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HINTS & SOLUTIONS (SET-A)

In question no. 1 to 40, only one option i s correct. Choose the correct option.

1. A lawn sprinkler shown in the figure has 1 cm diameter nozzleat the end of a rotating arm and discharges water at the rateof 8 m/s. Determine the torque required to hold the rotatingarm stationary. (Neglect friction.)(A) 1.5 Nm (B) 2 Nm(C) 4.2 Nm (D) 2.6 Nm

8m/s

16cm 14cm 8m/s

Sol. Torque = time rate of change of moment of momentum

T = Q (v1 r 1 + v2 r 2)

= 1000 2

18 0.14 8 0.16

4 100

= 1.5 Nm 8m/s

16cm 14cm 8m/s

2. A pulley–block arrangement is shown in the adjacent figure, ifaccelerations of m1, m2 and m3 are a1, a2 and a3, respectively, thenrelation between a1, a2 and a3 is(All strings and pulleys are ideal.)

(A) a2 a3 + 10a1 = 0 (B) a3 a2 + 10a1 = 0(C) a2 2a3 + 10a1 = 0 (D) 2a2 + a3 + 5a1 = 0

m3

m2 a2

a3

m1 a1

Sol. Length of string remain constant

3. One mole of an ideal diatomic gas is expanded. During expansion, volume and temperature of

gas vary such that 2V

T constant. Find heat transfer during expansion if temperature of the gas

increases by 50 K.(A) 225 R (B) 125 R(C) 100 R (D) None of the above

Sol. v

Q Pdv nC dT

4. A hollow sphere is released from rest on a rough inclined plane asshown in the figure. Find the velocity of point of contact as a functionof time.(m = mass of hollow sphere, R = radius of sphere)(A) 2 t (B) 4 t

(C) 6 t (D) 8 t 37

=0.2

Sol. Minimum for pure rolling,

min 2

tan 2tan 37 0.3

MR 51

I

So, the hollow sphere starts slipping on the inclined plane.

F ma mg sin mg cos = ma a = 4.4 m/s2

370

a

mg sin

f

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I =2.4

R

P cm ttV v R

= 4.4 t – 2.4 t= 2t

5. The track shown in the figure is frictionless. The block B of mass 4 kgis lying at rest and block A of mass 2 kg is pushed along the track withsome speed. The collision between A and B is perfectly elastic. Withwhat velocity should the block A start such that block B just reaches atpoint P?(A) 10 m/s (B) 5 m/s(C) 20 m/s (D) 15 m/s

P

4 m 4 m

A

B

Sol. Apply conservation of momentum and conservation of energy

6. Potential energy function U(x) for a system inwhich a particle is in one-dimensional motion is asshown in the figure. The plot of force vs x is givenby.

x1 x

2

x3 x4 O

U

x

(A)

x1 x2

x3

x4 x

F

(B)

x1

x2

x3 x4

x

F

(C)

x4

x2 x3

x1 x

F

(D)

x4 x2 x3

x1

x

F

Sol. dU

Fdx

7. A heavy uniform rope of length 7.5 m is suspended from a ceiling. A particle is dropped from theceiling at the instant when the bottom end of the rope is given a jerk. Find the time taken by theparticle to reach a point where it will be at same level of pulse.(A) 1 sec (B) 2 sec(C) 4 sec (D) 6 sec

Sol. p

ˆa g( j)

w

dV ga (V gx)

dt 2

2w / p1

L a t2

, 4L

t3g

8. A body of mass m falls from a height h on a pan (of negligible mass) of springbalance as shown. The spring also possesses negligible mass and has a spring

constant K. Just after striking the pan, the body sticks with the pan and startsoscillatory motion in vertical direction. The velocity of the body at mean position

will be (A) 22mg k 2gh (B)k2gh

m

(C) 2mg

2ghK

(D) 2mg

4ghk

m

h

Sol. mg (h + y) = 21

KY2

mg mg 2kh

Y 1k k mg

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mg 2kh A ' y A 1

k mg

V = A

9. A tuning fork vibrating with sonometer, having tension in the wire as T, produces 4 beats per sec.The beat frequency does not change when tension in the wire changes to 1.21T. Find thefrequency of the tuning fork (assuming that the sonometer wire is vibrating in the same mode inboth cases).(A) 80 Hz (B) 84 Hz(C) 88 Hz (D) 76 Hz

