quick revision test single correct choice type …

QUICK REVISION TEST

SINGLE CORRECT CHOICE TYPE QUESTIONS

1 If time taken by projectile to reach Q is T, then PQ =

P

v

090

Q

A. sinTv B. cosTv C. secTv D. tanTv

Answer :D

Solution : 2cosVT

g

2

22 2

1 sin 4 sin 20 sin2 2 cos cos cos

g V V VPQ g Tg g

tanPQ TV

2 A body is projected from ground with a velocity 140 2 ms at 045 to horizontal. After four seconds, it is stopped and again projected horizontally with a speed 140ms . Its horizontal displacement from point of projection (by the time it strikes ground) can be ____

A. 400 m B. 394 m C. 344 m D. 294 m Answer :D

Solution : 4 ˆ40ˆ 0V i j 40 410at s . After four seconds its at max height and its horizontal

displacement is 40 4 160x m . From the highest point in its trajectory its again projects horizontally with 140ms . Take that point as centre and assume a horizontal circle. The radius vectors of that circle are horizontal. In which ever direction you throw it the time of descent is 4 sec and from that point the horizontal distance is 160 m, since horizontal velocity is 140ms however this 160m can be in any direction hence total horizontal displacement is vertical addition of 1 160R and 2 160R . Range of

1 2R R will be between zero and 320 m. Hence D is correct option.

3 From a height 100m, a body is dropped and at the same time another body is projected vertically up with

Sai charan

Typewritten text

SRIGAYATRI EDUCATIONAL INSTITUTIONS ADVANCE MATERIAL

Quick Revision Test Single

velocity 110ms from the ground. The time after which they meet in air is _______ 2[ 10 ]g ms

A. 10s B. 8s C. 6s D. They do not meet in air

Answer :D

Solution : The time of descent of falling body is approximately 4.5 seconds. Time of flight of the body projected up is only 2 seconds. Hence they wont meet in air.

4 Velocity of ‘A’ increases at rate of 11ms every second, and velocity of ‘B’ decreases at rate of 11ms

every second. Their relative acceleration is ________ 2ms

A. 2 B. 4 C. 6 D. Zero Answer :D Solution : 2 21 1A Ba ms a ms 1 1 0ra

5 The acceleration experienced by a moving motor boat after its engine is cut off is 3a Kv . If 0V is speed at cut off, magnitude of velocity at time ‘t’ after cut off is _______

A. 02

02 1V

V kt B. 0

ktV e C. 0

2V D. 0V

Answer :A

Solution : 3dv kvdt

0

30

v t

v

dv k dtv

0

2 2 20

1 1 1 22

v

v

kt ktv v v

0202 1VV

v kt

6 A particle moves with a velocity 1ˆ ˆ3 4i j ms from origin. The displacement of particle along line x=y after two seconds will be ______ A. 10 m

B. 72

m C. 7 2m D. 5m

Answer :C Solution : | | 6ˆ ˆ8S vt i j

Unit vector along ˆ

2

ˆ

2i jx y is

Component of S along the unit vector will be 7 2m 7 Among the four graphs (figure), there is only one graph for which average velocity over the time

interval (0,T) can vanish for a suitably chosen T. Which one is it ? x is the position of a particle moving along x-axis.


A.

x

t

B.

x

t

C.

x

t D.

x

t

Answer :B Solution : Average velocity becomes zero when total displacement is zero. It happens if x at t = 0 is equal to x at t = T. It is possible with graph (B) only. x

tT

8 The co-ordinates of an insect crawling on a vertical wall changes as tx e and 2loge y t . Find the equation of trajectory of the insect. A. 2y x B. 2y x C. 2x y D. 2x y Answer :A Solution : tx e

2 2 22log log te ey t y t y e y x

9 A balloon ascends from rest vertically up relative to air with a constant vertical acceleration of 20.4 /m s whereas the air flows due east with a velocity 5 m/s. An insect moves up along the vertical wire fixed with the balloon, with a velocity of 0.1 m/s relative to the wire. Find the magnitude of velocity of the insect w.r.t ground at t = 6 s. Take necessary assumptions.

A. 3 52

m/s B. 5 52

m/s C. 5 32

m/s D. 3 32

m/s

Answer :B Solution : secballon air in tv v v v

5 0.1 0.4 6 0.1ˆ ˆ ˆ ˆ ˆ ˆ5 5 2.5at j i j j i i j

25 125 5 525 /4 2 2

v m s

10 In the arrangement shown, if 1 2,f f and T be the frictional forces on 2 kg block , 3kg block & tension


in the string respectively, then their values are:

2kg 3kg1N 8N1 0.1

2 0.2

A. 2 N , 6 N, 3.2 N B. 2 N, 6 N, 0 N C. 1 N, 6 N , 2 N D. 1N,2 N, 3.2 N Answer :C

Solution : 2kg1N T

3kgT

8

Fmax=2N fmax=6N

Net force without friction on system is ‘7N’ in right side hence first maximum friction will come on

3 kgblock.

2kg 81

1 2 2kg26

So 1 21, 6 , 2f f N T N

11 Three blocks A, B and C of masses 2 kg, 3kg and 4 kg are placed as shown . Coefficient of friction between A and B is 0.5 and that between B and C is 0.1. Maximum force Fthat can be applied horizontally on to A such that the three blocks move together is____ 210g ms

ABC

F 0.5AB

Frictionless

0.1BC 0C

A. 12.22 N B. 13 N C. 11.25 N D. 15 N Answer :C

Solution : When the three blocks move together, acceleration of the system= 9F

This acceleration on 9F

should be less than or equal to maximum possible accelerations of the blocks

for 4kg: 20.1(5)10max 1.254 4

a BCf ms

for (3 4) ;kg 20.5(20)max 1.47 7

a ABf ms


Now 1.259F

11.25F N 12 A small body starts sliding down from rest on a fixed inclined plane of inclination , where base

length is equal to l . The coefficient of friction between the body and the surface is . If the angle is varied keeping l constant, the time of sliding from top to bottom of inclined plane will be least for the relation

A. 1tan 2

B. 1tan

C. 1tan 22

D. tan

Answer :A Solution : sin cosa g g

2 sec 2 secsin cos

ta g g

10 tan 2dtd

13 The elevator has a mass M and the counter weight at A has a mass m . The motor supplies a constant

force F on the cable at B. Neglecting the mass of the pulley and cable, the speed of the elevator at time t after starting from rest is

M

MotorA Bm

A.

F m M gt

M m

B.

F Mg tM m

C.

F M m gt

M m

D. F i

M m

Answer :A Solution : T F Mg Ma

mg T ma

( )F mg Mg a

m m

( )F m M gv tM m

14 A particle moves in the X-Y plane under the influence of a force such that its linear momentum is


( ) cos sin ,p t A i kt j kt where A and k are constants of appropriate dimensions. The angle

between the force and the momentum is A. 00 B. 030 C. 045 D. 090 Answer :D

Solution : dpFdt

. 0and F P

15 In the system shown in figure, the friction coefficient between ground and bigger block is . There is no friction between the two blocks. The string connecting the two blocks is light. All three pulleys are light and frictionless. Then the minimum limiting value of so that the system remains in equilibrium is

A. 1

2 B. 1

3 C. 2

3 D. 3

2

Answer :C Solution : In equilibrium

T = mg

N = 3 mg

& f = 2T = 2mg

In limiting case maxf f

2mg N

2 3mg mg


2 .3

Ans

16 Statement: 1) Static friction can be equal to kinetic friction

Statement: 2) Static friction can be less than kinetic friction

Statement: 3) Static friction can be greater than kinetic friction A. Statement (1) alone is correct B. Statement (2) alone is correct C. Statement (3) alone is correct D. Statement 1,2,3 all are correct Answer :D Solution : Conceptual

m

T

mg

2T2m

T = mgf

N

2mg 17 In the arrangement shown in the figure,masses of the blocks B and A are 2 m , 8 m respectively .

Surface between B and floor is smooth . The block B is connected to block C by means of a pulley. If the whole system is released from rest, then the minimum value of mass of the block C so that the block A remains stationary with respect to B is : ( Co-efficient of friction between A and B is .)

A

B

C

A. m

B. 2

1m

C. 10

1m

D. 101

m

Answer :D Solution : FBD of A

C a

mg

T

8mA

N

N

a

8mg


If the acceleration of ‘C’ is a

For block ‘A’ N = 8 ma …………(1)

8 8mg N ma ………….(2)

And acceleration a can be written by the equation of system ( )A B C

1 110m g m m a

1

1

8 810

m gmg mm m

1 110m m m 110

1mm

18 Two particles are located on a horizontal plane at a distance 60m. At t = 0 both the particle are simultaneously projected at angle 045 with speeds 12ms and 114ms respectively as shown in figure. Find the time at which the separation between them is minimum?

11u 2ms

12u 14ms

045 045A B

1 260 m

B

A. 12 25

s B. 12 27

s C. 12 35

s D. 12 33

s

Answer :A Solution : In relative motion, from observer considers himself at rest and observes the motion of object. Graphically, we can drawn the direction of motion of particle 2 w.r.t particle 1

Both the particles are moving in gravitational field with same acceleration ‘g’. Hence, relative acceleration of particle 2 as seen from particle 1 will be zero. It means the relative velocity of particle 2 w.r.t. particle 1 will be constant and will be equal to initial relative velocity. Graphically we can draw the situation as shown.

y

1A l

N

CB

2

450

Minimum separation during motion

21v

12| u | 14 ms

11| u | 2ms

xB450


AN is the minimum separation between the particles and BN is the relative separation between the particle when the distance between 1 and 2 is shortest: From figure we can write

0 012 cos 14 cos 45 2 cos 45v ………………….(i)

0 012 sin 14sin 45 2sin 45v …………………..(ii)

From (i) and (ii) 112 10 2v ms

3cos 4 / 5sin5

as 037

Hence, minimum separation between the particles is

3sin 60 365

AN AB m

The time when separation between the particles in minimum

0

12

60cos37 12 2510 2

BNt t sv

19 A particle moves in the x-y plane whose co-ordinates vary with time t as given by cos sinx a t t t and (sin cos )y a t t t where ‘a’ is a constant of appropriate dimensions . Find the distance travelled by the particle in first t seconds.

A. 2at B.

2

2at C.

2

3at D.

3at

Answer :B

Solution : sin cos sin cosxdxv a t t t t at tdt

cos sin cos sinydyv a t t t t at tdt

Distance travelled: 2 2

0 0

t t

x yD vdt v y dt

22 2

0

( cos ) ( sin )2

t atD at t at t dt at dt

20 Figure given below shows the variation of velocity (v) of a particle moving along straight line, with displacement(s). Which of the following graphs best represent the variation in acceleration with displacement?


v0

v

s

A.

a

s B.

a

s

C.

a

s D.

a

s Answer :D Solution : From the velocity versus displacement graph, 0v ms v [where m is the slope of the straight line]

Differentiating eqn. (i) w.r.t. time t, we get

dv mdsadt dt

, i.e., a = 0[ ]mv m ms v

Or 20a m s mv

Which is of the from y = mx + C

A straight line with slope 2' 'm and y-intercept 0' 'mv .

21 A particle moves uniformly with speed v along a parabolic path 2y kx where k is a positive constant of appropriate dimensions. Magnitude of acceleration of the particle at x = 0 is A. 2kv B. 2

2k v C. 22kv D. 23kv

Answer :C Solution : 2ˆ ˆ ˆ ˆr xi yj xi kx j

2 ˆ ˆ ˆ( ) 2x xdr dx dv i kx j v i kxv jdt dt dt

At x = 0, 0(ˆ ˆ) ˆx xv v i j v i only

As speed is constant, tangential acceleration is zero. Particle possesses normal acceleration only i.e., a


is always perpendicular to v . Hence, at x = 0, ˆxv v i only ˆ

ya a j only

i.e. 0xa at x = 0.

22 [ˆ ˆ]x x xdva a i k xa v jdt

At x = 0, xv v and 22 ˆ0xa a kv j

22a kv 22 If a particle is projected from ground with an initial velocity 3 ˆ /ˆu i j m s

, find the time up to

which vertical displacement will be greater than horizontal displacement after projection. Take horizontal direction as x-axis and vertical direction as y-axis and 210 .g ms

y

x

u

o

A. 35

s B. 15

s C. 3 15

s D. 3 15

s

Answer :D Solution : 0[ 3 1 2, 60 ]y x u

21sin cos2

u t gt u t

3 10 1 3 12 2 2 3 2 102 2 2 5

t t t s

23 Given that 2siny A ct x

, where y and x are measured in meters. Which of the following

statements is true A. The unit of is same as that of x and A B. The unit of is same as that of x but not of A

C. The unit of c is same as that of 2

D. The unit of (ct-x) is same as that of 2

Answer :A Solution : Conceptual


24 For a particle moving along circular path, the radial acceleration ra is proportional to time t. If ta is the tangential acceleration, then which of the following will be independent of time t ? A. ta B. .r ta a C. /r ta a D. 2( )r ta a Answer :D

Solution : 2

rva tr

---------(1)

2 . tantv a cons tr

1/2ta t ---------(2)

2 0.r ta a t .

25 A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposite to the motion. The work

done by the cycle on the road is A. Zero B. – 200 J C. + 2000 J D. – 2000 J Answer :A Solution : As the displacement of the road is zero with respect to the ground, work done by the force of cycle on the road is zero.

26 A vertical circular frame start from rest and moves with a constant acceleration a.

A smooth sliding collar A is initially at rest in the bottom position 0 . Find the

maximum angular position max reached by the collar.

a

A

A. 1tan ag

B. 1 2tan ag

C. 12 tan ag

D. 1tan2ag

Answer :C Solution : From work-energy theorem w.r.t frame

g N pseudoW W W K

1 cos sin 0mgR O maR

2(2sin ) 2 sin cos 02 2 2

g a


2

aTang

12 aTan

g

27 The ratio of lengths of smooth and rough part of a fixed inclined plane is 1:2. A small body starts moving from top and again comes to rest at the bottom point. Then the co-efficient of friction between body and rough surface is

u = 0

v = 0Rough

Smooth

1

2

A. 3 tan

2 B. 2 tan

3 C. 3tan D. 1 tan

3

Answer :A Solution : From work energy theorem

g N f K

2sin 0 cos 03

mg mg

2sin ( cos )3

32

Tan

28 A platform P is moving with a velocity p over hemispherical shell. A vertical rod AB passing through a hole in the platform is moving on the shell and remains vertical. There is sufficient friction between rod and shell to stop the slip. C is the crown of the shell and O is its centre BOC at any instant. Find the velocity of point B in dowanward motion at that instant.


A

B

C

O

Platform

VP

A. sinp B. cosp C. tanp D. cotp Answer :C

Solution :

CB

O

pV

pV

BV

tan B

P

VV

tanB PV V hence option (C) is correct.

29 Two bodies P and Q connected through light rigid rod of length L are

placed as shown, in the figure. Body P moves toward O with a

constant speed V. The velocity of Q is

LQ

O Px

y

300

A. 3v B.

3v C. 2 v D. 3

v

Answer :A Solution : Let 1v is the velocity of Q.


1 0cot 30v v 0cot 30 v

1 3v v .

2 2 2x y L

LQ

OP

x

y

300

1v

v

2 2 0dx dyx ydt dt

dx dyx ydt dt

1xv yv

1x v vy

1 3v v 30 A block of mass m is on the smooth horizontal surface of a plank of mass M. The plank is on smooth

horizontal surface. Now, a constant horizontal force F acts on M. Now, for a person standing on the ground :

FM

m

E

N

A. The acceleration of m is FM

towards west

B. The acceleration of m is zero

C. The acceleration of m is F/m towards east


D. The acceleration of m is FM m

towards east

Answer :B Solution : No horizontal force (external) acts only on the body of mass M

Hence acceleration of m is zero, hence option (B)

31 A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by 1( )x t after time t and that of the second body 2 ( )x t after the same time interval. Which of the following graphs correctly describes 1 2( )x x as a function of time t ?

A. t

1 2x x

O B. t

1 2x x

O C.

t

1 2x x

O

D. t

1 2x x

O

Answer :B

Solution : 21 2

12

x x ut at , hence option (B) is correct

32 A particle is ejected from the tube at A with a velocity ‘v’ at an angle with the vertical y-axis. A strong horizontal wind gives the particle a constant horizontal acceleration a in the x-direction. If the particle strikes the ground at a point directly under its released position and the downward y-acceleration is take

as g then,

h

A

a

A. 22 sin cosvh

a

B. 22 sin cosvh

g


C. 22 sin cos sinv ah

g g

D. 22 sin cos sinv gh

a a

Answer :D

Solution : Since 21 2 sin0 sin2

vv t a t ta

Also, 21cos2

h v t gt

22 sin cos sinv gh

a a

33 A 2m wide truck is moving with a uniform speed 0 8 /v m s along a

straight horizontal road. A pedestrian starts to cross the road with a uniform speed v when the truck is 4m away from him. The minimum value of v so that he can cross the road safely is

v0v

Man

2m

4m

Truck

A. 2.62 m/s B. 4.6 m/s C. 3.57 m/s D. 1.414 m/s Answer :C

Solution :

v0v2m

4m

Truck

CA

B

Let the man starts crossing the road at an angle as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 + AC or 4 2cot .

vocs t AC

8 4t AC

8 4 cost v t

8 cos 4v t


48 cos

tv

2sin 2sin

v t tv

2 4 8 cos 2 sinsin 8 cos

v vv

2sin cos 8v

max2sin cos 5

min85

v

34 Find the tension T needed to hold the cart equilibrium, if there is no

friction.

T 30

W

A. 34

W B. 22

W C. 23

W D. 43

W

Answer :A Solution : cos30W N ,

And sin 30T N

3cos30 sin 304

T W W

35 In the diagram shown. The blocks are the same mass M. A force F is

applied on the lower block and both the blocks start moving together

without any relative motion. Suddenly, the lower block hits a fixed

obstacle and comes to rest. The upper block continues to slide on the


lower block. The upper block just manages to reach the opposite end

of the lower block. The ground is smooth. What is the coefficient of

friction between the two blocks ?

MFM

0.5 m 1 m

obstacle

A. FMg

B. 2FMg

C. 2

FMg

D. None

Answer :A Solution : Before striking the obstacle the speed of the block

2 0 2 1 2 1 /2Fv a F mm

Now by work-energy theorems, we have

210.5 02

f mv

10.52

Fmg mm

Fmg

36 A train of mass M is moving on a circular track of radius R with constant speed V. The length of the

train is half of the perimeter of the track. The magnitude of linear momentum of the train will be A. 0 B. 2MV

C. MVR D. MV

Answer: B

Solution: If we treat as a ring of mass M, then its centre of mass will be at a distance of 2R

from the

centre of the circle.

Velocity of centre of mass CM CMV R


2 2R v VR

2CM CM

MVP MV

37 Two blocks of masses m1 and m2 are connected by an inextensible light string. The string is passing over a pulley attached with movable wedge. All the fixed surfaces are smooth and the inclined surface of the wedge is rough. The system is released from rest. Consider the two blocks and the wedge as the system and m2 does not slide on the wedge.

1m

2m

Rough

Smooth

Smooth

A. The centre of mass of system moves towards right. B. The centre of mass of the system moves towards left. C. The centre of mass of the system moves downwards. D. The centre of mass of the system does not move at all. Answer: D Solution: The centre of mass of the system does not move in horizontal direction as net horizontal force is zero and as the mass m2 does not slide. There will be no vertical displacement.

38 A sphere B of mass m is moving towards a bigger fixed sphere A with velocity v on a smooth horizontal surface, as shown. Sphere B moves and returns back after making an elastic collision and

being in contact with sphere A exerts a contact force of magnitude 48 .3

mvt

Find the angle between

contact force and the horizontal at the point of contact. . t Time of contact between two spheres.

AB

A. 045 B. 030 C. 060 D. Zero Answer :B Solution : Since, collision is elastic the kinetic energy of ball B before collision is equal to kinetic energy of ball A after collision.

Hence, speed of ball B before and after collision would be same.

From impulse momentum theorem – Linear impulse = change in momentum.


cos 2F t mv mv mv

2 48 4cos 3 3mv mv mvF

t t t

3cos2

or 30

F V

F

39 A ball collides elastically with another ball of same mass. The collision is oblique and initially one of

the ball was at rest. After the collision, the two balls move with same speeds. What will be the angle between the velocities of the balls after the collision? A. 035 B. 045 C. 060 D. 090 Answer: D Solution: The initially stationary ball will move along the line of impact after collision. In elastic collision velocities gets interchanged along the line of impact.

m

m

0v

m0v

1v

0v sin

After collisionBefore collision

If v1 = 0, then velocity of 1st ball is v0 sin perpendicular to L0I and of 2nd ball is v0 cos along L0I. So, the required angle is 090 .

40 A small body A of mass m and B of mass 3m and same size as A move towards each other with speeds V and 2V respectively from the positions as shown, along a smooth horizontal circular track of radius r. After the first elastic collision, they will collide again after the time :


090

A. 2 r

V B.

2r

V C. r

V D. 2

3r

V

Answer :D

Solution :

12

Velocity of separationeV V

Velocity of separation = 3V

Required time = 23

rV

41 As shown in the figure, a block A moving with speed 10 m/s on a horizontal surface collides with another block B at rest initially. The coefficient of restitution is ½. Neglect friction everywhere. The distance between the blocks at 5 s after the collision takes place is

A B10m / s

A. 20 m B. 10 m C. 25 m D. cannot be determined because masses are not

given Answer :C

Solution :

A B10m / s

A B1v 2v

1 10 52rel relv e u

5 5 25rel relS v t m

42 Mass 1m collides elastically with mass m2 at rest. Out of the four ratios (of mass 1m and 2m ) given in


four options, which ratio ensures the second collision between 1m and 2m . Collision between m1 and wall is perfectly elastic.

1m2m

v

A. 12

B. 1 C. 1

4

D. 2

Answer :C

Solution : 1 2 11 1

1 2

.m m u

v vm m

should be 1 2ve m m

2 1 11

1 2

m m uv

m m

1 12

1 2

2m uvm m

Next collision takes place if 2 1 1 1 11 2

2 1 1 2

2m m u m uv vm m m m

2 1 12m m m

12 1

2

133

mm mm

Among the given values, 14

is only valid ratio

43 Two balls are dropped from the top of a tall tower of height ‘h’ with a time gap interval of t0 second,

where 02htg

. The floor below is inelastic. Let the second ball be dropped at time t = 0. The distance

‘s’ between the balls at time t is best represented by the graph : A.

tO

s

B.

tO

s


C.

tO

s

D.

tO

s

Answer :A

Solution : Separation till first ball reaches ground 20 0

12

s g t g t t (straight line)

Afterwards: 212

s h gt (parabola)

44 Two blocks of masses m1 and m2 connected by a non-deformed light spring rest on a rough horizontal surface of coefficient of friction . The minimum constant horizontal force to be applied to the block of mass m1 in order to shift the other block is

A. 122

m m g

B. 1 212

m m g C. 1 2m m g D. 21 2

mm g

Answer :D

Solution : work done by 21

12

F Fx kx m gx

Where 2m gxk

2 112

F m g m g

45 A particle moves along a circle of radius R such that its kinetic energy (E) on moving through a distance s is given by 2E s , where is a constant. The force acting on the particle is

A. 22 s

R B. 4 4

2

2 s s RR

C. 2 ( 1)ssR

D. 2 22 s s RR

Answer :D

Solution : 2 21 1 .2 . 22 2

dvmv s m v sds

2tF s (Tangential force)

2 22c

mv sFR R

(Radial force)

1/22 2 2 22t c

sF F F R sR


46 Two bars of masses 1m and 2m connected by a weight less spring of stiffness k, rest on a smooth horizontal plane. Bar 2 is shifted by a small distance 0x to the left and released. The speed of the centre of mass of the system when bar 1 breaks off the wall is

1m 2m1 2

A. 20

1 2

k mxm m

B. 02

1 2

x k mm m

C. 1 20

2

m mx km D. 1

01 2( )k m

xm m

Answer :B Solution : At the instant block-I, breaks of the wall its velocity is zero

According to conservation energy

2 20 2 2

1 1 02 2

K x m v

2 02

Kv xm

1 1 2 2

2 2cm

m v m vVm v

2 02

1 2

0 km xm

m m

02

1 2( )x Km

m m

47 A large circular table with smooth horizontal surface is rotating at a constant angular speed about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance l from the centre. Find the speed of the particle with respect to the table as its distance from the centre becomes L.

A. 1

2 2 2v L l B. 1

2 2 2 2v l L C. 1

2 2 2v l L D. 3

2 2 2v L l Answer :A Solution : With respect to observer on platform

2a x 2dvv x

dx


Simplifying 1

2 2 2v L l

48 In the figure, the block A moves downwards with velocity 1v and the wedge B moves rightwards with velocity 2v . Correct relation between 1v and 2v is

B

1v

2v

A

A. 2 1v v B. 2 1 sinv v C. 2 12 sinv v D. 2 11 sinv v Answer :D Solution :

1l

1v

l

2l2v

x

y

Total length of string 1 2L l l l

1 2dl dldL dldt dt dt dt

But 2 2l x y and 22

dldx vdt dt

11

dl vdt

1 2 2sin ( ) 0O v v v

1 2[1 sin ]v v

49 The coefficients of friction and masses of various blocks are shown. A constant horizontal force 2N is


applied on block B. What will be the acceleration of block B after 2 seconds? (g = 10m/s2)

2N BA(5kg)

1.6m

C2kg 3kg

0

0

0.1

A. 20.20 m / s B. 20.25 m / s C. 20.15 m / s D. 20.67 m / s Answer :B Solution : Block C does not move with A and B as there is no friction between C and A

Minimum horizontal force, required on B to stip on A is

3 240.1 10 3 5 4.85 5

mF g m M NM

. As the applied force is 2N, B does not slip on

A. Hence A and B move with common acceleration 22 2 0.253 5 8

a ms

50 Two blocks of masses M and m are connected with either end of a massless string which passes over a light smooth pulley (M > m). The pulley is connected with the roof of an elevator moving with

acceleration a upwards. The magnitude of acceleration of block M relative to block m is 2a. Ratio Mm

is

equal to

A. 1 ag

B. 1 ag

C. 2a1g

D. 2a1g

Answer :C Solution : Relative acceleration of M relative to m = 2a =a –(-a)

a

a

aM

TT

m

Since, M > m, M is moving downward with acceleration a M(g + a)-T = Ma or T = Mg

T – ma – mg = ma or T = m(2a + g)

Comparing Mg = m(2a + g) = Mm

= 2a gg = 1+ 2a

g

51 Find the minimum value of F to hold the system at equilibrium. Coefficient of friction between all the contact surfaces is 0.4. 2( 10 )g ms


F15kg25kg

A. 125 N B. 62.5 N C. 31.25 N D. 250 N Answer :C

Solution :

f

150Nf

2T

250Nf

FF F F

T

If all the surfaces are assumed as frictionless, it can be proved that 15 kg moves down and 25 kg moves up. Directions of frictions are taken opposite to the relative velocities of blocks as shown.

