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1. MEASUREMENTS AND EXPERIMENTATION 1—32(Includes Points to Remember, Checkpoint Questions and Problems)

2. CHEMICAL CHANGES AND REACTIONS 33—74(Includes Points to Remember, Checkpoint Questions and Problems)

3. LAWS OF MOTION 75—91(Includes Points to Remember, Checkpoint Questions and Problems)

4. GRAVITATION 92—113(Includes Points to Remember, Checkpoint Questions and Problems)

5. FLUIDS 114—139(Includes Points to Remember, Checkpoint Questions and Problems)

6. ARCHIMEDES’ PRINCIPLE AND FLOATATION 140—169(Includes Points to Remember, Checkpoint Questions and Problems)

7. HEAT AND ENERGY 170—177(Includes Points to Remember, Checkpoint Questions and Problems)

8. ENERGY SOURCES 178—198(Includes Points to Remember, Checkpoint Questions and Problems)

9. LIGHT 199—243(Includes Points to Remember, Checkpoint Questions and Problems)

10. SOUND 244—263(Includes Points to Remember, Checkpoint Questions and Problems)

11. CURRENT ELECTRICITY 264—276(Includes Points to Remember, Checkpoint Questions and Problems)

12. MAGNETISM 277—294(Includes Points to Remember, Checkpoint Questions and Problems)

CONTENTS

1Arundeep’s ICSE Can. Physics – 9th

Points to Remember1. Physical quantity : “Any quantity in terms of which some laws of physics can be stated

and which can be measured directly or indirectly.” Velocity, force, acceleration, work arephysical quantities.

2. Measurement of physical quantity :(i) Select a unit to measure that physical quantity.(ii) How many times that unit is contained in that physical quantity : P = nu.

3. Fundamental physical quantities : Three fundamental quantities in mechanics are mass,length and time and cannot be further resolved into simpler physical quantities. Total ofseven fundamental quantities in physics. Four more are temperature current, luminousintensity and quantity of matter. e.g., volume area etc. are fundamental quantities. Theycannot be derived from other physical quantities.

4. Derived physical quantities : “Which can be obtained from fundamental physical quantitiese.g., velocity, force, acceleration, work etc.

5. Derived units : Units of derived physical quantities.6. S.I. unit of mass is Kg.7. S.I. unit of length is metre (m).8. S.I. unit of time is second (s).

Vernier Calliper can measure length or depth or diameter upto 0.01 cm. SCREW GAUGEor MICROMETER upto 0.001 cm. Meter can measure correct upto 0.1 cm or 1 mm.

9. Zero error : If in a vernier calliper the zeros of the main and thevernier scale do not coincide when the jaws are placed in contact,then the vernier has an error called ZERO ERROR. Zero error isalways algebraically subtracted from observed reading.

10. Positive zero error : “When the zero of the vernier scale lies tothe right of zero of main scale, when jaws are in contact.”

11. Negative zero error : “When the zero of the vernier scale liesto the left of zero of main scale, when jaws are in contact.”

Measurements andExperimentation1

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2 Arundeep’s ICSE Can. Physics – 9th

12. Principle of vernier : “Is based on vernier scale. i.e., when vernier scale slides over mainscale M. n vernier divisions of vernier scale coincide with (n – 1) divisions of main scale.”

13. Least count of vernier scale : “The difference between the value of one main scaledivision and one vernier scale division.” To find L.C. MathematicallyLC = 1 MSD – 1 VSD.Let n vernier scale divisions coincide with (n – 1) main S.D.

\ nVSD = (n – 1) MSD

1 VSD = ÷øöç

èæ -

nn 1

MSD

\ LC = 1 MSD – ÷øöç

èæ -

nn 1

MSD =n1

MSD

LC = V.S.theondivisionsofNumberscalemaintheondivisionSmallest

14. Method of using vernier scale :(i) Take the main scale reading (N) before on the left of the zero of the vernier scale.(ii) Find the number (n) of the vernier scale division which coincides with any division on

the main scale. Multiply this number with the vernier constant. i.e., n × VC(iii) Final observation N + n × VC.

15. Vernier Calliper :

16. Illustration to find LC : An instrument has 20 divisions in 1 cm and a VS with 25 VSdivisions coinciding with 24 divisions on the MS. find the LC of instrument.

Solution : 1 MSD =201

= 0.05 cm

LC = 1 MS – 1 VSD, Also 25 VSD = 24 MSD


\ 1 VSD =2524

MSD

LC = 1 MSD –2524

MSD =25

24MSDMSD25 - =

251

MSD =251

MSD × 0.05 = 0.002 cm

17. Zero error of SCREW GAUGE : “If the zero mark of circular scale does not coincidewith the zero of the pitch scale when the faces are touching each other. Such instrumentspossess zero error or zero correction.”

18. Positive zero error : “If zero mark of the circular scale lies below the reference line onthe MS then error is said to be a positive error.” It is subtracted.

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10

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19. Negative zero error of MICROMETRE SCREW : “If the zero of the circular scaleadvances beyond the reference line or is above the reference line, the zero error is negativeand zero correction is positive i.e., it is added.”

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40

20. Pitch : “The pitch of the screw gauge is the distance moved by the screw in one completerotation of its head.”

OR“The distance between two consecutive threads of screw measured along the axis of

screw.”

PITCH = thimbleofrotationsofNumberMSonthimblethebymovedDistance

.

21. LC (Least Count) : “The least count of screw gauge is the distance moved by its screwin rotating the circular scale by one division.”

LC = scaleheadondivisionsofnumberTotalscrewtheofPitch

.


22. Neat well labelled diagram of screw gauge :

23. Purpose of ratchet in screw gauge : Good instruments are provided with a ratchetattached to the screw by a spring.

The screw is moved by turning the ratchet. In the position when the flat end B of thescrew is in contact with the end A, further movement of ratchet does not press B againstA.

24. Back lash error : Some times due to wear and tear of the threads of screw, it is seen thaton reversing the direction of rotation of the thimble, the tip of the screw does not startmoving in the opposite direction immediately, but remains stationary for a part of rotation.This is called ‘back lash error’.