Sol. Let frequency of tuning fork be f T.f T – f = 4 (I)

f f T = 4 (II)

f '

1.21 1.1f

(III)

From Eqs. (I), (II) and (III),f T = 84 Hz

10. A hemisphere of radius R and mass 4m is free to slide with its baseon a smooth horizontal table. A particle of mass m is placed on thetop of the hemisphere. The angular velocity of the particle relative

to the hemisphere at an angular displacement , when the velocityof the hemisphere is v, is 4m

m

(A)

5v

Rcos (B)

2v

Rcos

(C)

3v

Rsin (D)

5v

Rsin

Sol. Let vr be the velocity of the particle relative to the

hemisphere and v be the linear velocity of thehemisphere at this moment. From conservation of linearmomentum, we have

4 mv = m (vr cos v)

5 mv = mvr cos or vr =

5v

cos

4m

m

vr

vr cos -

vr sin

or = r v

R =

5v

Rcos

11. A 1 kg mass is attached to a spring of force constant 100 N/m and rests on a smooth horizontalsurface. A second mass of 1 kg slides along the surface towards the first at a speed of 2 m/s. Ifelastic collision occurs at time t = 0, find the velocity as a function of time for the mass attached tothe spring before its speed becomes zero for the first time. (The other end of the spring is fixed to

the wall.)(A) 2 cos 10t (B) 4 cos 10t(C) 8 sin 10t (D) 4 sin 10t

Sol. After collision, block ‘A’ comes to rest and block B moves withvelocity 2 m/s. Applying the conservation of energy,

2 21 1

mv KA2 2

m

A vk

A

1kg 1kg

2m/s

B

1

2 0.2m100

x = A sin t

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v = A cos t

= k

10m

or Vt = 2 cos 10t

12. A particle is moving in a plane with a velocity given by

ˆ ˆ ˆu 2i 3 cos(3 t) j 2k.

If the particle is at origin at t = 0. Then the magnitude of displacement of the particle from origin at

1

t2

sec will be

(A) 1 m (B) 2 m

(C) 3 m (D) 5 m

Sol. x = 1 m

vy = 3 cos(3t) dy

dt

y 1/ 2

0 0

dy 3 cos(3 t)dt

y = 1 m

1Z 2 1m

2

2 2 2r 1 1 1 3 m

13. Two particles having masses m and 2m start moving towards each other from rest at infiniteseparation. The kinetic energy of smaller body when the separation between them is R equals

(A)24Gm

3R (B)

23Gm

2R

(C)28Gm

5R (D)

27Gm

3R

Sol. Apply conservation of energy and conservation of momentum.

14. A closed cylinder of length L0 containing liquid of variable density (x) = 0 (1 +x2) rotates with constant angular velocity as shown in the figure. Find the netforce exerted on the liquid volume (take cylinder to be massless and A = crosssectional area of the cylinder)

(A)

2 2

0 0 0

1 1 A L L

3 2 (B)

2 2 0

0 0

L1 A L

2 3

(C)

2

2 2 0

0 0

L1 A L

2 4 (D) None of the above

Sol. 0L

2

0

dF d m x 0L

2 2

0

0

(1 x )Ax dt

15. Water flows out of a big tank along a horizontal tube AB of length Land radius R and bends at right angle at the other end as shown in thefigure. Find moment of force exerted by water on the tube about theend A. (Assume that the radius of the tube is small.)

