

or -ibComparing.H) U)

^t*id Sln'^xw^(3)

/ W ' ll - - ' Of ( >)

i/,.f (, hf ;» } ,( fringe ' me h known / < -ro order Irnag* * formed fiy th» - gr.iMng

If v/)- in- / 0 on nither *ld«- of till'. dim'lion, -i v.ilue of 0 //ill b<- -ifnved

,,t whi) h / 'I Mn f / i»nd / /• again K‘» bright image If |> * » f » diffi-n?nt * - 1'«

the iHO- j/rtf mulnpli - of / ., / / < • / /ill get *.n).oncJ# third order Hr , image**f |.. i - l - .t « in |'< [ i * ' •' ! / / < ' if . / /ill" .

< J -.ino / 1/

Iiik, i'. / 'll- -'! tli- - -'I'l.itioii i / f granting and 'n' f. ' ailed die order of diffraction

n - o, I-. if ,«. r ) *ritr «il rri.i/im.i, n I I*. Ihe lif '.t order m-i/lm.i arid v> or, fhe

u-paraf )’ Ifn»ip.< •% - if * obt.ilfM-d f f /fre',pondlflg fo « af ft v/a'/flviiglli or « oloor .

I) )0

, , 1 li' <>" , tfw -ri path different « o , -,o / / < • / /illr>.m * O.-ri’ A tmUu

1/ rn,) / irn,) fo tfi) *, / /,, y / / « • mm / olil -iiii • r » « • - . -,f » ‘a '* IO on

00* ',-., |, |yryju)4||0/ «M»IftiiDniifiirtxr Infurmoiion

Wfit-r < * n 0,1,?,3,tj 'l Wh*t it diffracHon grating and obtain the grating aquation to find thewavelength of light / .) > *- *'•Tf«a t r y

illIt- •> i. . i t i r J - "D) »

f >iffr «jr tion Oroting"/• dl j f r tJr Uf j f i t jml lhf j ' orr.i' j of // /// / / * -. i i la le on whl' h \/f ry JUv• yuirJ l i' in l f / f j / f i l l' l hut -. (\r / r ih hr % ) - / / . f lmwn hy rnmn o/ ni l lmj r t iy i f f rs/ l th f i rn fJhj t rut / i f l f j r j l t i i ih t hnmitorpnl - inn imj hr iwrrn ihr z / r r i t thmnn / fi* y/«4 *. f j lut* i v i i l l t i"/ * /pi- .< 1 dlff r #' gf .Olf / g bat ^ f ) < ii/ »|/t ) /) / to '/ /) / ) ) Horn |< tif ). nrillrnata) . |Oratirig alamant

I hr dlituw • hr I / / e r / i lh* nuilar. nf ( wo )»//)/) aril Un»* h ) ullp' l (JtrilInfJt/frrvarrf "

fti i/alo* it obtainad ») / dividing tbt Ungth i / / f tfm grating by th* "dalnornbar tl of » > *•llna » ruled on It

V) the grating •lament d - I /NIf //«- » ontlde) wrill i*r«gtli of tt.« dlffrat.tlort grating than - l - l/f «Working ar»d thao) / >r ) / f »*liJa» »ha parallel l#a»m i-»f ntor Hror au. light IHumlii -dlrig Iliagrating at nonnal Intldaflta ft » e pan « the wav« front that pall

through the illti hahavai %ourt *» of >•, • try wave U > «< « ordlftl hiHu /gar/t print ipla( ontldof tha pataff^ r ** /d " i. me' - an afigl« 0 with All after

dlffrertlon fhay «ra than t' *i •» to fotui >n Ota uraart at p by ronvgk

|glM > If the path dlffaren< «hei irl t ay ta i r llona Wfvglaftgth X, they

y/ dl r a i n f a l l a ««•h Other at P A ‘ha Iniidani beam contliti of the

0,10 Dotcrlbe the diffraction of K rayi through cryitah? Alio daicrlbatlif % Braggii- fpJMtlon and Its different uieir _

m/"III M), -,/ fti)|I)>1 Diffraction of X»Rayi by Cryitulft

X rayi h type ol elactromagnwtlc. radiation of much shorter wavalangth.of the order nl 10 m.In order to nli-vnivB the affect of diffraction, tha graving spacing must b«of the order of iliu wavelength of tha radiation uiad. The regular array of

•torni In a uy»tnl forma a natural diffraction grating.Tha Itudy uf nt.oml< Strut,turn of crystal* by X ray* was Initiated In 1914 by

W H, itrogg and W.l Uragg wllh remarkable achlavemenu Ihey foundThat « monuQhramall' of X ray* was raflettad from « cryatnl plane

•l If II acted like mirror To umleritand thl* effect, a *erlei of atomic

(dories of cunateni Idler planar ipaijlng 'd' parallel to a tryital h» » B

"hown by lines |V. IM* / l*,lb 1 shown In fig, and so un.fTragg'iaquttlont‘hippo** «n •• ray bagmW incident at an angle on orte of the uUne* 0|r^t||OII uf K ,.y.n<.rtTha beam rafle< lad from that lowai plana travgll »orn< •**'•d,*tfrn« piene.ofprVat«l

0»t ( IV) aa ccin.para tu Ihi beam reflet tad from the upper ulane rhu*ttffev tlvp path differnn » baiwenn the ' wo * fflectr*«i !*« i,|V* 1 • 111 1 ^

» gfLAPA •lONtIT))) ) " •- »" ''pAmUfl..

rwtuwm i>- flg|H l* fvTi«.r») CVt ItKKI

w< jt«i*livn i- - v. , M i|nsmallt' inn m ww

Mil, „M,;,I . • i'U-l- ()•«•" W'*"-'1 inirf »*

lull'll’ 1.11« lli(''""'ll,,,M If

uwvUf-

bA fiirfi^„„ cnmprt* I " ,.*1, a " 1

dim-ciior. u» all‘»W- -Tt,||„ 11,0 Itlft"""1' ,I«-VJ, tail”" •*' I,.,'11

nr ll»M Mlfl- e- Jion l» a

roll) ((»' •

. VS

332Chapter 9 [physir^i Q Scholar's PHYSICS- XI (Subjective)From triangle ABC, we have Ptitn]

333,n9 IIIBC •On

|n transverse mechanical waves, the vibration can be oriented alongvertical, horizontal or any other direction. In each of these! cases thewave is said to be polarized.The plane of polarization is the plane containing the direction of vibrationof the particles of the medium and the direction of propagation of wave.A light wave produced by oscillating charge consists of a periodicvariation of electric field vector along with magnetic field vector at rightangle to each other. The direction of polarization in a plane polarizedlight wave is taken as the direction of electric field vector.Unpolarized light:A beam of ordinary I ght consisting of large number of planes of vibrationis called unpolarized light *polarized light: IIThe bear of light in which all vibrations are confined to a single plane ofvibration s called polarized light.Production and Detection of plane polarized lightThe light emitted by an ordinary incandescent bulb is unpolarized,because its vibrations are randomly oriented in space. It is possible toobtain plane polarized beam of light from un-polarized light by removingall wave , from the beam except those having vibrations along oneparticular direction. This can be achieved by various method as givenbelow

-— = sinOACBC = AC sin f)BC = d sin ft

Similarly from triangle ACB', we have

= sinO

*Or<r *Or As AC=d

CB

AACOr A"

multi-.iportur«» ditfrnctlonpattenThis •• piclufb of a wtvita-Ugtopoint vutc.o *hol through a PTOCAof oghtiy woven doth,An unpolarized light,due to incandescentbulb, has vibrationsin all directions.

CB = AC sinOOr

Tidbit*CB' = d sin 0Thus, path difference = BC+CB/

= d sinO d sinO= 2d sin 0

Potarlod-2

Plane Po-(a)

* urpoianzed

(D OifTraction p.«tte<n oihair urvjor laPath differenceBoth reflected beams //ill reinforce (constructive interference) ifdifference is equal to integral multiple of wavelength,constructive interference

a single humiinilfuminaUon.-.or

" yrHence for

Path difference = n>. (2 )Comparing 1 and 2 , we get2d sin 0 = n). where n = 1,2,3,where n is the order of reflection. This is known as Bragg's equation, or Bragg'slaw

Fig. 9.15Experimental »rr»< jomont to V>owthill light WIVM arc tranavefee. Thalines »ntn arrows incftoatea eieoJncvibration* of light «ava*.

(1) Selective absorptionReflection from different surfacesScattering by small particlesRefraction through crystals

Selective absorption method:Select ,ve abortion method is the most common method to obtain planepolarized light by using certain types of materials called dkhari su"“These transmit only those waves, whose vibration are parallel t o t h e a r u adirection and will absorb those waves whose vibration are in other d.recbo .One such commercial polarizingmaterial is Polaroid. _ , ,d the« the un-polarized light is made incident on the shee of IP° roidjhetransmitted light will be plane polarized. If a second sheet osuch a way that the axes of the Polaroid, as shown by the 1hn«dr«* onthem, are parallel, the light is transmitted through second Poiaroid^lso

^the second Polaroid Is slowly rotated about the beam o

^ andthe light emerging out of the seconcI o a f The |jght reappearsd,sappears when the axes become mutually P Pparallel to^ further rotation and become brightest when the axes are aga

each other.Transverse Nature of Light:

waves. If the light waves were longitudinal they wouldTh‘s experiment proves that light waves aremulually perpendicular . Sunlight also becomes vnever disappear even if the two P° ar ,he E4rths atmosphere or by reflection we can obtamPolarized because of scattering by air moleculi

partially polarized light instead of glare o ign •

(2)(3)Uses:(4)Bragg's equation can be used to determine the Interplaner spacing

between similar parallel planes

1.

of a crystal i.e., d = n/'

2Sin0X-rays diffraction is very useful In determining the structure ^ofbiologically important molecules such as hemoglobin, which is anImportant constituent of blood, and double helix structure of DNABragg's equation can be used to determine the wave length of light.{ \ jt. by >.-

2.

tw3 -

2d Sine j ;<•> oiQ.11 Explain tha phenomenon of polarization. How plane polarized light Isproduced and detected? A

EXPLANATION:Little light is due topolarization

;/

<wPolarizationThe phenomenon of Interference end diffraction eve proved that light has wavenature, but these phenomenon do not show whether light waves arelongitudinal or transverse. The phenomenon of polarization shows that lightwaves are transverse .

V*l 0°*

asstfsgsJ

334

pHVSICS — XI (Subjective)Reflection from different surfaces: >cho\*l 335• Reflection of light from water, glass, snow and rough road surfa<>, for - g..angles of incidences, produces glare, since the reflect ••a light p.,rt jpolarized, glare can considerably he reduced by u.mg polarized .un;1 • . ,

Scattering by small particles:

— — — — -—Multiple Choice Questions iSunlight also becomes partially polarized du" ' »

earth's atmosphere. This effect can be observed bT loo* ng d uttfy up through ipair of sunglasses made of polarizing glass i cert i n da. • - , of t- « , \ vlight passes through than at others.

Polaroid:

f our possible answers each statement an- given below. Tick (S) ,/ie correct answer:

siThe wave theory of light w(a) Galileo(c) Kepler

The locus of all

I(b) Huygen(d) Newton

A synthetic doubly refracting substance that frc -i*;, ab - o r j. ,- j |ltfl » ,one plane, while easily putting polarized fight in another plane of rub:

* • pOMKbase of vibration is:l.

(b) Interference(d) Polarization

(a)Q.12 What Is meant by optical rotation >ffrac(c)

gs experiment the Plano convex lens used should be of:(b) Large focal length(d) None of these

In|Uf. [gthOptical Rotation M

The aif between lens and the plate in Newton's rings experiment Is replaced by water. The rir

pattern(a) Contracts

Remain same

focal lengthWhen .i plane i < 1 lightplane of polarization Quart! and - odium c> orate . y > t.iwhich are termed optic ally active crystalsA few millimeter thicknesses of such try t .* * - ret «u* tby many degrees Certain organic \ubst .»ncev such

Are examples

(b) Expands(d) Becomes darkerpolar *:at*on

W (C•s sugar and tartaric tod.show opticil rotation when they erg s property cf c‘'Ubstancc. can r

Fringe spacing In young's double slit experiment Increases due to increase In:(b) Wavelength(d) Frequency of source

• T'« «4 vtxar* 5(a) 5* »t separation

(c) Order of fringe

Fringe width In Interference pattern will be more, when coherent sources are:(b) Too much apart

(d) Inclined at 45°

F O R M U L A E! 6.(a) Dose together(c) incited at same angleWhich Is not optically active?(•) Sugar(c) Water

Path difference for constructiveInterference In light d $ n « s m/.^nere m« 01,2.3,

7.d $ R 0 •[2nv*l]^ (b) Tartarlcacld

(d) Sodium chloratePath difference for destructive

Interference In lightd Smd » Imt — JX,where m* 0,1,2 3 • where

8. Longitudinal waves do not exhibit:(•) Reflection(c) Diffraction

^y/positlon of mth bright fringe ALyma,T (b) Refraction

(d) Polarization

When light falls on a ball than its shadow has bright centre, It Is due to:(b) Diffraction(d) Refraction

The danger signals are red while eye 1* more sensitive to yellow because:

Scattering In yellow colour Is less than that of red

In wavalangth than yellow colour

yj/*Position of mth dark fringe aV v (m — 1—Q 9.

(a) interference(c) Polarization

/Av TFringe width /spacing

10.da~ (whenl lm/

Grating fitment yL (•)N fo) Red light is longer

^ Scattering in r«d Is less than In yellow fd)

Which one of the following Is nearest to monochromatic light?(b) Ught from fluorescent tube

21Displacement of mirror in

Mlchelson InterferometerNone of these

L*m- mM.

/ 2d am 0 •nkWhere n» 1,2 3.

Bragg's Law (•) Ught from neon lamp

%

oifes eiusv ' IOIA

ks P H / S I C S - X I ^Subjcctr.Z ' -' /' y s' r" jf*• < ** ^ *o o#.#**>«#* #.*

t>* •+> -*?i,' ; v' S,*-* •' X *

produced t r f t>**» k**^^ *" * ’ * <v **'* *'-"«* * <**> t«" *j*r**rt * **(M ft**

U

Short Questions of ExerciseM***

t

' 4

******fl

Oft* Unde/ wftM cofrtiiom, tfw two scuff of , r> ***,**sIL^sourcfe$?^ 7///-^2005-200^,Ifer 299t, _

i*r2y/,„J-x 20 2OT

s*t «'Ptrfx,^* > «v*s>

Cor Jitiom for coherent source;te/0 so-.rces 3'* S3 'I to "0> \ f

(> ) netOorce^ rnyjl-e(» ) rh£u'*/perr

•% id) 1k&s/u4 ff 0 ,J -,., vrtto+fp x < v*ri ' tt&l •flcse*4*'*2) MO CtattgA

A/4-<*hx. wmthoffclr

f t) «vo * /*d

*44** / '/Ifc/ f/x

' «, //Vy* *•' Y/f

'/ :cmf //< /e fengr " o"o o''c~ < c" .:*,'< /e constant '' *ce - ce

/W<

fh) A'T’ptoud*(d) fr«r' i ‘ i*rfS, f" < ***#'**'.' p.” «r- '/ *'.«> -J * <->.**** wk*n y.ll/yw llg*.if rtpUctd by Wu< 1,3*7

(b) Widerfdj Bright

// ' y> '/ fr * <V V« 'its r-*n r Of pfffditt* tcAoxKt #/ rth v/htU D^M?(b) Interferencefd) D»$p*f«*Of>

n 4 •/ '> s o n f&*rUf (jff { Uff 0 / mo/«r»f the mirrc r through -» dUteivce of r , the p * '

0.3-2 Mow »'. tr^ d'.st : ;ceyoong'; experiment? Can fringes disappear?

Ant, & / >ncra*,!ngrne se sargt'or bet //een s;its fringe spdc;'- g Js decreased sno /i’ce versa.

Interference fringes affected 6 / »ne *.eparatio" aetweer. the --'its o-?

M**tfsn*+ spac ng ' g- /en by,ry •

A /,7.A/ * —

*/Tr.ip equation shows that the distance between fringes(Ay) is inversely proport.onai to the sepa ^ a:orbetween the slits (d).Fringes can disappearWhen separation between the slits ismade /oroe enough, the fringes will be so dose that trey cannotbe distinguished from one another and pattern will disappear.

'v ’'*'4n */.'fCj / X. '//

%

by

r* > ' (b) >.0-9 *3 Can the visible light produce interference fringes? Explain.

/

(Mtn 2003-2006. Bwp 2006,Ptd 2005. Sgd 2005. Gnw 2009-2011)(< > 4

On d />/' ng th** k'lgth of th® grating 'L' by total number #N' of the lln? s ruled on

(b) Total length of slits

(d) None of these

( d ) 2/.Am. /es. visible light (white light) can produce the interference fringes.

Explanation

White light consists of seven colours. Each spectral colour produces its own mterference fringepattern. These patterns overlap to give rise a resultant diffus' d coloured interference pattern.

1

fa, Or atir g •bern^ r . t{'. ) rota he ght of slits

F'olari/ation of light shows that light wavts are

(a / lor < gftud>na! //ay«*s

(c) sound waves

S-9-4 In Young's experiment, one of the slits Is covered with blue filter and other with red filter . Whatwould be the pattern pf light Intensity on the screen?

Ans’ No Interference pattern of bright and dark fringes is formed on

ReasonWe know that for detectable interference, the two sources must be monochromatic havmg constantPhase deference Since the blue and red lights have So, the cond.tion lorinterference Is not satisfied.

v

(! j V < r werse waves screen.(d) all of these

AN^vVEf|0. c

9. b,k 2. a H. dX b 4. a 5.b 7. c6. 20.19. »br 12. d 18. a13. a 14. a 19.b 17. c16. c

* 3 VV. V ' i ni. S*M_» 'm

nifp.Q musw IOJ I\yk338 Chapter

Scholar’* PHYSICS XI (Subjective)

Q 9.5 Explain whether Young's experiment is an experiment for studying interfeof light.Basically, it is an experiment to study the Interference of light though it involves diffExplanation

As the light passes through the slits it bends around the slit (diffraction) . Then th *superpose each other to produce the interference pattern effect of light But in thit *only study the interference effect of light.

rence or diffract,on *ff*n16 100», 01 Q.9.9 How would you manaS* t0 &eX more orders of spectra using a grating?(federal200s, Gr <* 2005,Mir Pur 2006, Rwp 2006,Ihr 2006,Bwp 2008,Ans.

faction. Mir Pur 2009, Ihr 2009,Grw 2010-2011,ihr 2010)

Orders of spectra (n)

The formula for diffraction grating is,Ans.

racted r;jy,experiment wt.

wd sin0 = n /

OR ftQ.9.6 An oil film spreading over a wet footpath shows colours? Explain how does it h

(Bwp 2003, D.G.Khan 2005-2006, Rwp 2005, fsd 2005. Bwp 2008. Grw 2009,

Ans. This happens due to the interference of light waves

ExplanationOil film spread over a wet foot path acts like a thin film .

• A light beam is incident on the upper surface.

• It is partly reflected from upper surface and partly reflected from the lower surf.ice of thin film of oil.

• The two reflected coherent beams superpose and an diffuse interference pattern of different colours isobtained.

appen?Ihr aounoii.Grw,*,, for maximum value

SoThis equal ion show'. tha^We can get more orders of if;Tor a given diffraction gratingIts grating element is constant, so

r 1Hence, by decreasing the wave length we can obtain more order of spectra.

Q.9.10 Why Polaroid sunglasses are better than ordinary sun glasses?

(Grw 2003, Sgd 2003- 2005, D.G.Khan ZOOS. Ihr 2005. Fsd 2005, Mir Pur 2009, thr 2009,Grw 2010)

Ans.Q.9.7 Could you obtain Newton's rings with transmitted light? If yes, would the pattern be different from

that obtained with reflected light? Polaroid sunglasses pass small light having specific planes of vibration.

* Polarized sunglasses reduce glare.They protect the eyes from bright rays of sun light.Polaroid sun glasses blot out only the harmful glares .

( Ihr 2006, Bwp iOo9.Mir Pur 2009)

Ans. Yes, the Newton's rings can be obtained by transmitted light.