To avoid back lash error while taking the measurements, the screw should be rotatedin one direction only.

25. Physics is a branch of science that deals with nature and laws governing it.26. Since least count is of order 10 m so the screw is called micrometre screw.27. The metre is the length of path travelled by light in vacuum during a time interval of

2997924561

s.

28. The time period of a pendulum does not depend upon mass of bob so long as length ofpendulum remains same.

29. An ideal simple pendulum consists of heavy point mass called bob tied to one end of aperfectly inextensible, flexible and weight less string.

30.


31. When pendulum moves from extreme position B to C and then back to B one completeoscillation is completed.

32. Length of pendulum : “Is the distance between the point of suspension and the point ofoscillation.” It is represented by ‘L’.

33. Time period of pendulum T = 2p gL

and frequency V =T1

=π2

1Lg

34. Time period of simple pendulum depends upon(i) Directly on square root of length.(ii) ‘g’ the acceleration of gravity. Since ‘g’ varies from place to place. Time period will

be different at different places.(iii) Time period does not depend on (a) mass of bob (b) amplitude of vibration. So long

as it remains small.35. ISOCHRONOUS VIBRATIONS are one which always has the same time period of

vibration for small amplitude.36. Screw is actually an inclined plane wound into a spiral around a straight rod. Winding is

called a thread.

Check Point

Q.1. What is the advantage of Vernier callipers over a regular scale graduated inmillimetres?

Ans. A Vernier calliper give correct reading upto 0.1 mm, where as regular graduated scalecannot measure length less than 1 mm.

Q.2. What is the principle of Vernier?Ans. Vernier is based on Vernier scale. i.e., (n – 1) main scale divisions coincide with n vernier

divisions.Q.3. What do you mean by least count of Vernier calliper?Ans. Least count of Vernier calliper is the difference between one M.S.D. and one (V.S.D.)

Vernier scale division.”Q.4. What is the use of lower jaws of a Vernier calliper?Ans. Vernier calliper is used to measure the length upto 0.01 cm.

It is used to measure the external diameter of spherical body upto 0.01 cm.Q.5. What is the use of upper jaws?Ans. Use of upper jaw is to measure the internal diameter of a hollow cylinder or pipe.Q.6. What is the use of the thin strip moving behind the main scale?Ans. Thin strip moving behind the M.S. is used to measure depth of a small beaker or bottle.


Q.7. Two different Vernier callipers have different number of equal divisions on theirVernier scales (i) 10 equal divisions (ii) 50 equal divisions. Which can measuremore accurately.

Ans. (i) L.C. = 1 M.S.D. – 1 V.S.D.

1 –109

M.S.D. =101

= 0.1 mm = 0.01 cm

(ii) L.C. of = 1 –5049

=501

mm =100

2mm = 0.02 mm = 0.002 cm

\ (ii) is more accurate.More division on V.S. more is accurate.

Q.8. What is the zero error of a Vernier calliper?Ans. Zero error in Vernier calliper : “Is equal to the distance between the zero of the main

scale and zero of the vernier scale.”OR

“If zero of V.S. does not coincide with the zero of M.S. when jaw B touches the jawA and the straight edge of D touches the straight edge of C, then instrument has an errorcalled zero error.” Zero error is always algebraically subtracted.

If zero of the V.S. lies to the right of zero of M.S. it is positive zero error and if zeroV.S. is to the left of zero of M.S., it is negative zero error.

Q.9. What is the formula of least count?Ans. Formula of LC is 1 M.S.D. – 1 V.S.D.

= scaleVerniertheondivisionsofNumberscalemiantheondivisionSmallest

.

Q.10. What is the formula to calculate total reading with a Vernier calliper?Ans. Formula is N + n + VC

Where N is the main scale reading before on the left of zero of V.S.n is V.S. division which coincides with any division on M.S.V.C. is the vernier constant.

Q.11. How many types of zero errors will be possible in Vernier calliper?Ans. Two types of zero errors are possible :

(i) Positive zero error : “If zero of V.S. lies to the right of the zero of M.S.(ii) Negative zero error : If zero of the V.S. lies to the left of zero of the M.S.

Q.12. What is Positive zero error, how do we correct such an error in a Vernier calliper?Ans. If zero of vernier scale lies to the right of zero of main scale, the error is positive zero error.

This error is subtracted from the observed reading.Q.13. What is Negative zeror error, how do we correct such an error in a Vernier calliper?Ans. If zero of vernier scale lies to the left of zero of main scale, the error is negative zero error.

It is to be added to the observed reading.


Check Point

Q.1. What is a screw?Ans. A screw is actually an inclined plane wound into a spiral around a straight rod. Winding is

called a thread.Q.2. What is meant by “guage”?Ans. Gauge means an instrument.Q.3. Name two main parts of a screw gauge.Ans. Two main parts of a screw gauge are :

(i) Pitch scale or linear scale.(ii) Head scale or circular scale.

Q.4. What is meant by pitch of a screw?Abs. Pitch : The distance moved by the screw in one complete rotation of its head.”

ORThe linear distance moved by the screw, when it is given one complete rotation is equal

to the distance between two consecutive threads as measured along the axis of the screw.”

Q.5. How is the pitch found?Ans. Pitch of a screw gauge is the distance moved by screw along its axis in one complete

rotation of its head.Q.6. What is the least count (LC) of the screw gauge?Ans. Least count is the distance moved by the screw on the main scale when screw is given one

complete rotation.Q.7. How the LC of a screw gauge is found?

Ans. Mathematically : LC = scalecircularondivisionsofNumberPitch

=100

mm0.5orm1 = 0.001 cm or 0.0005 cm

Q.8. What is meant by zero error of a screw gauge?Ans. Zero error : If the zero mark of circular scale does not coincide with the zero of the pitch

scale when the faces are just touching each other the instrument posses zero error and zerocorrection.

Q.9. When is the zero error is positive?Ans. Positive zeror error is the zero mark of the circular scale lies below the reference line of


the main scale, then the error is positive zero error.” It is always subtracted from theobserved reading.

Q.10. What is the degree of accuracy of the screw gauge?Ans. Degree of accuracy of screw gauge is 0.001 cm.