(A) 5R2gHL (B) 10 R2gHL(C) 15 R2gHL (D) none of the above

H

A

2 H

3H

2

B H/4

Sol. A

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16. A rigid body is made of three identical thin rods, each of length L,fastened together in the form of letter H. The body is free to rotateabout a horizontal axis that runs along the length of the legs of theletter H. The body is allowed to fall from rest from a position inwhich the plane H is horizontal. What is the angular speed of thebody when the plane of H is vertical ?

(A)g

L (B)

1 g

2 L

(C)3 g

2 L (D) 2

g

L

Sol. Moment of inertia about the axis of rotation,I = I A + IB + IC

= 0 +2ML

3 + ML

2 = 2

4ML

3

If is the angular speed in the vertical position

K.E.=

2 2 21 1 4

I ML2 2 3 = loss in P.E. of the system

2 22 L

ML 0 Mg MgL3 2

or =3 g

2 L

AB

C

17. A sphere is in pure rolling on a horizontal surface. How many points of sphere will have a verticalvelocity?(A) 1 (B) 2(C) Data insufficient(D) There is no point on the sphere which has vertical velocity

Sol. D

18. There are two identical small holes of area of cross-section ‘a’ on

the opposite sides of a tank containing a liquid of density . Thedifference in height between the holes is h. The tank is resting on asmooth horizontal surface. Horizontal force required to keep thetank in equilibrium is

(A) gha (B)

2gh

Q

(C) 2agh (D)gha

h1

h2

h

Sol. Thrust force F = F1 – F2 = a 21v a2

2v

a(2gh1) a(2gh2) = 2ag(h1 – h2)= 2agh

19. A cyclic process ABCD is shown in the given P–V diagram. Which ofthe following curves represents the same process?BC and DA are isothermal processes.

V

P

C

D

B A

(A)

T

P

A B

C

D

(B)

T

V

D C

B

A

(C)

T

P

A

B

CD

(D)

T

V

A B

CD

Sol. AB constant P, T will be increasing with increasing V.BC constant T, P will be decreasing with increasing V.


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CD constant V, decreasing P, hence decreases T.DA constant T, decreasing V, increasing P. Also, BC is at a higher temperature than AD.

20. A pendulum has period T for small oscillations. An obstruction is placeddirectly below the pivot as shown in the figure so that only the lowestone-quarter of the string can follow the bob when it swings to the left of

its equilibrium position. The pendulum is released from rest at a certainpoint A. The time taken by it to return to that point ‘A’ is(A) T (B) T/2(C) 3T/4 (D) T/4

L

m

O

B

A

3L/4

Sol. Time taken by the bob to move from A to B and from B to A is T/2.

T LLet T1 be the time period of the pendulum when it is in the contact with the obstructions.

1

T L

T L

4

1T 1

T 2 T1 = T/2

Time taken by the bob to return to A = T/2 + T1/2

= T/2 + T/4 = 3T/421. A block of weight W produces an extension of 9 cm when

it is hung by an elastic spring of length 60 cm and is inequilibrium. The spring is cut into two parts, one of length40 cm and the other of length 20 cm. The same load Whangs in equilibrium supported by both parts as shown inthe figure. The extension now is(A) 9 cm (B) 6 cm(C) 3 cm (D) 2 cm

60 cm

40 cm

20 cm

Sol. Let k be the spring constant of the original spring

W = 9k k =W

9

Spring constant is inversely proportional to the length of the spring. Spring constant for the 40 cmpiece

k1 = k 60 3

k40 2

, k2 and = k 60

3k20

When connected in parallel, the equivalent spring constant k = k1 + k2

or k = 3

k 3k2

= 41

k2

Let x be the new extension. Then kx = W = 9k

9

k x2

= 9k x = 9k 2

9k = 2 cm.