Pattern of transmitted light:The pattern obtained from transmitted light is exactly opposite to that of reflected light . There isflflphase change in this case. Every dark ring is converted into Bright ring. So the centre of Newton s rings

is bright.Q.9.8 In white light spectrum obtained with diffraction grating, the third order image of wave kn^ave

coincides with the fourth order image of second wave length. Calculate the ratio of the two

lengths?

Ans. Ratio of two wavelengthsGrating equation is given by

d sinO = nA.For first wavelength n = 3

d sinO = 3\,For second wavelength n = 4

d sinO = 4A.,Comparing equations (1) and ( 2 ) we have

3A.1 = 4X2

OR AL =1K 3

Hence, the ratio is 4; 3.

Q.9.11 How would you distinguish between unpolarized and plane polarized light ?

(Bwp 2004, Mir Pur 2004,Rwp 2004-2005,Min 2005,Grw 2.006)

Ans. Plane polarized unpolarized and light

Suppose light is incident on a Polaroid. The Polaroid is slowly rotated about an axis which is along the

direction of incident ray.If the transmitted light is plane polarized, it becomes dimmer and dimmer and disappears at certain

orientation.

If the transmitted light is unpolarized, it becomes dim but not completely blocked at any orientatingQ-9-12 Fill In the blanks.

According towavelets .

In Young s experiment, the distance between two

smaller than that for green light

The distance between bright fringes n the

the light used increases.

source of secondaryHaugen' s principle, each point on a wave front acts as a

So .0) (i)

adjacent bright fringes for violet light is

interference pattern incr^jj££ « th* wavelength ofSo («)W )

(iii)

‘

r. lfpc PIMCW ‘tnJ -J

, PHYSICS XI rs141* : ^.-v* ^*-> 4 p«ter/ V V ..'-4 *9*r<*f.'^»w r* « -««s «• >» artvttrm- r>v v ^'• =•;*'<*-«s**«»8» '/ i'V*'- M**V y ,£" *>** •.•*V f--- ", <Wruos* fr -QfUz*

V***6 U a m't4t>'.Ci f o e «*»* i #* (r/JawJ x •*<;ojrf w _T.***J>jiante between 1-o bright friogn » fringe ipac;

( jlemWw-Af fiamsola for fringe tpacmg a

,y ?|r! .

XLLys T

p.jfo-g vaJ -jci, ?«rSolved Examples71M

*** ' < K^f' ioca 4*59r. er. Oa \<

» y^f n^fMMX M 4.251^« ^ Tk 4stMa of tW ikinf** wavefcugfli a# fW Mcidesi hgkf.fraf a

' MMjt rje ..ver > .*| * 4 0 25or. 2 5 / Iff *mf> Mr:**',i ygjstrr.\ y.r- shU L 1 I9f> cm i r,tt4a«eof ted4irk In^•jr•6^Wcm - 5.9 / JO 4mror « rlr. ird rfe/V fnr. jc rr. 2

jjgh » of wavelength 450 nm in incident on a diffraction grating on which 5000 ImeWcm have been

(0 How mao? order* of ipectra can he observed on either tide of the direct beam?( ii ) btttrrr ne the angle corresponding to each other.

6«ra|Data:h ; /clength of light = X 450nm - 450 / JO Yn

Line'; ruled on diffraction grating *S* 5000 lines per cm

cm *

t* FiafcVarveiengfh of vncUienf fight X~ 7

f . alcHbtNMKA the fomv,.3 fr*d*rk fringe it

, 1 Aly (m * - )6

1IIOrating element d — -N7 Find;

m5000005000

Orders of Spectra n - ?Angle = 9 ^ 7

(i)P’ / ungva we get

50 / 10 4(ii )

Calculation:(i) For maximum number of order of spectra, SinO “ I

As m a x i m u m possible value of angle of diffraction is 0- 90°, therefore sin 90° - 1Using the formula

I X / la > ’ > ( Y >2 5 / 10

fflfcfcr)V> / 10 4

d SinO * nXHutting values, wc get

2 / 5.0 / 10 4/ 2 5 / 10 1

«•/. -X 5.9 / I 0'7mX - 590 / 10 °m£ 590 nm)

Or5

\ 500000 J 1 * n / 450 * 10 /y

(* 10

500000 / 450 x 10-9Or109 _L x IO3'4

n *5 * 4 5 * 1 0*

n = 0.0044 x 103n•4.4

Hence, maximum order of spectrum * 4

225Yellow sodium light of wavelength *89 nm, 'milted by a single source passe* throu

^ ^ f9r $ p&slits J .00 mm apart. The interference f atter i is obs ved on a screen 225 cm away*

are two adjacent bright fringes?

EHGiven Data:Wavelength of sodium light-X = 589 m - 589 x 0 9mSeparation between the slits d - 1.00 nm 00 x lO^mDistance of screen from slits* J , 225cm - 2 25m

For first order spectrum, n - Id sinO = nX

(ii)Using

Ir\ i foo Pi l low •miJ

34a ChaptersPHYSICS- XI (Subjective)y*»caj

Scb°^ar s 343Putting values, we get

I Exercise ProblemssinO = 1 x 450 x 10"9

500000sin 0 = (500000) x (450 x 10'9)sin 0 = 225 x 106 x 10"9 Light of wavelength 546 nm is allowed to illuminate the slits of Young’s experiment. The

separation between the slits is 0.10 mm and the distance of screen from the slits where interferenceeffects are observed is 200cm. At what angle the first minimum will fall? What will be the lineardistance on the screen between adjacent maxima?

9-1sin 0 = 225 x 10 3

sin 0 = 0.2250 = s i n 1 (0.225)lQ = n°

For second order spectrum, 0 * 2d sin 0 = nX

Given data:Wavelength of light =* X = 546nm = 546 x

Separation between spts = d = 0.10 mm = 0.10 x10°mslits = L = 20cm = 20 * 10 2 mfroDistance o

Putting values we get To find:i = 0 = ?stance between adjacent maxima = Ay =?

Angle fc1 mm,iisin 0 -(2) (450 x 10"9)* lin<500000 i ringc spacmj

Calculation : fT As the formul.VII

sin6 = (500000) (2 x 450 x ( O'*)sinO “ 450 x 10* x 10"* la for minima issinO * 0.45 1=(m +-)X0 = Sin” 1 (0.45)

For first minima, m = 0|Q - 26,7°For third order spectrum f n 3 Putt mu values, we get

546»10'2 »0.10* I 0J

sin0= 2730 x 10“*

d Sin 0- nXsin &=Putting values wc get

1 sinO - (3) (450 x 10-9)500000

sinO = 0.002738= sin '(0.00273)

sinO- (500000) (3 x 450 x l O'9)sinO 675 x 10A x 10’9

sinO - 0.675 0= 0.16°0 = Sin 1 (0.675) For fringe spacing, using the expression19 - 42-5«

UAy =-For fourth order spectrum , n « X

d SinO * nX Putting values, we getPutting values, wc get20* IQ 2*546* 10 °

0.10xI0 J~

Ay =1.092* l0JmAy = 1.092mm

1 Av =sinO - (4) (450 x 10-9)500000sin9 - (500000) (4 x 450 x 10"*)sin9 = 900 x i0‘x 10-9sinO = 0.9 |A y m m approx

Calculate the wavelength of lightscreen placed 200

from the central bright image.

- twhich illuminates two slits 0.5 mm apart and produces an

away from the slits. The first bright fringe is0 = sin (0.9)b = 62.2 j cminterference pattern on a

observed at a distance of 2.40 mmwen data:

= d = 0.5mm =0.5*10"'mDistance between slits

• ?JP,

• 1 -I

Ir7 i

r u f o o o n io u I A I *344

Chapters«i PHYSICS- XI (Subjective)fj Stb**L 345Distance of screen from slits * L 200cm 2m

Distance of first bright fringe °yB 2.40mm * 2.40* 10 * mFor first order maximum m I

A monochromatic light of /1 =588 nm is allowed tMichebon interferometer . If mirrorobserved to shift?

1 wWavelength of light - X = 588nm =588 x lO’mDistance through which mirror AY, is moved = L =0.233mm = 0.233x1(r’mirv

Number of fringes shifted m = ?C alculation: As for Michclson interferometer, the expression is

ILJffi

I on the half silvered glass plate (7, , in theid through 0.233 mm, how many fringes will be

9.4i

To find:Wavelength of light A 9 Given data:

Calculation:Using the formula

XLTo Hod:y=n,T

x--£mLPutting values, we get

2.40* l 0‘l*0.5*10 >

As for first fiingcXH .20*0.5*10*

X 0.6* 106m

or

/2LmW2

Putting we getm 1o, >2*0.233*10

588*10’m = 792!

X 600 ntnA second order spectrum is formed at an angle of 38,0° when light fulls normally on a diffraction

atiog having 5400 lines per centimeter. Determine wavelength of the light used.9.5

Give n data:In a double slit experiment the second order maximum occurs at 0-0.25" Th< wavelength is 8C

nm. Determine the slit separation.9.3 Angle of diffraction = 0 = 38°

Number of lines on grating = N = 5400 lines per cm =540000 lines per meterGiven data:

Order of spectrum = n = 2For second order maximum ~ m 2To find:

Angle “ 0 0.25P

Wavelength of light = k = ?Wavelength of light X = 650nm = 650 x 10 Calculation:

To find: Using the formulaSlit separation d ? dsin9 = nX

Calculation:, dsinQX=Using the formula ndsin0 = mX ld «—mX NOr d = sinOsinO x =—Putting values, we get N*n

Putting values, we get2*650*10'9d = sin38°X ~

540000*2sin(0.25°)

d =2.979 x 1O'4X =5.70*10’

d =0.3*10°« X =570*10*11»

|Xa 570nmjm = 3.3mm

346Chapter 9 [phL5ic«i (V..

S.hnlar’s PHYSICS- X!(Subjective)j9.6 A light is incident normally on a grating which has 2500 lines per centimwavelength of a spectral line for which the deviation in second order is 15.0°f' ^0lnPuteGiven data:

347

Blue light of wavelength 480 nm illuminates a diffraction grating. The second order image isformed at an angle of 30° from the central image. Slow many lines in a centimeter of the gratinghave been ruled?

Wavelength of light = X = 480nm = 480x1O^rnFor second order image = n = 2Angle of diffraction = 0= 30°Number of line per cm = N = ?

9.8Number of liner on grating = N = 2500 lines per cm = 250000 lines per meterAngle of deviation = 0= 15°Order of spectrum = n = 2

Wavelength of light = X = ?

Given data:To find:

Calculation:To find:As grating equation is

dsin0= nXCalculation:1 1 Using the formula

dsin0=nX

— sin0= nXN„. sin0

— sin0= nXOr As d -—N iV_ sineN*n

Putting values, we get

sin15°

1OrAs d=—N

nXOr X =Putting the values, we get250000*2

X = 5.176*10‘7

X = 517.6*1O'*m

sin30°2*480*10-’

Or

0.5N = 2*480* I O'9

N =5.2*105 linesper meterX = 518nm

Sodium light ( A = 589nm) is incident normally on a grating having 3000 lines per centimeter.What is the highest order of the spectrum obtained with this grating?

9.7N = 5.2*103lincs perem

x-rays of wavelength 0.150 nm are observed to undergo a first order reflection at a Bragg angle of13.3° from a quartz (Si02) crystal. What Is the interplanar spacing of the reflecting planes in the

9.9Given data:Wavelength of sodium light = X = 589 nm = 589 x 1CT9 m

crystal?Number of lines on grating = N = 3000 lines per cm = 300000 lines per meterGiven dpta:

Wavelength of x-rays = X = 0.150nm -0.150x10 mAngle of highest order 0

To find:For l #t order reflection “ m = 1Order of the spectrum = n = ?Bragg ,s angle » 0* 13. 3°Calculation:

To find:As grating equation isInterplanar spacing d ?

Calculation:As dsin0- nXUsing Bragg’s equation

2dsin6“ mX1or — ilne- nXN

mXd*sinG2*in0NX

Putting values, we getPutting the values, we get1*0.150*104

d «2*iinl3.3*8W90°,n »

0.150M04300000*589*10d- 2*0.2310

300000*589" / * d•0.326*104md «0.326 nmn-5.659

n-5.66P- 5t MHence

m

*5* nmile%\_g •(348__Chapter 9 fuh

An X-ray beam of wavelength A undergoes a first order reflection from a crvof incidence to a crystal face is 26.5°, and an X-ray beam of wavelength ^11 its anK|third order reflection when its angle of incidence to that face is 60.0°. Asnni Unt*er8oe$beams reflect from the same family of planes calculate (a) interplanar spacin'! '?8 thal the to*(b) the wavelength A. R ° f *lc PUncs anjj

ysicalQPii^i 1H ICS ~ X1(Subjective)9.10349

Chapter10Given data:For first w avelength A1

Angle of incidence = 0|= 26.5°For Ist order reflection = /?, =lFor second w avclcngth A^Wavelength of x-ray beam = 0.097nm

Aj = 0.097x lO ^ mFor 3rd order reflection = n2 = 3Angle of incidence = = 60°

•Inter plane spacing = d = ?Wavelength of first beam = A, = ?

OPTICAL INSTRUMENTS

Learning Objectives ]1t

Recognize the term of least distance of distinct vision.

2. Understand the terms magnifying power and resolving power.j. Derive expressions for magnifying power of simple microscope, compound microscope and

astronomical telescope.

To find:

Calculation: Understand the working of spectrometer.Using Bragg's equation

Describe Michelson rotating mirror method to find the speed of light.2dsinO=nXUnderstand the principles of optical fibre.For 2nd wavelengthidentify the types of optical fibres.2dsinGj=n,A..

Putting values, we get Appreciate the applications of optical fibres.2dsin60° = 3*0.097* 1O'’

2d*0.866 s= 0.291x l O *0.291* 10’d * 2*0.866

d » 0.168* 10 #m

d-0.168ntn

For calculating wavelength Ax2 dsin0, v n,X,

2dsinQfn

Putting values, wc get

X - 2N0- 168* IO *sin:6.3°1

X, - 2*<XI68M0 **0 1 ft2X, -0.140*10*

X,«01$nm]

«•» *«» f

150 Chapter 10[Opticali= ]PHYSICS- XI (Subjective) 351

$cho,sir

Chapter No. 10 Optical Instruments

....««,o,— w34lfc amBiearecaWe usemicroscope, compound microscope telescope, spectrometer, optical fibers etc..= ?

X

Visual angle

The angle made by on object at the eye is called visual angle.

Least Distance of Distinct Vision (Near Point)

n‘"'‘TnTmjj";nCefrZ*'r #A » be distinct is I *called the least distance ofdistinct vision or near point. It is denoted by d „

>•:c_ u

V 27 v * zUiV r J-i fIu cV Fig. 10.2

When the seme object Is viewed at•ehorter distance, the image on theretina of the eye is greater; so theobject appears larger and moredetails can be seen. The angle 0the object subtends In (a) isgreaterthanS’In (b).

*2 ExplanationFor a human eye, a sharp mage is formed when the object is placed at a distance

beyond a minimum limit. But v/hen the object is within this limit, the image

appears to tfe blurred and fuzzy. Jhe distance of distinct vision is about 25

from the eye.(for young The location of near point increases with age.

r I I I7. S' X*i >• -Vi

A §y x_,

£ C §I5 ?|~ *- c -J —-z ~ .Ec c

SO -

e e • cmc.-J

—Linear MagnificationThe size of the image goes onpoint to the focus of the lens. So the magnification takes place.

of the size of the image to the size of the object is called magnification

XLJ a far offincreasing, when the object brought from

/.5

&•A -— -

r- £=0 V

The ratioor linear magnification .Le.0 o£••-# 2a 22 ^u — C

z, ~T

c g.fi,SisSi.sizeof imagerj JC Magnification =2 - SO O—u sizeof object22 ~ a. 26a u «0

I c Ic r*-,

I (1)|6 - J.1 • •

M = —08. £ ORQ -5.I ? "u m.ie.1I ?w * £ e

v.•x Another DefinitionThe rath of distance of image from the lens to the distance of the object from

lens is called magnification.Magnification

Viu- £c 5 Sa.1cu I image distance

" object distance3''I

mm PM (2)siB ORii

2- 13 j P£ ll Thu* by comparing (1) and (2), wa have

I.a0 p

*sJi M *s n I TK«U**r w,«nlfl«‘lon

he ^Unifying powar or angular magnification can be defined a* the ratio of

afi8les subtended by the Image as seen through the optical device to that

blended by the object at the unaided eye.a a the angle subtended by the object form near point at unaided eye and

3 9 angle subtendedby the object with optical dcvlee at near point.*4 till 214

Thenangle subtendedb;

Magnifying Power

M = — Diagramae,

OR M — 0.Resolving PowerThe resolving power of an instrument is it ability to reveal the minor details ofthe object under examination.

ORThe resolving power of an instrument is it ability to separate the image aftwovery close objects.Mathematically Working

int d. The angle subtended by the objectThe resolving power is the reciprocal of minimum angle of resolution Consider an within focal length of convex lens, then a

is obtained at near point d making an angle p at eye.Raleigh showed that for light of wave length X used to see two points separatelyby using a lens of diameter D, _ at eye is a

magnified and virtuPower =:2^=— P'Angle of minimum resolution * n„_ - 1.22 — L°gnificat on is

By using the figure (I)

tana =

The angulD.0) ( >1 I)OR Resolving Power — — »

amim 1-22XResolving power of plane diffraction gratingThe resolving power of plane diffraction grating is

a

sizeof objectrit row ftnd n to reW

wnai print nuke t pinho*# *a p*ete of paper and hold* mfront o< your rye do* to \ -page . You - *<*r the prwclearty.

distanceof the objectq m dX oX

.

R = tana = -7dXrX, AXwhere X* average of two wave length X,and X,)And Xi * Xa * XAlso AX = X,-If N is the total number of lines ruled on grating then

R N * mWhere m is the order of diffraction.

r%Q.10xaimp**For small value of a , tan a a

OSo a = —

propagation of I gfit through

th* p.nhole wtiic* mskti th«dear image at n a pinholacamera.

It H tt*r dBy using the figure (2)

tan(5 =Sizeof imagei

distanceof image

ltan|3 = -

For small value of p, tan(i * P. Also q - d

S°Putting the values of a and P from equation (2 ) and (3) 'n cqustion ( )

l/d

M.iO

l Size of image

O‘Site of object' DuUtnce ofobjecl P

Q>1 What Is a simple mlcroicopt? Calculate Its magnifying power.(3)

MicroscopeA microscope is a device which U used to see the magnified imag of very malland near object. —Simple MicroscopeA convex lens can be used for magnificationWorking principleWhen the object l* placed bet' «n foe poim md op if center of biconvexlent then an erect, virtue! and anlarged image It ov ned.ConstructionIt consists of a magnifying glass (l.t. blcorivi 4 lens}

Distance of image^

<1Also

T*«> ,1 *

-A •.•v \\ >•-

K ‘t ' :-v,‘*.<• * •

T E*Vv.f •**§.’{£ -

.

4 -+ 1 M>i.t*

354 — r— —Chapterio [C), K< ;i||

s PHYSICS XI (Subjective) 355"Mr,,5*?»bM * -~

/>DiagramRayTherefore, M= -

PNow, using the lens formula

(4 ) q = dj A

i i »h

5i*

I «y« P*f p qSince the image Is virtual, so q = -d

r

1ft - t«v«iiiu*nlh oanlury mteruvcofiaKtiloh could ba tno.»d up and -Itnt .i in

Ha kupport ring {Cb»>Ua,vt llw (*v.«um

of m» hlalriry of Osianta, tlormcs)

— C — —f p (1Multiplying both sides by d, we get

d d df p d< = <- >f Pd , cl— =I i -

Ray diagram of a Compound Microscop*

WorkingImage formed by objectiveThe object of height li is placed just beyond the principle focus of objective. A

rted and magnified image of height h, is formed. This image acts as an

's

real, Inveobject for eyepiece.Image formed by eye piece

formed by objective lies within the focal length of the eyepiece. Eye-

virtual and magnified image of height h2 at near point.

ORrp

The imagepiece makes aMagnifying PowerIfP = angle subtended by the final image of height h2 at the eye

or- angle subtended by the object of height h at the eye.

( IBut — = MP

M = 1 + —Therefore,f

Which is the formula for magnification; it shows that focal length shouldbe small for high angular magnification.i hen

P _ tanpa tana

M = -Q.2 What Is compound microscope? Describe its construction and working.