Q.11. What is back lash error?Ans. Some times due to wear and tear of the threads of the screw, it is seen that on reversing

the direction of rotation of the thimble, the tip of the screw does not start moving in theopposite direction immediately, but remains stationary for a part of rotation. This is called‘back lash error’.

Q.12. How back lash error is avoided?Ans. To avoid back lash error while taking the measurements, the screw should be rotated in one

direction only.

Check Point

Q.1. Define simple pendulum.Ans. ‘A simple pendulum is a simple device for measuring time.”

OR“A simple pendulum consists of a heavy point mass (called bob) suspended from a rigid

support by a perfectly inextensible, flexible and weightless string.”Q.2. Define the following terms : (a) amplitude (b) oscillation (c) frequency (d) time

period.Ans. (a) Amplitude : The maximum displacement of the pendulum from the mean position on

either side.OR

“Distance between the mean and extreme position is called amplitude.” AB orAC = amplitude it is represented by ‘a’.

A

B

Effectivelength

Point of suspensionO

BOB

C

l

(b) Oscillation : One complete to and fro motion of the pendulum is called one oscillation.OR


If the pendulum moves from one extreme position (B) to other extreme position(C) and back to initial position (B). It is one oscillation. B to C then C to B is oneoscillation.

(c) Frequency : It is the number of oscillations or vibrations made in one second. It isdenoted by “f ” or :n”.

(d) Time period : The time taken to complete one oscillation is called time period. It isdenoted by ‘T’.

Q.3. Time period will increase or decrease if we use a heavier bob.Ans. Heavier bob means more massive. Mass does not effect the time period.Q.4. Can you replace the thread by a metallic wire or an elastic band?Ans. No, length does not remain same. The string made of metal is not weightless. wt. of string

increases. We cannot use elastic band as length may stretch.Q.5. From where the length of the pendulum is measure?Ans. From the point of suspension.Q.6. How would the period of a simple pendulum be affected if it were located on the

moon instead of the earth?

Ans. T = 2p gL

on the surface of moon g is61

the of g on moon.

Hence, time period will increase.Q.7. What effect would the temperature have on the time kept by a pendulum clock if

the pendulum rod increases in length with an increase in temperature?Ans. Time period T of pendulum is directly proportional to the square-root of length. Length of

pendulum increases with increase in temperature. Hence, time period increases and clockruns slower.

Q.8. What kind of graph would result if the period T were graphed as a function of thesquare root of the length, L .

Ans. T µ L

\ T2 = L or 2TL

is constant.

Graph showing the variation of T with l.

If we plot a graph for time period (T) taken on y-axis and square-root of length )( l takenon x-axis, it comes out to be a straight line.

l(0, 0)

CT

A

B x

y


Q.9. What effect does the mass of the bob have on the period of a simple pendulum?What would be the effect of replacing the steel ball with a wooden ball, a lead ball,and a ping pong ball of the same size?

Ans. Time period of a pendulum does not depend upon mass of the bob. When steel ball isreplaced by wooden ball, still there will be no effect on time period so long as length ofpendulum remains the same.

Q.10. A pendulum with a string of length 2 m has a period of about 2.8 s. What would bethe period of the pendulum if the length of the string were increased to 8 m?

Ans. T = 2p gL

= 2p g2

when l = 2 m

2.8 = 2p g2

...(i)

When l = 8 m

T¢ = 2p g8

= 2p g22

= 2 úúû

ù

êêë

ég2π2

= 2 × 2.8 [From (i)]T¢ = 5.6 s

Q.11. A simple pendulum swings with a period of 1.5 s. What would the period of thependulum be if the length of its string were doubled, the mass of its bob were cutin half, and the force of gravity were doubled?

Ans. T = 1.5 s

T = 2p gl

1.5 = 2p gl

...(i)

T is independent of mass of bob.When l¢ = 2l g¢ = 2g

\ T¢ = 2p gl

22

= 2p gl

T¢ = 1.5 s from (i)\ Time period will remain the same.


Problems1. QUESTIONS

Q.1. What do you understand by the term scientific notation?Ans. Scientific notation is a way of expressing numbers that are too small or too big to be written

in decimal form. e.g., 1 year = 365 days.1 year = 365 × 86400 s = 31536000 s = 3.156 × 107 s.

Q.2. What do you understand by the term measurement?Ans. “Measurement” is the process of comparison of the given physical quantity with the known

standard quantity of the same nature.Q.3. What is the need for measurement?Ans. Without measurement, you cannot make a correct judgement about the length, area, volume,

mass and temperature of a given object. A guess or rough estimate may give an entirelywrong answer.

Q.4. What is a physical quantity?Ans. Force, velocity, speed, acceleration, work are physical quantities. A physical quantity is one

which can be measured directly or indirectly and in terms of which some laws of physicscan be stated.

Q.5. What is a unit of measurement?Ans. A unit is quantity of constant magnitude which is used to measure the magnitudes of other

quantities of the same nature.OR

A unit of a physical quantity is defined as a reference standard of same kind used tomeasure it.

P = nu.Measure of a physical quantity = Numerical value × Size of unit.

Q.6. What steps are involved in measuring a physical quantity?Ans. Following two steps are involved in measuring a physical quantity.

(i) Selecting of a unit to measure that physical quantity. Let it be (u).(ii) Finding how many times that unit is contained in that physical quantity. Let it be n.\ Physical quantity P = nu.

Q.7. What are fundamental and derived physical quantities?Ans. Fundamental physical quantities are those quantities which cannot be derived from other

physical quantities, neither from themselves and cannot be further resolved into simplerphysical quantities. e.g., mass, length time, temperature, current, luminous, intensity andquantity of matter.

Derived physical quantities are the physical quantities which can be derived or obtainedfrom fundamental physical quantities.


Examples are velocity =time

distance. It is derived from distance and time both fundamental

quantities.