150 = T + f = T + 0.4F

T = 150 – 0.4 F ______ (1)

250 + 2f = 2T 250 + 0.8F = 2T = 2[150 – 0.4F] = 300 – 0.8F

1.6F = 50

500 125 31.2516 4

F N

52 A is a fixed point at a height 23l above a perfectly inelastic smooth horizontal plane. A light

inextensible string of length l has one end attached to A and other to a heavy particle. The particle is held at the level of A with string just taut and released from rest. The speed of ball just after striking the plane is


Al

23l

A. 43gl B. 2

3gl C. 2

3 3gl D. 4

3 3gl

Answer :D Solution : 2v gH

223lg

l

l

23l

A

v

vcos

line of impact

As the collision is perfectly inelastic, the component of velocity perpendicular to the surface becomes zero.

' cosv v

4 2( )3 3gl

43 3

gl

53 Fig shows two parallel rays incident on a mirror. They are reflected as parallel rays as shown in the same fig. What is the nature of the mirror?


incident rays

reflected rays

A. Plane B. Concave C. Convex D. Parabolic Answer :A Solution : Parallel rays can be reflected as parallel rays only from a plane mirror.

54 A small bob suspended by a thread swings in a vertical plane so that angle made by thread in extreme positions with the vertical is 1sin 4 / 5 . The ratio of acceleration of bob in extreme position to its acceleration in lowest position is : A. 1/2 B. 2 C. 1 D. 1/3 Answer :C

Solution : 21 1 cos2 Bmv mgr

2

22 1 1 sinBv gr

BA

= 2g 161 125

= 45g

= g. 45

=g sin ….(1)

At A. 0, sinr Ta a g So, aA = g sin So, 1.0A

B

aa

55 The moment of inertia of a thin uniform semicircular wire of mass ‘M’ and radius ‘r’ about a line perpendicular to the plane of the wire through the centre is: A. 2Mr B. 21

2Mr C. 21

4Mr D. 22

5Mr

Answer :A Solution : 2I MR


56 A string of negligible thickness is wrapped several times around a cylinder kept on a rough horizontal surface. A man standing at a distance ‘l’ from the cylinder holds one end of the string and pulls the cylinder towards him as shown in fig. There is no slipping anywhere. The length of the string passed through the hand of the man while the cylinder reaches his hands is :

A. l B. 2 l C. 3 l D. 4 l Answer :B Solution : Since velocity of top end is twice the center so it will travel twice the distance traveled by the centre

57 A thin spherical shell of radius ‘R’ lying on a rough horizontal surface is hit sharply and horizontally by a cue. At what height from the ground should it be hit so that the shell does not slip on the surface

A. 23

R B. 54

R C. 53

R D. 32

R

Answer :C Solution : Fdt h I

Fh

C.M

cmFdt mv

From above relations: 23

h R

Height from ground 53

R

58 A sphere of mass ‘m’ is given some angular velocity about a horizontal axis through the center, and gently placed on a plank of mass ‘m’. The coefficient of friction between the two is . The plank rests on a smooth horizontal surface. The initial acceleration of the sphere relative to the plank will be:


A. Zero B. g C. 7

5g D. 2 g

Answer :D Solution : Acceleration of sphere = g (right wards)

Acceleration of plank = g (left ward)

So relative acceleration = 2 g 59 A circular platform is mounted on a vertical frictionless axle. Its radius is r = 2m and its moment of

inertia is I = 200 kg-m2. It is initially at rest. A 70 kg man stands on the edge of the platform and begins to walk along the edge at speed 0 1.0 /v m s relative to the ground. The angular velocity of the platform is: A. 1.2 rad/s B. 0.4 rad/s C. 2.0 rad/s D. 0.7 rad/s Answer :D Solution : Net external torque is zero. Therefore angular momentum of system will remain conserved i.e.,

Li fL

Initial angular momentum 0iL

Final angular momentum should also be zero.

Or angular momentum of man = angular momentum of platform in opposite direction

or 0mv r I

0 70 1.0 2200

mv rI

0.7 /rad s

60 A wheel of radius ‘R’ rolls on the ground without slipping with a uniform velocity ‘v’ The relative acceleration of topmost point of the wheel with respect to the bottommost point is:

A. 2v

R B.

22vR

C. 2

2vR

D. 24v

R

Answer :B Solution : As 0 ;CM CMa v const tangential acceleration of each point is zero


2relV V

222V

aR

61 A particle moves under the action of a force of magnitude F such that the angle made by the force with the instantaneous velocity varies with the distance ‘s’ as = ks where, k is a positive constant. Find the work done by the force in covering a distance ‘s’.

A. sinF ksks B. sinF ks

k C. sinF ks D. sink ks

Answer :B Solution : ks

F

cosF

d kds

cosdw F ds

cos sin sind F FW dw F kxk k k

62 A particle initially at the origin moves in xy plane with a velocity V = ai + bxj. Where a and b are constants. The radius of curvature of the trajectory is _____

A.

32 2

1a bxb a

B.

2

1a bxb a

C.

12 2

1a bxb a

D.

12 2

1a bxb a

Answer :A

Solution : .dy b xdx a

2

2

d y bdx a


2

2

3 32 22 2

1

1 1

d y bdx a

dy bxdx a

or

32 2

1a bxb a

63 A particle of mass m strikes elastically with a disc of radius R, and centre c, with a velocity ‘v’ as shown in the figure. If the mass of the disc is equal to that of the particle and the surface of contact is smooth, the speed of the disc just after collision is

m

R/2

C

A. 2

3v B. 3

2v C.

2v D. 2v

Answer :B Solution : No external force along the normal line hence we can conserve the linear momentum of the system of the normal.

Then velocity of disc = V1= ( cos )v j

/ 2 1sin2

RR

64 Four rectangular blocks A, B, C, D are placed one above the other in such away that each block projects a little beyond the block below it. If length of each block is ‘L’, the maximum projections of blocks B

1( )x and C 2( )x are......

line of impact

v cos

v sin

x

y


AB

CD

1x2x

A. ,

2 4L L B. ,

4 6L L C. ,

6 4L L D. ,

2 6L L

Answer :B

Solution :

AB

C L/2L/4

L/6D

65 A block of mass M is connected with a particle of mass m by a light inextensible string as shown in the

figure. Assuming all contacting surfaces as smooth, the acceleration of the block after releasing the system is.

Mm

A. 17

m gM m

B. 217

m gM m

C. 417

M gM m

D. 417

m gM m

Answer :D Solution : x- horizontal distance covered by block ‘M’ with acceleration a

y- vertical distance covered by particle ‘m’ with acceleration a1

4 4 'x y a a

a

Nm

N = ma

M


4T N Ma

4T m M a …….. (1)

T

mg

'am

'mg T ma

( 4 )T m g a ……… (2)

Solving we get 417

mgaM m

66 A particle A of mass 107

kg is moving in the positive direction of ‘x’. Its initial position is x = 0 and

initial velocity is 1 m/s. The velocity at x = 10 is : (use the graph given)

Power(in watts)

2

4

x10 (in m)

A. 4 m/s B. 2 m/s C. 3 2 /m s D. 100/3 m/s

Answer :A

Solution : From graph : area = 1 10 4 2 2 10 302

Area under graph = Pdx

32 3

1 1 1

10 13 7 3

VV V dV VF V dx m V dx m Vdx


3 31030 1 64, 4 /21

V V V m s

67 The blocks A and B shown in the fig. have masses 5 kgAM and 4 kg.BM The system is released

from rest. The speed of B after A has travelled a distance 1 m along the incline is

A

B

5 m

037

A. 32

g B. 34

g C. 2 3

g D.

2g

Answer :C

Solution : If A moves down the incline by 1m, B will move up by 1 .2

m If speed of B is ‘V’ then speed

of A will be 2V.

Gain in K.E = loss in P.E, 2 21 1 3 122 2 5 2A B A Bm V m V m g m g

12 3

gV

68 Water flows through a frictionless horizontal duct with the cross-section varying as shown in figure. Pressure p at points along the axis from left end is best represented by:

A.

P

X B.

P

X


C.

P

X D.

P

X Answer :A Solution : As cross-sectional area decreases V increases and hence P decreases.

69 A bent tube is lowered into a water stream as shown in fig. The velocity of the stream relative to the tube is equal to v = 2.5 m/s. The closed upper end of the tube located at the height 0h = 12 cm has a small orifice. To what height ‘h’ will the water jet rise after coming out of orifice?

A. 0.1 m B. 0.44 m C. 0.2 m D. 0.55 m Answer :C Solution : Let H is the depth of horizontal part of tube from top level of stream. Applying Bernoulli’s theorem from entrance of liquid into the tube upto the highest point of water jet,

20

12

P H g v O 0 0P O g H h h

202v g h h

2

02vh hg

22.50.12

2 9.8

0.2m

70 Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a light string. They are fully immersed in a fluid of density dF. They get arranged into an equilibrium state as shown in the figure with some tension in the string. The arrangement is not possible if

A

B

A. A Fd d B. B Fd d C. A Fd d D. 2dA B Fd d Answer :C


Solution : Equilibrium: 1 2 1 2B B W W

F g A Bd V V g V d d

2 (1)F A Bd d d

If B Fd d , the string becomes slack

To keep the string tight,

A Fd d and B Fd d

71 The gas inside a soap bubble expands uniformly and slowly so that its radius increases from R to 2R. The atmospheric pressure is Po and surface tension is s. The work done by the gas in the process is

A. 3

228 243

oR P sR B.

3225 24

3oR P sR

C. 3 225 23

3 2oR P sR

D. 223 sR

Answer :A

Solution : 2 3

2 2284 4 243

Ro

gas oR

R PsW P r dr R sr

72 A small opening near the bottom of the vessel shown in figure has area ‘A’. A disk is held against the opening to keep the liquid from running out. Let 1F be the net force on the disk applied by liquid and air in this case. Now the disk is moved away from the opening a short distance. The liquid comes out and strikes the disk inelastically. Let 2F be the force exerted by the liquid in this condition just after the

disc is moved. Then 1

2

FF

is

A. 1

2

B. 1 C. 21

D. 14

Answer :A Solution : Let ‘ ’ be the density of liquid. Then


1F P A gh A …………………(1)

In the second case

2F rate of change of momentum

22 2Av A gh

Or 2 2F gh A ……………(2)

From equations (1) and (2)

1

2

12

FF

73 A soap bubble having surface tension T and radius R is formed on a ring of radius b(b<<R). Air of density is blown into the bubble with velocity v as shown. The air molecules collide perpendicularly with the wall of the bubble and stop. Find the radius R at which the bubble separates from the ring.

Rv b

A. 2

Tv

B. 24Tv

C. 2

2Tv

D. 2

4Tv

Answer :D

Solution : Excess pressure inside a bubble 4TR

Let area of bubble at wall where air strikes be A.

Force due to excess pressure 4TAR

Let density of air, force due to striking air 2Av For bubble to separate from te ring,

2 4TAAvR

or 2 4Av R TA or 2

4TRv

.

74 A beaker containing water is placed on the platform of a spring balance. The balance reads 15 N. A stone of mass 0.5kg and density 500 kg/m3 is kept fully immersed in water without touching the walls of beaker and held at rest. What will be the balance reading now? (g = 10 m/ 2s ) A. 20 N B. 25 N C. 10 N D. 15 N Answer :B Solution : Reading of weighing machine = weight of the liquid + buoyant force on body

75 A smooth sphere ‘A’ of mass ‘m’ is moving with a constant speed ‘V’ on the smooth horizontal surface collides elastically with an identical sphere at ‘B’ at rest. After elastic collision speed of sphere ‘A’ is V/2 then the speed of sphere of ‘B’ is

A. 2V B. 3

2V C. 3

2V D. 5

2V


Answer :C Solution : The given elastic collision is oblique. Hence we apply conservation of K.E

2 2 21 1 1( )2 2 2 2 B

VmV m mV

32BVV

76 Two particles in S.H.M having same time period, same amplitude ‘A’ and same mean position x = 0

these particles always crossing opposite to each other at 2Ax then the minimum phase difference

between the two particles (in rad) is

A. B. 2 C.

6 D. 2

3

Answer :D Solution : sinx A

sin2A A

1 6

256

23

77 A tunnel is dug in the earth across one of its diameter. Two masses ‘m’ & ‘2m’ are dropped from the ends of the tunnel. The masses collide and stick to each other and perform S.H.M. Then amplitude of S.H.M. will be: [R = radius of the earth] A. R B. R / 2 C. R / 3 D. 2R / 3 Answer :C Solution : They collide at the centre of the earth

2.

2cm gR m gR

V Am m

0gAR

3RA

78 A quarter section of a stationary acoustic wave between consecutive node A and anti node B has the

graphic profiles as shown in the diagram at two instants of time t = t1 and t = t2, where 1 2~2Tt t ‘T’

is the periodic time of the wave. The stationary wave is formed by the superposition of two identical waves. Then the amplitude A and the wavelength of each wave are given by :


a

2a

2a

2t t

1t t

x b0x A

a

x

y

B

A. 2 , 2A a b B. , 2

2aA b C. , 4

2aA b D. 2 , 4A a b

Answer :C

Solution : If 1 2 ,2Tt t it is quarter section of a full wave 4b amplitude maximum at the antinode

22aA a A

Hence 3 is correct option 79 When the plane progressive sinusoidal transverse wave travels on a stretched string along the positive

x-axis. At a given instant, the shape of string is as shown in fig. The tangent at the point ‘P’ on the stretched string makes an angle 60. Then the ratio of the kinetic energy of a string element at ‘P’ to the elastic potential energy stored in the string element at ‘P’ is

P 60

X

A. 2 0tan 6 :1 B. 2 0sin 6 :1 C. 2 0cos 6 :1 D. 1 : 1 Answer :D Solution : K.E., P.E are maximum at the mean position and K.E, P.E are zero at highest position

80 A wall is moving with velocity ‘u’ and a source of sound moves with velocity u/2 in the same direction as shown in the fig. Assuming that the sound travels with velocity 10 u, the ratio of incident sound wavelength on the wall to the reflected sound wavelength by the wall is equal to


u

wall

u/2S

A. 19 : 22 B. 9 : 11 C. 4 : 5 D. None Answer :B

Solution : . 10 / 2 192

R VI

V u u uf f f

1110 18

19rr

uu u ff

911

I

r

81 A loop of a string of mass per unit length and radius ‘R’ is rotated about an axis passing through centre perpendicular to the plane with an angular velocity . A small disturbance is created at a point in the loop. The disturbance travels along string in both sides. The linear speed part of the disturbance which moves in the same sense of rotation as string for a stationary observer is A. R B. 2 R C. 3 R D. Zero Answer :B

Solution : pulseTV R

The velocity of disturbance w.r.t ground 2R R R 82 An acoustic wave given by sin( )iy A t kx is sent down a string. Upon reflection from one end, the

wave becomes sin( ).2rAy t kx

The resultant wave in the string will be

A. A pure standing wave with amplitude A/2 B. A combination of a standing wave with amplitude 2A and a traveling wave of amplitude A

C. A traveling wave with amplitude 32A

D. A combination of a standing wave with amplitude A and a traveling wave of amplitude A/2 Answer :D Solution : i ry y y

sin( ) sin cos2A t kx A kx t

83 A current carrying loop is placed in a uniform magnetic field pointing negative z direction. Branch


PQRS is a three quarter circle, while branch PS is straight. If force on branch PS is F, force on branch PQR is

B

Q

R

Y

XS

P

A. 2F B. 2

F C. 2F D. 2 F

Answer :A Solution : Force on ( 2 )PS F I R B ____(1) & Force on (2 )PQR I R B ______(2) From (1) & (2)

Force on 2 22FPQR F

84 Two particles A and B of same mass and having charges of same magnitude but of opposite nature are thrown into a region of magnetic field (as shown) with speeds 1v and 2v 1 2( )v v . At the time particle A escapes out of the magnetic field, angular momentum of particle B w.r.t particle A is proportional to (assume both the particles escape the region after traversing half circle).

BA2V1V

A. 1 2v v B. 1 2v v C. 1

2 22v v D.

1

2 22v v

Answer :C

Solution : 1 2&A BmV mVR RBq Bq

1 2 1 2 1 2BA A BL M R R V V V V V V

85 The figure shows two infinite semi-cylindrical thin shells. Shell-1and shell-2. Shell-1 carries current 1i in inward direction normal to the plane of paper, while shell-2 carries same current 1i , in opposite direction. A long straight conductor lying along the common axis of the shells is carrying current 2i in direction same as that of current in shell-1.Force per unit length on the wire is


rInsulation

Shell-1

Insulation

Shell-2 A. Zero B. 0 1 2

2i ir

C. 0 1 22 i ir

D. 0 1 22

2 i ir

Answer :D

Solution :

R

O

Note : 0

2I SinBR

For 090 (semi circular) 02

IBR

0 11 2 2 2

2Bnet

IB B BR

Force per unit length 0 1 22

2 I IR

86 A conducting slider EF of mass m and length L is placed on two parallel long conducting rails. The generator (G ) maintains constant current I in the circuit. The coefficient of friction between the slider (EF) and the conducting rails is 0.50. If a magnetic field 0 0

ˆˆB B i B k

exists in entire region, the acceleration of the slider will be (Neglect the gravity)

G

E

F

Y

X

I

A. 0

2ILB

m B. 0ILB

m C. 03

2ILB

m

D. Zero

Answer :A Solution : F along x – axis = 0.ILB

Normal reaction 0.ILB


Frictional force 0

2ILBf

0

2ILBa

m

87 Circular regions (1) and (2) have current densities J and –J respectively normal to the plane of paper , such that their region of intersection carries no current .Magnetic field in the region of inter section is

1r 2r

(1) (2)

d A. Uniform, proportional to 1 2r r -d B. Uniform, proportional to d

C. Non – uniform D. Zero Answer :B

Solution : Since 0

2B J r

0

r

0 01 22 2netB J r r J d

Hence B is uniform and proportional to d 88 A disc (of radius r )carrying positive charge q, distributed uniformly, is rotating with angular speed

(clock wise) in a uniform magnetic field B about a fixed axis ( as shown in figure), such that angle made by axis of disc with magnetic field is . Torque applied by axis on the disc is

Disc

Fixed axis

B

A. 2 sin2

q r B B. 2 cos4

q r B C. 2 sin2

q r B D. 2 sin4

q r B


Answer :D Solution : M B

sin2qL Bm

2

. sin2 2q mR w Bm

2

sin4

qR wB

As the axis is fixed and angular momentum is constant, this much amount of torque is also applied by axis on the disc in opposite direction.

89 A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by M, the wire resonates with the same tuning fork forming three antinodes for the same position of bridges. The value of M is A. 25 kg B. 5 kg C. 12.5 kg D. 1/25 kg Answer :A

Solution : 2p Tfl

5 9 32 2

g Mgl l

5(3) 3 M

25M kg 90 A cone made up of material of density 3

2 is fully submerged in two liquids as shown in figure. 1V is

volume of cone submerged in liquid 1 and 2V is volume of cone submerged in liquid 2. The ratio 1 2:V V is

1

2r

45°

2

A. 1 : 1 B. 2 : 1 C. 3 :1

2

D. 1 : 2

Answer :A


Solution : 1 1 2 2 1 232

V g V g V V g

1 2: 1:1V V

91 A dipole having dipole moment p is placed in front of a solid neutral conducting sphere as shown in

figure. The net potential at point A on the surface of sphere is

O

RA

R

r

P

A. 2

coskPr

B. 2

2

coskPr

C. 2

2

2 coskPr

D. 2

kPr

Answer :B Solution : Due to the placing of dipole in front of conducting sphere the charge gets redistribute on the surface of sphere but the total induced charge on the sphere would be zero. Potential at centre of sphere is equal to sum of potential at centre due to dipole and due to induced charge.

argcentre induced ch e dipoleV V V

2

2 2

cos0 kP kPrOP

arg[ 0,induced ch eV because the total induced charge is on the surface of sphere and is zero] As conductor is an equipotential surface, the potential at all points of conductor would be same. So, potential at

2

2

cos .kPAr

92 A thin bar of mass m and length L can freely rotate about a horizontal axis passing through the point O. The bar is deflected from the vertical by an angle and released. The speed of lower end of the bar when it makes an angle ( ) from the vertical

O

L

A. 3 cos cosgL B. cos cosgL


C. cos cos3

gL D. cos cosgL

Answer :A Solution : According to conservation of energy

21 2

12

mgh mgh I

1 cos 1 cos2 2L Lmg

2

212 3

mL

O

L

2h1h

2[cos cos ]3

mLmg or 3 (cos cos )gL

3 (cos cos )V L gL

93 A ball is rising towards the surface with a constant velocity in a liquid whose density is four times that of the material of the ball. How many times of its weight is the drag force acting on the rising ball A. 2 B. 3 C. 4 D. 5 Answer :B Solution : Let be the density of the material of the ball, the equation of motion is given by

3 34 4 .4 03 3

r g r g F ; F = force of friction & r = radius of the ball

34 .33

F r g

= 3 times the weight of the ball 94 A lead ring of radius r rotates about a vertical axis 1OO passing through its centre and perpendicular to

the plane of the ring. Find the number of rotations per second at which the ring just breaks. The ultimate strength of lead is m and its density is .


O

O1

A. 18

m

r

B. 14 2

m

r C. 1

2m

r

D. 12 2

m

r

Answer :C Solution : Consider a part of the ring subtending an angle at the centre.

Mass of the element ;m r ml

mass of the ring.

Tcos / 2/ 2

T cos / 2/ 2

Tsin / 2 Tsin / 2T

r

Let T be the tension on the ring. Of the two components only T sin / 2 is operational. The net force is

22 sin2

mT T rl

or 2

2 2 2 2

2 2m m mrT r rl r

Mass of the ring 2m rA where A is the area of cross-section of the wire.

22 22 22

rT A r r A n

m = ultimate strength 2 2 2maxmax4T r n

A

12

mnr

95 Two cars A and B are moving towards each other with a speed of 30 m/s. A person sitting in car A fires shots after every 3 seconds and person sitting in car B observes them. What will be the time difference recorded by him between two consecutive shots ? Velocity of sound 330 /m s . A. 3s B. 3.6 s C. 2 s D. 2.5 s


Answer :D Solution : The apparent frequency as heard by the person sitting in car B is given by

30 330 30 6' .30 330 30 5

nn n n

Let t and 1t be the time intervals between two shots as recorded by the persons in car A and B respectively.

Then ' . 5' 3 2.5' ' 6

t n t nor t st n n

96 A material of wire having density 1.4 /g cc is not wetted by water of surface tension70 dyne/cm. Find the maximum radius of the wire which can float on the surface of water A. 0.18 cm B. 0.9 cm C. 0.54 cm D. 0.3 cm Answer :A Solution : 22LT L r g ; L = length of the wire

T = surface tension of water

r = radius of wire

density of the material of wire

1/22Tr

g

On putting the values of the various terms we have

1/22 70 0.183.14 1.4 980

r cm

97 In a region where there is no gravitation field, a small sphere of mass ‘m’ moving with an initial velocity 0V in a viscous medium experience a drag force R bV . Determine the velocity of sphere as a function of time A. /

0bt mV V e B. /

0bt mV V e

C. /0

m btV V e D. /0

m btV V e Answer :B Solution : Equation of motion of sphere is given by

dVm bVdt

dV b dtV m


logebtV Cm

At t = 0, 0V V

0logC V or /0

bt mV V e

98 What will the percentage error be in calculating the atmospheric pressure equal to 760 mm of mercury according to the height of a mercury column if the diameter of the meniscus of the liquid surface is 5 mm. Surface tension of mercury = 0.5 N/m A. 0.2 B. 0.7 C. 0.4 D. 0.9 Answer :C

Solution : 2T h gr

2Thr g

3 3

2 0.52.5 10 13.6 10 9.8

= 3.00 mm

% error = 100 3 0.4760

99 The water in a reservoir is 20 m deep. A horizontal pipe 6 cm in diameter passes through the reservoir 11 m below the water surface as shown in figure. A plug secures the pipe opening. (Take g = 10 m/s2). Find the force of friction between the plug and pipe wall

20 m11 m

plug

A. 280 N B. 310.86 N C. 300 N D. 320 N Answer :B Solution : F.B.D of plug

PlugFPf

Force of friction = Force due to pressure difference

= Pressure difference on the sides of the plug × Area of cross section of the plug

( )gh A


3 2 210 10 11 (3 10 ) 3 410 10 11 3.14 9 10

1110 3.14 9 10 1990 3.14 10 99 3.14 310.86N

Force of friction is 310.86 N 100 A water drop falls in air with a uniform velocity. Find the difference between the curvature radii of

the drop’s surface at the upper most and lower most points of the drop which are separated by the distance h is (take proper approximation regarding radii of the surfaces of drop if required i.e., difference in the radii of the surfaces is very small).

A. 3

8ghT

B. 33

8ghT

C. 3

4ghT

D. 3

2ghT

Answer :A Solution : Suppose 1R and 2R be the radii of curvatures at the upper point and lower point of the drop respectively. The pressure inside the drop at the upper end,

01

2A

TP PR

and pressure at the lower end

02

2B

TP PR

Where 0P is the atmospheric pressure.