22. A uniform rod AB of mass m and length 2a is falling freely without rotation under gravity with ABhorizontal. Suddenly the end A of the rod is fixed when the speed of the rod is v. The angularspeed with which the rod begins to rotate is

(A)v

2a (B)

4v

3a

(C)v

3a (D)

3v

4a

Sol. Angular momentum about A will be conserved

Li = Lf or mva = I = 2m(2a)

3

v

A Baa

A

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= mva2

3

4ma =

3v

4a

23. The coefficient of friction between an insect and a hemispherical bowl of radius r is . Themaximum height to which the insect can crawl in the bowl is

(A) 2r

1 (B) r

2

111

(C) r 21 (D) r

21 1

Sol. The insect can crawl up the bowl till the component of itsweight along the bowl is balanced by the limiting frictionforce

R = mg cos f L = mg sin mg cos = mg sin tan =

h = r – y = r

2

11

1.

mgcos

y

mg

mgsin

fr

h

24. A block of mass m is placed in contact with another block of mass M as shown

in the figure. The coefficient of friction between the two blocks is . With whatacceleration should the block M move so that the block m does not slide down?

(A) g (B) g/ (C)2g (D) g/2

M

m

Sol. The FBD of the block m is shown for equilibrium in they-direction.

f = mg ma = mg

or a =g

m ma

mg

f

y

x

25. The space between two large plane surfaces is 2.5 cm and it is filled with glycerine. What force isrequired to drag a very thin plate, 1 m

2 in area, between the surfaces at a speed of 1 m/s? (The

plate is at a distance of 1 cm from one of the surfaces.) ( = 0.75 Ns/m2)(A) 100 N (B) 75 N(C) 50 N (D) 125 N

Sol. F1 = force on the upper side of the plate

du

dyarea

10.75 1

1.5

100

1 cm

1.5 cm

F2 = force on the bottom side of the plate

10.75 1

1

100

F = F1 + F2 = 125 N


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8II26. A satellite of mass m is revolving round the earth at a height R above the surface of the earth. If g

is the gravitational intensity at the surface of earth and R is the radius of earth, find the bindingenergy of the satellite.(A) mgR/2 (B) mgR/4(C) mgR (D) mgR/6

Sol.

2

2mv GmM2R 2R

GM

v2R

K.E. 21

mv2

3R

R

mgR

4

P.E. GMm mgR

2R 2

Total Energy = mgR

4

Binding energy mgR

4

27. Same spring is attached with 2 kg, 3 kg and 1 kg block in threedifferent cases as shown in the figure. If x1, x2 and x3 be theextensions in the spring in these three cases, then (Assume theaccelerations of the blocks of the respective system to behaving equal in magnitude and opposite in direction.)(A) x1 = 0, x3 > x2 (B) x2 > x1 > x3 (C) x3 > x1 > x2 (D) x1 > x2 > x3

2kg 2kg

k

x1

3kg 2kg

k

x2

1kg 2kg

k

x3

Sol. kx1 = 2g or x1 = 2gk

2

3g kx 3g 2g

3 5 x2 =

12 g

5 k

3

kx g 2g g

1 3 x3 =

4g

3k

x2 > x1 > x3.

28. A boy throws a ball upward with velocity 20 m/s. The wind imparts a horizontal acceleration of

4 m/s2 to the left. The angle (to the vertical) at which the ball must be thrown so that the ball

returns to the boy’s hand is

(A) tan1(1.2) (B) tan

1(0.2)

(C) tan1

(2) (D) tan1

(0.4)

Sol. vy = v0 cos = 20 cos vx = v0 sin = 20 sin

Time of flight of the ball is T =y

2v

g =

40cos4cos

10 . . . (i)

In this time, the horizontal displacement of the ball should be zero.

vxT 2

x

1a T 0

2 T =

x

x

2v (20sin 2

a 4 = 10 sin . . . (ii)

From Eqs. (i) and (ii), 4 cos = 10 sin

tan = 4

0.4

10

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29. An ice cube of side a = 10 cm is floating in a tank (of base area = 50 cm 50 cm) partly filledwith water. The change in gravitational potential energy when ice melts completely is (density ofice = 900 kg/m