Force is mass × acceleration. Acc. istime

velocity =

timetime

distance = 2time

distance

\ F = 2tdm´

Here, distance, mass, time all are fundamental quantities.Q.8. What are fundamental and derived units?Ans. Fundamental unit or Basic unit is that which is independent of any other unit or which

can neither be changed nor can be related to any other fundamental unit. e.g. unit of mass,unit of length, time and temperature.Derived units are those which can be expressed in terms of fundamental units.Example : Force = Mass × Acceleration

= mass ×time

velocity

= mass ×timetime

distance

Force = mass × 2(time)length

Q.9. Name the three fundamental quantities used in mechanics.Ans. Fundamental quantities used in mechanics are :

(i) Mass (ii) Length(iii) Time.

Q.10. What is a system of units?Ans. System of units : International bureau of weights and measures, the members of this treaty

have met over the years to define the standard units more precisely and achieve higherlevels of accuracy.There are four systems of units.(i) CGS system or french system : The unit of length is centimeter, the unit of mass

is gram and the unit of time is second.(ii) FPS system or British system : The unit of length is foot, unit of mass is pound and

unit of time is second.(iii) MKS system or metric system : The units of fundamental quantities of mass length

and time are kilogram (Kg) metre (m) and second (s) respectively.


(iv) The MKSA system : This is the modified MKS system of units in which another unitfor measurement of current was introduced. In this system the units of the fundamentalquantities of mass, length, time and current are kilogram (Kg), metre (m), second (s)and ampere (A) respectively.

Q.11. Name the system of units being used worldwide.Ans. System international or SI (System International d’ unites) : It is also a french system

of units and is an extended form of the MKS system. This system consists of seven basicand two supplementary units.

Q.12. Name the seven fundamental units of SI. Also, write their symbol.Ans. Seven fundamental units :

Basic physical quantity Name of unit Symbol of unit

1. Mass Kilogram Kg2. Length Metre m3. Time Second s4. Temperature Kelvin K5. Electric current Ampere A6. Luminous intensity Candela Cd7. Quantity of matter Mole mol

Q.13. Define one metre.Ans. It is SI unit of length.

One metre : The metre is the length of the path travelled by light in vacuum during the

time interval of299792458

1 of a second.

Q.14. Give the SI definition of a kilogram.Ans. Kilogram : It is equal to the mass of the international prototype of the kilogram.

OROne kilogram is the mass of a specific platinum–iridium alloy cylinder (90% platinum

and 10% of iridium), 3.9 cm in height and 3.9 cm in diameter, kept at the international Bureauof weights and measures at severs, France.

Q.15. Define time. What is its SI unit?Ans. SI unit of time is second (s). The second is the duration of 9 192 631 770 periods of the

radiation corresponding to the transition between two hyperfine levels of the ground stateof the caesium 133 atom.

Q.16. Name a convenient unit you will use to measure (i) width of your textbook (ii) massof a pencil (iii) mass of a cricket ball (iv) height of your desk (v) width of a chair.

Ans. (i) width of your textbook — cm(ii) mass of a pencil — gram(iii) mass of a cricket ball — gram


(iv) height of your desk — cm(v) width of a chair — cm

Q.17. Mention some points which you keep in mind while representing a unit and itssymbol.

Ans. The following points should be kept in mind while writing the symbols and units of a physicalquantity.(i) The units named after a scientist are written or should begin with a small initial letter.

For example, the unit of energy is written as joule and not Joule.(ii) The units do not have plurals, i.e., the unit of measurement of force is newton and not

newtons. Similarly, the unit of mass is gram and not grams. The symbol for gram isg and not gms.

(iii) The symbols of units named after scientists have initial capital letters.(iv) If the unit is not named after a scientist, then smaller letters are used for its abbreviation.(v) Always index of notation is used to express a unit i.e., metre per second should be

written as m s–1 and not as m/s.(vi) No full stop is placed after a unit.(vii) Space should be left between the numerical value and the unit.

e.g., Proper : 10 m s–1, Improper : 10ms–1.Q.18. Explain the term least count of an instrument.Ans. It is the minimum correct measurement we can make with an instrument.

With Vernier scale it is 0.1 mm.With screw gauge it is 0.01 mm.

Q.19. Explain how you will measure the length of an object correctly with the help of ameter scale?

Ans. We can measure the length of an object with the metre scale correctly upto one mm.Place the metre scale along the object, such that one and (A) of the object coincides

with the zero of the metre mark of the metre scale. This can be achieved with a fair degreeof accuracy by keeping the line of sight vertical.

01 2 3 4

metre scale

Object

To avoid the error due to parallax, some metre scales are made very thin or havebevelled edges.


Now shift your line of sight to the other end (B) of the object. Looking vertically downfind graduation on the metre scale, which coincides with the edge of object. This graduationgives the length of the object AB.

Q.20. What is the principle of a Vernier?Ans. Principal of Vernier : Auxiliary scale or Vernier scale slides along main scale and n VSD

coincide with (n – 1) MSD.Q.21. What is the Vernier constant or least count of a Vernier? How will you find it?Ans. The LC of the Vernier is defined as the difference in the value of main scale division and

one vernier division.To find it : LC = 1 MSD – 1 VSDn VSD = (n – 1) MSD

1 VSD = ÷øöç

èæ -

nn 1

MSD

\ Least count is LC = 1 MSD – 1 VSD

= 1 MSD – ÷øöç

èæ -

nn 1

MSD =n1

MSD

\ Least count = scaleVerniertheondivisionsofNumberscalemiantheondivisionSmallest

.

Q.22. What is a Vernier calliper? Why is it named so?Ans. The Vernier calliper is an instrument which is used to measure lengths or depths upto a

minimum of 0.01 cm.It is named so as it derives its name from French mathematician who devised the

Vernier scale. It is called calliper because it has two jaws or hinged arms.Q.23. Draw a labelled diagram of a Vernier calliper, showing its main parts.Ans.

Q.24. Describe in steps how you will use a Vernier calliper to measure the thickness ofa piece of chalk.


Ans. Measurement of thickness of a piece of chalk : To measure the thickness of chalk thefollowing procedure is adopted :(i) Find the LC and zero error of the Vernier.(ii) Fix the chalk between the jaws A and B to find its thickness.(iii) Note the main scale reading.(iv) Note the Vernier division which coincides with any division of the main scale. Multiply

this number Vernier division with the least count. This is the Vernier scale reading.(v) Add the main scale and Vernier scale readings. From this addition, subtract the zero

error with its proper sign to obtain the correct measurement of the thickness of chalk.Observed thickness = Main scale reading + (VD coinciding with any division on

the MS) × LC.Thickness = Observed thickness – Zero error with sign.