As the drop is falling with uniform velocity, so

B AP P gh

or 0 02 1

2 2T TP P ghR R

or 2 1

1 12T ghR R


or 1 2

1 2

2R R

ghR R

It can be assumed that 2R R and 1 2R R =h

so 1 2 / 2.R R h Therefore, we get

1 2 1 22T R R gh R R

2 2h hgh

3

1 2 8ghR RT

101 A conical glass capillary tube of length 0.1 m has diameters 3 410 & 5 10 m at the ends. When it is

just immersed in a liquid at 00 C with larger diameter in contact with it, the liquid rises to 28 10 m in the tube. The density of the liquid is 4 31/14 10 /kg m and angle of contact is zero. Effect of temperature on density of liquid and glass is negligible. A. Surface tension at 00 C is 26.6 10 /N m B. Surface tension at 00 C is 28.4 10 /N m C. Surface tension at 00 C is 27.7 10 /N m D. Surface tension at 00 C is 25.4 10 /N m Answer :B Solution : If r is the radius of the meniscus in the conical tube, then from the geometry of figure we

have 1 2 1tan r r r rL h L

L

A B

r1

r2

h


i.e., 44 5 2.5 102.5 100.1 0.08 0.1

r

4. ., 3 10i e r m ;

0

2 4 4 3 2

20

18 10 3 10 10 / 9.8 /14 8.4 10 /

2 2C

m m kg m m shr gT N m

102 A closed organ pipe has length ‘l’. The air in it is vibrating in 3rd overtone with maximum amplitude ‘a’. The amplitude at a distance of l/7 from closed end of the pipe is equal to

A. A B. a/2 C. 32

a D. Zero

Answer :A Solution : The figure shows variation of displacement of particle in a closed organ pipe for 3rd overtone.

For third overtone

74

l or 4

7l or

4 7l

Hence the amplitude at P at a distance / 7l from closed end is ‘a’ because there is an antinode at that point.

Alternate : Because there is node at x=0 the displacement amplitude as function of x can be written as

2sin sinA a kx a x

For third overtone

74

l or 4

7l

7sin sin2 7 2

lA a a al

At 7lx A a


103 A uniform rectangular lamina of mass ‘m’ and size

2bb

is resting on a smooth horizontal table.

A force F is applied at point ‘C’ perpendicular to side BC.

A

B C

D

Y

F

b/2

X

b

A. angular acceleration ' ' about centre of mass of the lamina is 12

5F

mb

B. acceleration of point ‘A’ w.r.t centre of mass is 65Fm

C. acceleration of point ‘A’ w.r.t ground is ˆ ˆ6 75 5

F Fa i jm m

D. angular acceleration ' ' about the centre of mass of the lamina is 45

Fmb

Answer :B

Solution :

Y

A D

/ 4b

B FX

/ 2b

comcoma

, /COM grounda F m

/ 2Fb

22 25[ ]

12 4 48m bI b mb

24 ;5

FI mb from fig : 1 2sin ;cos

5 5

acceleration of ‘A’ w.r.t. COM is ,65A COMFam


acceleration of ‘A’ w.r.t ground, 6 175

ˆ5

ˆF Fa i jm m

104 Inside a satellite orbiting around the earth, water does not fall out of an inverted glass. This is best explained by the fact that A. The earth’s force of attraction on the water is negligibly small at the height of the satellite B. The satellite and the earth exert equal and opposite forces on the water C. The gravitational attraction between the glass and the water balances the earth’s attraction on the water D. The water and the glass have the same acceleration towards the centre of the earth. Answer :D Solution : Water and glass have common acceleration. i.e. both are freely falling bodies. Hence one will be relatively at rest with respect to other.

105 A wire of length L and six identical cells of negligible internal resistance are connected in series. The temperature of the wire is raised by T in time t due to the current. N similar cells are now connected in series with a wire of the same material and cross-section, but of length 3L. The temperature of the wire is raised by the same amount T in the same time t. The value of N is A. 24 B. 18 C. 12 D. 9 Answer :B

Solution :

2

2

6

33

EL s T tm L L

R L NEL s T t

L

2

1 36 3 6 3 183

NN

106 A particle with charge +q and mass m enters a magnetic field of magnitude B, existing only to the

right of the boundary YZ. The direction of motion of the particle is perpendicular to the direction of

B. Let 2 mTqB

. The time spent by the particle in the field will be


mq

Y

Z

x

x

x

x

x

x

x

x

A. 2 T

B. 22

T

C. 22

T

D. T

Answer :D

Solution :

2 22

t T T

107 The conductor AD in the figure moves to the right in a uniform magnetic field directed into the paper. Then which of the following statements is/are correct? I. D will acquire a higher potential with respect to A II. The free electrons in AD will move towards A when A and D are connected externally. III. The current in AD flows from lower to higher potential when A and D are connected externally.

A

D

v

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

xx

x

A. I and II only B. I and III only C. II only D. I, II and III Answer :D Solution : Apply Fleming’s right hand rule.

108 A parallel plate capacitor, filled with a dielectric of dielectric constant K, is charged to a potential 0V . It is now disconnected from the cell and the dielectric slab is removed. If it now discharges, with time constant through a resistance, the potential difference across it will be 0V after a time


A. 1ln 1K

B. ln1

KK

C. ln K D. Zero

Answer :C Solution : /

0 0t tCV kCV e

/0

1 lntV V e t kk

109 The diagram shows a uniformly charged hemisphere of radius R. It has volume charge density . If the electric field at a point 2R distance above its centre is E, then what is the electric field at the point which is at 2R distance below its centre ?

A

B

(CA = CB = 2R)C

A.

0

ρR6ε

E B. 0

ρR12ε

E C. 0

ρR6ε

E D.

0

ρR24ε

E

Answer :B Solution : Apply principle of superposition

Electric field due to a uniformly charged sphere

+ρ+ρ +-ρ-ρ

=

= 0

ρR12ε

Resultant0

ρR12ε

E E

110 The linear charge density on a dielectric ring of radius R is varying with as 0 2

cos

where

0 is a +ve constant of appropriate dimensions. The potential at the centre of the ring is

θ 0θ=0

A. 0 B. 0

02

C. 0

04

D. 0

0

Answer :A


Solution : Potential is a scalar quantity. The charge density function suggests that net charge is zero. Net charge on the ring

2π

0

λRdθQ

= 2π

00

.

cos(θ/2)dθ 0R

0

1 ( ) 04πε

QVR

111 Two infinite dielectric sheets having charge densities 1 and 2 (charge per unit area) are placed in two perpendicular planes whose two cross sectional view is shown in the figure. The charges are distributed uniformly on the sheets in electrostatic equilibrium condition. Four points are marked as I, II, III and IV. The electric field intensities at these points are 1 2 3, ,E E E

and 4E

respectively. The

correct expression for electric field intensities is

II I

IVIII

1σ

2σ

A. 2 21 2

1 2 4

02E E E

B. 2 21 2

2 402ε

E E

C. 2 21 2

1 2 3 402ε

E E E E

D. none of the above

Answer :C Solution : Using principle of superposition E

due to infinite plane sheet having surface charge

density on its one face is, 02ε

1σ

2σ

y

x

So, 1 21

0 0

ˆ2ε 2

ˆε

E i j


1 22

0 0

ˆ ˆ2ε 2ε

E i j

1 23

0 0

ˆ ˆ2ε 2ε

E i j

1 24

0 0

ˆ2ε 2

ˆε

E i j

112 1 C charge is uniformly distributed on a thin spherical shell given by equation 2 2 2 25x y z . What will be intensity of electric field at a point (1,1, 2) ? (All the quantities are in S.I units) A. 5 N/C B. 45 N/C

C. 5 3 N/C2

D. Zero

Answer :D Solution : 1 1 4 6 5r

in 0E

Fcosθ

Fsin θ

Fcosθ

Fsin θ

P

F

Qθ

113 The density of the core of a planet is 1 and that of the outer shell is 2 . The radii of the core and

that of the planet are R and 2R respectively. If the gravitational acceleration at the surface of the

planet is same at a depth R, then 1

2

is

A. 74

B. 47

C. 38

D. 73

Answer :D

Solution :


114 A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to 1F

on a particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force 2F on the same particle. The ratio 1 2/ FF is

RR

A. 12

B. 34

C. 78

D. 97

Answer :D Solution : Mass of cavity = M/8 if mass of sphere = M as volume of cavity is 1/8 of sphere.

115 In the arrangement shown in the figure, a pulley of mass M is suspended from a light spring of

stiffness constant k and a mass m is suspended form a string tightly wound on the pulley. If the mass m is given a small downward displacement and released, the time period of its small oscillations is (Assume that the string does not slip on pulley)

ROR

R/2

M


R

M

m

A. 8 322

m Mk

B. 3 822

m Mk

C. 3 82 M mk

D. 8 32 M mk

Answer :A Solution : Let the initial elongation of the spring in the equilibrium state be e. Then 02Ke Mg T

0T mg

2Ke Mg mg

When the mass m is pulled and released, let the instantaneous displacement of m be x and that of pulley be x/2.

Since the total energy during SHM remains conserved.V=speed of m at that instant

M

Ke

T0

Mg

Mg

mT0

R

2 22 21 1 1 1

2 2 2 2 2 2 2v x xmv M I K e Mg mgx

constant

2

2MRI and for no slipping

2V R v

2VWR

2

2 2 21 12 4 8 2 2 2

M M x xmv v v K e Mg mgx constant

Differentiating w.r.t. ‘t’. we get


3 1. 08 2 2 2

dv x dx Mg dx dxm M v K e mgdt dt dt dt

(2)

Using equation (1) and (2) and then rearranging we get

2

8 32

dv K x xm Mdt

8 322

m MTK

116 A uniform ring of m with outer radius 2r is fitted tightly on a shaft of radius 1r . The shaft is rotated

about its axis with a constant angular acceleration . The torque due to elastic forces in the ring as a function of the radial distance r from the axis is 1 2(r r r and the ring does not slip on the shaft).

A.

4 4

2

2 22 1

m r r

r r

B.

4 42

2 22 12

m r r

r r

C.

4 42

2 22 1

m r r

r r

D.

4 42

2 22 12

m r r

r r

Answer :B Solution : Consider a ring element of radius r and thickness dr

2 2 2 22 1 2 1

22m mr drdm r drr r r r

M.I of the part of ring between r and 2r is

2 2

2 3 4 422 2 2 2

2 1 2 1

22( )

r r

r r

m mI dm r r dr r rr r r r

Torque due to tension at a distance r from axis is the torque on part of ring between r and 2r

4 42

2 22 12

m r rI

r r

117 A coin is placed on a rough horizontal platform, which undergoes horizontal simple harmonic motion about a mean position O. The coin does not slip on the platform. The force of friction acting on the coin is F. A. F is always opposite to restoring force acting on coin. B. F is directed towards O when the coin is moving away from O and away from O when the coin moves towards O.


C. F=0 when the coin and platform come to rest momentarily at the extreme position of the simple harmonic motion. D. F is maximum when the coin and platform come to rest momentarily at the extreme position of the simple harmonic motion. Answer :D Solution : Conceptual

118 A body of mass M is attached to the lower end of a vertical metal wire, whose upper end is fixed. The elongation of the wire when M is in equilibrium is .

1) Loss in gravitational potential energy of M is Mg .

2) Elastic potential energy stored in the wire is Mg .

3) Elastic potential energy stored in the wire is 2

Mg .

4) Heat produced may be 2

Mg .

A. only 1,2 are true B. only 2,3 are true C. only 1,4 are true D. only 1,3,4 are true Answer :D Solution : Conceptual

119 A hemispherical shell of mass M and radius R is made to execute small oscillations about the symmetric point at the surface as shown in figure. Find its time period of oscillation.

RC

A. 23Rg

B. 324

Rg

C. 423

Rg

D. 2 Rg

Answer :C Solution : Centre of mass of the hemispherical shell is at a distance R/2 form the centre of hemisphere.

The torque in displaced position is sin2RMg

Oc

M.I. of the hemispherical shell about the axis passing through O is 22 .

3MR (same as the M.I. about a

diameter of the base as they are equidistant form the centre of mass)


sin2RI Mg (Torque and oppositely directed)

22 .3 2

RMR mg is small

34

gR

2 34

gR

42 43 3

R RTg g

120 Angular frequency of SHM of a particle is . There is a point P at a distance ‘x’ from the mean position ‘O’. When the particle passes P, it has velocity is v towards OP. Find the time in which it returns to P again.

A. 11 tan vx

B. 12 tan vx

C. 11 sin ( )vx

D. 12 sin ( )vx

Answer :B Solution : Let the particle is at P at an instant t starting from mean position and it returns to p again after interval t’

Then sin ...(1)x a t

Also sin ( ')...(2)x a t t

Form equation (1)

cosdx a t vdt

cos / ...(3)t v a

From equation (2)

[sin 'cos cos '.sin ]x a t t t t sin ' cos 'v xa t ta a

(using equation (1) and (3))

sin ' cos 'vx t x t


1 cos ' sin 'vx t t

2 ' ' '.2.sin .2.sin cos2 2 2t v t tx

12' tan vt

x

121 The maximum internal pressure (in the absence of an external pressure) that can be sustained by a glass spherical flask of radius 25mm and wall thickness 1mm (Tensile strength of glass is 0.05 Gpa) is nearly A. 40 atm B. 20 atm C. 10 atm D. 25 atm Answer :A

Solution : Tensile strength 2

2 2F rr r r r

52 40 10 40rP pa atmr

122 A horizontally oriented thin copper rod of length is rotated about vertical axis passing through its middle. The breaking stress of copper is and density is . The frequency of rotation at which this rod just ruptures is

A. 2 2 rpm

B. 2 22rps

C. 2 2

2 rps

D. 2 2

32

rpm

Answer :C

Solution : 2

8mamT A

2

8m A

2

8 2 n

2 2

2n rps

123 A bar of mass m, length l is in pure translatory motion with its centre of mass velocity v. It collides with and sticks to another identical bar at rest as shown in figure. Assuming that after collision it becomes one composite bar of length 2l, the angular velocity of the composite bar will be


l

lvC.M.

A. 3 ,

4vl

anticlockwise B. 4 ,3

vl

anticlockwise C. 3 ,4

vl

clockwise D. 4 ,3

vl

clockwise

Answer :A Solution : By Law of Conservation of Angular Momentum ( )systemmvr I

2 22 42 22 12 12

mmmv

34v anticlockwise

124 A bead of mass m slides without friction on a vertical hoop of radius R. The bead moves under the combined influence of gravity and a spring of spring constant k attached to the bottom of the hoop. For simplicity assume, the equilibrium length of the spring to be zero. The bead is released at the top of the hoop with negligible speed as shown. The bead, on passing the bottom point will have a velocity of

2R

A. 2 gR

B. 222 kRgR

m C.

2

2 kRgRm

D. 2

2 kRgRm

Answer :C

Solution : . .

intint int

Loss inelastic Loss inGainin

potential gravitationalK E of the

energy of potentialbead at

spring at energy of beadlowest po

lowest po at lowest po

2 21 12 22 2

k R mg R mv

2

2 kRv gRm


125 A rocket of mass m exhaust fuel of average density , with a speed v relative to the rocket. The area of cross-section of the opening of the exhaust is A. The minimum value of v to lift the rocket is

A. mgA

B. mAg C. 2mg

A D. 2m

Ag

Answer :A Solution : 2mg Av

mgvA

126 A sinusoidal wave with amplitude my is travelling with speed V on a string with linear density . The angular frequency of the wave is . Mark the one which is correct. A. doubling the frequency doubles the rate at which energy is carried along the string B. if the amplitude were doubled, the rate at which energy is carried would be halved C. if the amplitude were doubled, the rate at which energy is carried would be doubled D. the rate at which energy is carried is directly proportional to the velocity of the wave Answer :D

Solution : Power 2 212

P PW A SV P V

wave P linear density S cross section

W angular frequency V velocity of wave

A amplitude 127 A machine gun is mounted on an armoured car moving with a speed of 20 ms-1. The gun can point

against the direction of motion of car. The muzzle speed of bullet is equal to speed of sound in air i.e., 340 ms-1. The time difference between bullets actually reaching and sound of firing reaching at a target 500 m away from car at the instant of firing is

A. 1.2 s B. 0.09 s C. 0.9 s D. 9 s Answer :B

Solution : Time taken by sound to reach target 1500340

t

Time taken by bullet to reach target 2500

340 20t

1 21 125 0.09sec.

16 17t t t

128 An interference is observed due to two coherent sources ‘A’&’B’ separated by a distance 4 along the y-axis where is the wavelength of the source. A detector D is moved on the positive x-axis. The number of points on the x-axis excluding the points, x=0 & x= at which maximum will be observed is


A

B D

Y

A. Three B. Four C. Two D. Infinite

Answer :A

Solution :

A

B

4

Dx

Path difference

2 216x x x 1/22

2

161x xx

2

2

81x xx

8xx

maxfor n n

8xn

8nx

As n is integer Total maximum = 2

129 A tube of diameter d and of length unit is open at both ends. Its fundamental frequency of resonance is found to be 1 . The velocity of sound in air is 330 m/sec.One end of tube is now closed.

The lowest frequency of resonance of tube is 2 . Taking into consideration the end correction, 2

1

is

d

A.

0.60.3

dd

B. ( 0.3 )2( 0.6 )

dd

C. ( 0.6 )2( 0.3 )

dd

D. ( 0.3 )2( 0.6 )

dd

Answer :C

Solution : for open pipe 1 2( 2 )lVf V

l e

for closed pipe 2 4( )cVf Vl e


2

1

4( )2

2( 2 ) 2( )

Vl e

V l eVV l e l e

0.62( 0.3 )

l dl d

But

0.3e d 130 Two identical sources moving parallel to each other at separation ‘d’ are producing sounds of

frequency ‘f’ and are moving with constant velocity v0. A stationary observer ‘o’ is on the line of motion of one of the sources. Then the variation of beat frequency heard by O with time is best represented by: (as they come from large distance and go to a large distance)

od

ov

ovi

f

f

2S

A. t

beat

freq

B. t

bea t

freq

C. t

beat

freq

D. t

bea t

freq

Answer :C

Solution : 2 11 1

cosf f f V

V VO V VO

0 90 f increases ; 090 f is maximum

090 180 decreases


131 In the figure shown an observer 1O floats (static) on water surface with ears in air while another

observer 2O is moving upwards with constant velocity 1 / 5V V in water . The source moves down with constant velocity / 5sV V and emits sound of frequency ‘f’. The velocity of sound in air is V and that is water is 4V. For the situation shown in figure :

1O

2O

sV V / 5

1V V / 5

A. The wavelength of the sound received by 1

45VO isf

B. The wavelength of the sound received by 1 /O isV f

C. The frequency of the sound received by 245vO is

D. The wavelength of the sound received by 2 5VO is

f

Answer :A

Solution : 1 ( )

5

vf fvv

54

f 11

45

v vf f

132 Velocity versus displacement of a particle moving in a straight line is 9 4v s , here v is in 1ms

and s in meter. Mass of the particle is 2kg. The time in seconds at which average power is 34

th of

the instantaneous power is A. 0.92 B. 0.67 C. 1.5 D. 3.0 Answer :C

Solution : 34av iP P Given 2 9 4V s . On comparing with 2 2 2V u as , we get

1 23 2u ms and a ms

2 2 2 21 3 [(3 2 ) 3 ] 3[ ] [ ] 2(3 2 )2 4 2 4

f iv v tm mav tt t

2 2 24 12 9 6 2 3 1.5t t t t t t t s 133 An in extensible string of length connects two masses m and 4m and the system rests on a smooth

horizontal floor. An impulse J is imparted to B as shown in figure. The tension in the string in the subsequent motion is


A. 2J

ml B.

2

5Jml

C. 24

5Jml

D. 2

20Jml

Answer :D Solution : 5 cmj m V . In the COM frame, particles will bein circuar motion

4mB

cmV

4 / 5

comVA

m

2 2 2

,( ) ( / 5 )4 / 5 4 / 20

A cmm v m J m JTl l m ml

20

00 2

L Lxdx m dm

134 A non-uniform rod OA of linear mass density 0 0( constantx where and x is the distance from the point O) and length L is suspended horizontally from the ceiling by a light string at one end A and is hinged at other end O as shown in the figure. The angular acceleration of the rod just after the string is cut will be

AO x A. 2 /g L B. /g L C. 4 / 3g L D. 3 / 4g L Answer :C Solution : Mass of the element, dm

X

CMdx

2 4

0 20 0

000

2 03 4

LL

cm L

x dxxdm LLx I dmxdm xdx

413cm

gmgxL


135 A uniform semi circular disc hanging freely about a fixed point as shown in fig(A) has frequency of

small oscillations of . When it is suspended from another point as shown in fig(B), the frequency of small oscillations will be approximately

R R

fig(A)fig(B)

2R

A. 3 of B. 00.2 f C. 1.8 of D. 0.9 of Answer :D

Solution : 2

0 01 8 1 2 16 9,& 0.9

2 3 2 2 9g gf f f fR R

136 A block of mass 'm' is performing SHM with an amplitude A, on a frictionless surface. When it is at

extreme position, a bullet of mass 'm' moving with a velocity 0v collides and gets embeded into it at time t=0 as shown in figure. The displacement 'x' measured from mean position at any time 't' will be ( 2 2

0 2mv kA )

m m

0x

extreme position0t mean position

A

0v

k

A. 2 sin2 4kx A tm

B. 2 sin

2kx A tm

C. 2 sin2 3kx A tm

D. sin kx A t

m

Answer :A

Solution : 2

2 '201 1 122 2 4 2

vKA M KA

(by energy and momentum conservation)

(or) 2 2 '2 1 10

1 522 4 2

kKA mv KA A A andm

' 1sin 2 sin[ ]2 4kx A i A tm

137 A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components of the acceleration are equal. If its speed at t = 0 is 0u , the time taken to complete the first revolution is

A. o

Ru

B. ouR

C. 2(1 )o

R eu

D. 2

o

R eu


Answer :C

Solution : 0

12

200

1 1 1v

u

dv v dv dt R tdt R v R u v

----------(1)

0

222

020

v R

u

dv v dv dsv dt v u edt R v R

----------(2)

And From (1) & (2) we get 2

0

(1 )Rt eu

138 In a region of space, the electric field is in the x-direction and proportional to x, i.e., 0 .Ê E xi

Consider an imaginary cubical volume of edge a, with its edges parallel to the coordinate axis. The charge inside this volume is

A. Zero B. 30 0E a C. 3

00

1 E a

D. 20 0

16

E a

Answer :B Solution :

D

C G

BH

FA E

Z

X

Y

x0

Oa

a

a

The field at the face 0 0 .ÂBCD E x i

flux over the face 20 0( ) .ABCD E x a

The negative sign arises as the field is directed into the cube. The field at the face 0 0( ) .ÊFGH E x a i

flux over the face 20 0( ) .EFGH E x a a

The flux over the other four faces is zero as the field is parallel to the surfaces.

total flux over the cube 30

0

1 ,E a q

where q is the total charge inside the cube.

30 0 .q E a

139 A large flat metal surface has a uniform charge density . An electron of mass m and charge e leaves the surface at point A with speed u, and returns to it at point B. Disregard gravity. The maximum value of AB is

A. 2

0u me

B.

20u e

m

C. 2

0

u em

D. 2

0

u em


Answer :A

Solution : The force on the electron is 0

e

and its acceleration towards the metal sheet is0

.em

The

electron will move as a projectile with an effective value of 0

.egm

Its maximum range will then be

220

0

u mue em

.

140 The two ends of a uniform conductor are joined to a cell of emf and some internal resistance. Starting from the midpoint P of the conductor, we move in the direction of the current and return to P. The potential V at every point on the path is plotted against the distance covered ( )x . Which of the following best represents the resulting curve?

A.

v

x

B.

v

x

<

C.

v

x

D.

v

x

<

Answer :B Solution : When we move in the direction of the current in a uniform conductor, the potential decrease linearly. When we pass through the cell, from its negative to its positive terminal, the potential increases by an amount equal to its potential difference. This is less emf, as there is some potential drop across its internal resistance when the cell is driving current.

141 A capacitor A with charge 0Q is connected through a resistance to another identical capacitor B, which has no charge. The charges on A and B after time t are AQ and BQ respectively, and they are plotted against time t. Find the correct curves.

A.

Q02

Q0

QB

QA

t B.

Q02

Q0

QB

QA

t C. t

QB

QA

Q02

Q0

D.

QB

QA

Q02

Q0

t Answer :A Solution : Conceptual

142 Let S be an imaginary closed surface enclosing mass m. Let dS

be an element of area on S, the direction

of dS

being outward from S. Let E

be the gravitational intensity at .dS

We define . ,SE dS

∮ the

integration being carried out over the entire surface S. A. Gm B. 4 Gm


C. 4Gm

D. no relation of the type (a), (b), or (c) can exist

Answer :B Solution : Follow the method used to prove Gauss’s law.

2. mE Gr

0. cos(180 ) cosE dS EdS EdS

2

2

. cos

cos. . 4 .

S S

S S

mE dS G dSr

dSGm Gm d Gmr

∮ ∮

∮ ∮

E

r

S

dS

m

143 A particle of charge per unit mass is released from origin with velocity 0ˆv v i

in a magnetic field

0ˆB B k

for 0

0

32

vxB

and 0B

for 0

0

32

vxB

. The x-coordinates of the particle at time

03t

B

would be

A. 00

0 0

3 32 2

v v tB B

B. 00

0 0

32 3

v v tB B

C. 0 0

0 0

32 2 3

v v tB B

D. 0 0

0

32 2

v v tB

Answer :C

Solution : 0 0

0 0

mv vrB q B

0

0

3 sin 602

6 3OA

xr

TtB

Therefore, x-coordinate of particle at any time 03

tB

will be


xO

A

x

xx

x

x

x

x y

x

x

xr

x

x

x

x

x

x

v0

v0

00

00 0

3 cos 602 3

vx v tB B

0 0

0 0

32 2 3

v v tB B

144 An infinite current carrying conductor, parallel to z-axis is situated at point P as shown in the figure. The

find . ?B

A

B d

0,a

0,0 ,0aA

B

YP

/ 3 , 0a

OX

A. 0

24i B. 0

16i C. 0

12i D. 0

8i

Answer :A Solution :

x

Y

X

a B

02 22iB

a x

02 2 2 2

3

. . .2

B x a

aA x

i aB d dxa x a x

0 02 2

3

2 24

x a

ax

ia idxx a


145 Two identical non relativistic Particles move at right angle to each other, possessing de Broglie wavelengths 1 and 2. The deBroglie wavelength of each Particle in their centre of mass frame is

A. 1 2

2 B.

2 21 2

2

C. 1 22 2

1 2

D. 1 2

2 21 2

2

Answer :D Solution :

V2

V1 1 2 1 2ˆ ˆ ˆ ˆ2 2cm

mv i mv j v i v jVm

1 2 1 21 1 1

ˆ ˆ2 2

ˆ ˆˆc c

v i v j v i v jV v v v i

1 21 2

h hmv mv

1 21 2 2 2 2

1 1 2 1 2

22c

c

h hmv m v v

Similarly 1 22 2 2

1 2

2c

146 The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from 1 2to . The work function of the metal is

A. 2 11 2

2hc

B. 1 2

1 2

(2 )hc

C. 1 2

1 2

( )hc

D. 1 2

1 2

( )hc

Answer :A

Solution : maxhcKE W

147 A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 watt of power, the force exerted by it on the surface is A. 83.2 10 N B. 73.2 10 N C. 75.12 10 N D. 85.12 10 N Answer :B Solution : Force is rate of change of momentum. Power absorbed is 0.4 P and power reflected is 0.6 P.

