COLLEGE

PHYSICS - IMechanics and Thermodynamics

As per Credit Based Semester and Grading System S.Y.B.Sc., Semester III,New Syllabus University of Mumbai

(w.e.f. 2017-2018)

Dr. Kailas R. Jagdeo(M.Sc., B.Ed., Ph.D., LL.B., GDC&A)Department of Physics,

DSPM’s K.V. Pendharkar College ofArts, Science and Commerce,

Dombivli (E) - 421 203.

Dr. Sopan A. Bhamare(M.Sc., Ph.D., MBA (HR))

Head, Department of Physics,B.N.N. College of Arts, Science and Commerce,

Bhiwandi, Thane - 421 302.

Dr. Suresh N. Kadam(M.Sc., B.Ed., M.Phil., Ph.D.)

Department of Physics,KET’s V.G. Vaze College of Arts, Science andCommerce, Mulund (E), Mumbai - 400 081.

Moharram Ali Khan(M.Sc., M.Phil.)

Department of Physics,Rizvi College of Arts, Science and Commerce,

Bandra (W), Mumbai - 400 050.

Vishwas V. Deshmukh(M.Sc., M.Phil., MBA (Edu. Mgt.))

Department of Physics,Rizvi College of Arts, Science and Commerce,

Bandra (W), Mumbai - 400 050.

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First Edition : 2018

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PREFACE

We are pleased to present this book “College Physics - I” on“Mechanics and Thermodynamics.” This book is prepared according tonew syllabus prescribed by the University of Mumbai for S.Y.B.Sc. -Semester III in Physics course.

We, the authors, were free to choose the methods and style ofpresentation. The explanation of the subject matter has been coupled withvariety of solved examples which have been introduced at appropriatestages.

We hope that this book will be extremely useful to the students inunderstanding the subject. We welcome the suggestion for improvement.

We thanks the Director, Shri. K.N. Pandey and the team of HimalayaPublishing House Pvt. Ltd. for bringing out this book.

Mumbai AUTHORS

SYLLABUS

USPH301: Mechanics and ThermodynamicsLearning Outcomes:

On successful completion of this course, students will be able to:(i) Understand the concepts of mechanics and properties of matter and to apply them to problems.

(ii) Comprehend the basic concepts of thermodynamics and its applications in physical situation.(iii) Learn about situations in low temperature.(iv) Demonstrate tentative problem-solving skills in all above areas.

UNIT I 15 Lectures(i) Compound pendulum: Expression for period, maximum and minimum time period, centres of

suspension and oscillations, reversible compound pendulum. Kater’s reversible pendulum,compound pendulum and simple pendulum – a relative study.

(ii) Center of Mass, Motion of the Center of Mass, Linear momentum of a Particle, Linearmomentum of a System of Particles, Linear momentum w.r.t. CM coordinate (i.e., shift oforigin from Lab to CM), Conservation of Linear Momentum, Some applications of theMomentum Principle, System of Variable Mass.Torque Acting on a Particle, Angular Momentum of a Particle, Angular Momentum of System ofParticles, Total angular momentum w.r.t. CM coordinate. Conservation of Angular Momentum.

(iii) Oscillations, The Simple Harmonic Oscillator, Relation between Simple Harmonic Motionand Uniform Circular Motion, Two Body Oscillations, Damped Harmonic Motion, ForcedOscillations and Resonance.

UNIT II 15 Lectures(Review of zeroth and first law of thermodynamics)

(i) Conversion of heat into work, heat engine, Carnot’s cycle: its efficiency.(ii) Second law of thermodynamics, Statements, Equivalence of Kelvin and Plank statement,

Carnot’s theorem, Reversible and irreversible process, Absolute scale of temperature.(iii) Clausius theorem, Entropy, Entropy of a cyclic process, Reversible process, Entropy change,

Reversible heat transfer, Principle of increase in entropy, generalized form of first and secondlaw, entropy change of an ideal gas, entropy of steam, entropy and unavailable energy,entropy and disorder, absolute entropy.