Q.25. What do you understand by the term zero error of a Vernier calliper. Illustrate thiswith the help of well labelled diagrams.

Ans. Zero error in Vernier calliper : “Is equal to the distance between the zero of the mainscale and zero of the vernier scale.”

OR“If zero of V.S. does not coincide with the zero of M.S. when jaw B touches the jaw

A and the straight edge of D touches the straight edge of C, then instrument has an errorcalled zero error.” Zero error is always algebraically subtracted from observed reading.

If zero of the V.S. lies to the right of zero of M.S. it is positive zero error and if zeroV.S. is to the left of zero of M.S., it is negative zero error.Illustration :Positive zero error. On bringing both the jaws together, if the zero mark of the vernierscale is on the right of the zero mark of the main scale, the zero error is said to be positive.It is equal to the distance between the zero of the vernier scale from the zero of the mainscale.Fig. shows a vernier callipers with positive zero error.

0

0 1 2 3 4 5 6 7 8 9 10

1cm

MAIN SCALE COINCIDING DIVISION

VERNIER SCALE

Positive zero error.How Zero Error is determined ?

To find this error, we note that division of the vernier scale which coincides with any divisionof the main scale. The number of this vernier division multiplied by the least count of the


vernier, gives the zero error.For example, for the scale shown in Fig. above, the least count is 0.01 cm and the 6th divisionof vernier scale coincides with a main scale division.

\ zero error = + 6 × L.C. = + 6 × 0·01 cm = + 0·06 cm.Negative zero error. On bringing both the jaws together, if the zero mark of the vernierscale is to the left of the zero mark of the main scale, the zero error is said to be negative.Fig. below shows a vernier callipers with negative zero error.

0

0 1 2 3 4 5 6 7 8 9 10

1cm

VERNIER SCALE

MAIN SCALE COINCIDING DIVISION

How Zero Negative Error is determined ?To find this error (i.e., the distance between the zero of main scale and zero of vernierscale), we note that division of the vernier scale which coincides with any division on themain scale. The number of this vernier division is subtracted from the total number of divisionson the vernier scale and then the difference is multiplied by the least count.In Fig. the least count is 0·01 cm and the sixth division of the vernier scale coincides with acertain division of the main scale. The total number of divisions on vernier are 10.

\ zero error = – (10 – 6) × L.C. = – 4 × 0·01 cm = – 0·04 cm.Q.26. State two uses of a Vernier calliper.Ans. Two uses of Vernier calliper :

(i) The lower jaws called the outside jaws are used to measure the length of rod, diameterof sphere or external diameter of a cylinder.

(ii) A thin long strip T attached to the Vernier strip at the back of the main scale is usedto measure the depth of beaker or bottle.

Q.27. What is the principle of a micrometre screw?Ans. It works on the principle of micrometre screw or that when one complete rotation is given

to the screw, the distance moved (linear distance) by the screw is equal to the distancebetween two consecutive threads.

Q.28. Explain the following terms with reference to a screw gauge (i) Pitch (ii) Leastcount.

Ans. (i) Pitch : The distance moved by the screw in one complete rotation of its head.(ii) Least count is the distance moved by the screw on the main scale when screw is

given one complete revolution.Q.29. Draw a well labelled diagram of a screw gauge.


Ans.

Q.30. What is the purpose of the spring and ratchet arrangement in a screw gauge?Ans. Ratchet is attached to the screw by a spring. Screw is moved by turning the ratchet.

In the position when the flat end B of the screw is in contact with the end A, furthermovement of the ratchet does not press B against A.

Q.31. What do you understand by the term zero error of a screw guage. Illustrate thiswith the help of well labelled diagrams.

Ans. Zero error : When the zero mark of circular scale does not coincide with the zero of thepitch scale when the faces are just touching each other, there is zero error or zero correctionoccurs in the instrument.

0

50

40

10

450

50

40

5

45

Nil zero error 2 positive zero error

0

50

40

10

45

–3 zero errorPositive zero error : If zero mark of the circular scale lies below the reference line onthe main scale, the error is said to be positive.Negative zero error : If the zero mark of the circular scale lies above the reference line,the zero error is negative.

Q.32. What is backlash error in a screw gauge? How is it removed?Ans. Some times due to wear and tear of the threads of the screw, it is seen that on reversing

the direction of rotation of the thimble, the tip of the screw does not start moving in theopposite direction immediately, but remains stationary for a part of rotation. This is called‘backlash error’.


To avoid backlash error while taking the measurements, the screw should be rotatedin one direction only.

Q.33. Describe the procedure you will adopt to measure the diameter of a wire by usinga screw gauge.

Ans. Measurement of diameter of a wire with a screw gauge.To measure the diameter of a wire with the help of a screw gauge, following procedureis adopted :(i) Find the least count and zero error of the screw gauge.(ii) Place the wire in between the stud A and the flat end B of the screw. The ratchet

is turned clockwise so as to hold the given wire gently between the stud A and theend B of the screw.

(iii) Note the main scale reading.(iv) Note the division of the circular scale which coincides with the base line of the main

scale. This circular scale division when multiplied by the least count, gives the circularscale reading.

(v) Add the circular scale reading to the main scale reading to obtain the total reading(i.e., the observed diameter of the wire).

(vi) From the total reading, subtract the zero error (if any) with its sign to get the correctmeasurement.Thus,Observed diameter = main scale reading + (circular scale division coincidingthe base line of main scale) × least count.True diameter = observed diameter – zero error (with sign).

Q.34. Explain with reference to a screw gauge (i) Linear or Pitch scale (ii) Circular orHead scale (iii) Stud (iv) Hub and (v) Reference line.

Ans. (i) Pitch scale : On the tabular hub along its axis, is drawn a line which is graduated inmm or half mm depending upon pitch of the screw is called reference line and thisscale is called linear or pitch scale.