Force 0.4 2 0.6P PFC C

Where C is the velocity of light.


148 A resistor and an inductor in series are connected to a battery through a switch. After the switch has been closed, what is the magnitude of current flowing when the rate of the increase of magnetic energy stored in the coil is at a maximum?

L R

V

A. 4VR

B. 3VR

C. 2VR

D. VR

Answer :C

Solution : The rate of increase of magnetic energy 2

2LIE

is the difference between the power output

of the battery and the power dissipated in the resistor. 2 2 2

2

2 4 4dE V V VVI RI R Idt R R R

. The rate of increase is maximum when 2VIR

149 Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii 1 2R and R respectively. The ratio of mass of X to the mass of Y is

A. 1/2

2

1

RR

B. 2

1

RR

C. 2

1

2

RR

D. 1

2

RR

Answer :C Solution : Let the masses of X and Y be 1 2m and m and let their velocities after being accelerated be

1 2and respectively. Since the particles have equal charges and have been accelerated through the same potential difference, their kinetic energies are equal,

i.e., 2 21 1 2 2

1 12 2

m m

In a uniform magnetic field B, the radii of the circular paths are given by 2

1 1 1 11

1 1

m mqB or qBR R

22 2 2 2

22 2

m mand qB or qBR R

Thereforce 2 2 2 2

1 1 2 2 1 1 2 22 2

1 2 1 2

m m m mand orR R R R

But 2 21 1 2 2m m . Therefore, we have


2

1 1

2 2

m Rm R

150 Three long, straight and parallel wires C, D and G carrying currents are arranged as shown in figure. The force experienced by a 25 cm length of wire C is

A. 0.4 N B. 0.04 N C. 34 10 N D. 44 10 N Answer :D Solution : The magnetic field due to wire D at wire C is

740 2 10 2 30 2 10

4 0.03DIB T

r

Which is directed into the page. Similarly, the field due to wire G at C is

7410 2 20 0.4 10

0.1GB T

Which is directed out of the page. Therefore, the field at the position of the wire C is

4 4 42 10 0.4 10 1.6 10D GB B B T And is directed into the page. The force on 25 cm of wire C is

0 4 4sin 90 1.6 10 10 0.25 4 10F BIl N

151 The system shown consist of two springs. If the temperature of the rod is increased by .T The coefficient of linear expansion of the material of rod is . If thermal stress in rod is zero, compression in left spring is

K 3KL

A. 4

L T B. 34

L T C. 3

L T D. Zero

Answer :B


Solution :

K 3K

1x 2x

1 2l x x

1 2l x x ______________(1) under Equilibrium of rol 1 23Kx Kx

2 1/3x x ____________________(2) From (1) & (2)

13

4lx

152 A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is A. virtual and at a distance of 16 cm from the mirror B. real and at a distance of 16 cm from the mirror C. virtual and at a distance of 20 cm from the mirror D. real and at a distance of 20 cm from the mirror Answer :B Solution : For biconvex lens, we have 1 1 1

u f

1 1 130 15

30cm For mirror 1 1 1

u f

20 ( )plane mirror Again for lens,

10u 6cm

from mirror it is at 16 cm Hence, option (b) is correct.

153 A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is A. 0.02 mm B. 0.05 mm C. 0.1 mm D. 0.2 mm Answer :D Solution : The least count Q is equal to 1 M.S.D – 1 V.S.D Here 20 V.S.D = 16 M.S.D


16 41 . . . . . .20 5

V S D M S D M S D

Hence, least count = 1 M.S.D- 4 . .5

M S D

Least count = (1mm) x 1 0.25

mm

154 The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is A. 0.9% B. 2.4% C. 3.1% D. 4.2% Answer :C

Solution : Least count of screw guage .

PitchNo of circular divisions

0.5 150 100mm mm

Diameter D = PSR + CSR 12.5 20 2.7

100mm

The density 343 2

mD

% error in density 100 3 100 %m Dm D

3.1%

155 A light ray traveling in glass medium is incident on glass-air interface at an angle of incidence . The reflected (R) and transmitted (T) intensities, both as function of , are plotted. The correct sketch is

A.

Inte

nsity

100%

0 090

T

R

B.

Int e

n sit y

100%

0 090

T

R

C.

Inte

nsi t y

100%

0 090

T

R

D. 0

100%

I nt e

n sit y

θ 90o

T

R


Answer :C Solution : The incident ray will undergo in TIR, when c .Hence no transmission take place.

Glass

Air

Hence for the range of

Transmission will be more0

Reflection willbe littlee

For Notransmissionwill take placee

156 5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. Taking the initial temperature to be 1,T the work done in the process is

A. 198

RT B. 132

RT C. 1158

RT D. 192

RT

Answer :A Solution : 1 1

1 1 2 2TV T V or 2/3 2/3

1 2(5.6) (0.7)T T 2 14T T

The work ( )1

nR TW

11

(3 ) 92 / 3 2

nR TW nRT

1 1 1PV nRT 14

n

19 .8

W R T

157 In a Young’s double slit experiment, the separation between the slits is 1.0 mm and the distance between the slits and the screen is 1.0 m. The light falling on the slits contains mainly two wavelengths 600 nm and 500 nm. The least distance from the centre of the fringe pattern where the intensity corresponding to one of these wavelengths is zero, is A. 0.30 mm B. 0.75 mm C. 0.25 mm D. 1.20 mm Answer :C

Solution : 2 2

oyd

4 6

3

1 500 10 102 1 10

6250 10 30.25 10 m 0.25mm


158 An ideal gas undergoes two successive process A and B. In the process A, the values of the increase Uin internal energy and the work W done by the gas are 72U J and 72W J respectively. For the process B, 0U A. Process A is adiabatic, process B is isochoric B. Process A is adiabatic, process B is isothermal C. Process A is isothermal, process B is adiabatic D. Process A is isobaric, process B is adiabatic Answer :B Solution : Conceptual dc=0 adiabatic, 4d =0 isothermal

159 Light is incident at an angle with the normal to a plane containing two slits of separation d. Select the expression that correctly describe the positions of the interference maxima in terms of the incoming angle and outgoing angle . (m = order of maxima)

d

A. Notransmissionwill take pla

since

sin e d

B. sind m

C. sin sin ( 1)md D. sin sin m

d

Answer :D

Solution : Path difference sin sind d for maxima, sin sin mx md

160 Figure shows two coherent point sources 1S and 2S vibrating in same phase. AB is an irregular wire lying

parallel to 1 2S S and at a far distance from the sources 1S and 2S . Let 310 0.12BOAd . Totally,

how many bright spots can be seen on the wire, including pointsA and B ? A

B

d

1S

2S

O

A. 2 B. 3 C. 4 D. more than 4 Answer :B

Solution : B) Angular width 310d (given)

No.of fringes within 0.12 will be

3

0.12 2 [2.09]360 10

n


The number of bright spots will be three including central bright

161 Samples of two radioactive nuclides, X andY , each have equal activity A0 at time 0t . X has a half-life of 24 years and Y a half-life of 16 years. The samples are mixed together. What will be the total activity of the mixture at 48t years?

A. 012

A B. 014

A C. 03

16A D. 0

38

A

Answer :D Solution : Given X has activity 0A at 0t and its half-life is 24 years Y has activity 0A at 0t and its half-life is 16 years

At 48t years, activity of 014

X A


Y A


Y A


Y A

Thus, total activity of the mixture of X and Y at 48t years is 0 0 01 1 34 8 8

A A A

162 A 45 10 Å photon produces an electron – positron pair in the vincinity of a heavy nucleus. Rest energy of electron is 0.511MeV . If they have the same kinetic energies, the kinetic energy of each particle is nearly A. 1.2MeV B. 12MeV C. 120MeV D. 1200MeV Answer :B Solution : If the kinetics energy of each particle is k, then

3

4

12.4 10 24.8 1.0222 2 0.51 11.95 10 2

hc MeVAk MeV k MeVA

163 A radioactive sample undergoes decay as per the following graph. At time 0t , the number of undecayed nuclei is 0N . Calculate the number of nuclei left after 1h .

6.93

N

N0

t (min)

0N4

A. 8

0 /N e B. 100 /N e C. 12

0 /N e D. 140 /N e

Answer :C Solution : 00,t N N 06.93, / 4t N N


0 / 4N is the sample left after two half-lives

1/22 6.93t 10.6932 6.93 0.2 min

60mint 0.2 60 0

0 0 12t NN N e N e

e

164 In the circuit shown in fig, when switch S is open, the potential difference across the capacitor is72

oP Q o R Q

VV V V and V V . If the switch ‘S’ is closed, then energy in the capacitor of capacitance

3C, when the current in the inductor is maximum is

2C

SL

3C

P

R

Q

A. Zero B. 2

032

CV C. 203CV D. 2

06CV

Answer :D

Solution : For maximum current di odt

Q

3C2C

SP

R

x 10CV x

Let x be the maximum charge on 3C. 10

3 2CV xxso

c c

6x CV

energy 2

262 3

x CVC

165 A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric

constant K = 2. The level of liquid is 3d initially. Suppose the liquid level decreases at a constant speed

v, the time constant as a function of time ' 't is


d Rd/3

\/\/\/\/

A. 6

5 3o R

d vt

B. 2 2 2

(15 )3 3 9

od q vt Rd dvt v t

C. 65 3

o Rd vt

D. 2 2 2

(15 )2 3 9

od q vt Rd dvt v t

Answer :A Solution : The liquid level falls by t in time ‘t’. The capacitor is a combination of two capacitors

connected in series with capacitances as , 1

3

K AC

d t

and 2 2

3

AC

d t

Here K = 2 and A = 1 unit

1 2

1 2

65 3e

C CCC C d t

time constant, 6

5 3R

CRd t

166 A stationary hydrogen atom of mass m in the ground state achieve minimum excitation energy after head- on, inelastic collision with a moving hydrogen atom. Find the velocity of moving hydrogen atom

A. 1/210.2( )ev

m

B. 1/240.8( )ev

m

C. 1/220.4( )ev

m

D. 1/2

40.8( )1.0078

evm

Answer :B

Solution : 22umU m . 2 2 21 1 1.2

2 2 4E mu m mu

Minimum excitation energy is to excite a hydrogen atom from ground state to first excited state 10.2ev

21 10.24

mu ev 1240.8( )evu

m

167 A magnetic field, confined in a cylindrical region of radius R, is changing at the rate of 14Ts . A

conducting rod PQ of length 32

R is placed in the region as shown. The induced emf across the rod will

be

PC

Q

60


A. 2

( 6)24R

B. 2

24R C.

2

6R D.

2

( 6)6

R

Answer :D

Solution : 0tan 60 PQCP

332

CP R

P

C

Q

T

S

2 23 2

2 22R R RCP and CQ R

01cos , 452 2

CP RCS R

total flux , B area of CPS CST

2 2 2

264 64 24 24 6

dB R R RRdt

168 In the diagram shown, a non – uniform magnetic field oB B x has been applied in the direction shown. A particle of mass m and charge – q is projected with a velocity ‘U’ from origin towards positive x-axis. The displacement of the charged particle along x-direction when its velocity becomes parallel to y – direction is

x

y

o

A. o

muqB

B. 2

o

muqB

C. 2 o

muqB

D. 4

o

muqB

Answer :B

Solution : Since the magnetic field is a linear function of ,2 2

o oare

O B R B Rx So B B

R = maximum horizontal distance 2

2o o

mu mu muR RB RBq B qq

169 In the circuit shown in fig, if both the bulbs 1B and 2B are identical


~

L = 10mH

1B

2B

500c F

220 ,50V Hz

A. Their brightness will be the same B. 2B will be brighter than 1B C. As frequency increases brightness of 1B will decrease D. Only 2B will glow because the capacitor has infinite impedance Answer :B Solution : 1 2Let i and i be currents through 1B and 2B then,

22 2 2

1 21220, 220i R X L i R

CW

2 22 2

6 22

2 12 2 22 2 21

1 1500 10 100 40

9.8710 100

R RCWi R i i

i RR Lw R

Bulb 2B will be brighter. As frequency increases, CX decreases, LX increases. 2I becomes less, 1i increases.

Brightness of 1B will increase and that of 2B decrease.

170 P-T diagram is shown below. Then choose the corresponding V-T diagram P

T

A

B

C

D

A.

VA

B C

T

D

B.

VA

B C

T

D

C.

V

A

B C

TD

D.

V

A

B C

T

D

Answer :D Solution : AB, CD are isothermal and BC is isochoric. , ,B A B C D CV V V V V V

171 Which of the following will have maximum total kinetic energy at temperature 300 K?

A. 21kg H B. 1kg He C. 21 12 2

kgH kg He D. 21 34 4

kg H kg He

Answer :A


Solution : Total kinetic energy = internal energy 2fU nRT

In case of 2H degrees of freedom is greatest (5) and number of moles n is highest. So, this is the case of maximum kinetic energy.

172 n moles of a gas filled in a container with a frictionless piston at temperature T is in thermodynamic equilibrium initially. If the gas is compressed slowly and isothermally to half of its initial volume, work done by the atmosphere on the piston is

A. 2

nRT B. 2

nRT C. 1ln 2

2nRT

D. ln 2nRT

Answer :A

Solution : Work done by atmosphere = atmP V 2atmVP …..(i)

Initially, gas in container is in thermodynamic equilibrium with its surroundings. Pressure inside cylinder atmP from PV nRT

atmP V nRT or atm

nRTVP

Putting in (i), 2

nRTW

173 A diatomic ideal gas is heated at constant volume until the pressure is doubled and then heated at constant pressure until volume is doubled. The average molar heat capacity for whole process is

A. 136

R B. 196

R C. 236

R D. 176

R

Answer :B Solution : Let intial presure, volume, temperature be 0 0 0, ,P V T indicated by state A in P-V diagram. The gas is then isochorically taken to state 0 0 0(2 , 2 )B P V T and then taken from state B to state 0 0 0(2 , 2 , 4 )C P V T isobarically.

P

V

02P

0P0T

02T 04T

A

B

02V0V Total heat absorbed by 1 mole of gas

0 0 0 0(2 ) (4 2 )v PQ C T T C T T 0 0 05 7 1922 2 2

RT R T RT

Total change in temperature from state A to C is 03T T

Molar heat capacity 0

0

19192

3 6

RTQ RT T

174 In the figure shown, AB is a rod of length 30 cm and area of cross-section 21.0cm and thermal


conductivity 336 SI units. The ends A and B are maintained at temperatures 020 C and 040 C respectively. A point C of this rod is connected to a box D , containing ice at 00 C , through a highly conducting wire of negligible heat capacity. The rate at which ice melts in the box is [Assume latent heat of fusion for ice, 180cal.fL g ]

Highly conducting wire

020 CA BC

040 C

00 CIce D

10cm 20cm

A. 184 .mg s B. 184 .g s C. 120 .mg s D. 140 .mg s Answer :D

Solution : Thermal resistance of LACKA

3

4

0.1 10 (suppose)336 1 10 336

R

Thermal resistance of 4

0.2 2336 10

BC R

Temperature of 00C C

1 220 40 20;

2H H

R R R

020 CA BC

040 C

00 CIce D

H

00 C

1H 2H

1 2 3 3

40 40 336 13440 13.4410 10

H H H WR

Rate of melting of ice 1 113.44 / 4.2 . 40 .80f

H g s mg sL

175 Three separate segments of equal area 1 2,A A and 3A are shown in the energy distribution curve of a blackbody radiation. If 1 2,n n and 3n are number of photons emitted per unit time corresponding to each area segment respectively, then

1A 2A 3A

E

A. 2 1 3n n n B. 3 1 2n n n C. 1 2 3n n n D. 3 2 1n n n Answer :D Solution : Area ( )A E d


But 1dQE ddt a

a=surface area of body ( ) / 1dn hcA

dt a

dn A a dndt hc dt

3 2 1 3 2 1n n n

176 A uniform magnetic field of induction B is confined to a cylindrical region of radius R. The magnetic field

is increasing at a constant rate of dBdt

(tesla/second). An electron of change q, placed at the point P on the

periphery of the field experiences an acceleration:

P

R

A. 1

2eR dBm dt

toward left B. 12

eR dBm dt

toward right

C. eR dBm dt

toward left D. Zero

Answer :A Solution : If we consider the cylindrical surface to be a ring of radius R, there will be an induced emf due to changing field.

R

P . d dBE dl A

dt dt

(2 )E R dBAdt

2 dBRdt

2R dBE

dt

Force on the electron.

2eR dBF Ee

dt 1

2eR dBm dt

As the field is increasing being directed inside the paper, hence there will be anticlockwise induced current (in order to oppose the cause) in the ring (assumed). Hence there will be a force towards left on the electron.

177 One of the lines in the emission spectrum of 2Li has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is: A. 4 2n n B. 8 2n n C. 8 4n n D. 12 6n n Answer :D Solution : For 2nd line of Balmer seires in hydrogen specturm


2 2

1 1 1 3(1)2 4 16

R R

For 2Li : Which is satisfied by only (D).

178 A ray of light is incident on a thin film. As shown in figure M,N are two reflected rays and P, Q are two transmitted rays. Rays N and Q undergo a phase change of . Correct ordering of the refracting indices is:

1n

2n

3nP

Q

M N

A. 2 3 1n n n B. 3 2 1n n n C. 3 1 2n n n D. None of these, the specified changes can not occur Answer :B Solution : Ray N undergoes reflection at surface II with phase change of 3 2n n Ray Q undergoes a phase-change of at II, but there is no phase change when it is reflected from surface I. 1 2n n

179 In which of the following process the number of protons in the nucleus increases. A. - decay B. - decay C. - decay D. k – capture Answer :B Solution : For - decay : 4

2y y

x xA B

For decay : 01 1

y yx xA B

For decay : 01 1

y yx xA B

For k – capture : there will be on change in the number of protons. Hence, only case in which no of protons increases is decay Hence (B).

QUICK REVISION TEST

MULTIPLE CORRECT CHOICE TYPE QUESTIONS

1 The motion of a body is given by the equation 6 3dV V

dt . If the body starts from rest,

A. its speed when the acceleration is zero is 12 ms

B. the speed varies with time as 3 12 tV e ms

C. the speed is 10.1ms when acceleration is half the initial value

D. the magnitude of initial acceleration is 26 ms Answer :A,D

Solution : acceleration = 0 0dvdt

6-3v=0 12v ms At t=0 v=0 26a ms

6 3dv vdt

0 06 3

v tdv dtv

32(1 )tv e C,D) initially, v = 0 6ia a = 6 – 3v

6 6 3 3 3 1 /2

v v v m s

2 A body falls starting from rest from a height 80 m. If 210g ms

A. Ratio of times taken to travel second metre and third metre are 2 1: 3 2

B. Magnitude of average velocity in the total journey 120ms

C. Magnitude of change in velocity from start of 2nd second to end of 4th second is 130 ms

D. If the gravity becomes horizontal at t=2s, magnitude of its velocity at t=4s is 120 2 ms

Answer :A,B,C,D

Solution :

Sai charan

Typewritten text


A) First meter Second metre Third metre

12 1t

g

22 2 2 1t

g g

33 2 2 2t

g g

B) 180 204avV ms

C) 110 3 30v gt ms

D) 1 ˆ20 0ˆ 2V j i

1| | 20 2.V m 3 Nail/nails are driven horizontally into a vertically placed drawing board. As shown in the figure, a small steel

ball is dropped from point ‘A’ and reaches point B by bouncing elastically on the protruding nail/nails, along the

edge DB [ which are not shown in the figure.]

10m

A

B

C

D

5m

A. Component of initial acceleration of ball along AB is / 5g

B. It is possible to arrange the nails so that ball bouncing on nails reaches ‘B’ quickly than a ball sliding without

friction in the straight path AB.

C. With the arrangement of the nails we can see that the ball reaches B in less than 0.4 s

D. Component of initial acceleration of ball along AB

is / 2g

Answer :A,B

Solution : A) 5cos

125 5gacceleration g g

10

5

B) Suppose the ball falls for 5m it takes 1s to reach the bottom and then reaches point ‘B’ by bouncing

practically horizontally along a row of closely spaced nails near the bottom edge of board. The speed with

which the ball reaches bottom is 110ms . Time taken to travel horizontal distance is 1s. Total time = 2s.

A ball sliding from A to B takes time ‘t’ 212

s at

D

ball

21 105 5 .2 5

t 25010

t 5t which is slightly more than 2s.

C) It will take a minimum time of 2s hence c is wrong.

4 Position vector of a particle varies with time t (in seconds) as 2 3 ˆ2 ˆr t i tj meters. Then

A. magnitude of average velocity in 3rd second is 109 /m s .

B. average acceleration is same in any time interval.

C. direction of average velocity in first three seconds makes an angle of 1 1tan2

with positive x-axis.

D. instantaneous velocity and acceleration are always in same direction.

Answer :A,B,C

Solution : A) at t = 3s, 23 2(3) 3(3) (18ˆ ˆ ˆ 9 )ˆr i j i j m

at t = 2s, 22 2(2) 3(2) (8ˆ ˆ ˆ ˆ6 )r i j i j m

displacement in 3rd second 3 2 10ˆ 3 ˆS r r i j

2 2| | 10 3 109S m

Average velocity 109 109 /1

m m ss

B) 4 ˆ ˆ3drv ti jdt

Acceleration 4ˆdva idt

. As a is time independent, average

acceleration will be same in any time interval

C) At t = 0, 0r

At t = 3s, 3ˆ18 9 ˆr i j

Displacement in first 3 seconds is 3 18ˆ 9 ˆS r r i j

Average velocity 6 3 /3

ˆ ˆa

Sv i j m s

Angle made by av with positive x-axis is 1 13 1tan tan6 2

5 Two particles A and B start simultaneously from the same point and move in a horizontal plane. A has an initial

Velocity 1u due east and acceleration 1a due north. B has an initial velocity 2u due north and acceleration 2a

due east.

A. Their paths must intersect at some point.

B. They must collide at some point.

C. They will collide only if 1 1 2 2a u a u

D. If 1 2u u and 1 2a a , the particles will have the same speed at some point of time.

Answer :A,C,D

Solution : A) 2 21 1 2 2 2 2

1 1,2

ˆ ˆ ˆ2

S u ti a t j S a t i u tj

2a

1a

O1u

2u

O

P

y

x

C) 2 21 1 2 2 1 1 2 2

1 1, , ,2 2

x u t x a t y a t y u t

They collide only if 1 2x x and 1 2y y simultaneously

2 11 2

2

21 (1)2

uu t a t ta

2 21 2

1

21 (2)2

ua t u t ta

(1) = (2) 1 21 1 2 2

2 1

2 2u u u a u aa a

D) 1 1 1 2 2 2,ˆ ˆ ˆ ˆv u i a tj v a ti u j

2 2 2 2 2 21 2 1 1 2 2v v u a t a t u

2 2

2 2 2 2 2 2 11 2 2 1 2 2

1 2

( ) u ut a a u u ta a

Clearly 2 1u u and 1 2a a or 1 2u u and 2 1a a .

6 A man is standing on a road and observes that rain is falling at angle 045 with the vertical. The man starts

running on the road with constant acceleration 20.5 /m s . After a certain time from the start of the motion, it

appears to him that rain is still falling at 450 with vertical, with speed 2 2 /m s . Motion of the man is in the

same vertical plane in which the rain is falling. Then which of the following statement (s) are true ?

A. It is not possible

B. Speed of the rain relative to the ground is 2m/s.

C. Speed of the man when he finds rain to be falling at angle 045 with the vertical, is 4m/s.

D. The man has travelled a distance 16m on the road by the time he again finds to be falling at angle 045 .

Answer :C,D

Solution : rg rm mgV V V

rm rg mgV V V

0 045 cos 45rm rgV cos V

2 2 /rm rgV m s V 0 0cos 45 cos 45rm mg rgV V V

1 12 2 2 2 4 /2 2mgV m s

Using 2 2 2v u as for the motion of man, s = 16m. 7 The measured values of two resistances 1 2&R R are 1 (100 0.3)R , and 2 (200 0.4)R . Then

A. The value of 1 2R R is (300 0.7)

B. The value of 2 1R R is (100 0.1)

C. The value of 2 1R R is (100 0.7)

D. The value of effective resistance PR when connected in parallel is given by1 2

1 1 1

PR R R . Then

(66.7 0.18)PR .

Answer :A,C,D

Solution : 1 2(100 0.3) , (200 0.4)R R

A) 1 2 (100 200) 300R R R

1 2 0.3 0.4 0.7R R R

(300 0.7)R R

C) 2 1 200 100 100R R R

1 2 0.3 0.4 0.7R R R

(100 0.7)R R

D) 1 2

1 1 1 1 1 1 200 66.7200 100 3

RR R R R

21 2 1 22 2 2 2 2

1 2 1 2

[ ]dR dR dR dRdR dR RR R R R R

2

2 2

200 0.3 0.4 4 1.60.3 0.1 0.183 100 200 9 9

dR

(66.7 0.18)R dR .

II method to find dR :

1 2 1 2 1 2 1 2 1 22

1 2 1 2

( ) ( ) ( ) ( )( )

R R R R d R R R R d R RR dRR R R R

1 2 1 2 2 1 1 2 1 22

1 2

( )( ) ( )( )( )

R R R dR R dR R R dR dRdRR R

2 2 2 2 2 2

2 1 1 2 2 1 2 1 2 12 2 2 2 2 2

1 2 1 2 1 2 2 1 2 1

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

dR R dR R dR R R dR R RR R R R R R R R R R

2 1 22 2

1 2

[ ]dR dRRR R

.