UNIT III 15 Lectures(i) Third law of thermodynamics, Nernst heat theorem, Consequences of the third law,

Maxwell’s thermodynamic relations, Clausius-Clapeyron equation, Thermal Expansion.(ii) Steam engine, Rankine cycle, Otto engine, Efficiency of Otto cycle, Diesel cycle, Efficiency

of Diesel cycle, Otto and diesel comparison.(iii) Low Temperature Physics: Different methods of liquefaction of gases, methods of freezing,

cooling by evaporation, cooling by adiabatic expansion.Joule-Thompson effect, JT effect of Van der Waal’s gas, Liquefaction of helium, propertiesand uses of liquid Helium.

CONTENTS

UNIT IChapter 1: Compound Pendulum 1 – 14Chapter 2: Conservation of Linear and Angular Momentum 15 – 46Chapter 3: Oscillations 47 – 68

UNIT IIChapter 4: Heat Engines 69 – 78Chapter 5: Second Law of Thermodynamics 79 – 90Chapter 6: Entropy 91 – 109

UNIT IIIChapter 7: Third Law of Thermodynamics 110 – 122Chapter 8: Combustion Engines 123 – 135Chapter 9: Low Temperature Physics 136 – 152

University Question Papers 153 – 164

(1)

Chapter 1

COMPOUND PENDULUM

1.1 IntroductionThe acceleration due to gravity which is denoted by g however, the value of g on the surface of

the earth is not the same everywhere. It is greatest at the pole and least at the equator and its value alsochanges with the latitudes. The pendulum can be used to calculate the absolute value of g. A pendulumis suspended body oscillating under the influence of the gravitational force acting on it. This is themost accurate method for determining of acceleration due to gravity. The simplest type of pendulum is

the simple pendulum whose period of oscillation isglπ T 2 , where, l is the length of pendulum.

In this chapter, we will discuss the real (or rigid body) pendulum which can be of any shape andusing such pendulum the accurate value of g can be studied.

1.2 Compound PendulumA rigid body capable of oscillating in a vertical plane freely about a horizontal axis passing

through any point of it is called physical pendulum or compound pendulum.Let us assume that the rigid body which is

suspended about a horizontal axis passing through thepoint O oscillates freely in a vertical plane. Let G be thecentre of gravity of the pendulum which is at a distance lfrom O and θ be the angular displacement of thependulum at any instant t (the displaced position isshown by solid line in Fig. 1.1). In the displacedposition, the weight mg is resolved into two componentsmg cos θ along OG and mg sin θ acting at right angle toOG. Thus the component mg sin θ contributes a couple(or a torque) and this tends to bring the pendulum backto its equilibrium position.

Moment of couple = – mgl sin θ ...(1.1)The negative sign indicates that couple is oppositely directed to the angular displacement.

Letdtdω be the angular acceleration by the couple, and I be the moment of inertia of the rigid

body about the horizontal axis passing through O, then

Moment of couple =dtdωI ...(1.2)

UNIT I

Point ofSuspension

Centre ofmass

O

G

P

l

Fig. 1.1: A Compound Pendulum

mg

2 College Physics - I (S.Y.B.Sc.)

Equate eqns. (1.1) and (1.2), we get,

θmgl–dtdωI sin

sin2

2mgl–

dtθdI

dtdθ

elocity,Anugular v

θI

mgl–dtθd2

2(when θ is small)

02

2 θ

Imgl

dtθd

0202

2 θω

dtθd

Imgl ω0where ...(1.3)

This is the differential equation of angular SHM showing that the motion of the compoundpendulum is simple harmonic. The time period of oscillation is given by,

0

2ωπ T

mglIπ T 2 ...(1.4)

Let Ig be the moment of inertia of the rigid body about an axis parallel to the axis of suspensionbut passing through its centre of gravity G. Then from the parallel axis theorem, we can write,