(ii) Circular or head scale : the bevelled surface of cap is divided into 50 or 100 equalparts. This is called circular or head scale.

(iii) Stud : Screw gauge having U shaped metal frame. At one end of it is fixed a metalpiece of gun metal and has a plane face. This is called stud.

(iv) Hub : The other end of U shaped metal frame is called hub.(v) Reference line : On the tabular hub along its axis is drawn a line known as reference

line.Q.35. Name the instruments you will use for measuring the following (i) The thickness

of a sheet of paper. (ii) Diameter of a pin. (iii) The diameter of a marble. (iv) Thediameter of a pencil.

Ans. (i) Screw gauge. (ii) Screw gauge(iii) Vernier calliper. (iv) Vernier calliper.

Q.36. What is an ideal simple pendulum?


Ans. An ideal simple pendulum consists of a heavy point mass (called bob) tied to one end of aperfectly inextensible, flexible and weightless string.

Q.37. Define and illustrate the following terms as used for a simple pendulum (i) pointof suspension (ii) point of oscillation (iii) Length of the pendulum and (iv) Oneoscillation.

Ans. (i) Point of Suspension is the point on rigid support from where the pendulum issuspended. ‘O’ is the point of suspension.

C

A

Effectivelength

Point of suspensionO

Point of oscillation

B

l

(ii) Point of oscillation : ‘C’ is the point of oscillation and it is the centre of the bob wherethe centre of gravity of bob lies.

(iii) Length of pendulum : ‘l’ the distance OC is called the length of pendulum.(iv) One oscillation : When pendulum moves one extreme A to other extreme B and

back to A is called one oscillation.Q.38. What is time period of a simple pendulum? What are the factors on which it

depends? Write an expression for it.Ans. Time period is the time taken to complete one oscillation.

Factors on which time period depends :(i) Square root of length of pendulum T µ L .

(ii) Inversely proportional to square root of acceleration due to gravity. T µ g1

.

Expression : T = 2p gl

.

Q.39. Explain how you will measure the time period of a simple pendulum.Ans. To measure the time period of a given pendulum, the bob is slightly displaced from its mean

position and is then released. It begins to move to-and-fro about its mean position in verticalplane along with the string. Now time (t) for 20 complete oscillations is noted with the helpof a stop watch. t is divided by 20 to obtain time period (T).

Q.40. Draw a graph showing the variation of time period of a simple pendulum with its


length. How will you find the value of g by using this graph?Ans. Time period (T) of a simple pendulum depends directly to the square root of length ( )l

i.e. T µ lor the square of time period T2 is proportional to the length l of the pendulum i.e. T2 µ l

or ÷øöç

èæ

2Tl

is constant

Graph Showing the variation of T with l :

l(0, 0)

CT

A

B x

y

Graph for T against lIf we plot a graph for time period (T) taken on y-axis and square root of length ( l ) takenon x-axis, it comes out to be a straight line.To determine the value of ‘g’ from graph : Take a pt. A on the line. Draw ^ AB andAC on x-axis and y-axis to find the corresponding values l and T..Calculate the value of ‘g’ the acceleration due to gravity from the relation

T = 2p lg or g l= 4 2p

T2

Q.41. A girl is swinging in a swing in the sitting position. How will the period of the swingbe affected if she stands up?

Ans. Is standing position centre of gravity rises up and effective length decreases.

Since T = 2p gl

.

\ T decreases and hence will swing slowly.Q.42. The bob of a simple pendulum is made of wood. How will its time period be affected

if the wooden bob is replaced by an identical bob of aluminium?Ans. Aluminium is denser than wood but time period of bob does not depend on mass of bob.

Hence, time period will not be affected.Q.43. The acceleration due to gravity decreases as we move away from the surface of the

earth. What will happen to the time period of a pendulum if the pendulum is taken


to the top of a mountain?

Ans. T = 2p gl

as ‘g’ decreases on the mountain T increases and hence time period increases.

Q.44. What happens to the time period of a simple pendulum if it is taken from the earthto the moon where the gravity is one-sixth that at the surface of the earth?

Ans. T = 2p gl

T is inversely proportional to square root of g.

\ T¢ = 2p6gl

= 2p gl6

= 6 T

Time period increases.Q.45. What happens to the time period of a simple pendulum if its length is decreased

to one-fourth of its original value?

Ans. T = 2p gl

T¢ = 2pg

l4 when length becomes

4l

T¢ = 2p gl

4 =21

÷÷ø

öççè

æglπ2 =

21

T

Time period reduces to half the original time period.Q.46. What is a second’s pendulum?Ans. Second’s pendulum is a pendulum whose time period is two seconds.

2. FILL IN THE BLANKS1. Physics is the branch of science which deals with and the governing

it.2. Measure of a physical quantity = × size of unit.3. A unit of a physical quantity is defined as a , of the same kind, used to

measure it.4. The metre is the length of the path travelled by light in vacuum during a time

interval of of a second.5. One kilogram is the mass of a specific iridium alloy cylinder, 3.9 cm in height

and 3.9 cm in diameter, kept at the International Bureau of Weights and Measures


at Sevres, France.6. In the common form the division of the Vernier are smaller in size than the smallest

division on the main scale M. The graduations are such that Vernier scaledivision (VSD) coincide with main scale divisions (MSD).

7. Zero error is positive when the zero of Vernier scale lies to the of the zerooof the main scale, when the two jaws of the Vernier are in contact.

8. Zero error is negative when the zero of Vernier scale lies to the of the zerooof the main scale, when the two jaws of the Vernier are in contact.

9. The linear distance moved by the screw, when it is given one complete rotation isequal to the distance between two consecutive threads, as measured along the axisof the screw. This distance is called the of the screw..

10. The least count of a screw is defined as the distance moved by the screw on themain scale when the screw is given complete revolution/s.

11. Since least count is of the order of so the screw is called a micrometrescrew.

12. If the zero mark of the circular scale of a screw gauge lies the reference lineon the main scale then the error is said to be a positive zero error.

13. If the zero of the circular scale of a screw gauge advances beyond the referenceline or is the reference line, the zero error is negative and zero correctionis positive.

14. An ideal simple pendulum consists of a (called bob) tied to one end of aperfectly inextensible, flexible and weightless string.