8 Three persons A,B and C lying at the vertices of an equilateral triangle of side length l are moving with constant

speed v. They are moving in such a way that velocity of A is always pointed towards B, velocity of B is always

pointed towards C and that of C is pointed towards A. Then

lvA B

C

vv

A. They never meet

B. They meet at the centroid of triangle in a time 23

lv

.

C. Average speeds of persons over the total time of motion is v.

D. Magnitudes of average velocities of persons over the total time of motion is32

v.

Answer :B,C,D Solution : B) Directions of velocities of particles change continuously but they always lie at the vertices of triangle whose side length decreases continuously. Finally they meet at centroid O.

Relative velocity between any two particles along line joining them 32 2v vv . This is constant throughout

the motion until they joint at O. Relative displacement = .

Time of meeting 23 / 2 3rel

Stv v v

.

O

060A B

C

v

v

v

C) As they are moving with constant speeds, average speed in any time interval = instantaneous speed v.

D) Displacement of any person S = AO = BO = CO

magnitude of average velocity / 3

2 / 3aSvt v

32

v

9 In the figure shown, the wedge is fixed and the masses are released from rest. The coefficient of friction between

4 kg and wedge is 0.8 and between 2kg and wedge is 0.6. Which of the following statement is / are correct?

2( 10 )g ms

24

037

A. Accelerations of blocks must be same

B. Friction on 4 kg is 24 N

C. Friction on 2 kg is 12 N

D. Normal reaction between blocks is non zero

Answer :B Solution : 2kg block accelerates down

2 0.6 2 10 cos37kgf

485

N

4 sin ( does not move down)kgf mg it

34 10 245

N

10 In the figure shown, all the surfaces are smooth. Blocks A and B are movable, whereas C is fixed. X- axis is Horizontal and y – axis is vertical as shown. Just after the system is released from the position as shown, pick up the correct options.

A

C

B

y

x

A. Acceleration of ‘A’ relative to ground is in negative y- direction B. Acceleration of ‘A’ relative to B is in positive x – direction C. The horizontal acceleration of “B’ relative to ground is in negative x – direction. D. None of the above Answer :A,B,C Solution : There is no horizontal force on block A, therefore it does not move in X-direction, whereas there is net downward force ( )mg N is acting on it, making its acceleration along negative y – direction. Block B moves downwards as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain , thus w. r. t B, A moves in positive x- direction

11 The value of mass m for which the 100 kg block can remain in static equilibrium is 2( 10 )g ms

100

m

0370.3

A. 35 kg B. 37 kg C. 83 kg D. 85 kg

Answer :B,C

Solution : For system remain in equilibrium net force is zero if m is greater than force of 100 kg along inclined

then system has tendency to move upward then friction will act downward.

0 0100 sin 37 100 cos37mg g g

3 3 4100 100 60 24 845 10 5

m

And if m is lesser then system has tendency to move downward, friction will act upward on 100 kg block .

0 0100 sin 37 100 cos 37 36g mg g m

So we got the range of m 36 84m In this range 37 and 83 lie

12 All the blocks shown in the figure are at rest. The pulley is smooth and the string is light. Coefficient of friction

at all the contacts is 0.2. A frictional force of 10 N acts between A and B. The block A is about to slide on

block B.

C

A

B5 kg

A. The normal reaction exerted by the ground on the block B is 110N

B. The normal reaction exerted by the ground on the block B is 50 N

C. The friction force exerted by the ground on the block B is 20 N

D. The friction force exerted by the ground on the block B is zero

Answer :A,D

Solution : The frictional force on block A is

2 150 10 110N N N

The net force on block B in vertical direction is zero

110N Normal reaction exerted by ground on block B is 10 10 0f

The net force on block B in horizontal direction is zero

frictional force exerted by ground on block B is zero

13 In the figure shown, all the contacts are smooth . Strings and spring are light. ‘A’ is held at rest by some means

and ‘B’ and ‘C’ are at rest and in equilibrium also .Find out the acceleration of each block just after the block

‘A’ is released. Masses of A, B and C are M, M and 2M respectively

A C

B

A. , ,2 2g g g B. , ,

2 2g gg

C. , , 0g g D. , , 0

2 2g g

Answer: D

Solution : 2mg

0ca

1T

Mg

1 2T MgB

2T

a2T

1T

Before block A was released, the system was at rest, and all blocks were in equilibrium. hence , tension in

both the strings is equal to 2Mg.

When block A is released , it will have an unbalanced force on it and hence the tension in string (2) will change

to say 2T . Now the arrangement is as shown in the figure since, tension is spring does not change

Instantaneously, hence, tension in string 1 will remain same i. e 2 mg . Thus , Block C will remain at rest and

0ca

Newton’s law along the string (2) , 2mg - mg = ma2ga

Hence acceleration of , , & 02 2g gA B C

14 The diagram given shows how the net interaction force between two particles A and B is related to the distance

Between them, when the distance between them varies from 1x to 4x . Then

O

Attraction

Repulsion

1x

2x 3x 4x

A. The potential energy of the system increases from 1x to 2x

B. Potential energy of the system increases from 2x to 3x

C. Potential energy of the system increase from 3x to 4x

D. Kinetic energy increases from 1x to 2x and decreases from 2x to 3x Answer :B,C,D Solution : If the interacting force is conservative

(1) If the nature of force is attractive on decrease of separation P.E decreases and vice versa.

(2) If the nature of the force is repulsive on decrease of separation P.E increases and vice versa

15 A man of mass m, standing at the bottom of the staircase of height L, climbs it and stands at its top. A. Work done by all forces on man is equal to the rise in potential energy mgL. B. Work done by all forces on man is zero C. Work done by the gravitational force on man is mgL D. The reaction force from a step does not do work because the point of application of the force does not move while the force exists Answer :B,D Solution: Work done by all the forces is equal to change in kinetic energy. Here 0initial finalKE KE

0 0netKE W

16 In a children’s park, there is a slide which has a total length of 10 m and a height of 8m. A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of his weight. Then, A. the work done by ladder on the boy as he goes up is zero B. the work done by ladder on the boy as he goes up is – 1600 J C. the work done by slide on the boy as he comes down is – 600 J D. the work done by slide on the boy as he comes down is 1600 J Answer :A,C

Solution : 10m 8m

Frictional force 3 200 60

10N work done by ladder on boy is zero because while ladder applies force

on boy, his point of application does not move.

Work done by slide = work done by friction = - 60 x 10 = - 600 J

17 There are two mass less springs A and B of spring constants AK and BK respectively and A BK K . If and

BW be denoted as work done on A and work done on B respectively, then

A. if they are compressed by same distance, A BW W

B. if they are compressed by same force (upto equilibrium state) A BW W

C. if they are compressed by same distance, A BW W

D. if they are compressed by same force (upto equilibrium state) A BW W Answer :A,B Solution : (A,B)

If the springs are compressed to same amount :

2 21 1;2 2A A B B A B A BW K x W K x K K W W

If the springs are compressed by same force.

2

2

2

2

1 .2; ; ;12

AA A B

A A B B A BA B B A

BB

FKW K KF FF K x K x x x

FK K W KKK

Hence, A BW W

18 A particle projected from horizontal ground at angle ‘ ’ with speed ‘u’. In same plane of motion a horizontal Acceleration ‘a’ exists so that projected particle returns back to point of projection. Find time of flight

A. 2 sinu

g

B. 3 sinu

g

C. 2 cosua

D. 3 cosua

Answer :A,C

Solution :

O

u

x

y

Initial position final position x = 0 ; y = 0 x = 0 ; y = 0

212x xx U T a T 21

2y yy U T a T

22 ( ) yx

x y

UUT or Ta a

Have cosxU u ; sinyU u

xa a ; ya g

19 A carom striker is given velocity on carom board has always friction, which causes constant retardation. Striker hits boundary of carom board and comes to rest at point from where it started. Take initial velocity direction is positive, choose correct graph depicting motion. ( -velocity, s-displacement, t-time)

A. t

s

B. t

s

C.

t

v

D.

t

v

Answer :B,C Solution : According to sign conversion

For s-t curve:

Forward motion:

Slope (velocity) +ve and decreased

Backward motion:

Slope (velocity) –ve and decreasing

(Option B)

For v-t curve

Forward motion:

Slope (accel) -ve and constant, velocity +ve

Backward motion:

Slope (accel) +ve (as per sign convertor and velocity -ve

20 A force F

(larger than the limiting friction force) is applied to the left to an object moving to the right on a rough horizontal surface. Then : A. the object would be slowing down initially B. for some time F

and friction force will act in same direction and for remaining time they act in opposite

directions C. the object comes to rest for a moment and after than its motion is accelerating in the direction of F

D. the object slows down and finally comes to rest. Answer :A,B,C Solution : A) F & f oppose motion. B,C) till body reverses its direction at motion. F & f are forwards left. Then body moves along f and friction acts towards right

21 A small block of mass of 0.1kg lies on a fixed inclined plane PQ

which makes an angle with the horizontal. A horizontal force of 1N acts on the block through its

center of mass as shown in the figure. The block remains stationary if (take g = 10 m/s2)

1N

Q

O P

A. 45 B. 45 and a frictional force acts on the block towards P C. 45 and a frictional force acts on the block towards Q D. 45 and a frictional force acts on the block towards Q Answer :A,C

Solution :

1N

Q

P1N

1cos

1sin

If = 45° then cos = sin hence, block will be at rest.

If plane is rough and > 45° then sin > cos so friction will act up the

Plane If plane is rough and <45° then cos >sin so friction will act down the

Plane so (a,c) are correct.

22 A particle of mass 1 kg is moving along positive x-axis with velocity of 3 m/s. Another particle of mass 2 kg is

moving along positive y-axis with a velocity of 6 m/s. At time t = 0, 1 kg is at (3m, 0) and 2 kg is at (0, 9m),

xy plane is horizontal plane. Then choose the correct options.

A. If the surface is smooth, the centre of mass of the particles is moving in a straight line, y = 4x + 2.

B. If surface is rough and both particles have the same value of coefficient of friction 0.2 , then the centre

of mass will stop after time t = 3.0 s.

C. If surface is rough and both particles have the same value of coefficient of friction 0.2 , then coordinate

of centre of mass where it will stop finally is (1.75 m, 12 m)

D. If surface is rough and both particles have the same value of coefficient of friction 0.2 , then the centre

of mass will stop after time t = 1.5 s.

Answer : A,B,C

Solution : At time t, 1 13 3 , 0x t y

2 20, 9 6x y t

1 1 2 2

1 2

1CMm x m xX t

m m

------------(1)

1 1 2 2

1 2

6 4CMm y m yY t

m m

----------(2)

From eqs. (1) and (2) y = 4x + 2

First particle will stop after time 1 1 v /t g = 1.5s and second particle will stop after a

time 22

6 30.2 10

vt sg

Centre of mass will stop when both particles stop.

Hence, CM will stop after 3s

Mass 1 kg will stop at

21

1 3 5.252vx m

g , 1 0Y

Mass 2 kg will stop at

22

2 9 182vy m

g , 2 0X

1.7512

CM

CM

X mY m

23 A ball A collides elastically with an another identical ball B at rest with velocity 10 m/s at an angle of 030 from the line joining their centers 1C and 2C . Select the correct alternative(s).

1C

2C

3010m / s

A

B A. Speed of ball A after collision is 5 m/s.

B. Speed of ball B after collision is 5 3 m/s. C. Both the balls move at right angle after collision. D. Kinetic energy will not be conserved here because collision is not head on. Answer :A,B,C Solution : Velocity of ball A along and perpendicular to line of impact after collision 0(0,10sin 30 ) Velocity of ball B along and perpendicular to line of impact before collision (0, 0) Velocity of ball B along and perpendicular to line of impact after collision 0(10cos30 ,0)

24 A pendulum bob of mass m connected at the end of an ideal string of length l is released from rest from

horizontal position as shown in figure. At the lowest point, the bob makes an elastic collision with a stationary

block of mass 5m, which is kept on a frictionless surface. Mark the correct statement(s) for the instant just after

the impact.

m I

5m

A. Tension in the string is 179

mg

B. Tension in the string is 3 mg

C. Magnitude of velocity of the block is 23gl

D. The maximum height attained by pendulum bob after the impact is (measured from lowest position) 49l

Answer :A,C,D

Solution : The velocity of bob just before the impact is 2gl along horizontal direction. From momentum

conservation,

1 2Mv mv 5mv .

From coefficient of restitution equation.

1 21 21 v v v v v

v

Solving above equation we get, 1 22 ,3 3v vv and v where 2v gl

For tension in string, 21mvT mg

l

179

T mg

Let maximum height attained by bob is h, then 21 4

2mv lmgh h

g

1v2v

v

5m

m

After collisionBefore collision

25 Three identical particles A, B and C (each of mass m) lie on a smooth horizontal table. Light inextensible strings which are just taut connect AB and BC and ABC is 0135 as shown in diagram. An impulse J is applied to the particle C in the direction BC for a very short time interval. Then just after applying impulse J,

A. Speed of particle A will be 2

7J

m B. Speed of particle B will be

107

Jm

C. Speed of particle C will be 37

Jm

D. Speed of particle C will be 107

Jm

Answer :A,B,C

Solution :

A B

C

2V1 x1xV

1y 1yV

45

v

The external impulse applied to C causes both strings to jerk exerting internal impulses 1J and 2J

2 1 v xV ----------(1) 0

1 2 1cos 45 J mv xJ ------------(3) 0

1 1cos45 = mv yJ ------------------(4)

1J J mv -----------------(5)

Also velocities of B&C along BC are equal i.e., 0 01 1cos45 v cos 45 vy xv -----(6)

After solving we get, 2 12

7xJv v

m

1 1 23 2 2 2 10 3 2; ; , ,7 7 7 7 7y A B

J J J J J Jv v V V J and Jm m m m m

26 In the projectile motion of a body from ground to ground , the power of the gravitational force

A. The average power is zero for the whole path B. The instantaneous power varies linearly with time

C. The instantaneous power is constant throughout

D. The instantaneous power is negative for the first half and positive for the rest half

Answer :A,B,D

Solution : .P F v

( sin )mg v gt

27 Two particles move on a circular path (one just inside and the other just outside the circle) with angular

velocities and 5 starting from the same position. They cross each other

A. At intervals of time 4

if their angular velocities are in the same sense

B. At successive points on the path subtending an angle of 60 at the centre if their angular velocities are

oppositely directed

C. At intervals of time 3

if their angular velocities are oppositely directed.

D. At successive points on the path subtending an angle of 45 at the center if their angular velocities are in the

same sense.

Answer :B,C

Solution : 12 ( )6 3

t C

1 ( )3 3

x B

5

22 ( )4 2

t A

is false

2 ( )2 2

x D

is false

5

28 The kinetic energy (K) vs time (t) graph for a particle is shown in the figure. The force vs time graph for the

particle may be

time t

kineticenergy K

A. time

force

B. time

force

C. time

force

D.

time

force

Answer :C,D

Solution : 1

2 212

mv t v t

1 12 2dv t F t

dt

29 One end of a light spring of force constant k is fixed to a wall and the other end is tied to a block placed on a

smooth horizontal surface. When the block undergoes a displacement, the work done by the spring is 212

kx . It is

possible that the spring was initially

A. Compressed by a distance x and finally was in its natural length.

B. In its natural length and finally compressed by a distance x.

C. In its natural length and finally stretched by a distance x

D. Stretched by a distance x and finally was in its natural length.

Answer :A,D

Solution : Work done is positive when the force and displacement are in the same direction

30 Two blocks A and B of equal mass, initially in contact on an inclined plane, are released from rest. The coefficient of friction of the inclined plane with A is 1 and that with B is 2 . It follows that

BA

A. If 1 2 , the blocks will slide down with different accelerations.

B. If 1 2 , there will be a force of interaction between the blocks. C. While the blocks slide, the work done by friction is zero D. If 1 2 , the blocks will always remains in contact. Answer :A,B,D Solution : 1 2 Blocks will slide separately

1 2 Blocks will slide together

31 In the arrangement as shown, block A of mass 3 kg has velocity 10 m/s towards left and it is at 100m at t = 0 from pulley on a smooth surface. Block B is of mass 2 kg. (g = 10 ms-2)

A

B100 m

A. At t = 1 s, velocity of A will be 6 m/s towards left. B. A will stop at t = 2.5 s C. Block A will be at a distance 108 m from the pulley at t = 4s. D. Block A will again be at a distance of 100 m from the pulley at t = 5s. Answer :A,B,C,D

Solution : Acceleration, 22 4 /(2 3)

ga m s

Initial velocity of A is towards left and acceleration is towards right. A t = 1 sec, v = -v + at = -10+4x1= -6m/s = 6 m/s (towards left) Block A will stop at t = 2.5second, as v = -10+4x2.5 = 0 Block A will be a distance given by (at t = 4 second)

212

S ut at

1 10 4 4 16 8m2

= 8m towards left So, distance from pulley = 108m. Time at which displacement will be zero.

2102

S ut at

T = 2 2 10 5sec.4

u xa Block A will be again at 100m at t = 5sec.

32 In a conservative force field, a particle of mass 1 kg starts moving from rest from the origin. The potential

energy at various x co-ordinates is shown in the figure. Find the correct option if 0U Uy z

x(m)

U(J)

-2-62 6

2060

6020

A. When the particle crosses – 4 m, the speed of the particle is 2 10 m/s

B. When the particle crosses – 4 m, the speed of the particle is 2 20 m/s C. The particle is moving along + x axis. D. The particle is moving along x axis. Answer :B,D Solution : Motion of particle is determined by

xUF ifx

Uy

=Ux

= 0.

From graph, Ux

= + 10N.

10xUF Nx

shows that particle is moving along (-) x-axis.

When particle crosses (-) 4 m dK= - dU

2

0 ( 40)2

mv

21 80v 2 20 / secv m

33 A block of mass 10 kg is placed in contact with the vertical wall of a box which is accelerating in horizontal

direction with constant acceleration of 20 ms-2 . Friction coefficient is 0.6. (g=10ms-2)

0.6

220ms

A. Magnitude of acceleration of the block as seen by an observer on ground is 20 ms-2.

B. Magnitude of friction on the block is 100 N

C. Magnitude of contact force between wall and block is 100 5N

D. Magnitude of contact force between wall and block is 200 N

Answer :A,B,C

Solution : N = ma = 10 × 20 = 200 N

Limiting friction 0.6 200 120f N N

Weight of block w = mg = 10 × 10 = 100 N

As w < flim, friction = w = 100 N only.

There will be no slipping of block i.e., block and box move with common acceleration 20 ms-2 w.r.t

ground. Net contact force 2 2 2 2100 200 100 5F f N F N

34 A particle is provided such a horizontal velocity at its bottom most position in a smooth spherical shell of radius

1.0 m so that it is just able to complete its circle.

A. Magnitude of acceleration of the particle at the instant when angular displacement becomes 600 is432

g

B. Magnitude of acceleration of the particle at the instant when angular displacement becomes 600 is672

g

C. Magnitude of acceleration of the particle when its velocity becomes vertical is 10g

D. Magnitude of acceleration of the particle when its velocity becomes vertical is 3g

Answer :B,C

Solution : At A, 2 5u gR

At B, 2 2 2 3 .v u gR gR

2 5u gR

mg

BCA

Tangential acceleration ( )Ta g

Centripetal acceleration 2 / 3Ca v R g

Acceleration = 2 2 10T Ca a g

At C, 2 2 02 ( cos 60 )v u g R R

= 5gR- gR = 4 gR 0sin 60 3 / 2Ta g g 2 / 4Ca v R g

Acceleration = 2 2T Ca a

= 2 23( ) 164

g g =672

g

35 A ball of mass m at A is given an initial velocity to slide down and collide with ball B of mass 2m so that ball B now has sufficient energy just to hit a ball at C. As the ball A slides down, its velocity just before collision

becomes twice of its initial velocity. The coefficient of restitution between A and B is12

. Neglect friction every

where.

hA

B

C

s

A. Speed of the ball B just after collision is 2 ( +s)g h

B. Speed of the ball B just after collision is 2g( +s)h

C. The initial speed of ball A is 2g( +s)h

D. The initial speed of ball A is 2 ( +s)g h Answer :B,C Solution : Let '

,A AV V and 1BV be the velocities of ball A as initial, before collision and after collision

respectively. 'BV is the velocity of the ball B after collision.

If AV = V, 'A

V = 2V and '21 22 BmV = (2 ) ( )m g h s 1 2 ( )BV g h s

From conservation of momentum,

' " '(2 )A A BmV mV m V 11 112 2 2 ( ) 2[ 2 2 ( )]A AV V g h s V V g h s 1 11

1 0B A

A

V VeV

2 ( ) 2[ 2 ( )]1 2 ( )[1 2] 22 2

g h s V g h s V g h s VV

3 3 2 ( ) 2 ( )V g h s V g h s

36 The torque acting on a body about a given point is found to be equal to A L

where A

is a constant ector

and L

is the angular momentum of the body about that point. From this, which of the following statements are correct?

A. /dL dt

is perpendicular to L

at all instants of time

B. The component of L

in the direction of A

does not change with time.

C. The magnitude of L does not change with time. D. L

does not change with time Answer :A,B,C

Solution : A L

i.e. dL A Ldt

This relation implies that dLdt

is perpendicular to A

and L

.

2.L L L

Differentiating with respect to time, we get . . 2 2 . 2dL dL dL dL dLL L L L Ldt dt dt dt dt

SincedLLdt

so, . 0dLLdt

Therefore 0dLdt

and then L does not change with time. So option C is correct.

Since L is not changing with time, therefore it is the case when direction of L

is changing but its magnitude is constant and is perpendicular to L

at all points.

This can be written as If ( cos ) ( si )ˆ ˆnL a i a j

(here a is a constant) ˆ( sin c ˆ) ( os )a i a j

So, that . 0L

and L

Now, A

is a constant vector and it will always perpendicular to . Thus A

can be written as

A

ˆA Ak

We can see that 0LA

i.e. L A

Thus we can say that the component L

along A

is zero or component of L

along A

is always constant. Finally we conclude that , A

and L

always constant.

37 A mass ‘m’ is attached to a light rod of negligible mass as shown in fig. The system is pivoted at point ‘O’ and

rotates about the indicated z-axis with angular velocity ,

maintaining a fixed angle , with the axis.

rO

P x

yz

A. Angular momentum L

of mass ‘m’ about pivot is parallel to vector

B. Angular momentum L

of mass ‘m’ about pivot is never parallel to

C. Direction of L about O changes continuously with time

D. Angular momentum of particle about ‘P’ is parallel to vector

Answer :B,C,D

Solution : Conceptual

38 A solid cone and a solid sphere is arranged as shown in the figure. The centre of mass of the system is (

:are the radii of solid sphere and solid cone 1 2& are the distance of solid sphere and solid cone,

1 2&m m are the masses of solid sphere and solid cone respectively.)

24R

2 2 2, ,R m

1 1 1, ,R m

A. At 3R from axis if 1 2m m and 1 2R R

B. At 2R from centre of mass of solid cone if 1 2 1 2and R R

C. If 1 2 1 22 , R R then distance from the centre of mass of solid cone is 11

3R

D. At R from centre of mass of solid cone if 1 2 1 2and R R

Answer :A,B

Solution : Position of CM of cone. = 4h

from base

22

44R R

2C

1m

24h R

2m

1R

axis

C1

R2

2m

Distance between C1 and C2

2 1x 3R R 1 12 1 2

1 2 1 2

. ( 3 )m mC M of system from C x R Rm m m m

If m1 = m2 and R1 = R2 then

y ( 3 ) 22m R R Rm

from axis, position of CM = R + 2R = 3R

C.M. of system from

3

12 1 2 1 2

3 21 2

43 4 [ , ]4 1 (4 )

3 3

RvolC x R R Rvol vol R R R

3

23 3 4 2R R R from CR R

39 A particle is projected from point ‘A’ at an angle ' ' . It passes through point ‘B’ while travelling upwards in a direction making an angle ' ' with horizontal. AB makes an angle ' ' with the horizontal. Then A. tan tan tan B. tan tan 2 tan

C. tan tancosgt

u

D. tan tan

2 cosgt

u

Answer :B,C

Solution :

21sin2tan

cos

u t gtyx u t

sinu

cosu

u

sinv

BV

cosv

y

xA

1tan tan2 cos

gtu

………. (1)

cos cos , sin sinu v v u gt

sin cos tangt u u …….. (2)

(2) in (1) tan tan 2 tan

From (2) tan tancosgt

u

40 Six identical elastic balls are suspended in a row on strings of equal length that the distances between adjacent

balls are very small as shown. How will the balls behave if two extreme balls are moved aside and released at

the same time?

A. Only one extreme left ball will bounce off, with the momentum double that of extreme right ball just

before the collision.

B. Two extreme left balls will bounce off, each with the momentum as that of extreme right ball just before

the collision.

C. Two colliding extreme right balls will rebounded again with the same momentum

D. All the four extreme right balls will rebounded again with the same momentum half that of extreme right ball

just before the collision.

Answer :B


41 An iron block and a wooden block are positioned in a vessel containing water as shown in the figure. The iron

block (1) hangs from a massless string with a rigid support from the top while the wooden block floats being

tied to the bottom through a massless string. Now, if the vessel starts accelerating upwards

A. tension in the string 1 will decrease B. tension in the string 1 will increase

C. tension in string 2 will decrease D. tension in the string 2 will increase

Answer :B,D

Solution : Since geff, increases tension in both strings will increase.

42 A light cylindrical vessel of radius r is kept on a rough horizontal surface so that it cannot slide but can topple. It is filled with water upto a height '2h' and a small hole of area 'a' is punched in its wall so that the water coming out of it falls at the maximum distance from a vertical plane through the hole, along horizontal surface. Water comes out horizontally from the hole. The value of h for which the cylinder topples is/are

2h

2r

Hole

A. 3

2ra

B.

32 ra

C. 33

2ra

D.