I = Ig + ml2

If k be the radius of gyration of the rigid body about an axis passing through the centre of gravityG, then Ig = mk2. Thus we have,

I = mk2 + ml2 = m(k2 + l2)Hence, the time period becomes

gl l kπ

mgl l kmπ T

2222

2)(2

...(1.5)

1.3 Length of Equivalent Simple PendulumThe time period of compound pendulum is,

g

ll

k

π gl

lkπ T

2

22

22 ...(1.6)

Comparing this time period of a compound pendulum as expressed in eqn. (1.6) with the time period

of an ideal simple pendulum of length L (given bygLπT 2 ), we find that the two time periods are

3Compound Pendulum

same if we take L = ll

k

2. This length L of a compound

pendulum is known as the length of equivalent simple pendulum.Thus the length of equivalent simple pendulum is,

ll

kL 2 ...(1.7)

If the whole mass of the compound pendulum is

concentrated at a point C at a distance

l

lkL

2 from the

point of suspension, as shown in Fig. 1.2, it will behave like asimple pendulum of same period. The point C is then the centreof oscillation.

1.4 Maximum and Minimum Time Periods of a Compound PendulumThe time period of compound pendulum is,

gllkπT

22

2

Squaring on both sides, we get,

llk

gπT

2222 4

lkl

gπT

222 4 ...(1.8)

Differentiating eqn. (1.8) with respect to l, we get,

2

22142

lk

gπ

dldTT ...(1.9)

(a) If l = 0, the R.H.S. of eqn. (1.9) becomes infinite. Hence, the time period of the pendulum ismaximum when the axis of suspension passes through the centre of gravity of the pendulum.The time period of the pendulum gradually increases as the point of suspension is shiftedslowly towards the centre of mass.

(b) If l = k, the R.H.S. of eqn. (1.9) is zero. Hence the time period of the pendulum is minimumwhen the distance between the centre of suspension and centre of gravity (c.g.) becomesequal the radius of gyration of the pendulum about the horizontal axis passing through c.g.,we then have,

glllπ

gllkπT

2222

min 22

glπT 22min (1.10)

O

G

L

C

Fig. 1.2: Length of equivalentsimple pendulum

4 College Physics - I (S.Y.B.Sc.)

1.5 Centre of Suspension and Centre of Oscillation are MutuallyInterchangeableIn compound pendulum the centre of suspension and the centre of oscillation are highly

important. The point O through which the horizontal axis of suspension passes is known as the centreof suspension. The length of equivalent simple pendulum is given as,

ll

kL 2

As k2 is always greater than zero, L is greater than l. Ifthe line OG is extended up to a point C, then the pointC is called the centre of oscillation.

Thus, OC = OG + GC

lklL

2

Let the distance of the c.g. from the centre ofoscillation C = l′. Then we have,

lkl'

2

2kll'

If the pendulum is suspended about the point O, then the time period becomes,

T =gl

lkπ22

2

=gl

lllπ2

2

=g

llπ 2 ...(1.11)

Again, when the pendulum is inverted and suspended about the point C, then the time period T′ is,

T′ =lglkπ 22

2

=lg

l llπ 2

2

=g

llπ2 ...(1.12)

T′ = T [using eqn. (1.11)]

Fig. 1.3: Centre of Suspension andOscillation are Mutually Interchangeable

O

G

C

l

5Compound Pendulum

Thus, the centre of suspension and the centre of oscillation are interchangeable. If the distancebetween the centre of suspension and the centre of oscillation is L and knowing the time period abouteither of them, we have the value of g as follows,

T =gLπ2

g = 2

24T

Lπ ...(1.13)

1.6 Reversible Compound PendulumA compound pendulum with two knife edges (i.e., the point of suspension and point of oscillation)

so placed that the periods of oscillation when suspended from either is the same is called a reversiblecompound pendulum. Their distance apart is then equal to the length of the equivalent simplependulum and if this length and the time period be accurately measured a precise determination of g ispossible.