Also calculate its magnifying power. Since tana = — and tanP =Cl

h,/ del

(jOg*) So M =h / d

Compound MicroscopeA compound microscope is used when high magnification is requiredConstruction

h2M = —h

Multiplying and dividing by h,h2 h,

M = — x-Lh h,It consists of two convex lenses.

ObjectiveIt is of short focal length and small aperture.Eye pieceIt is of large local length and large aperture

(0h, h,

= -!- X —i-h h(ii) 1

=Mlinear magnification of the objective

-y = linear magnification of the age piece -M,h,

(1)

Principle i

hd MicroscoP4'When the image formed by the objective of small focal length is within the focal

length of the eye piece of large focal length then a virtual, inverted andmagnified image is obtained.

A Compoun

i

ThusNow M,can be written as,

M = M,x M2i

.: t>.Vf *«9A.-'

rt v‘

ppm

Scb0Ur^PHYSICS- X1(Subjective)

Jk w .

356Chapter 10tOptioii '

M,=i (2) -PprincipleA real, Inerted and diminished image formed bv .

ct for eyepiece which is at the focal point of bo h rh 'ect've

nified image is formed at infinity. ' the er'les then a vi*ual

As eye piece acts asexpressed as.

a simple microscope, hence its magnification M2 serves as ancan beobjeMj =l+— (3) and magWorking

ige formed by Objective lens' 0ar3"el bea ' Hght rays coming from distinct object forms an image A’ B' at

the objective. The image formed is a real, inverted and

f.So, equation (i) becomes

M.aL-PL f« JFor higher magnificationWe should use the eye piece of shorter focal lengths.

imaThe pafocal P°int of

diminished.(4)

formed by eye-pieceImage

The real image A'B' acts as

focus of the eye-piece. The eye piece

image formed is virtual, inverted and magnified.Focusing for IniinityWhen the image

eye piece. Then theLength of Telescope:(innormal adjustment)

In normal adjustment of telescope, the distanceobjective is called the length of telescope.So

L = f0 + fe

object for eye piece. Image A' B' is formed at theforms the final image at infinity. The finalFor higher resolving power

With wider objective we u$e blue light (of shorter wave length) to produce lessdiffraction and to give more detail of the object. as well asMB — —

formed by the objective is at the focus of objective

telescope is in its normal adjustment or focused for infinity.

between eye piece andQ.3 What is an astronomical telescope? Find Its magnifying Power.

Astronomical TelescopeTelescopeTelescope is an optical Instrument used for viewing the distant objects. It Is ofmany kinds.Astronomical TelescopeThe telescope used to see the distinct image of distant heavenly objects likeplanets or moon, is called astronomical telescope.

Magnifying PowerFor a telescope, it canimage at the eye as seensubtended by the object as seen by unaided eye. So

be defined as angle subtended by thethrough telescope to the angle For Your Intorm.itiu"

PM = - (1)a

In right triangle OA 'B'

tafi a =A'B' A'B'

f.OBConstructionA'B'

tan a = a =A simple astronomical telescope consists of two parts. tto(i) Objective: It Is of large focal length fQand large aperture. in right triangle O'A'B'00 A'B'Eye piece: It is of short focal length f,and short aperture. A'B'tanp = P =Ray Diagram f.O'B'

Putting values of a and p in equation (1), we getSo, equation (1) becomes

A'B' / f.M =A'B' / f0 Reflecting Tel'

f.A'B'M = x

A'B'f.fOR Ms —f.

focal length of objectiveM = —locallengthof eye P1CCC

368

Notes PHYSICS \l (Subjective)A good telescope has an objective of large focal length and largeimage depends upon intensity of incidentuse the objective of large aperture for thismagnify as well as make image to be bright.

359brightness of finalobjective. We

aperture. Thelight on the

purpose. ItThe telescope should focus so that the parallel rays enter!

wire near the eye piece.of Spectrometer

are focused at themay cross

UsesIt is used to;. Study the spectra of different light sources.. Study the deviation o{.light by glass prism. Calculate the refractive index of material ofprill

Measure the wave lengthof light by grating

Q.4 What is a :write down Its

spectrometer? Describe itsuses. construction and working; also

SpectrometerThe optical device which is used to study the spectrum of variis called a spectrometer. Spectrum of light beamrefraction by prism or by diffraction f

Construction

Q 5 Describe the Mkhelson\ experiment to calculate the speed of light.ous sources of lightobtained either by

can berom a grating.

Speed of LightLight travels so rapidly that it is very difficult to measure its speed. Galileo

was the first who tried to measure the speed of light. Although he did not

succeed in measurement of speed of light yet he was convinced that ittakes some time to travel form one place to another.In 1926, Michelson made observations for the calculating of speed oflight in air The apparatus and experimental setup is shown in figure.Experimental arrangementAn octagonal (eight - sided) polished mirror M is mounted on the shaft ofa motor. The speed of motor is adjustable. Suppose that the mirror instationary position. The light from source S falls on face 1 of mirror M.The reflected light from M falls on plane mirror m. The light reflectedfrom mirror m falls again on face 3 of M. On reflection through face 3 itenters the telescope TWorkingWhen the mirror rotates anti - clockwise, initially the source of light willnot be visible but when mirror M gains a certain speed, the source Sbecomes visible. This happen when the time taken by light in movingfrom M to m and back to M is equal to the time taken by face 2 to moveto the position of face 3.Expression for speed of lightThe angle subtended by any side of eight sided mirror at the centerradian.

Thus

It has three main components:(i) Collimator77;e function of is to make theend of the tube aprovided.

67/^a 0 «rays coming fromconvex lens is fixed and >When slit is just at the focus of convex lens then light rays enteringfrom slit become parallel after passing through the lens.(ii) Turn Table

bv source parallel. Atan adjustable slit is

a nearon the other end. one

Telescope

Turntable is capable to rotate about the fixed verticalattached at the bottom of turn table,to observe the spectra.

Adaxis. A circular scale is alsoA prism or grating is place on the turntable Sourceof lighl

m

Michelson’s method formeasurement of speed of light.

is 2n/ SSch.nutl.'dl»Q.Spectrometer

0 = — radian .(i)S(iii) Telescope

A telescope fixed on a stand and is rotatable about the r-turntable. A vernier scale is also attached along with the teleWorkingBefore using the spectrometer, one should «horizontally by leveling screws. The collimator isof light.

2TTAlso 0 = (nt

0 = ( 27tf) t0 = 27rft

Comparing equations (i) and (il), we get

— = 2 Tift

OR (u=—T'^me axis as that ofU>= 27lfscope.

00carefully adjust the turn tableadjusted to get the parallel rays

= f tormunication system based on optical fibre has large capabilities, it

mit thousands of telephone conversation, T V program8

, The comcan be used to trans

and data transferring.

’ ^e'f'CienCy ^ „and

. Fiber optic system consists of much smaller and „ght weieht cah|o

. A fiber optic with its protective case is about 6ngh* Mbles'

can replace a 7.62 cm diameter bundle of m •dWeter< wh'ch

amount of signals.'^copper the same

q.7 What are the basic principlesMe optics? Explainn,.m

1t = —or

8f Optical fibre imagt.

The distance covered by the light in this time is 2d i.e., S = 2ddistanceSoeed of light =As

timeSc = —t

wPutting values of S and t in it, we get A•jvxsr issu'd v**^* if.

*r *****'rnJ* *'crt v;r.kr fjfA-

The Wrrr -i »2d

C "1/8f FXIM.ANATION

A typical glass fibre is

aboui 100-250fzm inOR c =16 fdHe calculated the value of c in vacuum and is

Fibre Optic PrincipleThe propagation of light in an optical fibre

fined within the fibre and cannot escape from it

For smallrequires that light should be totally

be done possiblediameterdiameter the phenomenonof total internal reflectionis observed and there willbe small chance of power

of light due to

Pom!to Ponderc = 2.99792458 * no* m/sed. This can

c = 3 * 10* m/secORNote

Total Internal Reflection(2) Continuous Refraction

discuss these two different ways in detail.

(1) Total Internal ReflectionWhen a light ray traveling from a denser medium towards a rare

medium, makes an angle of incidence greater then critical angle

of ihe medium, then the ray is totally reflected back into the same

dtuser medium. This phenomenon is called total internal

reflection.Critical angleThe angle of incidence in denser medium for which it

corresponding angle of refraction is 90 is called critical angle.It is denoted by OcRefractive Index:The transparent medium has particular value of refractive Index.

The ratio of speed of light in vacuum to the speed of light in

transparent medium is called index of refraction of that medium._ . . Speed of light in vacuumbQ > Refractive index of material = —Speed of light in the medium

The speed of light in other materials is always less than c.• The speed of light depends upon the nature of medium (refractive index)• However the speed of light in vacuum and air is nearly equal.

lossesdispersion

(1) ^Now tie :

e,Reflected RayAir nQ.6 Write down a short note on optical fibers. Also discuss its advantages?

••

:Eacft o'm* 9m optc*V ae n i o Si trough n

eye v'a /.by *r»82»'J 'c<e«TWSU<TP

iFiber Optics x / nFor hundreds of years man has communicated using flashes of reflected sun lightby day and lanterns by night. Navy signalmen still use powerful blinker lights totransmit coded message to other ships during periods of the radio-silence.Graham Bell invented photo phone after the invention of telephone. His Photophone used a modulated beam of reflected sunlight focused upon a seleniumdetector placed at a distance of several hundred meters away.In this way bell succeeded to transmit a voice message through a beam of lightIn recent past the idea of transmission of light through thin fibers optic has beendeveloped. In these days it use in communicational technology.Advantages of Fibre optics:It has large advantages due to the use of optical signal (light signal instead ofradio signal).• It has wider band width of capability and safe from electromagnetic

interference.Optical fiber is used to transmit light around the corners and intoinaccessible places.

/ / jWaicr.Gave*or Plwilc

If the angle of reflection in the air is

90" the angle of incidence is called

the critical angle.

//

I*

£f Wafer.Otuat«

For angles of incidence greater

tne critical angle, all tne lights la

reflected; none is refracted into the air.

*%

cn = -OR x

X4

I UWhere the value of 'n' (refraction Index) depends upon

tedium. Being ratio It has no unit.When light incident from one optical medium to the other then at

th« boundary, a part of Incident light may be reflected back to the

remaining part may be refracted into the

nature ofmen

• It is possible to study the interior of lungs and other parts of human body. same medium while theFor this purpose, a flexible fibre optic . nsertt nto the body. other medium.

i he amount of light reflected or refracted at the boundary dependsvalues of refractive indices of two optical media. It also dependsof incidence.upon the

upon the angle(2) Continuous Refraction

**** m°def Pr°Pagat,0n°f 'ight through optical flhIf 0, is the angle of incidence in medium T and 02 imedium '2'. Then by Snell's law. the angle of refraction in

n,sin0 = n 2 sin02_ sin0:or value

0f less index of refraction. Then such a fibre is called multimode graded indexsin0n2If right comes from denser medium fibre.

If the cladding is of constant index of refractive and less than that the core then -It is called multimode step index fibre.

and enter into rare medium then 0,must begreater than 0,.The particular value of angle of incidence 0,becomes 90® is called critical angle 0C.When 0*= 0t then 02 = 90?

n,sin0c = n2 singo^sioOc = —

. for which angle of refraction 02 As light ray passing from a denser medium to a rare medium, it bends away fromnormal and vice

In multimode graded index fibre, the light signal has confined within the optical

fibre by continuous refraction and then refraction from the boundary and thenagain 'efraction, so light remains inside it and cannot leave or escape from it.

MS

verse.

So Fia 10.11Croaa *eciion« l »1trm of(a) MuW-modo atop Indo* fit**(b) MuiU^rwxH graded Index fib#*n

^

sin0c = L2 cr .-.r~=yon?1.5

where n2 = refractive index of airrefractive index of glass

sin 0t = 0.667

= 1.0Hi = = 1.5 ight propagation with a hypothetical

multi layer fibre.Or Be = sin'1 (0.667)= 41.8°It is the critical angle of glass.

Condition for total Internal reflectionFor that two conditions are required(I) Light should travel from denser toII) The angle of Incidence should be g

Or Q.8 Explain the different types of optical fibres? Jackal0C

Qlaaa claddingTypes of Optical FibersOptical fibers may be classified Into three types.

(I) Single mode step Index fibre.(II) Multimode step Index fibre.

Multimode graded Index fibre.

Qlaaarare medium.raater then critical angle. (•)

1 (HI)Outgoing NotesThe term mode Is the method by which light Is propagated within the fibre.0) Single Mode Step Index Fibre1 It his very thin core about 3 pm diameter.a It has a relatively large cladding.1 Monochromatic light sourct Is raqulrad to sand light signals through it. That

li why wa use •laser source.* It cen cerry more than 14TV chinnals or 14.000 phone calls.(H) Multimode Step Index Fibre1 It Is the optical fibre In which central cora has •

diameter such as 50pm and high refractlva Index.1 The central core hes •constant rafractlve Index n, such as

i.sa which steps down up to Ml at the boundary with thecladding.

1 It Is useful for short distances only

Cladding

oran.Inoomli4

(t»ixial ray 4 * •ingla •moda alap •Indaa hbn.a n. •1.1

n, •1.0

Propagation of light within a glaaa rodsSSSS*3®rod from air at an an|la I... than e !'. ' r,y'T",n< '"«>,h*I1"1

boundary of lurfaet of rod juit Ilka ray 1 which l/lneldanTit th.r°Ug.h theWhlla tht ray a and j maxing any, 4J, 'k#t 'n,l,of »° •raflaetlon md 10 thay propagata Inilda tha >|IU r*OdVh\ t0t*1 l'ilt,rn*1

ssssr * * £STmEsz

rig. 10.10 (h)Light propagation within afoKlblaglaaaflbra.

21

Light pmpdflSpUon (hrou^Mul..ml*

365364 rHv^iCS- XI (Subjective)^EifUltOptica

Optical fiber(iii) Multimode Graded Index Fibre• t s g"i ssi-cs • £ *• " ch cent' - core nas h.gh retractive

3 :r - r. • r =:'e 5;= 5 :z .-. z' is is o ^ :e ~

• 7**e:'

s-t;e r"e co'e -arges from. 50|um to 1,000pm.

• ~-iere s -- parties ai- aoundap. setweer. core and cladding.

~" £ pt is co^ftir jcHxsfy refected with > the fibre optics. It is_ s £ *_ for org distances.

jy\icropbone ElectricalsignalElectrical

(sound) t— >signal

losses in oQ-9 Explain how the signal is transmitted and converted into sound?

losses of power due to givenLosses of Power

light s^al travel

(1)Power loss by scattering and absorption

When the ligh: travels along ribers by multiple reflections, some of light energy is

zbsorbed by trie glass medium. It is due to the impurity of the glass medium.

Some part of energy of light signals is scattered by the group of atoms such as

ugh fiber, it suffersSignal Transmission & Conversation to Sound~ " e * c'r optsc co— cat on system, consists of three — ajor components.

(I) T arvs— tter

(if) Optica fiber

Whenfactors.

(ifi) Receiver(i) Transmitter

transit tier converts :re electr cal signal into light signal which is obtained‘ror- microphone.

_ve ght so-'ce n the transmitter is either a LED (light

em/tr.-.g dope or ' a-se'. got signa is invisible infrared of typical wave‘3 JT A * c" ~o.es *‘aster than visible or ultrav det light. The lasers and

LECH -sed ' tr s app cation are tiny un.ts (less than naif the size of thetn - orde’’to mate* the size o* the fibres.~ * e r * s g'« s S'e modu ated to transt the information, it flashed on and off.D i c« mod- at o' s e /c'essec r b ts of megabits per second where the bit ise re' * ( on state) o' o (off state).

(ii) Optical Fibre"re modu- ated p - se tra/e through the optical fibre by total interna reflections'5 co i'-o-s r refracton w»th wery faster speed despite the ultra purity (99 -59% g ass; o* tr.e opt cs f per. Tr,e gnt s gna's while passing through the optical

* p'e become d rr and ~ - s t c e fegenerated by a device call repeater Repeatersg'e typ zc y p aced 30km apart, but -n the newer system this separation is about'OOvm

(iii) ReceiverReceiver captures the gnt sgrais at the ore' end, and - converts then toe ectr ca pfgnals by means of photo-bode y commur cation can be'ep'esented by a particular pattern or coo c- f V* anc - by using computer typeec - pment The rece.ver is pre *mme to c *. it to . nd os, and converts itr to sound, P‘Ctur t w d*« »J reou ?d

joints.^v^H

Reduction of loss

Care- - manufacturingcanand scattering.

losses by absorptionreduce the power

loss due to dispersion

If he light signal is not perfectly monochromatic, then

band of wave lengths are refracted in different directions.

So they cover different lengths of paths inside the fibre and

produce phase difference and reach at different time, as shown in

fig- (a), having different wavelengths Xu X2 and X3. So the signal ^received is distorted and faulty.Reduction of lossDisadvantage of the step-index fibre canfusing a graded index fibre, as shown in figure.

The different wavelengths still take different paths and suffer total internal

reflection at different layers, but still they are focused at the same point like x

an<* y etc. It is possible, because the speed is inversely proportional to the

refractive index. So the wavelength Xt travels a longer path than X* or X3 but at a

greater speed in a low density portion of fibre.

Time differenceIn step-index fibre, the over all time differencemay be about 33 ns per km length of fibre . But using

time difference Is reduced to about 1 ns per km

(2) Power a narrow

Light paths in (a) step-indexand (b) graded-index fibre.

considerably be reduced

wavelengthsfibre, thebetween different

a graded index

'r

366

Hll PtorlO [( >Pticai|SpHVSlCS XI (Subjective) 367FORMULAE)

Multiple Choice QuestionsLinear magnification IM = —O

ible answers to each statement arc given below. Tick ( S) the correct answer:P four possiM *’Angular magnification

M = “1uThe objective of telescope has large aperture to:

(a) Reduce spherical aberration

(C) Have high resolution

2 The sky appears blue because:

(a) Real light is absorbed(c) Blue light is absorbed

3 The magnifyingpower of telescope

(a) Increasing the length of telescope (b) Increasing the focal length of objective

(c) Increasing the diameter of objective (d) Increasing the length of eyepiece

4. When light ray travels from one medium to another medium, the characteristic which does not

change is:(a) „ Velocity (b) Wavelength

(c) Frequency (d) Amplitude

5. The magnifying power of an astronomical telescope is 10. If the focal length of objective is too cm.

What is focal length of eyepiece?

(a) 10 cm(c) 1000 cmFor normal adjustment, what Is length of astronomical te\escope

eyepiece are100 cm and 20 cm respectively?

IOO cm (b) 20 cm

(c) 5 cm7* Two convex lenses of focal lengths 10 cm and 5 cm

lengths are:(a) 15 cm

0(i

1.22 —y(b) Increase span of observation(d) Have low dispersion

Resolving power of an optical device < / I)R.P. = . 1Olid

D a 1 - 22X"MilX XV Resolving power of an diffraction grating = .(b) Blue light is scattered most

(d) It is sky's natural colour

R = N x mX. -X AXi

dMagnifying power of simple microscope

Magnifying power of simple microscopewhen focused for infinity

Magnification of objective in compoundmicroscope

L Magnification of eye piece in compound^1 microscope

M = -vM =I+i can be increased by:P r

dM = -f

M,= 5LPdM2 = l-f —

MM = - l+ —Magnification of compound microscope M = M,x M2

(b) 100 cm(d) ' 5 cm

fP e -Length of telescope under normal adjustment L = f0 + feIf focal lengths of objective andf

M = — 6.Magnification of telescope

Speed of light , (a)c =16 fd(d) 120 cm

are placed in contact, then their combined focalcx/ Refractive Index n = —v

sinQjsin0,

n(b) 5 cmSnell's law n , sinO, = n 2 sinO 2 n2 3(d) — cm10

10(c) — cm

The Image formed by eyepiece of compound microscope Is:8.(b) Real and diminished(d) Virtual and diminished

tlty Is not related to the wavelength of incident light?

(b) Radius of curvature

(c) Power (d) Chromatic aberrationl0, Which of the following Is used to obtain a virtual and diminished image? %

lb) Convex lens

(a) Real and magnified(c) Virtual and enlarge

Which of the following quan(a) Focal length

9.