343

ra

Answer :B,C,D Solution : For toppling, 2 2 2av h r h g r

Þ 22 2a gh h r h gr

Þ3rh

a

43 A disc is performing pure rolling on a smooth stationary surface with constant angular velocity and linear velocity of centre of mass as shown in fig. Find the correct statements (Radius of the disc is R) at the given nstant of time

A

B

C

V/RD V

A. Acceleration of the lowest point ‘C’ of the disc is 2V

R

B. The radius of curvature of highest of point ‘A’ of the disc is 4 R C. The relative acceleration between the highest point A and lowest point ‘C’ is

D. The magnitude of relative velocity between the points B, C is 2V Answer :A,B,C,D Solution : In pure rolling motion lowest point ‘C’ velocity is zero, but acceleration is V2/R

Radius of curvature 2 2

2

(2 ) 4/N

V V Ra V R

44 A rod of length ‘l ’ is sliding between the smooth vertical wall and on the smooth horizontal surface. At a

particular instant of time, the lower end has velocity ‘v’ as shown and rod makes an angle with horizontal.

‘A’ is point on the rod. Find the correct statements

V

Al

A. The angular velocity of the rod about it’s centre of mass is sin

vl

B. The component of velocity of point ‘A’ along the length of the rod is cosv

C. The velocity of the other end of rod (contact with wall) is tanv

D. The velocity of the centre of mass of the rod is 2sin

v

Answer :A,B,D

Solution : sinI AORV

l

V other end = cotV

2 2( ) ( cot )2 2 2sinCMV V VV

45 A particle of mass ‘m’ is projected with a initial velocity ‘u’ at an angle with the horizontal ground at t = 0 sec. as shown in fig. Then find the correct statements

A. The instantaneous angular momentum of projectile after ‘t’ sec. from the projection about the projection

point is 2cos .

2mg u t

B. The instantaneous torque experienced by the projectile after ‘t’ sec. from the projection about the projection point is mg cos .u t

C. The average torque experienced by the projectile between the projection point and striking point about the

projection point is 2 sin cosmu

D. The average angular momentum of the projectile between the projection point and striking point about the

projection is 2cos

4mg u t

Answer :A,B,C Solution : Instant torque = F perpendicular distance

cos .mg x mg u t

2 sin cosAverage

dtmu

dt

dLdt

dL dt 2

tan. cos .

2ins tmg u tL

2. cos .6average

L dt mg u tLdt

46 A train A crosses a station with a speed of 40 m/s and whistles a short pulse of natural frequency 0 596n Hz. Another train B is approaching towards the same station with the same speed along a parallel track. Two tracks are 99d m apart. When train A whistles, train B is 152 m away from the station as shown in Figure.

Station

A 40 m/s

152 mB

99 m

40 m/s

If velocity of sound in air is v = 330 m/s. A. Time taken by sound pulse to travel from A to B is 0.5 s B. Time taken by sound pulse to travel from A to B is 1.5 s C. Frequency of pulse heard by driver of train B is 596 Hz D. Frequency of pulse heard by driver of train B is 724 Hz Answer :A,D Solution : When train A whistles, sound pulse starts to travel in air from train A to train B. During this interval train B moves some distance towards the station. Let sound pulse take time t to travel from train A to train B. Distance moved by train B during this interval is 40t. Therefore, the distance of train B from station when its

driver hears the pulse is 152 – 40t. Hence, the distance travelled by the pulse is 2 2(152 40 ) (99) .t But it

is equal to 330 .vt t

(i) 2 2(152 40 ) (99) 330t t or 0.5t s Therefore, driver of train B hears the pulse when his train is 152 40 132t m from the station. Hence, path of pulse will be as shown in the figure. Its inclination with track is (ii) will be as shown in the figure. Its inclination with track is given by

47 A straight conductor AB lies along the axis of a hollow metal cylinder L which is connected to Earth through a conductor C. A quantity of charge will flow through C :

A

B

C

A. If a current begins to flow through AB

B. If the current through AB is reversed C. If AB is removed and a beam of electrons flows in its place D. If AB is removed and a beam of protons flows in its place Answer :C,D Solution : When the current flows through the conductor AB, it remains electrically neutral. Therefore, no charges are induce by it on the cylinder . On the other hand , a beam of electrons or protons has net negative or positive charge. They will induce bound and free charges on L. The free charges will flow through conductor C to the earth.

48 A cylindrical tank having cross-sectional area A is filled with two liquids of density 1 2and each to a height

h as shown in figure. A small hole having area a is made at the bottom of container. If the surface on which vessel is places is smooth, then the force (horizontal) required, to keep the cylinder at rest just after uncovering the hole is F, then

h

h 2

1

A. 1 22 ( )F a gh

B. If the surface on which vessel is kept is rough having coefficient of friction then the minimum horizontal

force required to keep the vessel at rest just after uncovering the hole is 1 2( )(2 )gh a A towards right

C. In option (b) maximum value of force is 1 2( )(2 )gh a A towards right

D. This situation is not possible Answer :A,B,C Solution : Let v is the velocity with which liquid comes out, them from Bernoulli’s theorem

22

0 1 2 0 2vp gh gh p

1 2

2

( )2v gh

The reaction force exerted by liquid on vessel is 22 1 22 ( )rF av gh a

For smooth surface, rF F

1 22 ( )F gha towards right

rF

'F

mg 1N

For rough surface: Limiting friction force is, 1Lf N mg

Where 1 2 1 2( )m Ah Ah Ah

2 1 2( )f Ahg

For maximum value of ', LF f must be along .rF

max 1 2 1 2' 2 ( ) ( )r LF F f agh Ah g

1 2( )(2 )gh a A

max 1 2' ( )(2 )r LF F f gh a A

49 A thin conducting rod AB is introduced in between the two point charges 1q due to 2q as shown in figure. For this situation mark the correct statement(s).

1q 2qBA

A. The total force experienced by 1q is vector sum of electric force experienced by 1q due to 2q and due to induced charges on rod. B. The end A will become negatively charged. C. The total force acting on 1q will be greater than as compared to the case without rod.

D. The total force acting on 2q will be greater than as compared to the case without rod. Answer :A,B,C,D Solution : Due to induction effect the situation is shown clearly in figure. Due to 1,q let induced charge is

1 'q at end A and 1 'q at end B while due to 2q induced charges are 2 'q and 2 'q at ends A and B respectively. Thus, the end A acquires negatively charged and B acquires +ve charge. Electric force experienced by 1q or 2q has to be computed by using principle of superposition.

1q 2qBA

2'q 2'q

1'q 1'q

For 1q due to 1 2q q towards right due to rod towards right. Hence, total force experienced

by 1q in present situation is greater than as compared to the case without rod. Same is the situation

for 2 .q

50 Two metallic bodies separated such that distance between two points A and B on the two bodies respectively 20 cm, are given equal and opposite charges of magnitude 0.88 .C The component of electric field along the

line AB, between the plates, varies as 23 0.4 / ,xE x N C where x (in metre) is the distance from one body towards the other body as shown. A. The capacitance of the system is 10 F B. The capacitance of the system is 20 F

C. The potential difference between A and C is 0.088 V D. The potential difference between A and C cannot be determined from the given data

X C

B20 cm

A

Answer :A,C Solution : 2(3 0.4) /xE x N C

0.22

0

(3 0.4)xV E dx x dx

3 0.20[ 0.4 ]x x

3(0.2) 0.4 0.2 0.088V volt 0.88 100.088

QC FV

51 A wire of density 3 39 10 /kg m is stretched between two clamps 1m apart and is stretched to an extension

of 44.9 10 m. Young’s modulus of material is 10 29 10 /N m . Then A. the lowest frequency of standing wave is 35 Hz B. the frequency of 1st overtone is 70 Hz C. the frequency of 1st overtone is 105 Hz D. the stress in the wire is 7 24.41 10 /N m . Answer :A,B

Solution : Speed of wave in wire 1T Y l Y lV AA l A

Minimum frequency means fundamental mode. 10 4

3

1 1 9 10 4.9 10 352 2 2 1 (1)9 10

V V Y lf Hzl l l

Frequency of first overtone = 70 .V Hz

52 A stationary wave given by equation

2 24

2 211.56 10d y d ydt dx

is established in L = 1.7m long pipe filled with a

gas and closed at both the ends. The permissible frequencies are A. 100 Hz B. 200 Hz C. 150 Hz Answer :A,B

Solution : Let L be the length of a loop and n be the number of loops then nL = 1.7 or 1.7L mn

Velocity 411.56 10 340 / secV m

N = frequency 340 100

2 1.7

Hz

The possible frequencies are 100, 200, 300 etc

53 A particle is subjected to two simple harmonic motions simultaneously along x and y directions according to

3sin100 ; 4sin100x t y t

A. Motion of the particle will be on a curved path

B. Motion of the particle will be on a straight line with slope 4/3

C. Motion will be a simple harmonic motion with amplitude 5 units

D. Phase difference between two motions is / 2

Answer :B,C

Solution : 3sin100x t 4sin100y t

Equation of path is

43

yx i.e.

43

y x

Which is equation of a straight line having slope 43

Equation of resulting motion is (3 4 )sin1ˆ ˆ 00ˆ ˆr xi yj i j t

Amplitude is 2 23 4 5

54 A cylinder as shown in the fig is filled with oil of viscosity . Within the cylinder, a thin disc of radius R is rotating with a constant angular velocity about its symmetrical, vertical axis. The separation between the disc and the horizontal plane of the cylinder is y. Viscous forces act on the disc. Choose the correct option (s). Due to viscous forces on the disc

R

A. the value of total torque is 4R

y

B. the value of power developed is 2 4R

y

C. total value of torque is 2 4

2Ry

D. the value of power developed is 2 4Ry

Answer :A,B Solution : Consider an element of radius r and thickness dr . Viscous force acting on it is given by

2( .2 )rdF rdry

[ Viscous force acts from both top and bottom contact layer of the disc)

3. 4d torque r dF r dr

y

Total torque 3

0

4 R

r dry

4Ry

Power developed =2 4R

y

55 A uniform sphere of mass m and radius r rolls without sliding over a horizontal plane, rotating about the

horizontal axis OA. In the process, centre of sphere moves with velocity along a circle of radius R. Total

KE of the sphere is : ( ,v vI are angular velocity and moment of inertia about vertical axis respectively)

R

AO

A. greater than 212

m

B. greater than 2710

m

C. greater than 2 2 21 2,2 5

I whereI mr mR andR

D. equal to 2 21 12 2

I m

Answer :A,B,C

Solution : The sphere has two types of rotational motion.

(i) About the horizontal axis OA. About this axis its angular velocity is equal to h r

. (ii)

About the vertical axis passing through O, angular velocity about it is R

. Moment of inertia of

the sphere about the horizontal and vertical axes are 225hI mr and 2 22( )

5I mr mR respectively.

Total kinetic energy of the sphere is equal to sum of kinetic energy associated with these two types of

motion.

Total kinetic energy

2 21 12 2h hE I I

or, 2

2 21 2 12 5 2

E mr Ir

2 21 15 2

m I

22 2 21 1 2

5 2 5m mr mR

R

2 27 110 5

m mr

Hence, options (a), (b) and (c) are correct and option d is wrong.

56 Figure shows a block A, held on a spring balance D and submerged in a liquid in beaker B. The beaker is kept

on a spring balance E. The mass of the beaker plus liquid is 2.5 kg. Balance D reads 2.5 kg and E reads 7.5 kg.

If the volume of the block is 0.0030 3m , then which of the following are correct ?

A. The density of the liquid is 350003

kg m

B. The mass of block A is 7.5 kg

C. Mass of spring balance is 2.5 kg

D. When half the volume of the block is pulled out of the liquid, E would read 5.0 kg

Answer :A,B,D

Solution : (a) Down thrust 7.5 2.5 5.0kg

This equals volume of A density

b) Weight-Up thrust 5kg = reading of D, 2.5 kg

So, weight = 7.5 kg

(d) Up thrust is halved = 2.5 kg

So, E should read = 7.5-2.5=5.0 kg

57 A string is fixed at both ends and transverse oscillations with amplitude 0a are excited. Which of the following

statements are correct ?

A. Energy of oscillations in the string is directly proportional to tension in the string

B. Energy of oscillations in nth overtone will be equal to 2n times of that in first overtone

C. Average kinetic energy of string (over an oscillation period) is half of the oscillation energy

D. Average KE of the string (over an oscillation period)is equal to oscillation energy

Answer :A,C

Solution : If a string of length l has cross-sectional area A, density of its material then its oscillation energy

is given by

2 2 20E A a lf

Where f is frequency of transverse stationary wave formed in the string.

But 1 Tf

m

Where is wavelength, T is tension in the string and m A

Since, string has a fixed length, therefore, wavelength of a tone excited in the string is constant. Hence,

energy E T .

Therefore, option (a) is correct.

If the frequency of fundamental tone is 0f , then frequency of nth overtone will be equal to 0( 1)n f

Hence, oscillation energy of the string will be equal to :

2 2 2 20 0 ( 1)nE A a lf n

Since, nE is not directly proportional to 2n , therefore, option (b) is wrong.

Since, every particle of the string performs SHM, therefore, rms speed of a particle

12

its maximum speed

Hence, average KE is half of maximum KE. But maximum KE is equal to oscillation energy of the

string. Therefore, option (c) is correct.

58 Choose the correct statements from the following in which k is a real, positive constant :

A. Function ( ) sin cosf t kt kt is simple harmonic having a period 2 / k

B. Function ( ) sin 2cos 2 3sin 3f t t t t is periodic but not simple harmonic having a period of 2s

C. Function 2( ) cos 2sinf t kt kt is simple harmonic having a period 2 / k

D. Function ( ) ktf t e is not periodic

Answer :A,C,D

Solution : Statement (a) is correct. The function ( ) sin cosf t kt kt can be written as

( ) 2 sin 2 cos4 4

f t kt or kt

both of which are simple harmonic. The coefficient of time t in the

argument of the sine or cosine function 2 / ,T where T is the period. Hence, 2kT

or 2 /T k .

Statement (b) is also correct. Each term represents simple harmonic motion. The period T of term

sin t is : 2T

or T=2s. The period of term 2cos 2 t is 1s, i.e., T/2 and the period of term

3sin 3 t is 2/3s, i.e., T/3. The sum of two or more simple harmonic motions of different periods is not

simple harmonic. The sum, however is periodic. By the time, the first term completes two cycles and

the third term completes three cycles. Thus, the sum has a period of 2s.

Statement (c) is incorrect. We can write

2( ) cos 2sinf t kt kt

( ) cos (1 cos 2 )f t kt kt

1 cos cos 2kt kt

The period of cos kt is 2Tk

and of cos 2kt is k , which is T/2. As explained above, the perod of

the two terms together is 2 /T k . The term 1 is a constant independent of time.

Statement (d) is correct. Function kte decreases monotonically to zero at t ; it never becomes

negative. Hence it is non-periodic

59 A rectangular vessel of dimension ( )l b h and mass M contains a liquid of density . The vessel has an orifice at its bottom at a distance c from the rear wall as shown in the figure.

l

h

c

F

A. The maximum volume of the water that can be stored when the vessel is accelerated is hcb/2. B. The maximum volume of the water that can be stored when the vessel is accelerated is hlb/2.

C. Force F that must be applied when maximum water is stored is [ ]2

hcb hgMc

D. Force F that must be applied when maximum water is stored is [ ]2

hcb lgMc

Answer :A,C Solution : A,C – The maximum amount of water that can be retained is shown in the figure. If is the angle made by the water surface with the horizontal, then

h

c

tan h a hgac g c

So the maximum volume that can be retained is (1/ 2) h c b and

[ ]2

hcb hgF Mc

60 For the situation shown in the figure (assume r > > length of dipole), select the correct statement(s) [ P = dipole moment, Q = charge on the particle which is on equatorial line of dipole)

Q rp (small dipole

kept vertically)

A. Force acting on the dipole is zero

B. Force acting on the dipole is approximately 30

PQ4πε r

and is acting upwards

C. Torque acting on the dipole is 20

PQ4πε r

in clockwise direction

D. Torque acting on the dipole is 20

PQ4πε r

in anti-clockwise direction

Answer :B,C Solution : The situation is shown in the figure below :

net 30

PQ2Fsin4πε

Fr

cos 2aF x

in clockwise direction = 20

PQ4πε r

61 Two stars of masses m and 2m are co–rotating about their centre of mass under the influence of mutual

gravitational attraction. Their centers are at a distance r apart. If r is much larger than the size of the stars,

then their

A. common period of revolution is proportional to 3/2r

B. orbital speeds are in the ratio 2 : 1

C. kinetic energies are in the ratio 2 : 1

D. angular momenta are in the ratio 1 : 4

Answer :A,B,C

Solution : For r > R ;

2

1 2

2 1

F rF r

For 3; GMr R F rR

1 1

2 2

F rF r

62 A satellite is orbiting the earth in a circular orbit of radius r. Its A. kinetic energy varies as 1/r

B. angular momentum varies as 1r

C. linear momentum varies as 1r

D. frequency of revolution varies as 3/2

1r

Answer :A,C,D

Solution :

11 24

22 8

2 6.67 10 6 102 93 10

GMR mmC

63 A long thin uniform rod of length L and mass m is free to rotate in a vertical plane about a horizontal fixed axis

through its one end “O”. A light spring of force constant k is connected vertically between the other end A of

the rod and ground. When the rod is in equilibrium, it is parallel to the ground.

O

A L

A. The period of small oscillations when the rod is rotated slightly and released is 23mk

B. The period of small oscillations when the rod is rotated slightly and released is 2 mk

C. The maximum speed of the displaced end of the rod is 03kLm

(the angular amplitude of small oscillations I 0. )

D. The maximum speed of the displaced end of the rod is 03kLm

(the angular amplitude of small oscillations is 0. )

Answer :A,C

Solution : (a) Restoring torque about “O” due to elastic force of the spring,

( )FL kyL F ky

2 ( )kL as y L

22

2

13

dl MLdt

22 2

2

13

dML kLdt

2

2

3d kdt M or

3kM

and 23MTK

(b) In angular SHM, maximum angular velocity

0 0max

3d kdt M

( / )v r d dt

So, max 0max

3d kv L Ldt M

64 Two identical simple pendulums each of length L and mass m are connected by a weightless spring as shown in

figure. The force constant of the spring is k. In equilibrium, the pendulums are vertical and the spring is

horizontal and un-deformed. The frequency of small oscillations of the linked pendulums, when they are

deflected from their equilibrium positions through equal displacements in the same vertical plane in the same

direction is 1f and when displaced in opposite directions the frequency is 2f . Then

L Lk

A. 11

2 2gfL

B. 11

2gfL

C. 21 2

2g kfL m

D. 21

2 2g kfL m

Answer :B,C

Solution : (a) When both the pendulums are displaced in the same direction by same amount, the spring will

neither compress or stretch, so the restoring torque on each pendulum about the point of suspension will be

due to its own weight only.

L LK

(A)

L Lk

k

mg mg

(B)

L L

ymgmg

ky

(C)

y

. . sini e mgL mgL [ as for small ,sin ]

But by definition 2

2 22

dl ML as l mLdt

65 A particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes

through a certain point P (AP < BP) at successive intervals of 0.5 s and 1.5 s with a speed of 3 m/s.

A. Maximum speed of the particle is 3 2 /m s B. The maximum speed of particle the is 2 /m s

C. The ratio APBP

is 2 12 1

D. The ratio APBP

is 12

Answer :A,C

Solution : Let the displacement equation of particles is sinx a t

Time period of particle would be ( ) ( )PA AP PB BPT t t t t

(0.5 ) (1.5 )s s 22s

POt = 0

AB

a a - xx

1s

sinx t and ( ) cosv a t

Let OPt t then 12OAPt t then 13 cos cos

2a t a t

cos sin

2a t a t

4t

Or max3 3 2 /

cos / 4a m s v

sin sin4 2

ax a t a

2 122 1

2

aaAP a xaBP a x

66 A thin copper rod of length is suspended vertically from the ceiling by one of its ends. Density of copper is

, Young’s modulus is Y and Poisson’s ratio is . Then which of the following are correct ?

A. The strain in the rod due to its own weight is 2gY

B. The elastic deformation energy stored in the rod due to its own weight is 2 2 3

6g A

Y

where A is area of cross

section of the rod.

C. Volume strain of the rod is 12gY

D. Volume strain of the rod is 1 22gY

Answer :A,B,D

Solution : At P,T= A xg

Elongation of element dx is ( )T dxdeAY

Px

dx

Deformation energy of the element is21 ( )

2 2T dxdU T de

AY

2 2 2 2( )2 2

A xg dx A g x dxdUAY Y

2 2 2 2 32

02 6A g A gu du x dx

y y

67 The torque on a body about a given point is found to be A L

where A

is a constant vector and L

is

Angular momentum of the body about that point. From this it follows that

A. dLdt


at all instants of time

B. the component of L

in the direction of A


C. the magnitude of L

does not change with time

D. L


Answer :A,B,C

Solution : Since dL A Ldt

dLdt


Since A L

dL A Ldt

dLdt

must be perpendicular to A

as well as Further component of L

along A

is . ( ).A L x say AlsoA

. . . 0d dL dAA L A Ldt dt dt

dL dAA and Odt dt

.A L

constant

.A LA

constant

Since ( )dL ordt


, hence it cannot change magnitude of L

but can surely change

direction of L

68 A sledge moving over a smooth horizontal surface of ice at a velocity 0 drives out on a horizontal road and

Comes to a halt as shown. The sledge has a length l, mass m and friction between runners and road is .

ROADICE

l

A. No work is done by the friction to switch the sledge from ice to the road

B. A work of 12

mgl is done against friction while sledge switches completely on to road.

C. The distance covered by the sledge on the road is 20 .

2 2l

g

D. Total distance moved by the sledge before stopping is 20 .

2 2l

g

Answer :B,C,D

Solution : Work done against friction on ice is zero and work done against friction on the road is ( ) .mg So,

average work done is ( ) .2 2

O mg mg

Thus indicating that the effective length of the sledge that has to be dragged so that it just gets

completely on the road is .2

Distance covered by the sledge on the road before coming to rest is 20

2v

g. 2 2

0 2O V g s

So that distance moved, by the sledge is 20 .

2 2v

g

69 A uniform rod kept vertically on the ground falls from rest. Its foot does not slip on the ground.

A. No part of the rod can have acceleration greater than g in any position

B. At any one position of the rod, different points on it have different accelerations

C. Any one particular point on the rod has different accelerations at different positions of the rod.

D. The maximum acceleration of any point on the rod, at any position, is 1.5g

Answer :B,C

Solution : A

C

PB

Mg

Taking torque about A, when the rod has fallen through an angle ,

21sin2 3lmg I ml

or

3 sin2gl

For any point P on the rod, at a distance r from A, the linear acceleration is

3 sin .2gra rl

70 A block of mass m is placed on a rough horizontal surface. The coefficient of friction between them is . An

external horizontal force is applied to the block and its magnitude is gradually increased. The force exerted by

the block on the surface is R.

A. The magnitude of R will gradually increase

B. 2 1.R mg

C. The angle made by R with the vertical will gradually increase

D. The angle made by R with the vertical 1tan

Answer :A,B,C,D

Solution : R is the resultant of the normal reaction, ,N mg and the force of friction, .F mg As P is

increased, ( )F P increase, while N is constant.

71 In which of the following cases the centre of mass of a rod is certainly not at its centre?

A. The density continuously increases from left to right

B. The density continuously decreases from left to right

C. The density decreases from the left to right upto centre and then increases

D. The density increases from left to right upto the centre and then decreases

Answer :A,B


72 The vibrations of a string of length 600 cm fixed at both ends are represented by the equation

4sin cos 9615xy t

Where x and y are in cm and t in seconds A. The maximum displacement of a point x=5 cm is 2 3 cm . B. The nodes located along the string are 15n where integer n varies from 0 to 40. C. The velocity of the particle at x=7.5 cm at t=0.25 sec is 0 D. The equation of the component waves whose superposition gives the above wave are

2sin 2 48 ,2sin 2 48 .30 30x xt t

Answer :A,B,C,D

Solution : (A) Displacement = 54 ( ) 4 2 3

15 5Sin Sin Cm

(B) 2 3015

xKx x

Nodes will form at 2 3 4 50, , , , , ..........2 2 2 2 2 0,15,30, 45........... ie 15 is where n = 0 to 40

C) 4 (96 ) 9615

dy xV Sin Sin tdt

at 17.5 ;4

x m t w

75 14 (96 ) 96 015 4

v Sin Sin

(D) 2 96 , 2 615 15

x xSin t Sin T t

2 2 48 , 2 2 3830 30x xSin t Sin t

73 The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length 1.5AL m and

one open end, contains a small sound source that sets up the standing wave with the second lowest resonant

frequency of that pipe. Sound from pipe A sets up resonance in pipe B, which has both ends open. The

resonance is at the second lowest resonant frequency of pipe B. The length of the pipe B is :

A B

AL BL

A. 1m B. 1.5m C. 2m D. 3m

Answer :C

Solution : 2

2nd

ndlowest resonance lowest resonance open pipeinclosed pipe

3 2 24 1.5 2 b

b

V V L mL

74 1S and 2S are two sources of sound emitting sine waves. The two sources are in phase. The sound emmited

by

the two sources interfere at point F. The waves of wavelength :

2m 4m1S 2S F

A. 1 m will result in constructive interference B. 2

3m will result in constructive interference

C. 2m will result in destructive interference D. 4m will result in destructive interference.

Answer :A,B,D

Solution : Path difference 2x

For constructive n l (or) 2n

in terferance

A) 1: :n l

B) 2 33

n

C) 2for destructive

xin terference

(2 1) 22

n

42 1n

4 1( 1)2

n

D) 4 n = 1

75 In a resonance tube experiment, a closed organ pipe of length 120 cm is used. Initially it is completely filled with water. It is vibrated with tuning fork of frequency 340 Hz. To achieve resonance the water level is

lowered then (given 340airV m/sec., neglect end correction) : A. maximum length of water column to have the resonance is 95 cm. B. the distance between two successive nodes is 50 cm. C. the minimum length of water column to create the resonance is 45 cm. D. the distance between two successive nodes is 25 cm. Answer :A,B,C Solution : Conceptual

76 A sound wave of frequency v travels horizontally to the right. It is reflected from a large vertical plane surface moving to left with a speed u. The speed of sound in medium is c.