3)

(A) 0.072 J (B) 0.24 J(C) 0.016 J (D) 0.045 J

Sol. Relative density of ice = 0.9, i.e. 90% of thevolume of ice is immersed in water. When icemelts completely, level of water does notchange.

h = GG = 0.5 cm.u = mgh = (0.1)3 900 10 0.5 102

= 0.045 Joule.5cm

4cm

1cm

G

4.5cm

G

30. A spring is held compressed so that its stored energy is 2.4 J. Its ends are in contact withmasses of 1 g and 48 g placed on a horizontal frictionless table. The speed of the heavier mass

when the spring acquires its natural length is(A)

2.4

49m/s (B)

2.4 4849

m/s

(C)310

cm/s7

(D)610

cm/s7

Sol. 2 21 1 2 2

1 1m v m v 2.4

2 2

2 21 1 2 2

m v m v = 4.8 . . . (i)

m1v1 = m2v2 v1 = 2 21

mv

m = 48 v2 . . . (ii)

From Eqs. (i) and (ii),

2 2 22 21 48

(48v ) v1000 1000

= 4.8 2248

(49v ) 4.81000

v2 =

4800

48 49 =

310 10m/ s cm/ s

7 7

31. Two identical balls A and B are releasedfrom the position shown in the figure. Theycollide elastically on horizontal position MN.The ratio of the height attained by A and Bafter collision will be (neglecting friction)(A) 1 : 4

(B) 2 : 1(C) 4 : 13(D) 2 : 5

M N

4h

A

B

60 h

Sol. After collision, the ballsexchange their velocities,i.e.

and v A = 2gh ,

vB = 2g(4h) = 2 2gh

Height attained by A willbe

h A =2

Av

2g

= hM

4h

A

N

B

60

h

h

hB

vp cos600 =vp/2

P


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10II The path of B will be straight line and then parabolic as shown.

Velocity of B at P may be found by conservation of energy.

2 2B P

1 1mv mv mgh

2 2

2 2B P

v v 2gh

2P

v 8gh 2gh 4gh

The height h above P to which the ball B rises is given by

2

2 P

p

v1 1mv m mgh

2 2 2

h = 2

pv3 3

6gh4 29 8g

=9

h4

hB = h + 9 13

h h4 4

A

B

h h

13hh

4

=4

13

32. A particle moves on a rough horizontal ground with initial velocity v0. If (3/4)th

of its kinetic energyis lost in friction in time t0, the coefficient of friction between the particle and the ground is

(A) 0

0

v

2gt (B) 0

0

v

4gt

(C) 0

0

3v

4gt (D) 0

0

v

gt

Sol. (3/4)th of energy is lost, therefore

(1/4)th K.E. is remaining.

Velocity becomes 0v

2 under a retardation of g in time t0.

i.e. 0 0 0v

v gt2

or gt0 = 0v

2

or = 0

0

v

2t g

33. In a circular motion of a particle, the tangential acceleration of the particle is given by at = 10t. Theradius of the circle is 2 m. The particle was initially at rest. Time after which total acceleration of

the particle makes an angle of 45 with the radial acceleration is

(A)6

5 (B)

1/ 34

5

(C)

1/ 33

5 (D)

1/ 38

5

Sol. tan 45 = 2

t

r

a V10 t

a R

1/ 3

4 425 t 20 t t5

34. A steel wire of negligible mass, length L0 and cross sectional area A0 is kept on a smoothhorizontal table with one end fixed. A ball of mass M is attached to the other end. The wire and

ball are rotating with angular velocity . If elongation in the wire is L, then the value of Young’smodulus of elasticity will be

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(A)

2 2

0

0

ML

A L (B)

2

0

2

0

MA

L L

(C)

3 2

0

2

0

ML

A ( L) (D) none of the above

Sol. Here,

20 00

LYA MLL

35. The speed of longitudinal wave is ten times the speed of transverse waves in a tight brass wire. IfYoung’s modulus of the wire is Y, then strain in the wire is

(A) 100 (B)1

100

(C)1

10 (D) 10

Sol.