(a) Concave mirror

\388

pHY3iC3- XI (Subjective)Chapter lo [Q 369SchoUi —(c) Convex mirror (d) Plane mirror11. Two convex lenses of equal focal length f are placed in contact, thecombination is:(a) Zero(c) 2f

12. Final image produced by the compound microscope is:(a) Real and inverted(c) Virtual and erect

13. For normal adjustment, length of telescope is:(a) f0 U

(C) ^

Short Questions of Exerciseresultant f0ca| ler>gth 0,«,(What do you understand by linear magnification and angular magnification? Explain how a convexlens Is used as a magnifier?

(b) f(d) i / 2

Q.10.1

(Mtn 2003,D.G.Khan 2005,Fsd 2005*2008.Bwp 2006,Mir Pur 2009,Grw 2009,Lhr 2009*2010*2011,Grwaoil)

(b) Real and erect(d) Virtual and inverted

Ans, Linear Magnification

The ratio of the slu.eflmOQt to the site ofobhet Is called linear magnification

Angular Magnification

i.a. M = 0 P

(b) f.- f,(d) i The ratio of the angle subtended bv the Image as seen through the optical device to the angle

<ubtendbd bv the oblect at naked eye is called angular magnification. M = jj*--r. f.14. A spectrometer is used to find:Wavelength of light

(c) Wavelength of different coloursLight rays coming from a distant object are considered to be:(a) Parallel to each other(c) Parallel to lens

16. Critical angle is that angle of incidence in denser mediummedium is:(a) 45°(c) 0°

Convex lens as a Magnifierplace the object within the focal length of a double convex lens then a magnified, erect and(a) When we

virtual image is obtained.(b) Refractive index of prism(d) All of above15.

angular magnification and resolving power of an optical instrument.

(Rwp 2005.Bwp 2007,Lhr 2008-2011)Q.10.2 Explain the difference between

What limits the magnification of an optical instrument?(b) Parallel to principal axis(d) Parallel to source

for which angle of refraction In rare difference between angular magnification and resolving power

Angular magnification increases apparent sae of intone of the object whereas the resolving power of

an instrument separates the images of two very close objects

LimitationThe chromatic and spherical aberrations are

magnification of optical instrument.

Ans.

(b) 900(d) 180° defects in lenses which limits the

the two main17. Mlchelson calculated the speed of light using:

(a) Spectrometer(c) InterferometerIn optical fibre transmission system(a) ~ Diode(c) Laser

fb) Galvanometer(d) None of these ^are used to regenerate the dim light signal.

(b) Repeaters

'0; TransformerThe least distance of distinct vision is 25 cm. The focal length of a convex lens Is 5 cm. It can act a. *timpie microscope of magnifying power:

18.Ans Reason

We know that resolving power,R = — — =

Since the blue light is of shorter wave length.It produces less diffrqction. Hence. «t increases thi

resolving power of compound microscope.

D11.222. •

a) 4 (b) .5(d) ^

In such a microscope

microscope for use by the children.The Images seenW 6Photodiode converts the ftght signals intoa) Electric u%r a;

W Ertfcer of these

Q.10.4 onecan buya cheaphave coloured edges.Why Is this so?28.

(D.G.Khan 100S,

(b) Sound signals An* Reasoncannot bring ifl

red edge.It Is due to fftf flfrffratlotL len*. Such tenjfj

which win five cclouA) Hone of these

after passing through thea single point (focalpointjThe white light will disperse

ANS VERS white light from the object tol.i 2. b 3. b 4. c ii 6. d 7. d

k 12 13.e 14. d 15.» 16.b 17. d

» .

370

^£!SLT° [Opuca,,=2*“**, PHYSICS - XI (Subjective). 371tcbo llQ 108 Identify the correct answer:

(|) The resolving power of a compound microscope depends on|a, The refractive index of the mediumIn which the objects placed(b) The diameter of the objective lens.

(c) The angle subtended by the objective lens at the object. ‘(d) The position of an observer s eye with regard to the eye lens.

(||) The resolving power of an astronomical telescope depends on:

(a) The focal length of the objective lens.(b) The least distance of distinct vision of the observer.(c) The focal length of the eye lens.

(d) The diameter of objective lens.(i) (b) Diameter of objective lens.

(li) (d) Diameter of objective lens

Draw .ketches shoeing the different light paths through a single-mode and

Why Is the single-mode fibre preferred in telecommunications?

The different light paths through single mode and multi mode fibre are shown below.

Chromatic aberration is a type of distortion in which J lcn> fails to fi,ru\ all c,,[lir j,> ,hcChromatic aberration \hows fringe* of color along boundaries ihai separate.J^ . v in,j br\ 'h* *

Reason It occurs because lensc*- hnv< a different refractive indo for differcm u , , n ;i ' ^index decreases HII /I ineregsum wavelemilh Since rhe focal length / « •! i, , . , .J, , . . i, , ,n. different uvivclcnglhsol light will be focused on diltcrcnr positions

P"int-"..b. TvT®"n the refrr

.s <CHOitr

Q.10.5 Describe with help of diagrams, how (a) a single biconvex lens can be used as abiconvex lenses can be arranged to form a microscope. magnifying gUn. lb)

IMtnJOOi. fvd VXA\ \an erect , virtual and magnified

Ans. For single biconvex lens:When object is placed within the focal length of the lens thenobtained.For biconvex lenses:In compound microscope, when the image formed by the objective is v. thmeyepiece then a virtual, inverted and magnified image is obtained

‘m*Rt it

Ans.»he fool length of th. a multi mode fibre.Q.10.9

Ans The path through a single-mode fibre.

Cladding

(b) 6 convex tenses used as(a) Kay diagram of single biconvexlens used as magnifying glass to form a microscope

Q.10.6 If a person were looking through a telescope at the full moon, how would the app < ranee of ;

moon be changed by covering half of the objective lens.( Sgd 2003.Fsd 2004.lt ' 2010 20111

The apparent size of image of moon does not chanoe It looks dim only

ReasonAns. " A.

Fig 10.14through Multi-mode step-index fibre

Light propagation

Preference of single-mode fibreSingle mode is preferred in telecommunication because

hromatic source is used in single mode fibre,

of light and hence no signal is lost.

Magnification of telescope depends upon the focal leriQth of the eye piece M - ^So only the intensity oflength.• When half of the objective lens is covered, it does not change its

the light reduces to halffind by ray

Q.10.7 A magnifying glass gives a five times enlarged image at a distance of 25 cm from the lens• A strong mono-c* There is nojiispersion

diagram, the focal length of the lens.Ans. Scale: (along x - axis)

5 cm = 1cm25 cm = S cm

From fig focal length = OF = 125 cmThus f = 1.25 x 5 = 6.2 cm

r

Q-io.,0 How the light signal Is transmitted through,h*2D08 0.G.Khan 2005. Fsd 2005, Grw 2010)M = 4

D = 25 cmf = ?

tical Fibre by:The signal is transmitted through the op

(0 Total internal reflection5 cm

of total internal reflection while in

Internal reflection andM = (ll) minupus rffrnrflpn

^mitirnode Step Ind** fihr, the signalgraded Index fibre.

is transmitted by meanthe signal Is transmitted by totalf

25 55 = 1+ — case of multimode.continuous refraction.25 cm

25 254 - — fa — = 6.2cmf 4

-

How the powar is lost in optical fl &rt through dispart o*' ? Exp •'!Uvi!>icMv ^ iVi4

*ns. Power Ion by diiptrslonWhen light vgnal is not perfectly monoch'O'’' tT *c. tra'light will d/iotrit on patsing through tht corf o( tr toptical fibre into different wavelengths >*, and /1eu asshown In figure

’ —| at 4— — — __—+r~jl U f M i f -Each wavelength will have different oath Itnoth

• Therefore, the signal received Is distorted or faulty least distance of distinct vision - d cmfrom cye-piecc -

DispersionWhan tha light p**»t» into a matanal at an angk t*a S«* t fcaww ,* oa t ©» 'afrwtad suwon- g toSnail's Law and tha indo* of rafroction of the * start* T>

_» rns *•** r|* h i at *slightly diffarant angla whan posing through a motana T * » ipts: 4 cart *N of Hg*» H

callad dltparsion or chromatic dtspartJon

of virtual image.

dl Pwe get

1± m l2 5

9

p i 25#

i - ± .J_p' 25 250

I , 250 -25p- (25X250)

Solved ExamplesnWIHTIlliHll

A microscope has an objective lens of 10 mm focal length, and an rye piece of 25.® >nn tad Ilength. VVhat iv the distance between the lcn %o and it % magnification, if the object »* in *rpta«|when It i* 10.5 nun from the objective?

I 275Given Data.p* 6250l ocal length of objective f0 1Ontm

Focal length of cyc-piccc fe 25mm

1 - 22.72rmijDistance between lens L * q P' .

Distance of object from objective lens p 10.5mm

To Find:Pulling values, we get

Distance between lens l q + p' 70) l - 210 + 22.72L- iy2.12snm[L * 233mm)•(Magnification of objective

(ii) Total magnification M ?

) (Magnification ofeyc-piece)Calculation:

(i) If wc consider the objective alone

1 - i + iu p

As q' » d --250mm

374

^PHYSICS - XI (Subjective

0/ = 9.4°

Again using Snell,s law (For air-core boundary)nSinO = n ,Sin0'

( l .O)Sin0 = (1.50) Sin 9.4°

Sin 0 = 1.50 x 0.163Sin 0 = 0.24450 = Sin'^O.pftJ9 = 14.2°1

52500M = -238.56

M = - 220Negative sign indicates that the image is virtual . n - refractive index of air = 1.0

Calculate the critical angle and angle of entry for1.50 and cladding of refractive index 1.48an optical fibre haviv,n8 core of refractivc indexGiven Data:

Refractive index of core = ni = 1.50Refractive index of cladding = n2 = 1.48

Exercise ProblemsTo Find.

(i) Critical angle = 0C = ?(ii) Angle of entry of light = 0 = 7

Calculation.A converging lens of focal length 5.0 cm is used as a magnifying glass. If the near point of the

observer is 25crn and the lens is held close to the eye, calculate (i) the distance of the object from

the lens (ii) the angular magnification. What is the angular magnification when the final image is

formed at infinity ?

10.1

Given data:Focal length = f = 5.0cm

AirNear point = least distinct vision - d - 25cmn = 1.0

To find:(TT = P = ?(i) Distance of the object from Jens

(ii) Angular magnification = M = ?

(iii) Angular magnification, when image is at infinity = = ?

(i) According to Snell 's law (For core-cladding interface)

niSinOi = n2Sin02When 0|= 0C then 02 = 90°Thus 1.5 x sin0c = l .48 x Sin90° = l

l .5 x sin0c = 1.48

(ii) From Figure

M JIr« g^PfMOtfw,

,4 PHYSICS-XI Subjective ) 377d

Asgolir nugmficagioo = M =! —CPE)25M*lS

M-l+5M=6 Putting the values, we get

4(in) Now- when image is at infinity, object must be at focus, hence p »f d =12 ’96

M-S|dc =0.5cmP

M - aA telescope is made o f an objective of focal length 20cra and an eye piece o f 5.0 cm, both convex

lenses. Find le angular magnification.

Focal length o f objective = f0 = 20 cm

PocaJ length of eye-piece = fe = 5.0cm

Angular magnification = M = ?

103Hence Gr ?n data:

Putting values, we get25M* T

M =5.0

Or To find:

Cak lotion:fo

Angle magnification = M = -pPutting values, we get

HPA telescope objective has focal length 96cm and diameter 12cm. calculate the focal length udminimum diameter of a simple eye piece lens for use with the telescope , if the linear magaificitioirequired is 24 times and all the light transmitted by the objective from a distance point on tbs

telescope axis is to fall on the eye piece.

10.2

§3Given data:

0(4 A simple astronomical telescope in noimal adjustment has an objective of focal length 100 cm and

an eye piece of focal length 5.0 cm.(i) where is the final image formed? (ii) calculate the angular

magnification.

Focal length of objective = f0 — 96cmDiameter of objective = d 0 =12cm

= M = 24Linear magnificationGiven data:To find:

Focal length of objective = fo-I 00cm

Focal length of eye-piece = fe =5.0cm

(i) Distance of the final image = q'=?

(ii) Angular magnification = M = ?

In normal adjustment, the image due to objective is formed at the focus of the eye

image acts as an object for eye- piece . thereforep'= ff =5.0cm

Focal length of the eye- piece = fc =?Diameter of the eye- piece = dc =? To find:

Calculation:- piece, thisf

As magnification = M = —f.= —* M

Putting the values, we get

f.-*’ 24f# •= 4cm

As focal leng is lirectly p. portumal to diameter, so the ratiois equal to the ratio of their fc< d lengths. Thus

calculation:(i)

Or11 l

f. P <1ii JL5 s q'

1-l.iq ^ ^

iivirtual)(As in*g*

f the f*®;0 of the diameter o

\X

l m

^i£!2H0[pptic PHYSICS- XI (Subjective)$cho 379

10.8q0.6q,= o

q -ooAngular Magnification

q =18cmthus final image is formed at infinity(ii) Thus the image A'B' is formed 18cm away froi

second lens , which is 26cm from the first lensThus

lens .This image will act as an object for theM= -^

M =I p'= L- q = 26 -18=8.0cmFor second lens

p-8cm,f -16cm, q -?

-5 r|M = 2p|Or

110.5 A point object is placed on the axis of and 3.6second thin convex lens of focal length 16.0on the side away from the object. Find

Distance of object = p = 3.6 cmFocal length of first lens = f = 3.0 cmFocal length of second lens = ( =16.0Distance between two lenses = L = 26 cm

cm from a thin convex lens of focal length 3.0cm . Ais placed coaxial with the first anil 26.0 cm from itthe position of the final image produced by the hvo lenses.

cmGiven data:

cm

To find:Position of the final image = q/= ? q'= -16cm

Calculation: Negative sign shows that image is virtual.Using the lens formula

10.6 \compound microscope has lenses of focal length 1.0cm and 3.0 cm. An object is placed 1.2 cmfrom the object lens. If a virtual image is formed, 25 cm from the eye, calculate the separation ofthe lenses and the magnification of the instrument.

Given data:Focal length of objective - f0 = 1.0 cm

Focal length of eye-piece = fc = 3.0 cmDistance of object = p = 1.2 cmDistance of image = q’® -25cm (virtual image)

T° find:Separation of lens = L =?Magnification = M ® ?(

“ R,culatlon:As separation of lens * length of microscope - L - q p

So first we have to calculate q and pAs lens formula is

Or •q 3 3.6

Putting values, we get1 3.6-3 I 1I— = — +_

f* P qq 3*3.60.610.8 .

»

MiK

W380 PHYSICS - XI (Subjective)Chapter 10 (Optical lmtm igr’s

Or i I I 28M=5x~3M 140M = 3M = 46.66M = 47 approx.

q p

I*1 X

q*

l"

l .21,

1.2-1q 1.21 0.2

Or

Sodium light of wavelength 589nm is used to \ ie\v an object under a microscope . If the aperture

of the objective is 0.90 cm, (i) find the limiting angle of resolution , (ii )using visible light of any

wavelength , what is the maximum limit of resolution for this microscope .— £3.-

10.7q 1.2q = 6cm

This image will act as an object for the cyc-pieccNow for eye-picce

q'= - 25 cm , fc = 3cm , p' = ?Applying lens formula

±=!+If. p’ q'

l l l

Given data:Wavelength of sodium light = >t =589 nm = 589xl0'*/w

Aperture of the objective = D = 0.90 cm = 0.90x 10 ’ />/

To find: -9= a,(i ) Limiting angle of resolution

limit of resolution = cc'rmm

-9min( ii ) Maximum

Calculation:(I) Limiting angleUsing the relation L _p' f. q’

for limiting angle of resolution.

1 I Ia =122 —mm p

i-ia. Putting values, we get589x1O'4 .

“ m..= 1 '22xo.90xia!p' 3 25718.58x10 ^'1 25+3

0.90p1 75=798.4x10’1 28=7.98x10-'p' 75

75 is used and it isMaximum limit of resolution elength of the visible spectrumP 28 (H) at , the shortest wav

For muximum limit of resolutionp-2.7cm400nm for violet colour light.Thus V =400nm=400x10 JmThus , separation between lens = L = q + p'

L = 6 + 2.7=1.22 —D/|L = 8.7 cm| So CC mm

q dMagnificatiqn= — ( 1+ — )L P fc

400xl0~loioxTo"=1.22x

iPutting values, we gel 4.88x10“ 0.90x10M= — -T^UTnidjan

1 mi*' — ,

M =1.2

i >

382 g^Pter lOTOpij1 «.R An astronomical telescope hasing magnifying power of 5 consists of two thinKind the focal length is of the lenses.

Magnifying power = M = 5Separation between the lens = L = 24cm

Focal length of objective = f =?Focal length of eye-piece = f =7

Magnification M is

M= -^

2l!*t , PHYSICS - XI (Subjective)383,enses 24

Sow using Snell’s law for glass-waier interfacer\\ sin 0, = n2 sin 0,

Where

'"••pmGiven data:

To find: refrective index of glass=1.50jn, = refractive index of waier = l.}3

, = angleof incidenceorcritical angle = 0' = ?

0,=90 — for total internal reflection

Thus above equation becomes1.59xsinG' =1.33xsin 90“

n , =

0Calculation:

fOr 5= —Or f, =5f.

Distance between two lens is

24 = 5fc + fcUsing equation ( I ) in n

24 = 6 fcr _ 24

T\\ = 4cmPulling this value offc in equation ( 1 ), wc getf„= Sx4f , =20 cm

0 )

The reflective index of the core and cladding of an optical fibre are 1.6 and 1.4 respectively.

Calculate ( i ) the critical angle for the interlace (li ) the maximum angle of incidence in the air , of a

ray, W hich enters the fibre and is incident at the critical angle on the interface.

Refractive index of core = n^ - 1.6

Refractive index of cladding = n2= 1.4

10.10

Given data:

To find : = 3 = ?(i) critical angle for core-cladding interface

(li) maximum angle of incidence =6' = ?

( I) Snell’s law for core-cladding interfacem Sin 0i = n 2 Sin 02nt = 1.6n 2 = 1.40 , = Gc = ?02 = 90°

Putting values, we get1.6 sin 0c = 14 Sin 90°

• ft1.4

sin Qc = —

Calculation:

wherenT* m Ulr Wln l0tUlly in,ernal|y reflect a light ray if its angle of incidence is ul leas'of water = 1

*33

* m,n,mum unK,t* ,or total Internal reflection if pipe is in water 7 ( refractive index

Given data: 1

Angle of incidence lor glass in air = 0 =39°To Find:

Angle of incidence lor water = 0' =?

For refractive index of glass light pipe using the formula1n = sin 0(

Putting the values, we gel

1.6Calculation: sin Bc = 0-875

(0.875)- 10c = sin

1

Maximum angle of incidence•From figure, for 0t = 61",

the angle of refraction 0 j inI

384 PHYSICS-XI (Subjective)£l22P!?MitOptica|, 385SchoJ^nstruithe core should be01=90^0Z C

e;=90°-6i° = 29°Snell ' s law ,

for air - core interfacen, sin0j=n2 sinO',IxsinOj =1.6xsin 29°

sin0[ =1.6x0.485sinO;=0.776

0J = sin* 1 (0.776)

Chapter .11

HEAT & THERMODYNAMICSFor air, n, = lFor core n2 =1.6O',=29° Learning Objectivesj

o; =5i°1. . state the basic postulates of kinetic theory of gases.

2. Explain how molecular movement causes the pressure exerted by a gas and derive the equation P = 2/3

N0,<i/2mv2>, where N0 is the number of molecules per unit volume of the gas.

3. Deduce that the average translational kinetic energy of molecules is proportional to temperature of

the gas.4- Derive gas

5- Describe that the internal ene

* %******* *

Scholar’slaws on the basis of kinetic theory.

rgy of an ideal gas is due to kinetic energy of its molecules.

rk and heat in thermodynamics.isothermal and adiabatic processes.

Alternative To PracticalUnderstand and use the terms wo6.

On Differentiate between

Explain the molar specific heats of a gas.

9. Apply first law of thermodynamics to derive Cr-Cv ,entropy.». Explain,0.second law.<— n.mic,and»— «•

I 1V Understand the concept of reversible and irreversible process •

7-

PHYSICS, CHEMISTRY,BIOLOGY & Computer Science

8.=R.