A. The number of waves striking the surface per second is( )c u v

c

B. The wavelength of reflected wave is ( )( )

c c uv c u

C. The frequency of the reflected wave as observed by the stationary observer is( )( )c uVc u

D. The number of beats heard by a stationary listener to the left of the reflecting surface is vuc u

Answer :A,B,C Solution : Conceptual

77

A B

2 2, r

+

+

1 1, r

Two cells of unequal emfs, 1 and 2 , and internal resistances 1r and 2r are joined as shown. AV and

BV and the potentials at A and B respectively

A. One cell will continuously supply energy to the other

B. The potential difference across both the cells will be equal.

C. The potential difference across one cell will be greater than its emf.

D. 1 2 2 1

1 2

( ) .( )A Br rV Vr r

Answer :A,B,C,D Solution : Let 1 2.

Current in the circuit = i = 1 2

1 2

.r r

2 2A BV V ir p.d. across each cell.

Here, 2 .A BV V

Current flows in the cell of emf 2 from the positive plate to the negative plate inside the cell and hence it absorbs energy

78 Consider an attractive central force of the form ( ) ,n

kF r kr

is a constant. For a stable circular orbit

to exist

A. 2n B. 3n C. 3n D. 1n

Answer :A,B,D

Solution : ( ) n

kF rr

1

1( ) ( )( 1) n

kU r F r drn r

If L is the angular momentum of the particle of mass m in an orbit of radius r then,

Kinetic Energy 2 2

2 ( )2 2L L k rI mr

Since, Total Energy ( ) ( ) ( )E r U r K r

2

1 2

1( )( 1) 2n

k LE rn r mr

The criterion that a circular orbit of radius 0r be stable is that ( )E r is MINIMUM. For ( )E r to be MINIMUM two conditions must be fulfilled.

0 0

2

2| 0 | 0r r r rE Eandr r

Using both these conditions, we get 2

(3 ) 0Lnm

This is possible only when n < 3 We also note that inverse Square Law belongs to this category. n = -1 also gives stable circular orbits (Law of Direct Distance) But n = 3 gives circular orbits which are unstable (Inverse Cube Law)

79 Three capacitor of capacitance C each are connected in series as shown in the figure. Initially switch S is open. Now capacitors are charged by a battery of emf V by connecting between terminal A and B.

After long time battery is disconnected and inductor of inductance L is connected between A and B at time t = 0 so that an oscillatory circuit is formed. Now at an instant 0t switch S is closed, then which of the following are correct.

S

C A

CB

L

C

A. The current in the circuit just after closing the switch S is (that is at 0t t ) 03sin

3CV tL CL

B. The frequency of oscillation of the charge in the circuit before closing the switch S is 1 32 LC

C. The frequency of or oscillation of the charge in the remaining circuit is (after closing the

switch S )

1 22 LC

D. The amplitude of charge oscillations of the remaining capacitor (after closing the switch S ) is

`

20

3cos1 16 3

tLC

CV

Answer :A,B,C,D

Solution : 1) At t 0 cosq q t where 03 ,

3q

LC at 0t t 0 0sindqq q t

dt

(before

closing the switch) . Just after closing the switch and just before closing the switch the current is same due electrical inertia of induction 0 0sini q t

2) After closing the switch frequency 11 1 ,2 2

CCLC

3) Det 1q be the charge on capacitor just after closing the switch

1 1q q diL

C C dL

1 2 1 2 11

2 222

q d q d q LCL qC dt dt

1

2LC

and 1 10 sin( )q q t

at t = 0 1 10 2 0cos , sinq q t i i q t

10 0 0

10 0 0

20

10

sin coscos sin

3cos1 16 3

q q tq q t

tLC

q cv

80 At given instant there are 25 % undecayed radio-active nuclei in a sample After 10 seconds the no. of un decayed nuclei reduces to 12.5 %. Which of the following statements are correct

A. The mean life of nuclei is 14.43 seconds

B. The time in which the no.of undecayed nuclei will further reduce to 6.25 % of the reduced number in 10 sec

C. The mean life of nuclei is 13.66 seconds

D. The half life of the nuclei is 10 seconds

Answer :A,D

Solution : 1/2

0.693T

and mean life time 1/21 100.693 0.693TT

From 12.5% for 6.25% to remain under cased 1/2 10sect T

81 An annular wire loop ABCD carries a current 1I as shown in figure. O is the common centre of the curved parts AB and CD of the loop. A straight wire passing through O and perpendicular to the plane of the loop carries a current 2I directed towards the reader. Then

A. The net force on the loop is zero

B. The net torque on the loop is zero

C. As seen from O the loop will rotate in clock wise sense about axis OP

D. As seen from O the loop will rotate in anticlockwise sense about axis OP

Answer :A,C Solution : The magnetic field B due to current 2I is tangential to the curved parts AB and CD of the loop. Hence every current element dl of parts AB and CD is parallel or antiparallel to B. The magnetic force on AB or CD is zero since 0 00 180or in the expression dF = BIdl sin . The magnetic force on straight parts AD and BC is not zero. The magnetic force on AD is directed towards the reader which is equal and opposite to the force on BC which is directed away from the reader. These equal and opposite forces cancel each other. Therefore, the net force on the loop ABCD is zero. Since these equal and opposite forces do not act at the same point, they will exert a net torque on the loop which will rotate it in the clockwise sense when viewed from O.

82 A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat ‘Q’ flows only from left to right through the blocks. Then in steady state

A B

C

D

2K

E1L

3L4L

heat 0 1L 5L 6L

6K

3k

4k

5k

A. heat flow through A and E slabs are same

B. heat flow through slab E is maximum

C. temperature difference across slab E is smallest.

D. heat flow through C = heat flow through B + heat flow through D.

Answer :A,B,C,D Solution : in steady state heat flow through A and E are definitely same

b) Heat current through slab (3 ). .4 4

BdQ A TB kdt L

Heat current through slabe 2(4 ). .4 4

CdQ A TC kdt L

Heat current through slab (5 ). .4 4

DdQ A TD kdt L

Thus B D CH H H

Hence (d) is correct.

THR of . .6

LE i eKA

least resistance

Hence maximum current thus(c) is also correct

( ) ( ) ( )E E TH ET H R

83 A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 600 (see figure) If the refractive index of the material of the prism is 3 , which of the following is (are) correct?

A

B

C

D

060

090 075

0135

O

P

A. The ray gets totally internally reflected at face CD

B. The ray comes out through face AD

C. The angle between the incident ray and the emergent ray is 090

D. The angle between the incident ray and the emergent ray is 0120

Answer :A,B,C Solution : The correct options are(a),(b) and(c)

A

B

C

D

090 075

P060

030

Q

R060

030

045045

045

060

At P,(1) sin 600= 3 sin r

030r

The critical angle 1sin (1/ )r 1sin (1 / 3)

Thus it get total internal reflection at face CD because

1sin3c

Further the diagram shown is self explanatory.

84 A solid rectangular parallelopiped has sides of lengths x,y and z, respectively. The solid is pulled along the Z – direction which produces an extension z in this direction. The relative lateral contractions in

the x and y directions are given by ,x y v zx y z

where v is a constant, The relative change in the

volume of the solid is given by

A. (1 2 )v zz

B. (1 )v zz

C. (1 2 )v zz

D. 3 zz

Answer :A Solution : Conceptual

85 A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 metre. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of ( , )u v values recorded by the student (in cm) are (42,56),(48,48),(60,40),(66,33)(78,39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is(are):

A. (42,56) B. (48,48) C. (66,33) D. (78,39)

Answer :C,D

Solution : 1 1 1v u f

Thus, .f uvu f

The values given can be checked and we find options(c) and(d) only correct

86 Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choices) given below

8

6

4

2

0

100 200 A

/B A

A. Fusion of two nuclei with mass numbers lying in the range of 1<A<50 will release energy

B. Fusion of two nuclei with mass numbers lying in the range of 51<A<100 will release energy.

C. Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments.

D. Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments.

Answer :B,D Solution : 1 < A < 50

On nuclear fusion mass number of resulting nucleus will be less than 100

On fusion, mass number of resulting nucleus is between 100 and 200. The B/A value increases, and hence energy will be released.

On fission for 100 < A < 200, mass number for fission nuclei, i.e, between 50 to 100 B/A value decreased hence no energy will be released.

On fission, for 200 < A < 260, the mass number for fission nuclei will be between 100 to 130 B/A value increases and hence energy will be released

Thus, Options(b)and(d) are correct

87 A parallel beam of light ( 500 )nm is incident at an angle 30 with the normal to the slit plane in a Young’s double-slit experiment. Assume that the intensity due to each slit at any point on the screen is 0I . Point O is equidistant from 1S and 2S . The distance between slits is1mm . Then

S

1S

2S

O

3 m

A. the intensity at O is 04I

B. the intensity at O is zero

C. the intensity at a point on the screen 1m below O is 04I

D. the intensity at a point on the screen 1m below O is zero

Answer :A,C

Solution : Path difference at point sin 0.5O d mm corresponding phase difference, 2 p

3

10

2 0.5 102000 2 1000

5000 10

O is a point corresponding to a maxima with the point at 1m below O corresponding to central maxima.

88 In Young’s double-slit experiment, two wavelengths of light are used simultaneously where 2 12 . In the fringe pattern observed on the screen,

A. maxima of wavelength 2 can coincide with minima of wavelength 1

B. fringe width of 2 will be double that of fringe width of 1 and nth order maxima of 2 will

coincide with 2nth order maxima of 1

C. nth order minima of 2 will coincide with 2nth order minima of 1

D. none of the above are true.

Answer :B

Solution : 2 1 12 12D l lD D

d d d

nth order maxima of 2 coincides with 2nth order maxima of 1

nth order minima of 2 does not coincides with 2nthorder maxima of 1

99 The minimum value of d so that there is a dark fringe at O is mind . For the value of mind , the distance at which the next bright fringe is formed is x. Then, (x<<D, d<<D)

xdB

P

O

D D

A1O

A. mind D B. min 2Dd

C. min3x d D. minx d

Answer :B,C

Solution : There is a dark fringe at O if the path difference 1

2ABO AO O

xd

BP

O

D D

A 1O

1

2 2 2 222 2

2 2

22 2 2 1 2 12 2 2

d d d dD d D D DD D D D

min 2Dd

The bright fringe is formed at P if the path difference 1 1AO P ABP

2 2 2 2 2 2( )D D x D d D x d 2 2 2 2( 2 )

2 2 2x d x d xdD D D

Given mind d

On solving, we get min3 32Dx d

90 Mark out the correct statement(s) A. In both fission and fusion processes, the mass of reactant nuclides is greater than the mass of product nuclide B. In fission process, BE per nucleon of reactant nuclides is less than the binding energy per nucleon of product nuclide C. In fusion process, BE per nucleon of reactant nuclides is less than the binding energy per nucleon of product nuclide D. In fusion process, BE per nucleon of reactant nuclide is greater than the binding energy per nucleon of product nuclide Answer :A,B,C Solution : In general fission and fusion processes are exothermic reactions, i.e., energy is released. Hence, mass of products must be less than mass of the reactant nuclide, and BE/A of reactants < BE/A

of product nuclides

91 23592 U is active. Then, in a larger quantity of the element :

A. the probability of a nucles disintegrating during one second is lower in the first half-life and greater in the fifth halflife

B. the probability of a nucleus disintegrating during one second remains constant for all time

C. every nucleus must disintegrate by the lapse of the average life of the population

D. quite an appreciable quantity of 235U will remain even after the average life

Answer :B,D Solution : Probability of a nucleus to disintegrate in unit time is decay constant

92 A non – conducting ring of radius R having uniformly distributed charge Q starts rotating about 1x x axis passing through the centre in the plane of the ring with an angular acceleration ' ' as shown in the figure. Another small conducting ring having radius ( )a a R is kept fixed at the centre of bigger

ring in such a way that the axis 1x x is passing through its centre and perpendicular to its plane. If the resistance of the small ring is 1r , find the induced current in it (in ampere).

R

X1X

(Given 2

216 10 , 1 , 0.1 , 8 /o

Q C R m a m rad s

)

Answer :8

Solution : 2,2 2 .2 4qd dq qwd qwdq di d

T

2 2

3 2

( sin ) sin2 2 4

o o

o

di R qwdB dB dR R

d

Rd

22 2.

16 16 16o o oqw qw qwaB B a a

R R R

induced emf; 2 88 8

16 1oqad V i A

dt R

93 The space between a pair of co-axial cylindrical conductors is evacuated. The radius of inner cylinder is a, and the inner radius of the outer cylinder is 3a, as shown in figure. A static homogeneous magnetic field B parallel to the cylinder axis, directed into plane of figure, is present. An electron of mass m, charge – e starts out with an initial velocity ov in

b B

a

Vo

the radial direction. The minimum value of magnetic field B in tesla such that the electron does not reach the outer cylinder is (take | | 4omv ae

)

Answer :3

Solution : 1 2 .C C b R In 2 2

2 21 2 ,

2b aC PC b R R a ie R

b

B

R

pa

1C

2C

oV

Also 2

min 2 2 2

.2 4 6 3( ) 8

o oo

m m b ae aev B B TR e b a e a

94 An ideal gas is taken from state A (pressure P , volumeV ) to state B (pressure / 2P , volume 2V ) along a straight line path in the P V diagram. Select the correct statements from the following

A. The work done by the gas in the process A to B exceeds the work done that would be done by it if the system were taken from A to B along an isotherm

B. In the T V diagram, the path AB becomes a part of parabola

C. In the P T diagram, the path AB becomes a part of hyperbola

D. In going from A to B , the temperature T of the gas first increases to a maximum value and then decreases

Answer: A,B,D Solution : Figure shows the straight line path along with the corresponding isothermal path. Since the work done by the gas is equal to area under the curve (such as shown in figure by the shaded portion for the isothermal path), it is obvious that the gas does more work along the straight line path as

compared with that for the isothermal path.

P

/ 2P

,A P V

/ 2,2B P VV2VV

P

As the volume is increased from V to 2V , the difference of pressure between the straight line path and isothermal path initially increases and then decreases after attaining a maximum volume. The same trend is observed in the case of temperature ( P T ., So V is constant).

Now, the slope of straight line path is

Slope = / 22 2

P P PV V V

2P V (slope)

Patting this in the ideal gas equation, PV nRT

[ 2V (solpe)] V nRT

2 (constant)2( )

nRV T Tslope

Which is the equation of a parabola.

Similarly, eliminating V from ideal gas equation, we get Similarly, eliminating V from ideal gas

equation, we get [ ]2( )

PP nRTslope

or 2P = (constant) T

95 An ideal gas undergoes an expansion from a state with temperature 1T and volume 1V to 2V through

three different polytrophic processes A, B and C as shown in the P-V diagram. If | |AE , | |BE and

| |CE be the magnitude of changes in internal energy along the three paths respectively, then:

ABC

P

V1 V2 V

A. | | | | | |A B CE E E if temperature in every process decreases

B. | | | | | |A B CE E E if temperature in every process decreases

C. | | | | | |A B CE E E if temperature in every process increases

D. | | | | | |B A CE E E if temperature in every process increases

Answer :A,C Solution : Initial state is same for all three processes (say initial internal energy = 0E )

In the final state, A B CV V V

And A B CP P P

A A B B C CP V P V P V A B CE E E

If 1 2T T i.e. if the temperature decreases in the process,

Then 0E E for all three processes And hence

0 0 0( ) ( ) ( ) | | | | | |A B A B CE E E E E E E E E

If 1 2T T , then 0 AE E for all three processes A and hence

0 0 0( ) ( ) ( ) | | | | | |A B C A B CE E E E E E E E E

96 A non-conducting vessel containing n moles of a diatomic gas is fitted with a conducting piston. The cross-sectional area, thickness and thermal conductivity of piston are A, l and K respectively. The right side of the piston is open to atmosphere at temperature 0T . Heat is supplied to the gas by means of an electric heater at a constant rate q :

0T

l

Atmosphere

A. the temperature of the gas as a function of time is (2 )/(7 )0 1 KAt nRlqlT T e

AK

B. the maximum temperature of the gas is max 0qlT TAK

C. the ratio of the maximum volume to the minimum volume is max

0 0

1V qlV KAT

D. the maximum temperature of gas is 02T Answer :A,B,C Solution : A) Let temperature of the gas in the vessel be T at any instant t . The net rate at which heat is absorbed by the gas

0 0( ) ( )KA T T KA T TdQ q qdt l l

0T

l

Atmosphere

Let heat supplied dQ will change temperature by dT . The process will be isobaric as piston is open to atmospheric pressure

Thus 72PdQ nC dT n R dT

From eqn. (1) and (2), we have

0( )72

KA T TnR dT qdt l

0

2( ) 7

dT dtql KA T T nRl

On integrating within limit, we get

0 0 0

2( ) 7

T t

T

dT dtql KA T T nRl

or 0

0 0

( )1 2ln( ) 7

ql KA T T tKA ql KA T T nRl

or 02ln[1 ( )]7

KA KAtT Tql nRl

or (2 )/(7 )0 [1 ]KAt nRlqlT T e

AK

B) At the maximum temperature maxT the rate of heat supplied by heater is equal to rate of heat lost

by the gas max 0( )KAq T Tl

or max 0qlT TAK

Same result can be obtained taking t in the expression of T

C) As the process is isobaric, from ideal gas equation we have

0 max

0 max

V VT T

max max

0 0 0

1V T qlV T KAT

97 Two spherical black-bodies A and B , having radii Ar and Br , where 2B Ar r emit radiations with

peak intensities at wavelengths 400 nm and 800 nm respectively. If their temperature are AT and BT

respectively in Kelvin scale, their emissive powers are AE and BE and energies emitted per second

are AP and BP then :

A. / 2A BT T B. / 4A BP P C. / 8A BE E D. / 4A BE E

Answer :A,B

Solution : 800 2400

A B

B A

TT

4

4 16A A

B B

E TE T

4 2

4 2

4 44

A A A

B B B

P T rP T r

98 In the circuit diagram shown

RL

R

A. time constant is L/R

B. time constant 2L/R

C. steady state current in inductor is 2 / R

D. steady state current in inductor is / R

Answer :A,D Solution : In the circuit diagram shown

RL

R

99 A vessel is partly filled with liquid. When the vessel is cooled to a lower temperature, the space in the vessel, unoccupied by the liquid remains constant. Then the volume of the liquid ( )LV , volume of the

vessel ( )vV , the coefficients of cubical expansion of the material of the vessel ( )v and of the liquid

( )L are related as

A. L V B. L V C. / /V L V LV V D. / /V L L VV V

Answer :A,D Solution : L VV V

L L V VV V (or) VL

V L

VV

But V LV V L V

100 Heat is supplied to a certain homogenous sample of mater of matter at a uniform rate. Its temperature is plotted against time as shown in the figure. Which of the following conclusion can be drawn?

Temp

Time

A. Its specific heat capacity is greater in the solid state than in the liquid state

B. Its specific heat capacity is grater in the liquid state than in the solid state

C. Its latent heat of vaporization is greater than is latent heat of fusion

D. Its latent heat of vaporization is smaller than is latent heat of fusion

Answer :A,C Solution : Slope of graph is greater in the solid state i.e., temperature is rising faster, hence lower heat capacity. The transition from solid to liquid state takes lesser time, hence latent heat is samller.

101 In the circuit shown, resistance 100R , inductance 2L

H and capacitance 8C F

are

connected in series with an ac source of 200 volt and frequency ‘f’. If the readings of the hot wire

voltmeters 1V and 2V same then:

R L

~

1V2V

A. 125f Hz B. 250f Hz

C. Current through R is 2A D. 1 2 1000V V volt

Answer : A,C,D Solution : 1 2V V L CX X

1 1252

f HzLC

00

200100

VIR

( 0 )X Z R

2A

1 2 LV V IX I . ( ) 2 2 125 2 /L

= 1000 volt Ans.

= 1000 volt Ans.

102 When an electron accelerated by potential difference U is bombarded on a specific metal, the emitted X-ray spectrum obtained is shown in adjoining graph. If the potential difference is reduced to / 3U , the correct spectrum is –

I

X-ray wavelengthm 2.5 m

A.

I

B.

(B)

C.

I

D.

D

Answer: B Solution: m will increase to 3 m due to decrease in the energy of bombarding electrons. Hence no characteristic x-rays will be visible, only continuous X-ray will be produced.

QUICK REVISION TEST

MATRIX- MATCHING TYPE QUESTIONS

1 ˆ ˆ ˆ3 4 ; 2 3 ;ˆ ˆ Â i j B i j C i j

Column I Column II

A. A B is rotated through 045 clockwise. The resulting vector is P. ˆ î j

B. 2 j is rotated through 045 clockwise. Resulting vector is Q. 2i

C.

2A B C

is rotated through 045 . The resulting vector is R. ˆ2

ˆ1 32

i j

D. C is rotated through 015 . The resulting vector can be S. ˆ

2ˆ3 1

2i j

Answer:

A-Q; B-P; C-Q; D-RS;

Solution: A) ˆ Â B i j

After rotation vector ˆ2i 045

B) 1 12 22 2

ˆ ˆ ˆ î j i j

C) 2 22 ˆ2 ˆ

2Â B C i j i

D) ˆ ˆ( ) 2i j

3 12 ˆ2

ˆ22

i j (Clockwise)

Sai charan

Typewritten text


1 32 ˆ2

ˆ22

i j (Anticlockwise)

2 MATCH THE FOLLOWING

Column I Column II

A. Particle moving with increasing acceleration P. Speed of the particle may not change

B. Particle moving with zero acceleration Q. Direction of velocity may be changing without change in

magnitude

C. Particle moving with acceleration of constant

magnitude

R. Speed of the particle may increase

D. Particle moving with uniform deceleration S. Velocity of particle will decrease

Answer:

A-R; B-P; C-PQR; D-S;

Solution: CONCEPTUAL

3 A man wants to cross a river. If mwv velocity of a man relative to water wv velocity of water relative to

ground, mv velocity of man relative to ground, match the following : [ angle between normal to the river

flow and mwv ]

Column I Column II

A. Minimum distance for mw wv v P. 1sin mw

w

vv

B. Minimum time for mw wv v Q. m wv v

C. Minimum distance for mw wv v R. mw wv v

D. Minimum time for mw wv v S. 1sin w

mw

vv

Answer:

A-QS; B-R; C-P; D-R;

Solution: Conceptual

4 A small body is projected up at an angle 090 to the horizontal. Column I contains variables to be taken on x and y-axes and column II contains corresponding graphs for this body, match the following.

Column I Column II

A. Slope of trajectory of the body on y-axis and time on x-axis

P. O x

y

B. Magnitude of tangential acceleration on y-axis and time on x-axis

Q.

O x

y

C. Magnitudes of normal acceleration on y-axis and altitude h on x-axis

R.

O x

y

D. Speed of projectile on y-axis and time on x-axis

S.

O x

y

Answer: A-S; B-R; C-Q; D-P;

Solution: A) Path equation is 2y Ax Bx

slope 2 2 ( )xdyS A Bx A B u tdx

2 xS A Bu t This is in the form of y = -mx + C. Hence S is the suitable graph.

taO

g

v

x

B) As the particle rises from point of projection O to highest point, decreases from to tO a decreases from sing to zero. angle of projection. After that, ta again increases from O to sing . This variation is best represented by (R) C) Normal acceleration cosNa g . Na increases from cosg to g and then decreases to cosg again. Thin is best represented by Q.

D) 2 2 2 2( cos ) ( sin )x yv v v u u gt

2 2 2 2 (sin )v u g t u gt With increase in t, v first decreases and then increases after reaching maximum height. But

minimum 0v . This is best represented by P.

5 Lift can move in y-axis as well as along x-axis. A ball of mass m is attached to ceiling of lift with inextensible light rope and box of mass m is placed against a wall as shown in figure. Neglect friction everywhere.

y

m

T

m

x

N

Column I Column II

A. In figure lift is moving along x-axis then value of T may be P. Zero

B. Lift moving towards right along x-axis with decreasing speed then value of N may be Q. >mg

C. Lift is moving in upward direction (y-axis) then value of T may be R. < mg

D. Lift is moving in downward direction with constant velocity then value of T may be S. = mg

Answer: A-QS; B-QRS; C-PQRS; D-S;

Solution: 1 2netF F F

Here 1F mg

2 pseudoF F

Constant velocity 0pseudoF

6 For a particle in one dimensional motion, the following columns :

Column I Column II

A. Zero speed but non-zero acceleration P. Body which is about to fall

B. Zero speed and non-zero velocity Q. Extreme position of oscillating body

C. Constant speed non-zero acceleration R. Possible

D. Positive acceleration and speeding up S. Not possible

Answer: A-PQR; B-S; C-R; D-PQR;

Solution: Consider possible in horizontal, vertical otions in one dimension and linear SHM.

7 Two identical blocks A and B, connected by a massless string , are placed on a frictionless horizontal plane. A bullet having same mass, moving with speed u, strikes block B from behind as shown. If the bullet gets embedded into block B, then match the following: Take direction of u shown in the figure as positive.

m uCmBA

m

Column I Column II

A. Speed A after collision P.

23mu

B. impulse on A due to tension in the string; Q.

3mu

C. Impulse of C due to force of collision R.

3u

D. Impulse of B due to force of collision S.

23mu

Answer: A-R; B-Q; C-S; D-P;

Solution: Common velocity,

3 3A B Cmu uV V Vm

Impulse on A due to tension in the string, 3A

muJ

Impulse on C due to normal force of collision, 2

3 3Cu muJ m u

Impulse on B due to normal force of collision 2

3B CmuJ J

8 In the arrangement shown in figure match the following :

All smooth

K 4N / m

2m / s

2kg 2kg

Column I Column II

A. Velocity of centre of mass P. 2 SI unit

B. Velocity of combined mass when compression in the spring is maximum Q. 1 SI unit

C. Maximum compression in the spring R. 4 SI unit

D. Maximum potential energy stored in the spring S. 0.5 SI unit

Answer: A-Q; B-Q; C-Q; D-P;

Solution: 1 1 2 2

1 2

1 /CMm v m vV m s

m m

During maximum compression also, velocity of combined mass is 1 m./s.

Now, 2 2max

1 12 (2) 4 (1) 22 2i fU K K x x x x J

From 2max

1 12

KX m

We have, max 1X m

9 A Horse is pulling a cart. Match the statements of the columns.

Column I Column II

A. The force with which horse pulls the cart is responsible for P. Motion of the system

B. The force which ground exerts on the horse, is responsible for Q. motion of the cart

C. The horse will move if force exerted by the ground on the horse exceeds the force with which

R. cart pulls the horse

D. If force exerted by the ground on the horse is greater than frictional force acting on cart, it is responsible for

S. motion of the horse.

Answer: A-Q; B-PQS; C-R; D-PQS;

Solution: CHF = force on cart due to pull of the horse

F = Frictional force on the cart.