T Y10

A

Y T

100 A

Strain 1

100

36. A body at temperature 310 K is kept in an atmosphere whose temperature is 300 K. Thetemperature versus time graph of the body will be(A)

TimeO

300 K

310 K

T (B)

Time

300 K

310 K

T (C)

300 K

310 K

T

Time

(D)none ofthe above

Sol. Applying Newton’s law of cooling

T = 300 + 10 ekt

37. Two blocks of masses M1 and M2 are connected with a string passingover a pulley as shown. The block M1 lies on a horizontal surface. Thesystem accelerates. What minimum additional mass m should beplaced on the block M1 so that the system does not accelerate? (The

coefficient of friction between M1 and horizontal surface is .)

(A)

2 1

M M (B)

2

1

MM M2

M1

m

(C) M2

1M

(D) (M2 – M1)

Sol. For the equilibrium of the mass M1 + mf = T

(m + M1)g = TFor the equilibrium of block of mass M2

T = M2g

(m + M1)g = M2g

or m =

2

1

MM


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12II38. A block of mass 2m is suspended vertically with a string of mass m and length L. The whole

system is placed on a planet whose mass is same as that of earth but radius is half of earth. Thevalue of wave pulse velocity at mid-point of the string will be

(A)gL

2 (B)

gL

4

(C)5gL

8 (D) none of the above

Sol. D

39. A moving mass of 8 kg collides elastically with a stationary mass of 2 kg. If E be the initial kineticenergy of the moving mass, then the kinetic energy left with it after the collision will be(A) 0.80E (B) 0.64E(C) 0.36E (D) 0.08E

Sol. In collision of two bodies of mass m1 and m2, velocity of the first mass after the collision is givenby

v1=

1 2 2

1 2

1 2 1 2

m m 2mu u

m m m m

= 1 1

8 2 6u 0 u8 2 10

22

1

2

1

(1/ 2)8vK.E. after collision 60.36

K.E. before collision (1/ 2)8u 10

K.E. of the moving mass after collision = 0.36E.

40. A particle is projected with a velocity of 20 m/s so that it just clears two walls of equal height 10 mat a horizontal separation of 20 m from each other. Then, find the angle of projection is

(A) Angle of projection is 30 with the vertical. (B) Angle of projection is 30 with the horizontal.(C) Angle of projection is 45 with the horizontal. (D) Angle of projection is 45 with the vertical.

Sol. A

In question no. 41 to 50, one or more options may be correct. Choose all the correct options.

41. A uniform rod of mass M and length L is kept on a horizontal surface. The rod receives animpulse of ‘J’ at its lowest point, normal to its length as shown in figure. Then, choose correctstatement(s) from the following.

(A) Velocity of point P just after impact is2J

M.

(B) Velocity of point P just after impact is zero.

(C) Kinetic energy after impact is23J

2M.

L

2

P

J

L

L

6

(D) Kinetic energy after impact is

22J

M .

Sol. J = M v

J

vM

2L MLJ

2 12

6J

ML

P cm

LV v 0

6

K.E. 2 21 1

Mv I2 2

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22J

M

42. An impulse 30 Ns is given to the block of mass 2 kg in downward directionas shown in the figure. As a result of the impulse,(A) both the blocks start moving with 5 m/s in opposite directions

(B) centre of mass of the system is moving upward(C) the net impulse acting on the block of mass 2 kg is 10 Ns(D) the net impulse acting on the block of mass 4 kg is 10 Ns 4 kg

2 kg

30 Ns

Sol. A, B, C

43. A solid ball rolls down a parabolic path ABC from a height of 10 m asshown in the figure. Portion AB of the path is rough while BC is smooth.Then, choose correct statement(s) from the following.(A) At maximum height along portion BC, K.E. of the particle is zero. (B) At maximum height along portion BC, K.E. of the particle is not zero.