15. Length of a simple pendulum is the distance between the point of and thepoint of oscillation.

16. The pendulum is said to have completed one oscillation if it moves from oneposition to the other and back.

17. The time period of a pendulum does not depend upon the of the bob so longthe length of the pendulum remains same.

18. A second’s pendulum is one, whose time period is second.Ans.

1. Physics is the branch of science which deals with nature and the laws governing it.2. Measure of a physical quantity = numerical value × size of unit.3. A unit of a physical quantity is defined as a reference standard, of the same kind, used

to measure it.4. The metre is the length of the path travelled by light in vacuum during a time interval of

65457922991

of a second.

5. One kilogram is the mass of a specific platinum iridium alloy cylinder, 3.9 cm in height and3.9 cm in diameter, kept at the International Bureau of Weights and Measures at Sevres,France.


6. In the common form the division of the Varnier are smaller in size than the smallest divisionon the main scale M. The graduations are such that n Vernier scale division (VSD) coincidewith (n – 1) main scale divisions (MSD).

7. Zero error is positive when the zero of Vernier scale lies to the right of the zero of the mainscale, when the two jaws of the Vernier are in contact.

8. Zero error is negative when the zero of Vernier scale lies to the left of the zero of the mainscale, when the two jaws of the Vernier are in contact.

9. The linear distance moved by the screw, when it is given one complete rotation is equal tothe distance between two consecutive threads, as measured along the axis of the screw.This distance is called the pitch of the screw.

10. The least count of a screw is defined as the distance moved by the screw on the main scalewhen the screw is given one complete revolution/s.

11. Since least count is of the order of 10 mm, so the screw is called a micrometre screw.12. If the zero mark of the circular scale of a screw gauge lies below the reference line on the

main scale then the error is said to be a positive zero error.13. If the zero of the circular scale of a screw gauge advances beyond the reference line or

is above the reference line, the zero error is negative and zero correction is positive.14. An ideal simple pendulum consists of a heavy point mass (called bob) tied to one end of

a perfectly inextensible, flexible and weightless string.15. Length of a simple pendulum is the distance between the point of suspension and the point

of oscillation.16. The pendulum is said to have completed one oscillation if it moves from one extreme

position to the other and back.17. The time period of a pendulum does nto depend upon the mass of the bob so long the length

of the pendulum remains same.18. A second’s pendulum is one, whose time period is two second.3. MULTIPLE CHOICE QUESTIONS1. Which is not a fundamental unit?

(a) metre (b) litre(c) kilogram (d) second

Ans. (b) litre2. The main scale of an instrument has 20 divisions in 1 cm and 24 divisions of main

scale coincide with 25 divisions of vernier. The least count of the instrument is :(a) 0.002 cm (b) 0.001 cm(c) 0.02 cm (d) 0.01 cm

Ans. (a) 0.002 cm3. A girl sitting on a swing stands up. The time period of the swing.

(a) Increases (b) remains same(c) Decreases (d) none of these

Ans. (c) Decreases4. Moon has no atmosphere and acceleration due to gravity at its surface is one-sixth


that at the earth’s surface. When a pendulum is taken from earth to moon surface,its time period will :(a) remains same (b) Decrease(c) Increase (d) Be infinite

Ans. (c) Increase5. Which of the following is a derived unit?

(a) mass (b) volume(c) length (d) time

Ans. (b) volume6. In Vernier calliper N divisions of Vernier scale coincide with (N – 1) divisions of

main scale in which length of a division is 1 mm. The least count of the instrumentin centimetre is :(a) N (b) N – 1

(c)10N

1(d)

N1

– 1

Ans. (c)10N

1

7. A boy is swinging on a swing. If another boy sits along with him without disturbinghis motion, then the time period of the swing will :(a) increase (b) decrease(c) remain same. (d) be doubled.

Ans. (c) remain same.4. NUMERICAL PROBLEMS1. A Vernier calliper has 20 Vernier divisions which coincide with 19 divisions on the

main scale. If one main scale division is 1 mm , find the least count and Vernierconstant of the calliper.

Ans. One main scale division = 1 mm

1 VSD =2019

MSD

LC = 1 MSD – 1 VSD

= 1 MSD –2019

MSD =20

1920- =

201

MSD

LC =201

× 1 = 0.05 mm

LC or vernier constant = VStheondivisionsofNumberMStheondivisionSmallest

=20mm1

= 0.05 mm


2. A Vernier calliper has 50 Vernier divisions which coincide with 49 divisions on themain scale. If one main scale division is 0.5 mm, find the least count and Vernierconstant of the calliper.

Ans. One main SD = 0.5 mm50 VD = 49 MSD

1 VD =5049

MSD

Vernier constant is 1 MSD – 1 VSD

1 MSD –5049

MSD =501

MSD

Vernier constant = L. constant =501

× 0.5 = 0.01 mm

LC = VStheondivisionsofNumberMStheondivisionSmallest

=500.5

= 0.01 mm

3. The main scale of a Vernier calliper reads 4.7 cm; the 3rd division on the Vernierscale coincides with a main scale division. The least count of the calliper is 0.1 mm.What is the reading of the Vernier calliper.

Ans. LC of calliper = 0.1 mm = 0.01 cmMain scale reading N = 4.7 cmn = 3rd

\ Reading of Vernier calliper = N + n × LC= 4.7 cm + 3 × 0.01 cn= 4.7 + 0.03 = 4.73 cm

4. The least count of a Vernier calliper is 0.01 cm and its zero error is – 0.03 cm. whilemeasuring the length of a rod, the main scale reading is 4.6 cm and the sixthdivision on the Vernier is in line with a marking on the main scale. Calculate thelength of the rod.

Ans. LC = 0.01 cmDivision coinciding n = 6thMain scale reading N = 4.6 cmLength of rod = N + n × LC= 4.6 + 6 × 0.01 = 4.66 cmZero correction = –0.03 cmCorrect length = 4.66 – (–0.03) = 4.69 cm

5. What is the ratio of the least count of an ordinary Vernier calliper to that of anordinary screw guage?

Ans. gaugescrewofLCVernierofLC

= cm0.001cm0.01


=100

1 ×

11000

=1

10 = 10 : 1

6. The pitch of a screw gauge is 0.5 mm. The number of divisions on the circular scaleis 100. What is the least constant?