The time period of the compound pendulum from one side is,

T =gl

lkπ22

2

squaring on both sides,

T2 = gl

lkπ22

24 ...(1.14)

T2 l = )(4 222

lkgπ

...(1.15)

and the time period of the compound pendulum from other side is,

lglkπT

)(2

22

squaring on both sides,

lglkπT

22

22 4 ...(1.16)

)(4 222

2 lkgπlT ...(1.17)

Let us put T′ = T, and subtract eqn. (1.17) from eqn. (1.15), we get,

T2 (l – l′) = )(4 222

l – lgπ

If l ≠ l′, then

T2 = )(44 22

llgπ

llll ll

gπ

6 College Physics - I (S.Y.B.Sc.)

gllπT

2

or g = 2

2 )(4T

llπ ...(1.18)

The distance between any two points on opposite sides of G and at unequal distances from G, the timeperiods about them being exactly equal, then the distance )( ll is equal to the length of theequivalent simple pendulum and is given by

L = l + l′ ...(1.19)Experimentally, we can measure the values of l, l′ and the time period T = T′. Hence, the accelerationdue to gravity can be found using eqn. (1.18).

But in actual practice, it is extremely difficult to find the positions of the axes for the time periodT and T′ to be exactly equal. However, they can be made very nearly equal by adjusting the weightscarried by the pendulum. According to Friedrich Wilhelm Bessel, it was not necessary to make thetime periods exactly equal, but it was sufficient if the two periods are nearly equal. Then the timeperiods of the compound pendulum of both sides (after squaring) using eqns. (1.14) and (1.16) can bewritten as,

222

2

4lk

πglT

...(1.20)

and 222

2

4lk

πTlg

...(1.21)

On subtracting, above equations, we get,

222224

llTllTπg

22

2224llTllT

gπ

)()(

22

ll llTllT

...(1.22)

It can be solved by using partial fractions as follow,

gπ24 =

llB

llA

...(1.23)

where A and B are undetermined constants.

gπ 24

=)()(

)()(ll ll

llBllA

gπ24 =

)ll llBAlBAl

()()()( ...(1.24)

Comparing the coefficient of l and l′ of eqns. (1.22) and (1.24), we get,A + B = T2 and A – B = T′2

7Compound Pendulum

2

22 TTA

and2

22 TTB

Substituting the values of A and B in eqn. (1.23), we get,

gπ24 =

)llTT

llTT

(2)(2

2222

)(2)(2

42222

2

llT – T

llTT

πg

...(1.25)

Equation (1.25) is called Bessel’s formula which is used to determine the acceleration due togravity g using reversible compound pendulum. The distance l + l′ is accurately measured as being thedistance between the two knife edges. As T = T′, (T2 – T′2) → 0 but (l – l′) is appreciable. As thesecond term in the denominator of eqn. (1.25) is very small as compared to the first term, henceignoring the second term and there is no need to measure the quantity (l – l′) accurately and it isenough to determine the position of centre of gravity by balancing the pendulum on a horizontal knifeedge and measuring l and l′. Thus,

22

2 )(8TT

llπg

...(1.26)

1.7 Kater’s Reversible PendulumIt is compound pendulum, designed to give an accurate value of g. The principle is based on

reversibility of centre of suspension and centre of oscillation. It is therefore called Kater’s reversiblependulum. It consists of a metal rod of uniform cross section, a heavy cylindrical weight W1 can befixed near its one end so that the centre of mass of the rod is shifted towards this end. Two weights W1

and W2, one heavy of metal and others light of wood can be made to slide along the length of the barand clamped at any position. Two knife edges k1 and k2 facing each other may be adjusted near the twoedges as shown in Fig. 1.4.