Define the term heat engine.12.is Available in Market 13- Understand and describe Carnot theorem.ics scale of temperature.

H. Describe the thermodynami

^ Describe the working of petrol and diesel e

16. Explain the term entropy.17‘ Explain the change in entropyAS -

^ntropY crisis‘

Appreciate environmental crisis as an e

9|.||YHie® X M'.HI.JM IKI-I

Chapter No. 11tllJ,vi, nlo,n between be,.,,,,,,, ' «W,

ot iutt, tramform^n hU/'*"«*' "'""""thi,*

g&tunial*ork- ^

l,„b I' l It" - '<Iy11.111,/ 1 _ ^ir.lyy

ntral role in iethnolog, ...tid In the formof heat ' & jr use *rt IK

II ,)!<* Vhlilies

ifltHt pOttUlatf t of l- lr.etjr111

<E>Kinetic Theory of Gases1MV behaviour of gases Is well described by

macroscopic properties ( T, P , and V etc) of gases to microscopic properties (K.Eetc.) It provides a mathematic!:al model to study the behavior of gases.Postulates

• A finite volume of gas consists of very large number of molecules.

The size of the molecules is much smaller than the separation between

molecules.• The gas molecules are in random motion and may change their direction

of motion after every collision.• Collisions between gas molecules themselves and with walls of container

are assumed to be perfectly elastic.• Molecules exert no force on each other except during a collision.

Q- 2 Derive the relations for pressure and temperature in terms of average

__ kinetic energy of the molecules?

the kinetic theory. !t relatesThe

<5-CLUu/

o0 > c-3 s q CL,u O u

merely

transferred to the wall of the container per Icollision of molecules of the gas.Expression for pressure of gasConsider a cubical box as shown in figure.LetLength of side of box = £Number of molecules of a gas = NMass of eachMass molecule of a BaS " m

ln‘bal momentum of molecules be o

^ rebound «°,f the collision Is elastic, the molesPeed. So

li «h 3E >= v«

Final momentum=-nW«Change in momentum = -mvu -mv, *2mv„

\

V"T - mm

JSIf v

389388 PHYSICS- XI (Subjective)

, m . . 4

value of — in eq. (1), we get

Chapter 11 [Heat & Tu —^> na lar’s[Jig sJJAfter recoiling the molecule travels to opposite face EFGHE and collides with it.

The molecules again rebounds back to the face ABCDA after covering a distancej.bstitut-ng the

2 C .Let, time between two successive collisions = At Px

2t 2Then [v S = vt]At =NVlx

The rtumber of collisions per second (or collision frequency) that the moleculemakes with this face is,

2 + v )*2 1 + VNl is called the mean of

NWhereI vi*f molecules moving along x-direction So equation (2) becomes

//> (3)

the faces perpendicular to y and z axes will be

f = —At I t velocity osquareRate of change of momentum of the molecules-2mvu

P„= P<V

Similarly, pressure on tne

Py = p<VyfllP,=P<v/>

As the motion of molecules is random, therefore mean

component velocities will be equal,i.e.<vx2> = <vv2> = <v,2><V2> = <v/> + <Vy

2>+<V22><v2> = <vx2>+ <vx2>+ <vx2>

f ®Kv2> = 3<VX2>

due to collision with face ABCDA=At

square velocity of all thePutting value of At, we getv

Rate of change of momentum --2mvu x — =Now, according to the 2nd law of motion in terms of momentum: . 21

— By second law of motion, rate of change of momentum of molecule is equal tothe force applied by the wall.From Newton's third law of motion Force Fix exerted by the molecule on theface ABCDA is equal but opposite.

~mv> x2

t

Since

1<v 2> = -<v2>

3Putting the value of <vx2> in eq.(3)/ we 8et

-p < v2 >3

According to Pascal Law,pressuthe same provided the gas is of uniform density.

So,2 \mV„

Kt Px = inside the vessel will be

every where= mv'»

2re of the gasF« t

Thus, the total force acting along x-axis due to N molecules of the gas movingwith velocities v1K, v^ v3x. So,

,vNx isP« = Py = p> = 3

< v J >F* = Ffx + p

2X+ ^3x2mvlx mv Thus in generalFx =OR (4)i p < v2 >

3 ,

This is the relation for pressure in te

Relation between pressure and average

t t P =As pressure is normal force per unit area velocity.

rms of mean squareP.= ^-f-Hence

A t 2

mNAsOR P V

Putting value of p in equation (4)

3 V

P-‘3I2 V

ORP= -

massDensity =Since VolumemN

OR3 V 2

OR

-JM

A,m ii -

391\] (Subjective)

I lr = - \ < - *m >3 ' 2 mV >T * <:

o.«»»"TTZLcr,rfi*»m

s— = N.= nymt>er of moiecuies p*r un-t volume in the average

IN.1r s constant < -m\ > 1 I - Ci 1 of gases.the basis ofthe|asJavrson 1P * < -mv >0«

o* P or <K £>|0yU*SUW

mt

Coockiwon

^wnrt eitrttJ by i% fa* « dirtxily pmport*mji to tht wragtK E of jgus molnyUx t- OS at constant temperature is inversely

totHkas1

Prove that. T<cm/>Q-3 l-O"<=>Interpretation of TemperatureAccording to ideal f

PV •nRTWhere n is the number of molesV is the volume. T is the absolute temperature andR Hthe u g a s convt t(R = 8 Jt4 Jmol ’K ’)If NA is Avogadro number thennumber of mole can be r ijxetved ai

0f the moleculeskinetK energyiHW

I<i®r >«k^r» rtmaMW constant

I a*XM V*,1<tc>rs *"*

pv * constant

| oa p « constantly )

‘elso constant. So

Nn * rrNA 1oa p * —Thus equation (i) can be written .» v

NRTV

***** proved it is the Boyle s law.

temperature when the pressure is kept constanfrom kinetic molecular theory of gastS-

o *Njj¥i

v,Ui<i«v' >3 P J

PV *the absolute

PV * NkT

•$ the Boltzmann's constant its value s 138 * K>'23‘

( 2)or

Where k =1 >-noconstant per molecule

Also we know thatOR

D :N I ,P = < — nn ^3 V 2 IS is also constant

j ** Ihe pressure is constant. —Thus,

PV * — <.~mv 2 >3 2

Comparing the equations(2) and (3). we have

NkT = -N < -- mv ‘ >

113)OR

IID* 5 >V ~ constant * ~

2

v « <-* mv >2

3 : OR„ 2 i *OR T = — <-mv‘>

3K 2 — = Constant 1Ah©•K

21— ms ' >2

T = constant <

392 Chapter 11 rHeat * Th PHYSICS- XI (Subjective)ermod lar’s$d>°So. V o c T.

Hence verified, it is Charles law. Work doneby the system (gas) on its environment is consolered as positive.

positiWork done

tjve work doneQ.5 Write a note on Internal Energy and show that it is independent of the

path.Nega

by the environment on the system is considered as negative.Work d°neamount of heat Q enters the system it either appears as an increase in

of the system or is used up in doing work by the system on itsInternal EnergyThe sum of all the forms of molecular energies (such as kinetic and potentialenergy) of a substance is called internal energy.ExplanationIn the study of thermodynamics, an ideal gas is usually considered as a workingsubstance. The molecules of an ideal gas are mere point mass which exerts noforce on one another. So the internal energy• of an ideal gas system is generallythe translational Al£ of its molecules. Since T oc <K.E >, thus the internal enemyot an ideal gas is directly proportional to its temperature.How can we Increase the internal Energy

1) By heatingWhen we heat a substance, energy associated with its atoms or molecules isincreased, i.e., heat is converted to internal energy'.

2) By doing mechanical workWhen two objects are rubbed together, their internal energy increases becauseof mechanical work. The increase in temperature of the object indicates anincrease in the internal energy.

NoteSimilarly, when an object slides over any surface and comes to rest because offrictiona forces, the mechanical work done on or by the system is partiallyconverted into internal energy.Internal energy is a state functionin thermodynamics, internal energy is function of state. Consequently, it does notdepend or the path out depends on initial and final states of the system.Explanation JConsider a system wh ch undergoes a pressure and volume change from and

to Dr and Vt. respectively, regardless of the process by which the systemchanges from r.it = : to final state. By experiment it has been seer that thechange in interna) energy is a . ways the same and is independent of the paths C1

and C2. ^ ^i'te— a ere'g, is sim ar to the gravitational P.E. So like the gravitational P.E wet2ke tne change in internal energy and not its absolute value, which is important.

If aninternal energy

environment.ression for Work in terms of directly measurable variables

Her a gas enclosed in the cylinder with a moveable, -frictionless piston ofC0P

ectional area 'A'. In equilibrium, the system occupies volume V, and

exerts^ pressure 'P' on the walls of the cylinder and its piston. As pressure is

defined, force per unit area i.e.,

p =I

\vFIV"

1; iv

fa! ,bT "iryA cas sealed iaq' "d-e' b> 3.veigHess. fcxtonatessTre corsta-.t <3o*Tvrard aec*edforce F eauais FA and n thepiston s aisc deed db^- jrafd

wor< s dc'e cn me ass

A d<alomsc gas molecule hastranslational and roUta^energy. It also has vfcratatfenergy assocaled w*h the sprirq

*e bond betv^en its atoms

both AF = PA

This is the foijce exerted b / gas on piston.

The gas expands through AV very slowly so that it remains in the equilibrium. As

the piston moves up through a small distance = d - Ay

Work done by the gas is

W = FAy

W = PAAy

Since AAy = AV (change in volume)

W = PAVis is the work done by gas on piston.

We can express work in terms of directly measurable quantities.

Sr»phlcal RepresentationWork done can be calculated from the area under P-V graph.

Bv the details of change In internal energy and the mechanical workc,n describe the general principles which deal with heat en«| gv' ,cs

mechanical energy. These principles are known as laws >

S2 Explain the first law of Thrminify— wig»nd »» eonseduencjL

or

Coratar< Prwsura> fBP

-4TP &V •;

A0 %V—p. jC,

C,.

l-ui Your lolornuiUon' "a";(P,. V„TJ -ec*v - aS

V,V, •- 01

SvremV- 0

Discuss transfer of energy into wo'k and heat Also calculate the workdone by a thermodynamic system d Tngtne volume change.

Q.6 Flrst Law of Thermodynamics benmWork

he woix. Wits [ foundings."•miticiiiyWork and Heat

Both hec: and >. ark correspond to ar fer of en*. - y by some meansTea was first applied to the sr m engine ere it was natural to transfer

neat in and get work out.

Q * A U * W

\

pHVSlCS Xl394 Clmptor 11 [||CII| A mHiiI2l«4y«.*Explanation

\s nen heat s added to a s\ stem; there Is an increase In the Internal energy fromt , to U: due to the rise in temperature and an increase in pressure or change inthe state, if 3t the same time, a substance is allowed to expand, then W is thework done on its environment

Q = (U2-U,) + WQ = AU + W

Thus, the change in internal energy AU = U2 - U,From equation (i)

atlonV. ' >"after

..Initial pressure and volume whert <5. j,

the isothermal change takes pt*> ThenP,V, = P,V2

I energy of an ideal gas depends or

constant .

the 'sxmur iVviir tr MII/I*il ?- > \

io\^e VHeal entering apositive and Waving a ivxvcmis negative lhe work dont y1by ihc system is pos»iivc a^\\ihc work done on the sy«etn\\is negative, as shownintt»\\figure.

’ k , */' O r *-internaAS theleasedTherefore

OR (D** • ' .O.ii

AU = 0“ f thermodynamics

Q = AU + WQ = O + WQ = W

ucesfirst law oAU = Q- W

Examples of First Law of ThermodynamicsBicycle PumpA bicycle pump provides a good example. When we pump on the handle rapidly,it becomes hot due to mechanical work done on the gas, in this way; it increasesits internal energy.NoteThe arrangement consists of;Bicycle pump with a blocked outlet. A thermocouple connectedthrough the blocked outlet to note the temperature of air Whenpiston is rapidly pushed, thermometer shows a temperature risedue to increase of internal energy of the air. The push force doeswork on the air, thereby increasing its internal energy, by theincrease in temperature of air.Human Metabolism

Human metabolism also provides an example of energy conservation. Humanbeing and other animals do work when they walk, run, or move heavy objects.Work requires energy. Energy is also needed for growth to make new cells and toreplace old cells that have died. Work done will result in decrease in internalenergy of the body.According toi'1 law of thermodynamics, to an organism of human body

AU = Q- WHence, the body temperature or internal energy is maintained by the food we

(2) Hence

thermal Expansionif a gas expands and does external work W, an amount of heat Q has tc be|

supplied to the gas in order to produce an isothermal change.

Since, transfer of heat from one place to another requires time; hence to keep

temperature of the gas constant, the expansion must take place slowly. |

IsothermThe curve representing an

(2) Adiabatic ProcessA process in which noprocess.ExplanationSince in adiabatic process no^ence the first law of thermodynamics becomes

Q = AU + W0 = AU + WW =-AU

Adiabatic Expansionlf gas expands and does external work, itenergy of its molecules and hence, the tempera

Adiabatic compression1 Sas is compressed, wor

I4U»

MllivoltmotorTrappod airin pumpThermocouple

isothermal process is called an Isotherm.Pistonpushed in

is called adiabaticleaves the systemheat enters or

i.e.,Q = oheat enters or leaves the system

tp

ORof internal

is does at the expenseture of the gas falls.

ses the temperature of the

i

k is done on the gas; it increaeatgasNote (Metabolism)

Energy transforming processes that occur within an organism are named asmetabolism.

Applications of First law of thermodynamics(1) Isothermal Process

-W = AUwressed rapidlyCondition for adiabatic change exponds or is con

Qf adiabaticMabatic change occurs »hen the cylinder. In caseParticularly when the gas is containe >

n()t rernain constachanges, as the temperature of the gasPV7 =constant tant pressure toA process in which the temperature of the system is constant is called

isothermal process.ift isothermal process the condition for the applicat on of Boyle's Law is fulfilled.

y Is the ratio of the molar specific heat of the gas at coosG ^olar specific heat at constant volume, i.e*.

Therefore, when gas expands or compresses Isother ally, the product of itspressure and volume during the p ocess emains t nstant.

1397 I

X aHH

396 Chapter 11 [Hct & Th p^i^SlCS M lSul) jci.iiv<)rCrn»od

AdiabatThe curve representing un adiabatic process is called an adiabatAn adiiibat is steeper than an Isotherm.Examples of Adiabatic ProcessThe examples of adiabatic processes are

The rapid escape of air from a burst tyre.The rapid expansion and compression of air through which a sound wave isCloud formation in the atmosphere.

Q.8 (a) Define the following terms:(I) molar specific heat(ii) molar soecific heat at constant volume (Cv)(iii) molar specific heat at constant preasure (Cp)

(b) Prove that Cp - Cy = R

Qv = cv Art thermodynamics.

Qv =AU + WCV.AT =AU + WCVAT =AU + PAV

remains constant (i.e AV= o),

first law of thermodynamics

CAT =AUAU=Cv AT

r A|constant pressureIif one mole of an Ideal gas is heated a

I risesby AT then the heat transferred Qpi

d)

first law oAPP'Ving

|[ v W =-

ork done by the system isi.yolurn^

Thus theSince2. passing. zero3HenceOR

slant pressure so that its temperature

given by(3)

J the internal energy increases

I me rise in temperature AT.

e amount as at constant volume for thethe s.(4)= CvAT

ep the pressure constant, so the work done by theSpecific heatThe amount of heat required to raise the temperature of one kilogram of a substance up to one Kelvin is called I S,nce' the g*specific heat. gas isOne kilogram of different substances contains different number of molecules. Sometimes it is preferred toconsider a quantity called mole. One mole of any substance contains same number of molecules.Molar Specific Heat of a Gas IMolar specific heat of the substance is defined as the heat required to raise the temperature of one mo oft jsubstance through IK.NoteIn case of solids and liquids the change of volume and hence work doneagainst external pressure during a change of temperature Is negligiblysmall.But gases suffer variation In pressure as well as in volume with the rise intemperature. Hence, to study the effect of heating the gases, eitherpressure or volume is kept constant. We can define molar specific heat ofa gas in two ways.(1) Molar specific heat at constant volume ,olli .The molar specific heat at constant volume is the amount of heat required to raise the temptrtWtj^mole of the gas though IK at constant volume.It is symbolized by C v . Its SI unit is J Mole ’ K'\

thuspands t<

(5)= PAV— — —

according to first law of thermodynamicsQp = AU W

CpAT = CvAT + PAV

%

[using equations(3l(4)(6)

«5)lAccording to general gas equation.

PV = nRTi one mole of an ideal gas,n = IPV = RT

At constant pressure P, amount of work done^by

mansion AV caused by the rise in temperature AT g

PAV = RATPutting value of PAV in equation (6), we get

C AT = C v AT + RAT

Cps Cv +R

t i t i j Imole of a gas due to

Ccntf*'1*prMWftAU C.AT

Centumvo*Vm«AU * C.OT

TL T 0.0. - C.AT

OR

= Rif > ^ p C y

^j^that Cp > Cv by an amount equal to

a note on Reversible and Irreversible Processes.universal gas constant^

(2) Molar specific heat at constant pressure I ,The molar specific heat at constant pressure is the amount of heal required to raise themole of the gas through Ik at constant pressure. .

It is represented by symbol C..(^Derivation of Cp- Cy * RAt constant volumeif one mole of an ideal gas Is hes .ed a constant slumt so that Its tempers;- 'ctransferred CL is given by

art of 0' *verstble Process

n be retroA reversible process is one 11 siirroundings‘wi(Aoitf producing any change nExP'anat\on

order,ced in exactly reverse

6 dlr*« process, but thermal and mechanical effeAT then th*^rises by

398 Chapter 11 [Heat A Th prtVSlCS-XI (Subjective)399 f^0(j

§cbol*r' s

tficiency o

rhe efficiencysupplied.It isdenoted by rj

W =Q,- Q/

exactly reversed. If heat is absorbed in the direct process, it will be given out inthe reverse process. If work is done by the substance in the direct process, workwill be done on the substance in the reverse process. Hence, the workingsubstance is restored to its original conditions.CycleA succession of events which bring ike system back to its initial condition is calleda cycle.

Examples of Reversible Process .The process of liquefaction and the evaporation of a substanceperformed slowly are reversible processes.Slow compression of a gas in a cylinder is reversible process as thecompression can be changed to expansion by decreasing the pressure onthe piston.

Irreversible Process

f heat engineof heat engine is defined as the ratio of wor

i

ne to the heat

HTR T,Q,

> t

Where Wefficiency = —

Q.Thus, .

i<V- Qi wEnginen= Q.(D

Q, Qj."" o r* J(2 ) OR

Q2(nil- II Q.IThis is the expression for e

Q.11 State and explain Seco

LTR T2

of heat engine.

ermodynamicsA reversible process is one which can not be retraced in exactly reverse order,without producing any change in the surroundings.ExplanationAll changes which occur suddenly or which involve friction or dissipation ofenergy through conduction, convection and radiation are irreversible.Examples

mSecond Law of ThermodynamicsAccording to Lord Kelvin's statement for working of heat engine,

It is impossible to make a heat engine which converts all the heat absorbedfrom a hot reservoir into work without rejecting any heat to sink0) Explosion is an example of highly irreversible process.

Work done against frictionCM) ORThere is no perfect heat engineExoir?nation

the heat source at

SSSTSSSTBSSSS:which the substance eventually returns to ItsInttrnil energy Is zero AU 0 rk dont |,Hsnce, according to First lew of thermodynamics, the

tv W- Qi - Q*

Q.10 Write a note on Heat Engine?