A

B

C10 m

(C) Maximum height reached by the particle is 10 m because work done by frictional force is zeroin portion AB.

(D) Maximum height reached by the particle is less than 10 m.

Sol. B, D

44. Two blocks of masses 2 kg and 8 kg are connected through pulleysand strings as shown in the figure. If all pulleys and strings are ideal,then

( = 0.6, take g = 10 m/sec2.)(A) friction force between the 2 kg block and the surface is 12 N(B) friction force between the 2 kg block and the surface is 10 N(C) acceleration of the 2 kg block is equal to zero(D) tension in the string connected to the 2 kg block is 10 N

2kg

8kg

Sol. f max > 10 N

45. The temperature versus pressure graph of one mole of an idealmonatomic gas undergoing a cyclic process is shown in the figure.Choose correct statement(s) from the following.

(A) Total net work done from the cycle is RT0(1 ln 2).(B) Work done for process CA is zero.(C) Heat is supplied in the process CA.(D) Heat is rejected during process BC.

2T0

T0

P0 2P0

A

P

B

C

Sol. For AB:

W AB = nRT0 ln 01

RT ln22

AB AB 0Q W RT ln2

For BC,

WBC = nRT0 = RT0

QBC =

0 0

5 5n R T RT

2 2

2T0

T0

P0 2P0

A

P

B

C

For CA (isochoric process),WCA = 0

QCA =

0

3RT

2

46. One mole of an ideal gas at pressure P1, volume V0 and temperature T1 is isothermally expandedto four times of its initial volume. Then, it is compressed to its original volume at constantpressure. Finally, the gas is compressed at constant volume to its original pressure. The correctdiagram indicating the entire process is


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14II (A)

1P

4

3 2

1

4 V0

P1

V0

(B)

P1/4

3

21

4V0

P1

V0

(C)

32

1

T1

P1

P1/4

T1/4

(D)

32

1P1

P1/4

T1 4T1

Sol. A, C

47. There are two spheres of same radius and material at same temperature, but one is solid whilethe other is hollow. If same amount of heat is supplied to both the spheres, then choose thecorrect statement(s) from the following.(A) Final volume of hollow sphere is greater than that of solid sphere.(B) Final volume of hollow sphere is same as that of solid sphere.(C) Final temperature of hollow sphere is same as that of solid sphere.(D) Final temperature of hollow sphere is greater than that of solid sphere.

Sol. A, D

48. Choose the correct option which could complete the statement: “When a body is submerged in aliquid, …”

(A) buoyant force acts through the centre of mass of the body(B) buoyant force act through the centre of mass of displaced liquid(C) for stable equilibrium the point of application of buoyant force must be above the centre of

mass(D) for stable equilibrium the point of application of buoyant force must be below the centre of

massSol. B, C

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49. Choose the correct statement(s) regarding the motion of a particle.(A) If instantaneous velocity of a particle is zero, then its instantaneous acceleration will

necessarily be zero.(B) If instantaneous velocity of a particle is zero, then it can have instantaneous acceleration.(C) A situation is possible in which instantaneous velocity of a particle is never zero but

average velocity in that interval is zero.

(D) A situation is possible in which the speed of a particle is never zero but average speed inthat interval is zero.

Sol. B, C

50. F1 is the gravitational force of attraction between two point massesm1 and m2 separated by distance r and F2 is the force of attractionbetween two rods of masses m1 and m2, having centre of massseparated by distance r as shown in the figure.Choose the correct statement(s) from the following.

(A) F1 = F2 (B) F1 F2

m1 m2 r

r

m1 m2

(C) Even if the two particles have different masses, they exert equal forces of gravitationalattraction on each other.

(D) If m1 and m2 are kept in water, then values of F1 and F2 do not change.

Sol. B, C, D

Phy 1 (3).pdf

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