The pendulum is suspended with a knife edge k1 on thehorizontal rigid support and time period T is measured. Thependulum is reversed and allowed to oscillate about the knifeedge k2. The time period T′ are again measured. The mass W1 isadjusted up or downwards to make the time periods from the twoknife edges nearly equal. Fix up this mass W1, the mass W2 ismoved by means of a screw attached with it to make the finaladjustment to equality of the time periods. The rod is thenbalanced on a sharp knife edge horizontally and its centre ofmass is marked. Distance of two knife edges from centre of massgive l and l′. Knowing l, l′, T and T′ we can calculate the value ofg using Bessel’s formula.

Fig. 1.4: Kater’s Pendulum

k1

k2

W1

W2

8 College Physics - I (S.Y.B.Sc.)

1.8 Advantages of a Compound Pendulum Over a Simple Pendulum(i) In compound pendulum the length being the distance between the knife edges can be

accurately measured. In the simple pendulum since the point of suspension and the centre ofmass of the bob are both indefinite, thus the effective length of the pendulum cannot bemeasured accurately.

(ii) In compound pendulum, being of large mass, can oscillate for a very long time beforecoming to rest and hence the period of oscillation can be determined with great accuracy ofmeasuring the time for 100 to 200 oscillations. In a simple pendulum the oscillation die outvery soon on account of small mass of the bob and the accuracy is limited.

(iii) In compound pendulum, being a rigid body, oscillate as a whole while in the simplependulum there is always a possibility of change of its length due to slackening of the thread.

(iv) An ideal simple pendulum cannot be realized in practice. But the compound pendulum canbe realized in practice.

SOLVED PROBLEMS1.1. A thin uniform bar, one meter long is allowed to oscillate under the influence of gravity about a

horizontal axis passing through its one end. Calculate: (a) the length of equivalent simplependulum, (b) the period of oscillation of the bar and (c) its angular velocity.

Solution: M.I. of uniform bar about the axis passing through its one end and perpendicular to its lengthis given by,

22

12MkMLI

12Lk

(a) The length of equivalent simple pendulum is,

lLl

lklL

12

22

12Lk

In this case, the distance between centre of suspension and centre of gravity2Ll = 0.5 m

m666705012

)1(502

..

.L

(b) The period of oscillation of the bar,

s634189

666701423222

2

..

..gLπ

gl

klπT

(c) The angular velocity of the bar is,

srad84536341

142322 / ..

.Tπ

9Compound Pendulum

1.2. A uniform circular disc of radius R oscillates in a vertical plane about a horizontal axis. Find thedistance of the axis of rotation from the centre for which the period is minimum. Also evaluatethe value of this period.

Solution: Assume circular disc acts as a compound pendulum, the time period is given by,

gl

klπT

2

2

where, l is the distance between the centre of suspension and the centre of mass of the disc,and k is the radius of gyration.The period of a compound pendulum is minimum when l = k

gkπT 22min

For the disc, the momentum of inertia about an axis parallel to the axis of suspension andpassing through its centre of mass, i.e., about its own axis is given by,

22

21 MkMRI

2Rk

Thus we have,

gR.π

gRπ

gRπT 41412222

22min

1.3. A circular disc of radius R oscillates in its own plane as a pendulum about a point in its owncircumference. Find the position of the centre of oscillation.

Solution: Let M and R be the mass and radius of the disc. Let the disc rotate about an axis through Pon its edge.

M.I. of the disc about an axis passing through G =2

2MR

From parallel axes theorem, M.I. of the disc about P will beI = IG + MR2

=2

32

22

2 MRMRMR

The time period of the disc is,

T =Mgl

Iπ2

Here, l = R , is the length of the c.g. from the centre of suspension.

P

R

G

Fig. 1.5

10 College Physics - I (S.Y.B.Sc.)

T =g

R/πMgR

πMR 2322 2

3 2

Compare it with the time period of compound pendulum as,

T =gLπ2

Length of equivalent simple pendulum,2

3RL

The distance between the point of suspension and the centre of oscillation is2

3R . Thus the

centre of oscillation2

3R is from the centre of suspension.