Do Yon Know/Heat EngineHtal engine It a device which converts heal energy Into mechanical workIntroductionThe earliest hett engine wes the steam engine. It was developed on the feet thatwhen water Is boiled In e vessel covered with « lid, the steam Inside trios to pushthe lid off,showing the eblllty to do work. This observetlon helped to develop nsteem engine.ConductionA heet engine consists of;

hot reservoir or source which cen supply heet it high temperature• e cold reservoir or sink Into which heet Is rejected at a lowertemperature.

working substance Is needed which can absorb hunt a from source,zsssi w* “• «

Q -AU + W0.1- Q, .o+ WQi- Qa W

w-a-o,end Q - Qi -Qiltherefore

OR

of ••cond law of thermodynemlci

i "ngl*hoat r *mnHa/f°r th$ conversionofheat Into work.0 p«rforrn

#it*rvo,r'no mattar how much energy It conttlm cennot b# mido

l0Urc* ofh#.tV H,nci for th# working of hut #ngln# th#r# muit bo a

»h* iJcp j. ** t#mp#r#tura and•link at low ttmparatura to which haot

irnoum* # J1^Ul,t: *•trui for ocaam andour atmoiphar# whichconttlma

of boat energy but cannot ba conv#rtad Into uiaful meebinleel

bodies at differenttwo

WorkingA hoat angln# li mad# cyclic to provide •contlnuoui aupply of work. Workingiub»t#nc# abiorbi haat Q, from lourca, conv#rti voma of It Into work W V#*p#nilon and rojacti th# r#it h«it Qju old r#i#rvolr or link.

Chapter 11 [Hcat Sc Tt,400 Intei'estyig Informationduring one cycle equals to the area enclosed by the

for our inability to utilize the heat contents of oceans & atmosphere

ib that there is no reservoir at a temperature lower than anyone of the two.The reason pV d»agrflrn -

net heat AQ absorbed in one cycle.NoteIn practice, the petrol engine of a motor car extract heat from the burning fuel

and converts a fraction of this energy to mechanical energy or work & expels the thermodynamics,„(1)AU + W _

rest to atmosphere.Petiol engine converts roughly 25% & diesel engine 35% 40/0 available heat AQ - we get

f AQand AU in equationvalue oPuttingenergy into work. = 0 + WQ, - Cb

w = Q» - QJWhat is Carnot's Engine? Explain its working and calculate its efficiency.

n (eta) of heat engi

Output (Q.12

Also state Carnot's theorem. According lo Kvr>.n•I !t« 2rxi Lwr yr irwrn r„_rv ircitctt Nv* a

A waterfall analogy forheat engine.

%Ui,H- it *rtjn* JKMC* <jt a singlecannot be convert* vftrtfy ^Carnot's Engine

A Carnot heat engine is a hypothetical engine that operates on the

jars-*.:»the link •«

reversible Carnot cycle. Sadi Carnot in 1824 proposed this ideal engine using only

isothermal and adiabatic process. heat i* inn to

running » heat engine-He showed that a heat engine operating in an ideal reversible cycle between two

neat reservoirs at different temperatures would be the most efficient engine. out to beor compression turns

isothermal expansionPrinciple The energy transfer in an iso

jroportional to Kelvin temperature

are proportional to Kelvin temper*..Carnot's engine works on the same principle as that of cyclic heat engine t takesreT,3n<tt,reSPr-c"ve,y

neat from not body, convert a part of it into work and reject the remaining part* G. 0,

Qito cold bed /

Hence T. QWorking 1

i Carnot cycle using an, ideal gas as a wor< *rg substance is s'owr PV- (2)becomest corsets v following four steps 1

f 1 f ;tsotherr^ai expansion1.r>t gas s aflowtd *0 eroartd iorherr-,ai. / at temperature aosoro '?-«ar Cn ' v '* * cr ' 9S4rjcr '*?pryjjj s '«preser:ed bycu've Ai. r'* ..usually taken in percent

/ H*» %( lA<flabtr;c2. , l'fiTA'Aff f ffiner,' /

^s s ' *- a* * s rxoard ad a^a-ICJi 1 a" t;•e'-peratyr#

& co'd•<j % s *ed jy $C Son* fif hotfbethrC-at vy“ywv<v•yi*1 depend*. 00

•uixtanci*/ ’if .; r A f f X A r , rijltijr» of d.« lht Ml.'.*" /

STJTSTH Hgat r rage s so*-vesstd $orfc#r~ar;y at ",

V> « **er- * ta* V, o e - vd " - e /sci(4 s ^r j . .

r . jtt.t*', the*<*> *0.4.

*>* „-.*11hy »»«"* ** *«''.r„,of h««l.rigm-b.I*"'47

IOCKH

/ *•<(>'•* ga’v s ad atsi 4# , 4 wf.v H * 444 »" jfevrAvt*4 a*r,v*W'foW’ a 4'*y+i+' -s>i y* ' , \ .* l >urW' 4 / f K 'jfrf O'S'« •# % ‘ \ e r * .< Mi- :*,u,r 'I

xv-A# *40 w v ' - r# \TK m, 4 *-v •' ** V '

f* K -tr & 4 W ’S

402

Do You Know?Carnot's TheoremStatement

Heat outA'o heat engine can be more efficient than a Carnot engine operating between

is used to Since this scale isdent o

the same two temperatures.f working substance. So it can be applied at very low temperature.Extended statement

;dePen thermometer and gas thermometer are basedresistanceAll Carnot’s engines operating between the same two temperatures have the PlatiniRn’ftaerato, I scale of temperature.odynamicasame efficiency, irrespective of the nature of working substance. Work tlieriflT = 273.16^-

Note(For platinum resistance thermometer )All real heat engines are less efficient than Carnot engine due to friction & heatlosses.

T = 273.16~ (For gas thermometer)mLow l*mp«»rnturBQ.13 Describe the thermo-dynamical temperature scale. compartintni

A rufngnmlof Irm^lors hoal f,ow-lomocnaiuto 'omscompartment t0 T ilTn the working of petrol engine and diesel engine.h hof -icmpwjitvjto surroundingswith thn holp of oxtemni woi*. It it«1boot engtno operating in roverwo«Tk»r

Q.UThermodynamical Scale of Temperature

mA temperature scale is defined by using thermometric property of certainworking substance. If the working substance is not pure then its thermometry Petrol EngineAlthough different engines may differ in their construction technology but they

ire based on the principle of a Carnot cycle.

A heat engine takes heatproperty Is changed and reading of temperature measuring instrument becomes from a source, converts aunreliable.pan of it into work andThermodynamic scale of temperature is Independent of nature of working rejects the remaining pan to

Constructionthe sink. But a refrigeratorsubstance.Petrol engine consists of pistons, crankshaft, sparking plug and valves. It is showntakes heat from a cold bodyQ help of externalwith theAccording to this scale, the ratio is equal to the ratio of temperature of n Figure.work and rejects the hcaL soQj

hot body. Thus a refrigeratorworks in reverse order of Workingsource and sink.

Atypical four stroke petrol engine also undergoes four successive processes inheat engineQ,= TSo, (A)

tach cycle.1 Intake stroke

Thermodynamic scale of temperature is defined by choosing 27316K as theRecycle starts on the intake stroke in which P .jn(jer from the carburetor*mixture is drawn through an inlet valve .nto the cy^atmosphere pressure.

absolute temperature of triple point of water as one fixed point (upper fixedFor Your Informationpoint) and absolute zero as the other (lower fixed point ).

CMKelvinthermomr irrbulb

1impression Stroke1

of the thermodynamic temperature of theOne Kelvin is defined as

closed and the mixture isithe273.16corn c°mpression stroke, the inlet valve is

Passed adiabatically by inward movement of the piston.i - Poty

triple point of water.Triple Point

On

#,*SSure ^ temper-0^ 8 Spar>( ^res mixture causing a rapid increase in

Plsl°n to mratUre* The burning mixture expands adiabatically and forces

3ft to dor °Ve 0utward- This is the stroke which delivers power to

f0J6s -"ward fv!0^6 out et va*ves opens. The gases are expelled & piston

ost motorbikes have one cylinder engine but cars usually have

Triple point of water is the temperature at which water, ice and water vapoursare in equilibrium state which is obtained at particular temperature andpressure.

kshA triple-po*11a = the heat absorbed or rejected by the system at unknown temperature T. •sr<^t NU

Q, = the heat absorbed or rejected by the system at temperature of triple point temperate £, t>een . constant-voW** *tcof water.

Let

Then, according to Carnot cycle

- XI (Subjective)sPHVSlCS404 Chapter 11 flleai * Thermodof EntropyEfficiency , ,e of increase

0Decrease of entropy of HTR = —

The actual efficiency of tuned engine is not more than 25 to 30% because of heat& friction losses. Cylinder block

SD»> rk plugvalv« • VAlvr-

The jel engine on this aircraftconverts thermal energy towork, but the visible exhaustclearly shows that aconsiderable amount ofthermal energy is lost aswaste heat.EXPLANATION:The remaining part of thermalenergy is rejected to the sink,according to second law ofthermodynamics.

.

0f entropy of LTR - —Increase o T: V

= — - — = positive

the sign of net change of enincreased.net increase of entropy due to a natural process (i.e.

lower temperature). This is also called another

in entropyNet change

tropy is +ve or we can say that netI AST,> T2 SO

I entropy of the system is

that there isExhaustThis provesflow of heat from higher to

nt of second law of thermodynamics.of Thermodynamics in terms of Entropy

natural process, it will go L

big end crankshaft FlywheelDiesel EngineNo spark plug is needed in the diesel engine. Diesel is sprayed into cylinder atmaximum compression. Because air is at high temperature after compressionthe, fuel mixture ignites on contact with air in cylinder & pushes the pistonoutward.

statemeSecond Law

lit the direction that the Do You Know?If a system undergoes nentropy of system plus the environment increases.For example, an irreversible heat flows from a hot body to a cold body to

increases the disorder. So wc can say that the entropy is increased.Addition of heat increases the disorder; hence the entropy is also increased

Entropy as Unavailability of Mechanical Work (i.e., Degradation of

Approximate•fl'etonctoxvxrWvjx davtca* -

E«etoneCMD*vic»Efficiency:

The efficiency of diesel engine is about 35% to 40%.70-9«50-03

Eto lrir. nwor

ener„vi.o.y <*it>««y

°' 1’

Domestic fias furnace

Let US Con 'H

Sloraoe battery

molecules ink 1*° °f different temPera‘ure, so the average K.E of' “

two water,n lg1er temPerature water is greater than lower temperature. The sw.->

“Crated,betweeTrh";^95T“^Sink°f 3^engl"e’wWch C0U'd be EETST JS

But if ^ ° them and mechanical work can be obtained. Noe****-

fl°wfrom ho^htanl<S 30 ronnected with a conducting rod then heat starts fo

*

S°n° tfecha • °^y towards cold body until thermal equilibrium is reached. 2000

UnaVailfl|5j|j. n,Ca* Wor^ 's done due to the absence of heat engine, which results G-W.*" >20

s0^ mecbaniral work. Hence, increase in entropy means (he 2^*°*****

Accord,* fener&'

• ^

lncreases af? Principle of increase of entropy, the entropy of the universe w,n-, ^

^th^orpmechanical work from heat. It would be called 'hear

lnS ,JP°f thermal energy.

I ^n ,^ o°ftUniverse^ at same t °Py of universe will reach at maximum value. ev

*odcand it •ernpGrature and there will be no way to convert

ls caHed heat death.

9070-85Q.15 Define and explain the term entropy.

72

)

EntropyEntropy is state variable of thermodynamically system. It was introduced bvRudolph Claudius in 1856.This gives quantitative basis or mathematical formulafor second law of thermodynamics.The physical significance of entropy is that it is a measure of disorder ofmolecules of a system. Change in entropy is denoted hy ASIf AO is the Quantity of heat absorbed hy the system at temperature Tchange in entropy (sta»e variable) of the system is,

AS*— ^

39

30

12-16Vterc*“

3. Then

( for reversible process)

'ust like internal energy and potential energy , it is change in entropy which ismore important than its absolute value.Sigh ConventionThe change in entropy is positive (means t.ha: ntropy mcre^scs) when heat isadded to a system.Change of entropy is negative (means that entrop decreases) when heat istaken out of this system.UnitThe Si unit of change of entropy or enfc - oy is joule/Kelvin (JK ’)

TH«atrJ

ervtbmg w»H

usefulheat ^to

I

, 4«

Describe environmental crisis as entropy crisis.Q.16

Diesel fuei il'Of,•v»Pontt,siov^tybecaEnvironmental Crisis As Entropy CrisisitAccording to 2nd law of thermodynamics, every real process causes to increase contains

ffiore'toon itomj,nthe disorder or entropy of the universe. Any increase in the disorder of a systemthan gasoline doesproduces and even greater increase in the disorder of the environment, which is (gasoline ,typically C9H20. while deseicalled "environmental crisis.

'S constant/gas k'* typ«caHy C14H30) goltx^30

nstant per N?The disorder producing activities due to all industries may result a great moleculeIt takes lessttfjnjQg to criatt £0increases of disorder which affect the overall life support system.T =. — <-mv * >l* 3K 2 __diesel fuel. of gayWhich IS Why * U|#(J TemperatureOur mechanical energy producing processes are not efficient. For example petrol to be cheaper than Qa»oltn«engine has its efficiency about 30% and diesel engine's about 40%. Hence most Diesel fuel has W = PAVof energy is transferred into the environment in form of heat, which causes to Work donedrniiry than gasokne On Q = A U + W Q = (Ua -U,) + Wyincrease the entropy of it.

First law of•*•'»9«. t gallon (3 8 L) 4The second law of thermodynamics impose limit on the efficiency of mechanical thermodynamica -escI fuel contamQ = Wenergy produces by engines, which says that thermal pollution is an inevitable Isothermal pr°cessapproximately 155*106 >oula* -W = AU (compression)

W =-AU (expansion)result of second law of thermodynamics. Due to the thermal pollution in (147.000 BTU). white 1 gallon of Adiabatic process9*environment, temperature change may occur. But a small change in gasotne contain* Relation between and Cp ? = CVRCCenvironment may serious effects on metabolic rate in plants and animals. This 132x106 joules (125.000 BTU)

and CvThts. comtyned wth thamay disturb ecological balance.Effidency of heat enginmproved efficiency of deseiThe imperative from thermodynamics is that whenever you do anything., you

engines, explains why dese1should be sure about its present and future impact on your environment. This iseng nes get better mileage tta' Efficiency of carnot'san ecological imperative that we must keep in view now if we are to prevent aequivalent gasolme enginesstrong degradation of life on our beautiful but fragile earth.

-Xi (Subjective) 409pHySlCS«04

The(b) Only potentialOnly kinetic

partlyMultiple Choice Questions (3)

kinetic and partly potential (d ) None of theseof heat between two bodies is determined by:

b) Kinetic energy»0>Which 0

of flowfl*directionfWr-nn.'Vr mm.rum v cjurk am/cment art given below. Tick the correct internal energy

w •spheric pressureitic process?

Total energy(C

f the following is a best container for gasduring(bj TheSjos flask( <$) GlaSs-vessel

1. ~>e a ««rage soeed 0*' •-'e gas — o- ecu es in a co~ *ainer is:Zero'CO

“'e -nea* CE. 0* a gas s ze-o at

1(b )(a ) Wood vessel

10K-^the Wblowing process, the system aiways return to original thermodynamic state?

a in which or i" |&) Adiabatic(a) l«>b3ric

i

273• d nfir, te

*

oX- (b; -273* C(d ) isothermal

of 50% when its sink temperature is at

(b) 32/C

«5) 273f'C

int and steam point. Its efficiency wi» be:

(b > 71.23%

26.81%

bVO <: (ri) *x>°C3as -c ec.kes exerts pressure or the w* Is of container because they:

Collide with ore anotherHone of these

(c) Cyclic

A carnot engine has

source is:

M(c) 373sCAcamrt engine works between ,cepoi

(a) 85.42%

27°C The temperature of3.ii

Cai ce •- >•

* s -creased - r point of ice:

s

: 'd4.

- fb) frtcreases/ d ) Uocte of Thesezn mr&:*r- 'esrper*? v-e. z pressure s -a sedtfcer ft* vo* um*

--"^ 'c ar •

(diWf 53-^%r efficiency of heat engine should be:

;i if the temperature of sink is absolute ze ,

^C'/s?ar -'SV

aV

(a; .00%

K o1 /rtven the temperature

id ) infinity

nd Sink IncreasesId) fioub d- -« rrr-Vr'.’SS1 e* -- *r ^ ^ gasstfb# PAxe

, the efficiency of heat «•*"«f

difference between source a* tecc

wll:(& J Decreaseesc v .-e- 'ec"'/* c. ' i'/ e c»f •' ece

(a) Increase(c) )s not effected(dj >/3/ increase/decreate d*P*nd up0"» The internal energy of »n W***"8

fa, Keeps ©rt Changingte) Zero

' rsr>Ar -tforv ng substance

Veers'. S'

v. • <-/ 'vlo«r - ? V ^ V fir c• «

r --

^ r V <

tCi

(b) Remaps constant‘ioneoftrese

*

4

VaC£ '4 w • ,

Al J&tfifc 10 *)

\ 9. bS. d7. «20. b5. d U, a \ **• »

, /. ^ 4. •2. b 17. d)< f 16. b<X V

111' '7'/ //

14. i13« »12. af#>S*^ j=*r * >i

V/ -yf /

>" A....,4f, «r^ ewtnWV ' ^ ‘‘‘Isf .'S> « /C / of 4 >

it, »/ ' *’ yV^/ir* ^fe

'' > - — r

r n-

rt

fCte.

IA

- XI (Subjective) 411Chapter 11 [Hear* Th410

IT^ou £Short Questions of Exercise t t

Q.11.1 Why the average velocity of the molecules in a gas container is zero but thevelocities is not zero? ave|,age of the squareo| V— * V— >V—

Flfl. 1t 13(«)(Fsd 2005,Rwp 2006,D.G.Kh F* M 1316) F10.11.1Kc)an 2°06,Lhr 201Ans. ReasonAverage velocity of the molecules

The change in internal energy of the system is zero in all cases

Aftf- ReasonAs all the processes are cyclic. So the system returns to

internal energy of the system does not change.

at constant pressure is greater than specific heat at constant volume.Why?(Mtn 2003, Lhr 2003,Rwp 2004,D.G.khan 2005.

Fsd 2008, Bwp 2007-2008,Mir Pur 2009,Lhr 2010-2011)

The motion of gas molecules is random. So if a molecule is moving in one directionmoving in opposite direction. Thus the vector sum of these velocities is zero. its initial state after each cycle. Hence, theanother will be

M-^-oi.e

Average of the square of velocitiesAs the square of negative velocity is positive, so the average of square of velocities is

v2 +(-v)2

Q,,5 specific heat of a gas

not equal to zero.i.e * 0 ReasonAns.2

is heated at constant pressure then;When gas(i) a part of heat is used to do work on piston

(ii) rest of heat is used to increase the temperature through 1K

When gas is heated at constant volume then all the heat absorbed is used to increase temperature

Q.11.2 Why does the pressure of gas in a car tyre increase when it is driven trough some distance?(Mir Pur 2004-2009,Mtn 2006, Bwp 2007-2008,Fsd 2008,Grw 2009-2011)

Ans. ReasonThis is due to the work done against friction between road and tyre.This work done appears as heatThis heat is absorbed by the gas molecule.It Increases the average kinetic energy of molecules.As pressure P oc <K,E.> ,so pressure of gas is increased

onstant pressure is greater than molar specific heat at constantthrough1K.That is why the molar specific heat at cvolume. (Cp > Cy).

0,11.6 Give an example of a process In which notemperature of the syst«m changes.

*. „„„„no heatMn i.ms«» ““ »'“•S“*

from the system but theheat Is transferred to or

(Grw 1009» 1009)

What will be the change In Internal changes.Q.11.3 A system undergoes from state P,V, to state P,Va as shown In fig.energy?

Ans. The change In Internal energy of the system Is taro.ReasonIt Is clear from figure that temperature of system Is constant• It means that Internal energy Is also constant as It depends upontemperatureSo; there will be no change In Internal energy

Q B At/+^For adiabatic process 0Mt\Pi

O- ACZ + W'(adiabatic expansion)

Or At/ -W (adiabatic compression)

So^doing so whole mechinlcel energy Is converted Into heat energy.So

t Con®l«n1ftrnpofiUtf*P

(MPf 4! Samples

* Bunturn flf ilr 1kurl* *vri1

[h which sound WIVI Is pissing,

ABCDA,*.<*’"* cioud^ffnatipn *" th* itmo|ph*rtIsItpouibli to eonwort Intirnilinir»y W»

****** »««».Hiloowool.v*.two.

Q.114 Verletlon of volume by pressure Is given In FIR . 11.13. •!•* ti^tin #,ongA to A. what will be the change In Internal energy?