Ans. LC = scalecircularondivisionsofNumberscrewofPitch

=100

mm0.5 = 0.005 mm

7. The zero reading of the circular scale of a screw gauge is 97. What is the zero errorin csd and in mm? What is the zero correction in csd and in mm? The least countis 0.01 mm.

Ans. Reading on circular scale = 97\ 0 of circular scale lies 3 points above the reference line.\ Zero error is –3 csd [circular scale divisions]

Number of divisions on circular scale = 100

LC =100mm1

= 0.01 mm (given)

\ Zero error in mm = –3 × 0.01 = –0.03 mmSince, zero error is always subtracted.

\ Zero correction is – (–3 csd) = +3 csd.Zero correction = –(–0.03 mm) = +0.03 mm

8. The observed reading of the thickness of a glass plate measure with a screw guageis 2.33 mm. The zero error is - 0.03 mm (negative). What is the correct thicknessof the plate?

Ans. Observed thickness = 2.33 mmZero correction = –0.03 mmCorrect thickness = 2.33 – (–0.03) = 2.36 mm

9. The thimble of screw gauge has 50 divisions. The spindle moves through 1 mm fortwo complete turns of the screw. (a) Find the pitch of the screw and the least countof the screw gauge. (b) Find the diameter of the wire if the reading on the sleeveis 0.5 mm and the reading on the thimble is 6 divisions.

Ans. (a) Pitch is the distance moved by the screw in one complete rotation of its head

=2mm1

= 0.5 mm

LC of screw =50

Pitch =

500.5

= 0.01 mm

=10

mm0.01 = 0.001 cm


(b) Reading on the sleeve = 0.5 mm = 0.05 cmCoinciding division = 6th

\ Diameter of wire = MS reading + LC × CSD coinciding= 0.05 + 6 × 0.001= 0.05 + 0.006 = 0.056 cm

10. In screw gauge, each division on the main scale is 0.5 mm and the pitch is 0.5 mm.There are 100 divisions on the circular scale. The position of the circular scale isas shown : (a) Calculate the least count of the instrument. (b) What is themeasurement of the object?

Ans. (a) Pitch of the screw = 0.5 mm

LC =n

Pitch =

1005.0

= 0.005 mm = 0.0005 cm

(b) Measurement of object = 4 mm – (76th × LC)= 4 – (76 × 0.005 mm)= 4 – 0.380 = 3.620 mm

=10620.3

= 0.3620 cm

11. The time periods of two pendulums are 2.0 s and 1.0 s respectively. If the lengthof the first pendulum is 1.0 m. What is the length of the second?

Ans. For first pendulum

T = 2p gl

2 = 2p gl

...(i)

l = 1 mFor second pendulum

1 = 2p gl¢


1 = 2p g1

· l¢ [From (i)]

1 = 2 l¢ squaring both sides

(1)2 = ( )22 l¢

1 = 4l¢

\ l¢ =41

l¢ is41

of l

l¢ is41

× 1 m

l¢ = 0.25 m12. What is a second’s pendulum? What is its frequency? What is its length at a place

where g = 9.87 m s–2? Take p2 = 9.87.Ans. Second’s pendulum is a pendulum whose time period is two seconds.

of a pendulum T = 2p gL

When T = 2 seconds

2 = 2p gL

squaring

1 = p2gL

1 = 9.8787.9L

\ L = 1 mFrequency of second’s pendulum

f =T1

=21

= 0.5 Hz

13. If two pendulums with time periods T1 and T2 have lengths 1.0 m and 4.0 m

respectively, what is the ratio2

1

TT

?


Ans.

g

g

22

11

Lπ2T

Lπ2T

=

=Þ

2

1

TT

=

g

g1·L

1·L

2

1

=41

=21

2

1

TT

=21

\ T1 : T2 = 1 : 214. How will time T of a simple pendulum change when anyone of the following changes

are made? (a) Its length made one fourth. (b)It is taken to a place where g is onefourth of its value at the first place. (c) Its amplitude is made one half. (d) Massof the bob is made one half.

Ans. (a) When length is made one fourth T = 2p gL

...(i)

T¢ = 2p g4L

= 2p·21

gL

=21

úúû

ù

êêë

égLπ2

T¢ =21

T

Time reduces to half.(b) When ‘g’ is one fourth

T = 2p gL

...(i)

T¢ = 2p gL4

= 2 úúû

ù

êêë

égLπ2 = 2T [From (i)]

T¢ Time period doubles.(c) Time period does not depend on amplitude. Hence, no effect on time period.(d) Time period does not depend an mass of bob.

15. A simple pendulum is suspended from a nail fixed on the wall of a room just belowthe roof. The bob is just above the floor. The period of oscillation is found to be6.28 s. Calculate the height of the room from the length of the pendulum. (Takep = 3.14, g = 9.8 m s–2)

Ans. T = 2p gL


T = 6.25 sg = 9.8 m s–2

p = 3.14

\ 6.28 = 2 × 3.148.9

L

8.9L

= 1 squaring

8.9L

= 1

\ L = 9.8 × 1 = 9.8 m16. The ratio of the lengths of two pendulum is 4, what is the ratio of their time

periods?Ans. Let T1 and T2 be the time periods of two pendulums having length L1 and L2.

Such that2

1

LL

= 4

\

g

g

22

11

Lπ2T

Lπ2T

=

=

2

1

TT

=2

1

LL

2

1

TT

= 4 = 2

17. Calculate the value of g at a place where a simple pendulum of length 2.25 m hastime period T = 3.0 s. Take p2 = 9.87.

Ans. T = 3.0 s; p2 = 9.87; L =100225

=49

m

T = 2p gL

Squaring both sides,


T2 = 4p2

2L

÷÷ø

öççè

æg

T2 = 4p2

2

49

÷÷ø

öççè

æg

T2 = 4p2 × g49

T2 = g

29π

g = 2

2

T9π

= 239

× 9.87

g = 9.87 m s–2

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