1.4. Show that there are four points collinear with the centre of gravity for which the time period ofangular SHM of a compound pendulum is the same.

Solution: We now draw two circles with centre of gravity

G as centre and l andl

k 2as radii. Let the circles intersect

the line OC at the points P and Q respectively (as shownin the Fig. 1.3). Then we have,

OG = GP = l

and GC = GQ =l

k 2

QP = QG + GP

= ll

k

2

= OCThus, there are four points O, Q, C and P collinear withcentre of gravity G, about which the time period of thecompound pendulum is same.

1.5. In a reversible pendulum, the periods about the two knife edges are t and t , where is a smallquantity. If the knife edges are distant l and l′ from the centre of mass of the pendulum, provethat,

τll

ltπgtll 2

4 2

Solution: The period of oscillation of a reversible pendulum about a horizontal axis passing through a pointof suspension, which is at a distance l from the centre of mass of the pendulum, is given by,

gl

πt lk 2

2

where, k is the radius of gyration of the reversible pendulum about the parallel axis passingthrough its centre of mass. On squaring, we get

P

Q

O

l

l

G

k2

l

k2

l

Fig. 1.6: Four collinear points withsame time period

C

11Compound Pendulum

gl

πt lk )(

42

22

lkl

πgt 2

2

2

4

l

πgtlk 2

22

4...(1.27)

For another point of suspension which is at a distance l′ from the centre of mass, is given by,

gl

πt lk

2

2

On squaring, we get,

lkl

πtg

2

2

2

4)(

ltτtπglk )2(

42

22 ( is small, 02 ) ...(1.28)

Equate eqns. (1.27) and (1.28), we get,

l

πgtlltt

πgl 2

22

2 4)2(

4

22

222

2 4)2(

4l

πlgtlltτt

πg

22222 ]2[

4llltltτlt

πg

)()(]2[4

22 llllltτlltπg

τll

ltπgtll 2

4 2

Hence proved.

QUESTIONS1. What is physical pendulum? Obtain an expression for the time period of angular SHM of a

physical pendulum.2. Define compound pendulum. Show that a compound pendulum executives SHM. Find its

periodic time.3. Explain the concept of length of equivalent simple pendulum.4. Derive the condition for maximum and minimum time period of the compound pendulum.5. Derive an expression for the time period of a compound pendulum and show that it is

minimum when the distance between the centre of gravity and suspension equals its radiusof gyration about a horizontal axis passing through the centre of gravity of the pendulum.

12 College Physics - I (S.Y.B.Sc.)

6. Show that there are four points collinear with the centre of gravity of the pendulum aboutwhich its time is the same.

7. Under what condition does the time period of a compound pendulum becomes maximum?What is the maximum value of the time period?

8. Define centre of suspension and centre of oscillation. Show that in compound pendulumthey are interchangeable.

9. What is reversible compound pendulum? Derive an expression for the acceleration due togravity in terms of two nearly equal time periods about the two parallel knife edges of thependulum.

10. Show that in reversible compound pendulum the Bessel’s formula to calculate theacceleration due to gravity g is given by,

llTT

llTT

πg

22

42222

2

11. Explain Kater’s reversible pendulum to measure the accurate value of acceleration due togravity g.

12. What are the advantages of a compound pendulum over a simple pendulum?

UNSOLVED PROBLEMS1. A uniform rod of mass M and length L , is pivoted about one end and oscillate in a vertical

plane. Find the period of oscillation if the amplitude of the motion is small.

gL

322:Ans.