(O.O.khan

412

Yes, it is possible to convert internal energy into mechanical energyExampleWhen a gas is allowed to expand adiabatically, the gas does work on theenergy, due to which its internal energy decreases, i.e W = ,\UGases cam be liquefied by this process.

Ans.

tropY of a sYstem increases or decreases due to friction?(Mtn 2004-2005,Bwp 2006,D.G.Khan 2006,Sgd 2005, Ihr 2006-200«,Rwp 2008)

Does en

y 5 the entropy of the system increases due to friction.

Reason

surrounding by usingil%n* Hfli-

TSince

Due towhich increases the entropy of 5*stem

Q.11-13 Give an

Ans. We know that entropy

Q.11.8 Is it possible to construct a heat engine that will not expel heat into the atmosphere?ISgd 2004, Mtn 2004,

friction, some mechanical energy is converted into heat (i.e heat is added up into the system )Gr* 200S|

Ans. No, it is not possible .

ReasonIf it is possible, then it w,ll be violation of second low of thermodynamics winch st tiespossible to construct a heat engine that will not expel heat into atmosphere with our i ' ' 'Snt"change on the working substance, '- , v|r'gany

example of a natural process that Involves an increase in entropy?(Sgd 2005,Grw 2005,Lhr 2006, Bwp 2006, Ihr 2009,Grw2011)

is measured by following equation.

AQAS = TQ.11.9 A thermos flask containing milk, as a system is shaken rapidly. Does the temperature of milk rise?

{Lhr 2005, Fsd 200S,Mir Pur 2009)

ExamplesThe melting of ice involves the increase in entropy

changes its state.(i.e from solid into liquid). Thus, entropy increases.

All natural processes in which friction is Involved, the entropy of the system increases.

. Ice absorbs the heat from its surrounding andAns. Yes, the temperature of the milk rises.ReasonWe know that

T «r <K.E>When we rapidly shake the thermos flask we do some work on it, this work done increases the K.E ::molecules of milk. Hence, the temperature of milk rises.

adiabatic change is the one in whichNo heat is added to or taken out of a system. (b) No change of temperature takes place ,

(d) Pressure and volume remains constant.

Q.11.1A An(a)(c) Boyle's law is applicable.

No heat is added to or taken out of a systemis irreversible?

I Ans. (a). Q 11.15 Which one of the following process

(a) Slow compressions of an elastic spring.

I (b) Slow evaporation of a substance in an

Q.11.10 What happens to the temperature of the room, when an air conditioner is left running on a tablethe middle of the room?(Bwp 2003, Fsd 2005,Mir Pur 2006-2009, Lhr 2005-2006-2009, Grw 20101

Ans. The temperature of the room will not decrease even it increases slightly . isolated vessel.(d) A chemical explosion.

(c) Slow compression of a gas.Ans. (d) A chemical explosion is irreversible.Q.11.16 An ideal reversible heat engine has.

ReasonAs air conditioner is running at th middle of room it absorbs as well as rejects heat in the same r00^Jthe same rote . So temperature of the room remains unchanged. But due to working of compresomee heat is produced due to friction. So temperature of the room will increase slightly. (b) highest efficiency,

nature of working substance.(a) 100% efficiency.

fficiency which depends on the(c) an e(d) none of them.(b) highest efficiency.

i example-(Mtn 2005,Grw L. 2004. S«d 2005, fsd 2005,M,n 2006)

Q.ll.ll Can the mechanical energy be converted completely Into heat energy? If so give an2005-2006)

(RwpAns.Ans. yes, it can be converted completely into heat energy.Reason

jDuring isothermal compression work done 'W' on the system is converted into heat Q •

Q =A U +WFor isothermal process A U =0

all theExample:During slow compressionheat rejected is used as

So -Q =0+(-W)Q =W

Or W =QSo by doing so whole mechanical energy is cor erted into heat energy.

r

414 Chapter 11 flj XI (Subjective),s PHYSICS£2iAThe 415£2<Kjy

Solved Examples -332 x 10m 6.022 xlO23W i l l Thus

32N'hat is the average translational Kinetic energy' of molecules in ma gas at temperature 27C°?6.022 xlO 26

5.31 x 10~26kgGiven Data.

Temperature = T = 27C°= 27 + 273 = 300K

mvalues in equ. ( 1), we gel

< II ylttflfrX

puttingTo f ind.

<\b> =Average translational kinetic energy = <K.E> = ^Calculation. 1130.22 103<v2> = - x

Where 5.31T = — <K.E >3 K

<K . F'.> = 3 KT2

K = Boltzman constant = 1.38 x 10“ 23JK|X 1.38 X 10'“ X 300

- x 1.38 x 10~1J2

<K.E > - 6.2I x 10~31 jj

A5<v2> = 212.84 x 10*<v2> = 212840Or A|

<v 2 > = 461ms'1=Where Or rm*- I

Thus <K.E> *

A gai is enclosed In a container fitted with a piston of cross-sectional area 0.10 m2. The pressure of

the gas is maintained at 8000 Nm"\ When heat is slowly transferred, the piston is pushed up

throu ;!' a distance of 4.0 cm. If 42 J heat is transferred to the system during the expansion, what is

the change In Internal energy of the system?

<K .E.> - x 3 x 102

SEEfBEBDE Given Data.Find theGiven Data.

At ST.P. , wc haveTemperature - T - 0°C - 0 + 273 - 273KPressure * P = 1.01 x 103 Nm' 2

- erectional area of the piston•A"0.10maverage speed of oxygen molecule in the air at S.T.P. CrossPressure of the gas P•8000 Nm‘2

Distance moved by the piston - Ay 4.0 cm “ 4 x 10 mHeat supplied to the system Q * 42J

To Find.To Find. Change in internal energy - AU * ?

Average speed of oxygen molecules - <v> - ?Calculation.

Calculation.As we know that

work done by the gas “ W •PAVW-PAAy

<v*>- —mWhere k - 1.38 x 10 ^JK* 1

To find mass m, of one molecule of oxygen, we use the formulam- Molecular miss of oxygen

Avogadro vnumher

As T «As AV - AAy

SoPutting values, we getOr

W- 8000 x 0.10 x 4 x 10-2

Now AW*32J

PPlying first law of thermodynamicso-AU + W

V/hCr.A U-Q-W

PUf,Crc ** change in internal energy,n8 values, we get

(1 )

OrOr

Mm - — ~NAAU - 42- 32

^u»TojlWhere Molecular mass of oxygen m 32g•32 a 10

_)kg

NA•Avogadro’i nurr.oer•6.022 x 10 'and

i

V _ xi (Subjective)417Chapter 11 rHea.A T||

416

Exercise ProblemsQQESnZllEIThe turbine in a steam power plant takes steam from a boiler at 427°Ctemperature reservoir at 77°C. What is the maximum possible efficiency?

and exhausts ite the average speed of nitrogen molecules in air under standard conditions of pressure andinto * lo* £stima' Ucrature-

Given Data.Temperature of hot body (boiler) = Tj = 427C° = 427 + 273 = 700KTemperature of cold body (reservoir) = T2 = 77C° = 77 + 273 = 350K

V •

tinder standard conditions= T = 0°C = 0 + 273 = 273/r

» P - latm. 1.01 x 105Pa

To Find.TemperatureEfficiency = r|= ?

Calculation. Pressuren = 0 - £- )

11As

foM5 *

Cilcutation:= <v> »7speed of nitrogen molecules

Using the formula2 i ,

T- — <-mv 3 >3k 2 "

3kT«<mvJ >, 3kT

<V >=

AveragePutting values, we get/1 350 X*1 = 0 - ZTT ) /

17001n = d - j)

i( i )= 0.5o = -

% n = 0.5 x 100m - i= 1.38x10’23JKWhere k = Boltzman constant

Now the mass of one molecules of nitrogen (ty) is

Molar massm= Avogadronumber

I%r) = 50°/<J

BfflBTTTECalculate the entropy change when 1.0kg icc at 0 'C melts into water at 0nC. Latent heat of »ionof icc Lf = 3.36 x l O 'J kg “ ' m = _i!i_=2M0^M.65K|0“ kg

6.02xl 0u 6.02*1023Given Data.Putting values in eqution (1), we 8elMass of icc* m - 1.0 kg

Temperature * T * 0°C - 273KLatent heat of fusion of icc* Lr = 3.36 x 105Jkg

3* l .38*10‘23 x273<v2 > = 4.65 x10 “

1.13022x 1020<v 2 > =To Find.

4.65 x 1026Change in entropy =* AS = ? <v 2> = 2.43058x10'Calculation.v = V<v 3 > *493ms 11mu * i

Change in entropy is given byuses at a certaindifferent K

'U Show that ratio of the root meanAS —T3onndCmpCnitUrC'* C<,ual t0 thC 8quarcHere

AQ Heat added to icc mLfV

i rms =?mLfSo AS - CMc^Ut|0n:T 3 rmsPutting values, wc get

Using the formulaAS - L° x 3,36 x 10 *

273AS- 0.01230 x 10* JK. »

Or

\n

For first gas1

d )T= — < — m. v \ >3k 2

1 or second gas

T= — < — m. v/ >3k 2 • (2)

odynamic system undergoes a process in which its internal energy decreases by 300,1

" * Hciunc time 110 J of work 1» dyt. ® ^ ^. H a t

Dividing equation (i ) by (ii )2 1— < — m , v . 2 >, 3 k 2 1 1

— < — m , v , 2 >3k 2

TGiven dat*1’.

in internal energy - AU=«300JT DecreaseWork done on the system = W5« 120J

3<m,v, >Tollndi j lent lost by the system<m 2v, >

J< V ' , > mC,ltul"l0nl

According to first law of thermodynamicsQs U+W

5

< v )* > . m ,

< v , J >. Putting the values, w gel

Qs =* 300-120< vi >

LLVL.ISLLVlr„, |Q»- 42QJ

Negative sign shows the lost of heat.This is the required proof .

C and the sink temperature \%

rCe and heat rejectedis 227°.Th... »11.3. A carnot engine utilize* an ideal gas127°C. Find the efficiency of the engine. Also findto the sink when 10000 J of work is done.

11.3. A sample of gas is compressed to one half of Its Initial volume at constant pressure of 1.25 x 10*Nm*\ During the compression, 100 J of work is done on the gas. Determine the final volume of thegas.

Given data:Given data:

°C=227 + 273= 500K.=Tt = 227Temperature of sourceSuppose initial volume of gas =V =Vi_ T2=127 °C = 127 + 273= 400K

Temperature of sinkVFinal volume of gas =V, -—Work done = W - 10000 J

find:Pressure =P = 1.25x 10'Nm :

Efficiency of Carnot engine n ?Work done = W = - 100 J (i )Heal input = Q,-?

To find:(ii)

Heat rejected to sink O 2

Final volume of gas =Vf =?(hi)Calculation:

Elation:W = PAVAs

© As wc knoww = P( Vf - v.)or

M

( vW=P V I2 J As V, = V and Vf =Putting the values, we get( V -2V

«122.xioow=p

2f y ^- i0O=* P

2 j

5» 421-

XI (Subjective)Chapter 11 fHeai JL T||420

T2(ii) To find Qj n == 1 *

TMoutput workefficiency =As input heatWn "

Q,0.268 = —Putting values, we get

20 _ lOOOO100

”

Q,10000*100

100

T(<= 373.13K.Q.- 20

Q,=50000JQ, » 5.0*104J Also

T,•Tj=100K

373.13 — T2 =100(iii) To find Qj OrT2 =373.13 - 100

T,=273.13KAs W = Q1 - Q2

C h- Q r - WOr

O rPutting values, we get

Q2 = 50000-1000002 = 40000fe = 4.Qx 1 Q4 J

327°C and 27°C claims to have11,7 A mechanical engineer develops an e n g i n e, b C t W C C n

. Does he claim correctly? Explain.

Temperature of source = Tl =327C° =327+273-600K

Temperature of sink = T2 =27C° =27+273-300K

Claimed efficiency = 52%

efficieucy of 52%Chen d?la:11.6 A reversible engine works between two temperatures whose difference is 100°C If it absorbs 746 J

of heat from the source and rejects 546 J to the sink, calculate the temperature of the source andthe sink.

Given data:Temperature difference = T,-T2 =100Co =100KAmount of heat absorbed = Q,= 746JAmount of heat rejected = Q2=546J

To find:

Actual efficiency = r}=?C,l'»UHon:Find data:

Temperature of source = T, =?Temperature of sink = T2 =?

AsT,

Putting values, we getCalculation:

„-i - 4 300As^ ~ 1 ~

600Q,Putting values, we get 1

" m l ~ 2l

^* 2

^ = 0.5Ventage efficiency - 0.5 * 100

546n - i -— •

746

746n-0.268. (1)

* Or

i

Chaptor 11 [ H422 21*1*.*, , PHYSICS XI (Subjective)

"4l - 50?ijSo the claim of 52% efficiency is not correct . lution:full'11

To Find 7/% V = 7 0 %V - o . 7N o w

To Hud A- 5 0 %1 1 .S. A heat engine performs100 J of work and at the same time rejects 400 J 0f|lCi|(resen olrs. NVhat is the efficiency of the engine?% ncnc,'g> to th n - o . 5N o w

c '0|,jGhen data:work done - W- 100 JHeat rejected * Q,-400J

TT" T T T

«1.1*.tl

P u t t i n g v a l u e s , w

0 . 5- 1 -T

P u t t i n g v a l u e s , w e g e t

To HndiEfficiency•rj— ? i

Calculation! 2 8 0 2800 , 7* 1 -A* W.Q,.QQ. - H' + f tQi * 100 * 400Q , S00J

T T/J t2 8 0 « 1 - 0 . 5 2 8 0

9*r TO r iJ 2 8 0 n .— .0 . 5 2 8 05 0 , 3

T,2 8 0H t r - *100 T ,N o w 2800 . 5 T/ -Q

0 . 3T , - 5 6 0 Kthe nines T,' » 9 3 3 . 3 3 K

100*4— xlOO Thus increase in temperature

500Tl -T1 =933.3-560=373K

4:r = 0.2x100% n = 20%

T/ -Tl =373K

T/ -T. =373C°

11.9 A Carnot engine whose low temperature reservoir is at 7 C has an efficiency of 50%. It is dcs luo A steam engine has a boiler that operator at 450K.The heat change water to steam, wh'cbdnv«to increase the efficiency to ” 0%. BN how manv degrees the t e m p e r a t u r e of t h e source be the pjston j^e exhaust temperature of the outside ai r i s about 300K. at « maxi

•creased? of this steam engine?

Temperature of boiler (source) T, 4_ 0K

Temperature of outside air (sink) = T2 =300K

Maximum efficiency**1=?

Given data:Gh« data:

1espemre of lov. resen. o n = T.=1C = 273-" = 280 Kefficiency =% r = 50%

effioeacy = %-' = 70%

To Tind: .Calculation:

JjLT# Fimd: 1 T;. nc of sours* foe 50% efilcscocy T, =?Temperature of source for "0% cfiicie y = T =?Then increase in temperstuve T - T *?

300 rx450

- . 5 ..

aSrC'-f -

424 Chapter 11 jllvat Ai)^ 425XI (Subjective)PHYSICSISOf )

450ip0, 33

percentageclIideney ( ) , M loo

lil- 33%|

Board Model paper^Model Paper Physics Objective

nndlato Part - I (11». Class) Examination Session 2013-2014 and onward

Tatal marks* 17 Paper Code Time Allowed: 20 minutes(D)(C)(B)(A)' QUESTIONS

ThiTunU of pressure in bascT"complete Equihbr.um jflMpdV

K(jmJ Sec 7K(jrn

? Sec 1iQ.l. T Kgm SecJfx 0jjfyjO

Kgm Sec11.11 336 J of energy Is required to inch 1 g of Icc at 0 C. what is the change Inat 0" C as it Is changed to Icc at 0° ( by a refrigerator?£F = 0

11= 0entropy of 30 i. = o

R 0fHit#r The2.Given (lulu; Implies that vAt highest point, the vertKid component

of velocity of Projectile be< (i(^Impulse has the s.» 'm* nmt •, th.it of

V,cos"MinimumZeroMaximumIlent of Atllon of Icc -I f * 336.1Mass of water - m•30gTemperature - T•0°C-273K

3.Linear

MomentumMassEnergyForce

4.MorsEarthTo find: MoonSundue to gravitationalThe Tidal Energy is

pull of theThe rotational K E. of a disc >sTorque per unit Moment of Inertia is

Equivalent toEscape velocity on surface of earth is 11.2km/Sec-1. The escape velocity on theSurface of another planet of same massas that of earth but of l/4tlmes the

5.Change in entropy - A.V ? 1/8 rm/1/6 m /A mi/A mvCalculation; 6. Radius ofGyration44.8 Km

sec 1

InertiaAngularAcceleration

11.2 Kmsec*

Angularvelocity

Since boat is transferred from thedenoted by negative sign

Change in entropy AS --~7.water to freeze it into icc, therefore entropy of icc decreases and ij 22.4 Km

sec' 15.6 Kmsec' 18.

T1lore AQ-heat energy token out of water mLr radius of earth is M3 sec 1

m2 sec' 2nrsec' 1mJ sec' 1ml The SI unit of flow rate of fluid isThus x = x0 cosJAS x = x0 sin'x = x0 COS «Xx - xQ sin nXFor a spring mass system arrangedT aXaXPutting values, we get instantaneousthehorizontally.displacement Is 2AA30*336 A/2A/4In the time required for the tuning fork toA S-- 273 make one complete vibration, :he wavein air will travel a distance equal to MediumA S - -36.92JK -i PressureDensilhofVelocity of sound is indepei '4'rrTwo tuning forks of frequencies 40Hz and243Hz respectively are sounded together,

second is

Zero

THE END the no. of beats producedIn young's Double slit experiment, the

position of Bright fringes are given by

mdLdAdv ALr~ ~m~d

Y- =ur- =mTy„ = m—m L

Virtual andVirtual andRea/ andRea! andinverted

>v- ! • i. vm .s , v

t

427XI (Subjective)426 l^oardp PHYSICSm sec-1 in a direction 309 above the horizontal. Determine the height to2. Write answers of any EIGHT questions. (8 x 2 = 16 )

Define dimension. Check the correctness of the equation v=f/.by the principle of Homogeneity 0f H-( .i) Briefly explain the two drawbacks to use the period of simple pendulum as a time standard.

Assess the total uncertainty in the final result of a timing experiment with the help of an(rv) Determine the dimensions of pressure and density.

(v) Under what condition would a vector have components that are equal in magnitude.(vi) Justify the statement "A body cannot rotate about its centre of gravity under the action of its o(vii) If A . B =0 , Can it be concluded that A and B are perpendicular to each other? Support your

• proof.(Viii) Why fog droplets appear to be suspended in air ?(ix) Discuss the sign of acceleration due to gravity for a cricket ball thrown upward, for its

motion.(x) Can the velocity of an object reverse the direction when acceleration is constant? Justify with an exa( xi) It is advisable to fasten the seat belts during a fast drive. Why is it?( xii) Explain how would a bouncing ball behave in each of an elastic and inelastic collision with floor of3. Write answers of any EIGHT questions. (8 x 2 = 16)(i) When a rocket enters the atmosphere, why does its nose cone

come from?(ii) State the work energy principle. Express it in equation.

While calculating the Absolute Gravitational potential energy, why is the distance between infinity and surfactof earth is divided into very small steps.( iv) What is meant by moment of Inertia? Give its significance.(v) How is artificial gravity created in an Artificial satellites.(vi) Centripetal force and centrifugal reaction are equal in magnitude but opposite In direction. Why these forcesde

not balance each other.What happens to the period of simple pendulum if(a) Its length Is doubled(b) Its suspended mass Is doubled.

(viii) Show that in SHM, the acceleration Is zero when velocity is greatest and the velocity Is zero when lacceleration is greatest?

(lx) Why can we not realize an Ideal simple pendulum.(x) What features do longitudinal waves have in common with transverse waves.

Why does sound travel faster in solids than in gases?Justify the statement "Velocity of sound In a gas is independent of pressure of the g a s '

4. Write answers of any SIX questions. (6 x 2 = 12)(I) Define coherent sources of light. How two light beams can be made coherent. ^ young'5(li) How is the distance between Interference fringes is affected by the separation between the s

double shit experiment?.How would you distinguish between unpolarized light and plane polarized light.Name and explain any two of major components of a fiber optic communication system.

(v) How the resolving power of a compound microscope can be increased.What happens to the temperature of the room, when an air conditioner isof the room.

t 0(vii) What is meant by tripple point of water. What is the value of Absolute temperature of tripple P°(viii) Can the efficiency of a carnot engine be 100%? Justify your answer with proof.(ix) Normal Human body temperature is 98.6 °F . Convert it into °C and K.

SECTIONII (EisayType)

with a speed of 30A pall is thrown

I*1 Which it rises.3{•)

tationary orbits. Derive an expression for orbital radius of a Geostationary orbit1+4

a force is required to accelerate an electron (m=9.1 x 1CT51 kg ) from rest to a speed of 2x10’3

's formula for speed of sound in air. How did Laplace correct it .

6 j3j vVhat aregeos

(£J| How largethrough a distance of 5.0 cm.

What is the limitation of Newton

(u.) example. -imsec

Wn weight".

answer1+4

7. (a)

A simple pendulum is 50cm long. What will be its frequency of vibration at a place where g

(bl \

Explain the principle, construction and Magnifying power of a compound microscope

with the help of a ray diagram

jb) A light is incident normally on a grating which has 2500 lines/cm. compute the wavelength of a spectral line for

which the deviation in 2nd order is 15°T 3

Explain the carnot cycle and calculate the efficiency of a carnot heat engine. 2+3 = 5

Water flows through a hose whose internal diameter is 1cm at a speed of lm sec-1. What should be the

diameter of the nozzle if the water is to emerge at 21m sec"1.

SECTION III (PRACTICAL)Give answers to any Four Questions.

How does the electronic timer measure time of free fall accurately.

dent measured the diameter of cylinder as 2.45 cm by a vernier calliper having least count +0.01

cm. But later on he observes a zero error in the instrument and finds zero of the vernier scale lies to the

right of the zero of principal scale and 4th division of vernier scale faces any division or the principal

scale . Find the correct value of diameter of cylinder.

The wire of sonometer is stretched with a load of 4kg wt Including the hanger and resonant length of

wire is found to be 11cm by using a tuning fork having frequency 512 Hz. If diameter of the wire Is

find the resonant frequency of this wire for the same resonating length and same load.

Find clockwise torque from diagram. 0 10cm 50 cm 70cm 90cm 100cm 12 N 4N 5NM#ttr Rod

=9.8m sec-2

3*i»ij1+2+2

8.(3)upward and do^Wi-mple

9.(a)3room.

become very hot? Where does this heater<b|

4 x 2 = 8Note:*

10.( a) (I)(iii)(II) - A stu

(III)

(vii) doubled(Iv)

"-•!

9 N4 N12 N

angle of deviation vary with the angle of Incidence In case of prism.rent material varies with the colour of light.

perlmental determination of mechanical equivalent of Heat(v) How does the

Does the critical angle ofatranspa

sources of error during the ex

(xi)(vl)( Xii)(Vii) What are the

by electrical method.(viii) Design a table of observ

string of sonometer.10.(b) Write down the brief procedure to show experimentally that time period of simple pen

amplitude . 3

of length by using the vlbr.tlons In theatlons/calculatlons to prove the law

dulum is Independent of

(Hi) ORfocal length of a convex lens by displacement

to determine experimentally the

the basisof graph drawn below.(iv)

Write down the Brief proceduremethod.left running on a table In the(vi)

following Ouestlon onfW3ter 10.(c) Answer the

> 2 4 )( 8 x 3Note:- Attempt any three questions. - .z.jtam5. (a) Define Rectangular component - of a vector How two vectors can be added by Rectangular comp ent ir1

1+4

C

IMS

5428 Paste I ,.rhnlar's PHYSICS XI (Subjective) 429

SECTION-1Attempt any eight parts.Give the drawbacks to use the period of a pendulum as a time standard.The period of simple pendulum is measured by a stopwatch.

What type of errors are possible in the time period?.How many seconds are there in I year?Write the dimensions of (i) Pressure

Can a vector have a component greater than the vector 's magnitude?

Can you add zero to a null ^Opr?

Can a body rotate about its centre of gravity under the action of its weight?

Can the velocity of an object reverse the direction when acceleration is constant?

If so. give an example.^At what point or points in its path does a projectile have its minimum speed, its maximum speed?

Find the velocity of a heavy body when it elastically collides with a stationary light body.Derive a relation between impulse and linear momentum.Explain, how the swing is produced in a fast moving cricket ball?

Attempt any eight parts.Calculate the work done in Kilojoules in lifting a mass of I()kg through a vertical height of 10m.

(ii) A girl drops a cup from a certain height, which breaks into pieces.What energy changes are involved?

(iii) What is Escape Velocity?(iv) What is meant by Moment of Inertia? Explain its significance.

(v ) Show that orbital angular momentum L( ) = mvr.

Explain Rotational K.E of a Disc and a Hoop.(vii) Does frequency depend on amplitude for Harmonic Oscillators?( viii) Can we realize an ideal simple pendulum?fix'.) What is Phase?( x ) Is it possible for two identical waves travelling in the same direction along a string to give rise to

stationary waves?Explain why sound travels faster in warm air than in cold air.

(xii) Define Doppler Effect.Attempt any six parts.

( > ) State Huygen's Principle.Under what conditions two or more sources of light behave as coherent sources?Why the Polaroid Sunglasses are better than ordinary sunglasses? .

How the light signal is transmitted through optical fibre?When object lie within principal focus of convex lens what is the nature of image and where is it formed ?

( vi ) Is it possible to convert energy into mechanical energy? Explain with an example.(vii) Why is the average velocity of the molecules in a gas zero but the average of the square of velocities is

not zero?( viii) State Carnot’s Theorem.

A thermos fiusk is shaken which eoma^r^ iJIk^^Does the temperature of milk rise?

8 x 2 = 1 6Q.NO-2

( i )(ii)

(iii) (ii) Density(iv )( v)

(vi)(vii)(viii )

I( ix )( x)( xi)(xii) 8 x 2 = 1 6

(J.No.3( i )ILLlil IJiJ

' 4

I.(i) What can you conclude from the graph 1(ii) Find the value of "g" from the graph 2(iii) Measure the length of second's pendulum from the graph 1

ORAnswer the following Question on the basis of graph drawn below.

( Vi )

\ - 4\ t \ | .null vj - 11,001cmAlong V - jtxi. |.mjll -Hj u onion

(xi)006 Q.No.4 6 x 2 = 12

0.05( ii)

0.04 (iii)

i0 03(iv )(v)V0 02 X0 0 1

NEx »00 01 0 02 0 03 0 04 (1.05 0 06

. 1/p >(i) What is value of "p" corresponding to 1/q = 0.5 cm'1

^ sm^ enafvalues of 1/p and 1/q from

(ix)

NOl E:- Attempt any three questions of the following;-5 (a) Define rectangular components of a vector. How two vectors can be added by rectangular component*

method?-1

evaluate focal length.(b) A football is thrown upward at an angle of 30° with respect to the horizontal To throu ii 4Om pass.

must be the initial speed of the ball?

und explain Work-Energy Principle.

1000 kc cur is turning round a comer at I0m#-J a* *"***[•circular pith Is 10m. hot large « force must be cxened b> the pieman, on rhe ,o 60*,*c*,**MncTmplc Pendulum und derive rel.' - P"** jLf '*4

MULTAN 20136.(a) State

(b) A jI

2013 (A) 3.1«Roll No:PHYSICS TIME ALLOWED:PAPER-I (NEW COURSE)

HoursSUBJECTIVE MAXIMUM MARKS: 83

Write same question nt mber ant its part number on answer book, as g*venpaper.

the 7.(a )

NOTE:-

•i

4314Jl) p««t B XI (Subjective)PHYSICS -2ardplar’sSch®A pipe has. a length ol I m. Deleunmo the frequencies ol the fundamental and i|,c (-

pipc is iIVliiu* Simple Micioscope I md Ms Magnification

( h) Graph B"si harm°nics ;.pen .11 holh ends, ( speed ol sound in ail 140ms’ ) Gr a ph be twe e n p+q and pxq

N (.il 3 i

mM tfi l lIn a double '‘In evpeniiienl ( he second oidei maximum occurs ai 0 - n.2Vf '|'|10I Vieimmc Hie slu separationI \plain Isolheinul and \diahaiic hocesscs\ wain hose with an iniem.il diameiei ol .0 mm ai ihe oullei discharges 3( ) |< p .

^ " waier in /.i>f.iU ulaie ihe waiei speed at lhe outlet . Assume the density ol water is 10( H ) ko/m3 . . ISccundi

SECTIQN-III (PRACTICAL PART) U ,ls n°w issl '

v *

5 i:. i n >-* iii> ) a11'wavelength ; ‘madi,s 65«> m„ Tm4- ^Oi' ) Ia ) 1 . \ ii-I .-irrihill( hi i5 i

! J *t"HJ4++HIfi. • 11 —-hrTTTT • : i— .

'

IO4 .V ) Write answers of any four parts.Ilow does the electrome timer measures time ol tree fall accurately?Why a sv. 1ew gauge is more .iccui.ite than a vernier callipers?How the weight ol a metiv uni is loinul when u is suspended to two spring balances?t he velocity ol sound at 0* V is VI2ms * What is its velocity at 2>*V?t he index ol ic fraction ol glass with respect to air is 3/2.What 1 s the index ol retraction ohm with res|vet to glass?\\ hat aie the conditions to Iotal Internal Reflection?What aie the sources ol enoi duimg the experimental verification of law of mass of vihiatinn• suing using.1 sonometer? ' LC 1e“

( \ m) Design a table ol observations to find acceleration due to gravity b\ oscillating mass spring systemWrite down the brief procedure to find experimentally the unknown weight of body by the method ofvector addition of forces.OR

W rite down the biicf procedure to find experimentally the refiactive index of thecritical angle method.Answer the follow ing questions on the basis of graph draw n below

Graph A

TT

tT::.-ngm: :4 \ 2( it = 8 n *» 1h• • f j ; I - ]irl:u( »1' m 1s I ; l : i( I I I ) i-1 ' I J -fi r

Hi: : n i;r( IN » j : [ ri.: :: FTTTTFT4 tSE: ; N T"|• t rmat 1

;Ft t I'M;Tn:( Ml( Mi l

pin 1

• Mt • [L ; ! * iIffl ihii •f 1-hrt 1 *

fltin y± m«=r n 1».i3 1 *-

R *

material of a prism by a: i -_to ..r ;m 44 M-SSSSSI TTJ

I X T '

isT n*i t:I !! 1i .1 , _: « h- -Hi rail nmtHi l 5 1Ts -r •: J

! IT 1 y r: hm .» 1 • 1 A1 1

»• LTJ1 :

11 hi? l7t ±h1it —nn! y T• i i - : I. 1

4 i -f _P.ii i :.; t:

From flie graph plotcd,ai point"p"

( i) Find iJ»o value o /‘-h;uidh-(ii)Find ihe value of local length of tens

i11 I: I.-Hi•• :h! hi I • 1:

ft* 4 • • ‘‘I l 1

I,if. 1 i:l ;l i t

.J }

GUJRANWALA 2013r r^: :r 1; r— - ^ » !R7^EAWCOURSEj11,nHoursMAXIMUM MARKSjM ^ ^^^1(<>n

_CLAS»5>)TIME ALLOWED: 3.10t

4 i PAPER-I4 L PHYSICSSUii^^par,number un

|

1rl question number fix * *» 16Write same••h i NOTE:-n\ 1 . T rvasiMiah/c umf riwinch eoald serve

Attempt any eight parts.Name several repetitive phenomenon occurring

Why do we find il useful tohave twounits for amount

Distinguish between base and derived units

How many nanoseconds in one year are there’

in nature and mole 'T0/|ifb*ranee, kilogramWNIT Q.N0.2

CO~ I . .«1 * 1|U\ Ml »L :l !i ' iS\ l I (ii)(iii)I I I W I.Ml .!*• Iwl.lllini I..n.yyH | 'C(iv)1 1 1 I " 'll lit.' k il||k |l| 111^ ^l.ll«tu.l '

ORAnswer ihe following queslion on ihe buiis of graph drawn below:-

past B XI (Subjective)43? PHYSICS -2?ra plar'sS*r, Scll«Define IIK* tarns unit vector and position vector.

( an you add zero to a null vector?( iive two conditions ol equilibrium.Define impulse and give its relation with momentum.Motion with constant velocity is a special case of motion with constant acv« * i,.r. •

true’*al *on.|s

Can velocity of an object reverse the direction when acceleration is constant? II s oExplain circumstances in which acceleration and velocity V p f a car ^, Ve un

( b) v is zero hut a is not zero

know about zero error?ieht and write its SI. unit .

( v )What do you

( vi ) ( i‘)Define wetgWhat is second pendulum9

What is the difference between

Define critical angle.Write down the brief procedure to show experimental

independent of amplitude. on

( vii )

( viii )( iii )

a real image and a virtual image?( iv )( v )( tx )

that time period of simple pendulum( v i )

( B)( x ) exanjp|c( X I I convex lens experimentally.the brief procedure to find out the focal length

What can you conclude from the graph?

Find the value of ‘g’ from the graph

a is zero but v is not zeroExplain bow swing is produced in a fast moving cricket ball.Attempt any eight parts.A girl drops a cup form certain height which breaks into pieces. What energy changes arcA boy uses a catapult to throw a stone which accidentally smashes a green housepossible energy changes.What is escape velocity and its mathematical expression?When the mud Mies off ( he tyre of a moving bicycle, in what direction does it fly?Why docs a diver Change his body position before and after diving in the pool?What are satellites and flow they move around the earth 9

Name two characteristics of simple harmonic motion.Can we realize an ideal simple pendulum?

( a )Write down

( xii )(CKOQ.No.3 « x 2

involved?window.

'16( i )( ii ) Li« Du( iii )( iv )( v )( vi )( vii )( viii )(ix ) Define the term resonance.

Differentiate between longitudinal waves and transverse waves.( xi ) Why docs sound travel faster In solids than in gases?(xii) How arc beats useful in tuning the musical instruments?

Attempt any six parts.( i ) Define wave front and spherical wave front.( ii ) How would you manage to get more orders of spectra using a diffraction grating?( iii ) Under what conditions two or more sources of light behave as coherent sources?( iv ) One can buy a cheap microscope for use by the children . The images seen in such a microscope ha *t

couloured edges. Why is this so?(v) Why would it be advantageous to use blue light with a compound microscope?

^fviK Derive Boyle’s law from kinetic theory of gases.( vii ) Define reversible process and irreversible process.( viii ) Docs entropy of a system increase or decrease due to friction9

Why does the pressure of a gas in a car lyre increase when it is driven through some distanceSECTION-11

5/a ) Define scalar product of two vectors and give its four characteristics.( h i A football is thrown along an angle of 30^ with respect to horizontal to throw 40 m

the initial speed of the ball?6 (a ) Describe work energy principle.

fb) A 1000 kg car travc ling with a speed of 144 km IT * round a curve of radius^lOO m.centripetal force. . 9

7 (a; What is drawback of Newton s formula for speed of sound in air? How Laplace correct 1j 4

where g s^

( x ) -

6 x 2 = 12Q.No.4

(I) What do you infer from the graph?

Find out the focal length of a lens from the graph.Graph B(ii)

( ix )

1.4

pass. Whaim* 1*35

Find nece**)

3

BAHAWALPUR 2013= — ^ — — — — j sUH

( b ) A simple pendulum is 50.0 cm long What will be its frequency of vibration at a P*ace

m/* ?3

TIME ALLOWED: JrINTERMEDIATE PAH(NEW COURSE)

HoursSUBJECTIVE MAXIMUM MARKS

number and Us part number on an

I-l (1 i in54/a ) Describe the Michel .son s experiment for calculation of speed of light.

fb > In a double sht experiment, the second .Mcr ma imurr, recurs at 0 = 0.25°. The waDetermine the slit separation.

Z/a / Explain isothermal process and adiabatic procc( b ) What gauge pressure is required in the city mains for a stream from a fire hose connc

reach a vertical height of 15 0m?

is 650 «*® /velength PAPER-1PHYSICS : &33 given in5 .ns *

c.ed .o ^ Write same questionNOTE:-paper.

SECTION III (PI ACTICAL) 4 x 2U A ) Writ* short answers to any four questi , n

What is vernier constant?( b

\t*

434435rHv*lCS- Xl (Subjective)

lar'sSECTION 11

5e Projectile Motion. Derive expression for Height of Projectile and Time of Flight .De inCt ons of two aeroplanes at any instant are represented by two points A (2. 3. 4) and B (5. 6. 1) in,b’ Tm. Calculate the distance between the two aeroplanes. 3

in interconversion of P H. and K..E. ma hill of heicht 10.0 m. If the disc starts from rest at the lop of the hill.

808- Intcr ( Part - I )

Session (2012-14 )

New Marks Schem.MarksHours

Note: It is compulsory lo attempt (8 ) parts each from Q. No.2 and 3 while attempt any 6 n- n ~lattempt any (03) questions from Part II . While attempt Pan III ( Practical pan )'

instructions.

Roll No.

5. (a )Inter-A -2013Physics (Subjective ) = S3 Time : 3. lo

Explain( bl A disc without slipping rolls down

what is its speed at the bottom.Wha, is Doppler's Effect - What changer* Frequency

away and towards the observe^A. Explain .

(b) A block of mass 4 kg is dropped from it height ciCfl.SOw

through Jvhich the spring will be.compressed.Write dftwn iis working and magnification.

6. (a )r°m Q.Jyj8 !o theaccordin

:y Rec

mop ,eiving by observer when source is movinggi'rr.

5Write same questions no. and us pan no. as given in the question paper 7. ( a )

lo a spring of spring constant K = 1960 N.nrfSECTION I3Q.No.2 Attempt any eight parts.

Define Physics, significant figuresDefine Radian. Stcradian.

I . find the maximum distanWhat is Compound Microsc

|th 0J 50 nm a, crystal . Wha

5Xx 2 = l 6( i )

first order reflection at a Bragg angle of 13.3°8. (a )( ii )observed to undergo a; the interplaner spacing of the reflecting planes in the crystal?( h ) X-rays of Wav _

froni a quart/ (Si02)( iii ) Can you add zero to a null vector?(iv ) Define Displacement , instantaneous velocity.( v ) Slate Newton’ s First and Third Law of Motion.( vi ) Define Impulse and Isolate System.( vii ) Write the Dimension of ( i ) Pressure ( ii ) Density( viii ) Give the drawbacks to use the period of a pendulum as a time standard( ix ) Can ( lie magnitude of a vector have a negative value?( x ) Can a vector have a component greater than the vector’s magnitude?

At what point or points in its path docs a projectile base it' minimum speed, its maximum speed?( xii ) Why Fog droplets appear to be suspended in air ’’Attempt any eight parts.Calculate the work done in kilo joules in lifting a mass of 10 kg through height of 10 m.When a rocket re-enters the atmosphere it ’ s nose cone becomes scry hot '* Where does this heait .comes from?What is meant by Moment of Inertia? Explain its significance?

( iv ) Explain the difference between Tangential Velocity and Angular Velocity( v) What are Geo-Stationary Satellites?

Name two characteristics of Simple I larmomc Motion.( vii ) Explain the terms Node and Antinode.

( viii ) Differentiate between free and Forced Oscillations.

What is total distance travelled by an object mo\ ing with SHM in aA A \|( x ) What features do longitudinal waves have in common with tran - '. erse waves ’

How astronomours use the Doppler effect to calculate the speeds of different stars?( xii ) How Pollution can be reduced?

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What function of sounding box ol sonometer is ?kind of vibration is executed b\ Sonometer?

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( vii ) Define Refraction of Light.( viii ) When Zero Error is positive and w hen it is negative?

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time equal to its period if ampli^6( ix )A?

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6 X 2 = 12Q.No.4 Attempt any six parts.Stale Huyge’s Principle9

How the power is lost in Optical Fibre?Under what conditions two or more sources ol ii jhi behave s C. iierent Sources .An oil film spreading over a wet foot th shows coh rs. How does it happen9

What is meant by Linear Magnification and Angular Magnification?( vi ) Derive Charles’ Law from Kinetic Theory of Gases. * '( vii ) Is it possible to construct a heat engine ihat will not expell heat into the atmosphere ,

( viii ) Whal happen to the temperate e of the m when an air conditioner is left running

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