2. What is the time period for a body suspended as its centre of gravity. [Ans.: infinite]3. The distance of the point of suspension from the centre of gravity for minimum time period

of a compound pendulum is 1.5 m. What will be the value of the radius of gyration of thecompound pendulum about an axis through the centre of gravity? Find the minimum timeperiod of a compound pendulum. [Ans.: k = 1.5 m, Tmin = 3.47s]

4. A uniform circular disc of radius 2 m oscillate in a vertical plane about a horizontal axis.Find the distance of the axis of rotation from the centre of gravity for which the time periodis minimum. What is the value of this time period? [Ans.: 2 m, 2.867s]

5. A ring of radius 10 cm is suspended from a peg so as to oscillate as a compound pendulumin vertical plane (i.e., about an axis perpendicular to its plane). Find its period of oscillation.

[Ans. 0.897s]6. A body of mass 1 kg oscillate about a horizontal axis at a distance of 0.5 m from the centre

of gravity. If the length of the equivalent pendulum is 0.3 m, find its M.I. about an axis ofsuspension. [Ans.: 0.15 kg m2]

7. Using Kater’s pendulum the following date are observed from one side of the c.g. as,l = 46.7 cm T = 1.7308 s

and from other side of the c.g. asl = 37.0 cm T = 1.7472 s

Calculate the acceleration due to gravity g using Bessel’s formula. [Ans.: 985.95 cm/s2]

13Compound Pendulum

MULTIPLE CHOICE QUESTIONS1. The time period of a compound pendulum executes angular SHM is ________.

(a)mgI

l2 (b)mgl

I2

(c)I

mgl2 (d)

lmgI

2

2. Which of the following is correct for the time period of a compound pendulum?

(a)gk

lk 22 (b)

gllk 2

2

(c)gk

lk 222 (d)

gllk 22

2

3. The length of equivalent simple pendulum corresponding to a physical pendulum is _______.(a) l (b) k

(c)l

kl2

(d)klk

2

4. The length of equivalent simple pendulum is ________.(a) the distance between the point of suspension and its centre of gravity(b) the distance between the point of suspension and the centre of oscillation(c) the distance between the centre of gravity and the centre of oscillation(d) none of them

5. A physical pendulum is made to oscillate about some axis through a point on it. Then thelength of the compound pendulum is equal to ________.(a) the distance between the point of suspension and its centre of gravity(b) the distance between the centre of suspension and the point within its gravity and point

of oscillation(c) the distance between the centre of suspension and the point beyond point of oscillation(d) the distance between the centre of suspension and the centre of oscillation

6. The length of equivalent simple pendulum corresponding to a compound pendulum is _____.(a) k (b) l

(c)mlI (d)

mglI

7. The distance between the centre of suspension and the centre of oscillation is equal to _____.(a) the radius of gyration(b) the length of compound pendulum(c) the length of equivalent simple pendulum corresponding to a compound pendulum(d) none of them

14 College Physics - I (S.Y.B.Sc.)

8. The time period of the compound pendulum is a maximum when ________.

(a) l = 0 (b)2Ll

(c) l = L (d) l = K9. The time period of the compound pendulum is a minimum when ________.

(a) l = 0 (b)2Ll

(c) l = L (d) l = K10. There are ________ points collinear with the centre of gravity for which the time period of

a compound pendulum is the same.(a) two (b) three(c) four (d) infinite

11. In order to determine the accurate value of acceleration due to gravity g using Kater’spendulum, it is necessary to make the two time periods about the two parallel knife edgesmust be ________.(a) nearly equal (b) exactly equal(c) one half the other (d) one double the other

12. If the two time periods about the two parallel knife edges of a Kater’s pendulum are exactlyequal then the acceleration due to gravity is ________.

(a) ll

TT

2224 (b) ll

TT

2228

(c) 22

224TT

ll

(d) 22

28TT

ll

13. Which of the following relations are true for a compound pendulum?(a) centre of suspension and oscillation are interchangeable(b) ll = k2

(c)l

kl2

(d) all are correct

[Ans.: 1. (b); 2. (d); 3. (c); 4. (b); 5. (a); 6. (c); 7. (c); 8. (a); 9. (d); 10. (c); 11. (a); 12. (d); 13. (d)]

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