1R2RCHF

Mg mgHGF1f

HCF

HCF = force on horse due to pull of cart.

HGF = Force on horse exerted by the ground.

CHF is responsible for motion of cart if

CHF > f1

HGF is responsible for motion of horse, The horse will move if

HGF > HGF

HGF is responsible for motion of system if HGF > 1f

1f

1R

CHF

Mg

CART

H CF

2R

H GF

mgHORSE

1f

1 2R R

HGF

M m g

SYSTEM

10 Block A slides down the wedge of same mass. All surfaces are smooth. The angle of inclination of wedge is with horizontal. Match the columns :

B

A

θ

Column I Column II

A. Relative acceleration of block A w.r.t. wedge P. 2 2g sin / 1 sin

B. Absolute acceleration of block A Q. Zero

C. Acceleration of centre of mass of block and wedge in horizontal direction. R. 22g sin / 1 sin

D. Acceleration of centre of mass of block and wedge S. g sinθ

Answer: A-R; B-S; C-Q; D-P;

Solution: Let a be the acceleration of wedge towards left and ra be the relative acceleration of the block down the plane.

mg

N

ma pseudo

So, cosra a a 2a = ar, cos For block A, N + masin = mg sin For wedge N sin = ma

sinN

N

N Cos From equns.(2) (3) and (1)

22 sin / (1 sin )ra g

ya = Acceleration of block in vertical downward direction

= 2

2

2 sinsin(1 sin )r

ga

02

ycom

maa

m

= 2

2

sin2 (1 sin )

ya g

Net force on system in horizontal direction is zero. So acceleration of COM in horizontal direction is zero.

11 In each situation of Column – I a system involving two bodies is given. All strings and pulleys are light and friction is absent everywhere. Initially each body of every system is at rest. Consider the system in all situations of Column – I from rest till any collision occurs. Then match the statements in Column – I with the corresponding results in Column – II

Column I Column II

A. The block plus wedge system is placed over smooth horizontal surface. After the system is released from rest, the centre of mass of system

M

m

L

P. Shifts towards right

B. The string connecting both the blocks of mass ‘m’ is horizontal. Left block is placed over smooth horizontal table as shown. After the two block system is released from rest, the centre of mass of system

Q. Shifts downwards

m m

C. The block and monkey have same mass. The monkey starts climbing up the rope. After the monkey starts climbing up, the centre of mass of monkey – block system

m

R. Shifts upwards

D. Both blocks of mass ‘m’ are initially at rest. The left block is given initial velocity ‘u’ downwards. Then, the centre of mass of two block system afterwards

m m

S. Does not shift

Answer: A-Q; B-P,Q; C-R; D-S;


12 Two blocks of same mass m = 10 kg are placed on rough horizontal surface as shown in fig. Initially tension in the massless string is zero and string is horizontal.

FB10 kg string A

2 0.3 1 0.2

10 kg

A horizontal force 40sin( )6

F t is applied as shown on the block A for a time interval t = 0 to t = 6s. Here ‘F’

is in Newton and‘t’ in second. Friction coefficient between block A and ground is 0.20 and between block B and ground is 0.30. (Take g = 10 m/sec2). Match the statements in Column – I with the time intervals (in seconds) in Column – II

Column I Column II

A. Friction force between block B and ground is zero in the time interval P. 0 < t < 1

B. Tension in the string is non zero in the time interval Q. 1 < t < 3

C. Acceleration of block A is zero in the time interval R. 3 < t < 5

D. Magnitude of friction force between A and ground is decreasing in the time interval S. 5 < t < 6

Answer: A-P,S; B-Q,R; C-P,Q,R,S; D-S;

Solution: max 40 (20 30)F N N So as long as force ‘F’ is in positive x-direction, both the blocks are at rest at t = 0, 6s, F = 0 at t = 1, 5s, F = 20 N at t = 3s, F = 40 N For 0 < t < 1, 5 < t < 6 0 < F < 20 N aA = 0, Tension = 0, aB = 0, fB = 0 for 1 < t < 3, 3 < t < 5 20 < F < 40

T AF

20N N

0, 0Aa T 20T F i.e. 0 < T < 20

30 NT aB = 0

a – p, s; b – q, r, c – p, q, r, s; d – s

13 Match the following Column – I with Column – II : All surfaces are rough and no slipping takes place.

Column I Column II

A. Ring

P. Body accelerates forward

B. Disc

Q. Rotation about centre of mass is clockwise

C. Solid cylinder

R. Friction force acts backward on the body

D. Solidsphere

S. No frictional force acts

Answer:

A-P,Q,S; B-P,Q,R; C-P,Q,S; D-P,Q,R;


14 In the given arrangements in Column – I, for small displacements of block from equilibrium position match the time periods in Column – II springs, threads and pulleys are massless

Column I Column II

A.

K

m

block is suddenly released from rest initially the spring is in it’s natural length block is in vertical SHM

P. 2 mTK

B.

K

m

block is slightly displaced from the equilibrium then released . the block is in vertical SHM

Q. 52 mTK

C.

K K

m

300300

block is slightly displaced from the equilibrium then released, the block is in vertical SHM

R. 24mTK

D.

K

m

block is slightly displaced from the equilibrium then released the block is in vertical SHM

S. 22 mTK

Answer: A-P; B-Q; C-S; D-R;

Solution: A) For single spring Keff = K

2 mTK

B) for two springs

2 5effeff

mT K KK

C) for two springs 22 sineffK K

2eff

mTK

= 2K

D) 4effK K

2 24eff

m mTK K

15 A charged particle passes through a region that could have electric field only or magnetic field only or both electric and magnetic field or none of the fields. Match Column – I with possible situations in Column – II

Column I Column II

A. Kinetic energy of the particle remains constant P. Under special conditions, this is possible when both electric and magnetic fields are present

B. Acceleration of the particle is zero Q. The region has electric field only

C. Kinetic energy of the particle changes and it also suffers deflection

R. The region has magnetic field only

D. Kinetic energy of the particle changes but it suffers no deflection

S. The region contains no field

Answer: A-P,R,S; B-P,R,S; C-P,Q; D-P,Q;


16 Match the situations in column I with the resulting torques and forces on loop with 2i current in column II.

Column I Column II

A.

2i1i

P. 0, 0F

B.

i1i2

Centers of two circular loops are same and they are coplanar

Q. 0, 0F

C.

1i2i

Centers of two circular loops are same and their planes are mutually perpendicular

R. 0, 0F

D.

1i

2i

Planes of the loops are perpendicular.

S. 0, 0F

Answer: A-S; B-R; C-P; D-Q;

Solution: & m eqM B F I l B

17 Match the following Column-I with Column-II

Column I Column II

A.

qHollow neutral conductor

P. E

inside the conductor is zero

B.

q Hollow neutral conductor

Q. | |E

inside the conductor is constant

C. 1q 2q

Hollow neutral conductor

R. | |E

inside the conductor is varying

D. 1q 2q

Hollow neutral conductor

S. Potential inside the conductor is same as that of conductor

T. Potential inside the conductor is varying

Answer: A-P,S; B-R,T; C-R,T; D-R,T;

Solution: For A, as no charge is there inside the conductor, so E

inside the conductor is zero and hence potential is constant and equal to that of potential of the conductor. For other cases, E

inside conductor is

non-zero and varying as we are going from centre to periphery and so potential inside the conductor is

varying.

18 Consider an incompressible and non-viscous liquid in a container. Density of liquid is and acceleration due to gravity is g and h represents the vertical separation between two points. All points considered in Column I are inside the liquid. Match the statements given in Column I with corresponding all possible conditions given in Column II.

Column I Column II

A. Pressure difference between two distinct points is gh where two points are at same vertical line at a separation h.

P. Container is stationary.

B. Pressure difference between a pair of two distinct points on same horizontal level is zero.

Q. Container is accelerating in horizontal direction.

C. Pressure difference between a pair of two distinct points on same horizontal level is non-zero.

R. Container is falling freely.

D. Pressure difference between any two distinct points on same vertical line is zero.

S. Container is accelerating up in vertical direction with an acceleration < g.

T. Container is rotating about a vertical axis passing through its symmetry.

Answer: A-P,Q,T; B-P,R,S,T; C-Q,T; D-R;

Solution: (P) (i) Two points in same horizontal level will have same pressure. (ii) gh (Q) (i) 0al along vertical. (ii) 0gh along vertical. ® Pressure at every point is zero. (S) (i) gh ah gh along vertical. (ii) 0 along horizontal. (T) (i) gh along vertical.

(ii) 2 212

r along horizontal.

19 A particle of mass 2 kg is moving on x-axis under the action of force (8 2 )F x N . The particle is released from rest from x = 6m. For the subsequent motion, match the following. (All the values in the right column are in their S.I units.)

Column I Column II

A. Equilibrium position is at x = P. 2

B. Amplitude of S.H.M is Q. / 2

C. Time taken to go directly from x = 2 to x = 4 R. 4

D. Energy of S.H.M is S. 6

Answer: A-R; B-P; C-Q; D-R;

Solution: F = 8 – 2x = -2(x – 4) At equilibrium position, 0 4F x m As particle is released at rest x = 6m, i.e. it is one of the extreme positions. Hence, Amplitude A = 2m. Here, force constant 1 22 2k Nm m

Or 1 time period, 2 2T

Time to go from x = 2 m to x = 4 m (i.e. from extreme position to mean position) 4 2T

Energy of S.H.M 21 1 2 4 42 2

kA J

As particle has started it’s motion from positive extreme

Phase constant 2

20 Match the statements in column – I with the statements in column - II

Column I Column II

A. A tight string is fixed at both ends and sustaining standing wave

P. At the middle, antinode is formed in odd harmonic

B. A tight string is fixed at one end and free at the other end

Q. At the middle, node is formed in even harmonic

C. A tight string is fixed at both ends and vibrating in four loops

R. The frequency of vibration is 300% more than its fundamental frequency

D. A tight string is fixed at one end and free at the other end, vibrating in nd 2 overtone

S. Phase difference between SHMs of any two vibrating particles will be either or zero

T. The frequency of vibration is 400% more than fundamental frequency

Answer: A-P,Q,S; B-S; C-Q,R,S; D-S,T;

Solution: A) Number of loops (of length / 2 ) will be even or odd node or antinode will respectively be formed at the middle. Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be . (B) and (D) Number of loops will not be integral. Hence neither a node nor an antinode will be formed in the middle

Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be .

21 The following figures shown different bodies which are either free to rotate or translate on smooth horizontal

surface. An impulse J is given to the bodies in the direction shown in figure. Match the columns :

Column I Column II

A.

M

M

J

P. Translation

B.

M

M

J

Q. Rotation occurs

C.

JL- shaped strip not

fixed anywhere

R. Angular momentum about CM increases

D.

J

hinge

S. Linear momentum increases.

Answer:

A-P,QR,S; B-P,S; C-P,Q,R,S; D-Q,R,S;

Solution: A,C – Dumbell experiences a force and net torque, so it has translation and rotation.

Also J P , so linear momentum increase.

,J L so angular momentum also increases.

B :- 0 , so it experiences translation and linear momentum increases

D :– Due to hinge strip will not move.

22 Figure shows a siphon. It is a long pipe which is used to drain water from the reservoir at higher level to a reservoir at lower level. Regarding with the siphon match the following columns :

Summit

2h

1huP

4Inlet leg

32

1

Column I Column II

A. Pressure is more than atmospheric pressure at P. 1

B. Pressure is Less than atmospheric pressure at Q. 2

C. Pressure is highest of all the five points at R. 3

D. Pressure is least of all the five points S. 4

T. 5

Answer: A-P; B-Q,RS; C-P; D-R;

Solution: For pipe of uniform cross-section 3 4 5 Applying Bernoulli’s equation between (1) and (5) , we have

22 2

12a aP P gh

2 22gh

Thus for 2 20, 0h Also

2 23 4 1

1 12 2aP P P gh

22

12aP gh

From the above equation following conclusion can be made 4 3 aP P P

1 2 4ag h h p p

41 2

ap ph hg

or 1 2aph hg

Summit

2h

1huP

4Inlet leg

32

1

23 A circuit consisting of five ideal cells, three resistors 1 2( , 20 )R R and and a capacitor of capacitance 1C F

is shown in the figure. Match the conditions given in Column I with their results given in column II.

30V 10 V 20 V

R1 R214 V20

A B1 F

dK1K2

12 V

c

Column I Column II

A. 1K is in position c and 2K is open P. potential at point A is greater than potential at point B

B. 1K is in position c and 2K is closed Q. current through 1R is downward

C. 1K is in position d and 2K is open R. current through 2R is upward

D. 1K is in position d and 2K is closed S. charge on capacitor is 10 C

Answer:A-P,Q,S; B-P,Q,S; C-P,Q,R,S; D-P,Q,R,S;

Solution: q = (20 10) 10C C

30V > 10 V, 20V > 10V 1Ri , A BV V

21 14 12 20 RK C i

24 Some electric circuits containing any two of the components – a resistor, an inductor, and/or a capacitor – supplied

with either a variable DC source or an AC source of frequency 50 Hz, are shown in column II below. When a

current I (steady state for DC or rms for AC) flows through the circuit, let the corresponding potential difference

1 2V and V across the components be related as shown in column I.

Match the items in column I with those in column II.

Column I Column II

A. 10,I V is proportional to I

P.

V1 V2

V

6mH 2

B. 2 10,I V V

Q.

V1 V2

V

6mH 3 F

C. 1 20,V V V

R.

V1 V2

V

2 F1k

D. 20,I V is proportional to I

S.

V1 V2

V

6mH2

T.

V1 V2

V

6mH 3 F

Answer:

A-P,R,T; B-P,R,T; C-Q,S; D-P,R,T;

Solution: A–PRT ; B–PRT ; C– QS ; D– PRT

25 A uniform solid cylinder of mass M and radius R is connected to light spring(s) of force constant K kept on rough surfaces shown in figures of column I. When it is displaced slightly and released, it executes SHM. Their time periods of oscillations are given in column II. Match the two columns. Assume that cylinders roll without slipping while oscillating.

Column I Column II

A.

K

P. 324

mK

B.

K K

Q.

328

mK

C.

K

R.

322

mK

D. 30o

K

S. 2 mK

Answer: A-R; B-P; C-Q; D-R;

Solution: Apply the concepts of angular SHM

26 A thin uniform rod of mass 1 kg and length 1 m is acted upon by different forces a shown in column I. If area of cross section of the rod is A and its Young modulus is Y. Then match column – I with elongation of the rod given in column – II.

Column I Column II

A. 1N

1N P. 2

AY

B. 1N

1N

Q. Zero

C. 3N

1N R. 4

AY

D. 2Nrigid

S.

1AY

Answer: A-Q; B-S; C-P; D-P;

Solution: Consider a small element of thickness dx at a distance x from B.

F2F1

A xB

Fax

2 1 1xx xT F F

2 1x xF F

2 11 ( ( ) ) )Txde F x F x dx

AY AY

2 10

1e F x dx F xdxAY

2 2

2 10 0

12 2xF x F

AY

2 2

2 11

2 2F F

AY

1 2( )2

F FAY

a) Zero. In the left half of the rod, stress is compressive and in the right half, stress is tensile. Compression in the left half and extension in the right half are equal. Hence net elongation of the rod is zero.

b) 1

AY c)

2eAY

d) 2

AY

27 The system shown below is initially in equilibrium Masses of the blocks A,B,C,D, and E are, respectively, 3 kg, 3kg, 2 kg , 2kg and 2 kg, Match the conditions in Column I with the effect in Column II.

A

B

C

D

E

spring 1

spring 2

Column I Column II

A. After spring 2 is cut, tension in string AB P. Increases

B. After spring 2 is cut, tension in string CD Q. Decreases

C. After string between C and Pulley is cut, tension in string AB R. Remain constant

D. After string between C and Pulley is cut, tension in string CD S. Zero

Answer: A-R; B-Q; C-QS; D-Q;

Solution: A,B After spring 2 is cut , tension in string AB will not change. ( ) 4CD iT mg

( ) . .A B C DCD f D D

A B C D

m m m mT m g m gm m m m

12 (1 ) 2.45

mg mg

Hence CDT decreases. C, D After string between C and pulley is cut, tension in string AB will becomes zero. ( ) ( ) 4CD i D ET m m g mg Acceleration of C and D blocks is ( ) ( ).C D E C Dm m g m g m m a

6 3 , ( )4 2 CD f c c

mga g T m g m amg

3( ) 2 22CD fT m g mg mg

The tension decrease,

28 A particle is projected with a speed u at angle with horizontal from point A. It strikes elastically with a

vertical wall at height h/2. It rebounds and reaches maximum height h and falls back on the ground at point B as

shown in Fig. Distances from A to wall and from wall to B are 1x and 2x , respectively, and time to cover 1x and

2x are 1t and 2 ,t respectively. Match the values in column 1 with the expressions in column II.

Wall

h/2h

u

B x1A

x2

Column I Column II

A. 2 P. 2 1 2 1

2 1 2 1

x x x xorx x x x

B.

12

Q. 2 1 2 1

2 1 2 1

t t t tort t t t

C. 1 R.

1 2

sin( )u

g t t

D. 1

2 S. 1 2

1 2

cos ( )( )

u t tx x

Answer:

A-PQ; B-PQ; C-S; D-R;

Solution: 1 1 2 2cos ; cosx u t x u t

1 2 2 12 2;2 2 2 2

t t t th T hg g

Dividing. 2 1

2 1

2t tt t

Also, 2 1 2 1

2 1 2 1

x x t tx x t t

1 22 sin 2 sin;u uT t t

g g

1 2

1 sin2 ( )

ug t t

Also, 1 2 1 2cos ( )x x u t t

1 2

1 2

cos ( ) 1u t tx x

29 Two blocks of masses 3 kg and 6 kg are connected by an ideal spring and are placed on a frictionless horizontal

surface. The 3 kg block is imparted a speed of 2 m/s towards left. (Consider left as positive direction)

3kg 6kg2m/s

Column I Column II

A. when the velocity of 3 kg block is 2 /

3m s P. velocity of centre of mass is 2 /

3m s

B. when the speed of 3 kg block is 2 /

3m s

Q. deformation of the spring is zero

C. when the speed of 3 kg block is minimum R. deformation of the spring is maximum

D. when the velocity of 6 kg block is maximum S. The blocks are at rest with respect to each other

Answer:

A-PRS; B-PQRS; C-P; D-PQ;

Solution: 3 2 6 0 2 /

3 6 3CMv m s

So velocity of centre of mass is always 2/3 m/s

If velocity of 3 kg block is 2/3 m/s then velocity of 6 kg block is also 2/3 m/s. In this situation, spring will be

maximum elongated. Their relative velocity will be zero.

When the speed is 2/3, velocity can be 2 / 3 m/s, if 3 2 / 3 /kgv m s

223 6

2 33 3 6CM

vv

24 /3

v m s

Let deformation in spring be x. Then 2 2

2 21 1 2 1 4 13(2) 3 62 2 3 2 3 2

kx

x=0

Minimum speed of 3 kg block is zero. At this speed, spring will have some deformation, but not maximum.

When the velocity of 6kg is maximum the spring will become unstretched. From conservation of linear

momentum 0relv

30 Match the following two columns.

Column I Column II

A. A ring of mass m is projected on a rough horizontal plane with a velocity 0v . The magnitude of work done by friction in setting pure rolling is

P. 20

13

mv

0v

B. Kinetic energy of rotation of the pivoted thin uniform rod of mass m is

0vC C = center of rod

Q. 20

18

mv

C. Kinetic energy of translation of a smooth thin rod of mass m is

0v

C C = center of rod0v2

R. 20

14

mv

D. KE of the thin uniform rod of mass m at the instant shown is 0v

m045

S. 20

23

mv

Answer: A-R; B-S; C-Q; D-P;

Solution: A)

Oinitially

after attainingpure rolling

vv0

About O, 2 00 2

vvmv R mvR mR vR

fW = loss in 22

2 2 20 00 0

1 1 1(1 1)2 2 2 2 4

mv vKE mv mv m mv

B) 22 2

2 2 200

1 1 22 2 3 6 / 2 3

vm mKE I mv

C) 2

2 200

1 1 12 2 2 8t cm

vKE mv m mv

D)

0v0450v

A

B

2

2

O

From constrained equation, velocity of end A will be 0v vertically

downwards. O is the ICR.

0 02/ 2v v

2 22 2

2 2 200 02

21 1 1 12 2 12 2 2 3 3

vm mKE I m mv

31 In Column I, some operation performed on capacitor are given, while in Column II are given some probable effects on capacitor. Match the entries of Column I with the entries of Column II.

Column I Column II

A. A dielectric slab is inserted into the capacitor slowly keeping the charge constant.

P. Work done by external agent in negative

B. The plates of capacitor are moved apart keeping the charge constant

Q. Work done by external agent is positive

C. A dielectric slab is inserted into the capacitor slowly keeping the voltage constant.

R. Electric potential energy stored in the electric field in between the capacitor plates is decreasing.

D. The plates of capacitor are moved apart keeping the voltage constant

S. Electric potential energy stored in the electric field in between the capacitor plates is increasing

Answer: A-PR; B-QS; C-PS; D-QR;

Solution: Conceptual.

32 A charged particle having specific charge is released from origin. There exists a uniform electric field

0ˆE E j

and 0

ˆB B k

is the magnetic field. After t seconds, the velocity of the charged particle is ˆˆ ˆ

x y zV i V j V k and position vector is ˆˆ ˆxi yj zk . Then match the following here 0( )w B

Column I Column II A. xV P. 0

0

sinE wtB

B. zV Q. 0

0

(1 cos )E wtB

R. 020

(1 cos )wtB

D. z S. 020

( sin )E wt wtB

T. zero Answer:

A-Q; B-T; C-R; D-T;

Solution: 0 0

0 0

ˆˆ ˆsin [1 cos ]( )E Ev t e t e BB B

0 0

0 00

0 02 20 0

. [1 cos ] sin

sin [1 co

ˆ ˆ

ˆ ˆs ]

t E Er v dt t i t jB B

E Er t t i t jB B

33 Match the following Column – I with Column – II :

Column I Column II

A. Radius of orbit is related with atomic number (z) P. Is proportional to z

B. Current associated due to orbital motion electron with atomic number (z)

Q. Is inversely proportional to z

C. Magnetic field at centre due to orbital motion of electron related with (z)

R. Is proportional to 3z

D. Velocity of an electron related with atomic number (z) S. Is proportional to 3z

T. Is proportional to 5z

Answer: A-Q; B-R; C-S; D-P;

Solution: 00.529nr A

z

62.2 10 /nV z m sn

2

3

1.06 zI mAn

3

5

12.5 zB teslan

34 Match the following Column – I with Column – II :

Column I Column II

A. In case of series L-C-R circuit, at resonance. P. Current in the circuit has same frequency

B. Only resistor in an a.c. circuit Q. Voltage lags the current by / 2

C. Only inductor in an a.c. circuit R. Current lags the voltage by / 2 .

D. Only capacitor in an a.c. circuit S. Reactance of the circuit is zero

T. Current is in phase with applied voltage

Answer: A-PST; B-PST; C-PR; D-PQ;

Solution: (a) R-LC circuit at resonance At resonance, L CX X Net reactance = 0.

2 2( )L CZ R X X R

If 0 0sinE E t then 00 0 0sin EI I t where I

R

Frequency of current is saem as frequency of alternating voltage source.

tan 0 0L CX XR

current is in phase with applied voltage. (b) Only resistor in an ac circuit Frequency of current is same as frequency of applied voltage. Reactance of the circuit is zero.

tan 0 0XR

So current is in phase with applied voltage. (c) Only inductor in ac circuit Let applied voltage 0 sinE E t

Current in the circuit, 00 0sin ,

2 L

EI I t where IX

Frequency of current is same as applied voltage. Current lags behind applied voltage.

(d) Only capacitor in an a.c. circuit Current 0 sin2

I I t

Frequency of current is same as applied voltage. Current leads the applied voltage by 2

.

35 In column I some circuits are given. In all the circuits except in A, switch S remains closed for a long time and then it is opened at 0t while for A, the situation is reversed.

Column II tells something about the circuit quantities. Match the entries of column-I with the entries of column-II

Column I Column II

A.

/\/\/\/\/\

/ \/\ /\

/ \

L

R

S

E P. Induced emf can be greater than E

B.

/\/\/\/\/\

/\/\/\/\

L

R

SE

Q. Induced emf would be less than E

C.

/\/\/\/\/\

L

R

S

E

/\/\/\

/\

R. Finally, energy stored in inductor is zero

D.

/\/\/\/\/\

/\/\/\/\

L

R

S

S. Finally, energy stored in inductor is non-zero

Answer:A-R; B-R; C-PR; D-PR;

Solution: A : current in inductor when switch is open, 0EiR

Initially induced emf will be equal to E and finally it is zero. So, energy stored will be zer

B : same as P C & D : Here current becomes zero suddenly. So, didt

is large

36 A circuit is shown in fig. R is a non-zero variable with finite resistance. ' 'e is some unknown emf with polarities as shown. Match the columns

/ \/\ /\

/ \ /\

/\ /\/ \

/\ /\

/ \ /\ / \

/\ /\

A

B C D

EF

R2

AV

4

6Ve

Column I Column II

A. Current passing through 4 resistance can be zero P. Possible if e = 6V

B. Current passing through 4 resistance can be from F to C direction

Q. Possible if e > 6V

C. Current passing through 4 resistance can be from C to F direction

R. Possible if e < 6V

D. Current passing through 2 resistance will be from B to A direction

S. Possible for any value of e from zero to infinity

Answer:A-Q; B-PQR; C-Q; D-PQRS;

Solution: From loop CDEFC, 1 26 4 1e i R i

From look ABCFA, 2 1 2 1 26 4 4 2 2 6 2 2i i i i i

/\ /\/\

/\/\

/\ /\/ \

/\ /\

/\ /\ /\

/\ /\

A

B C D

EF

R

1 2i i

2

4V

4

6V2i

1i

e

on solving we get, 1 23 14 6,3 4 3 4e R ei iR R

A : 2 0 6. 0, 6i e R As R e V B : For current from F to C direction , 2 0 6i e . Possible for any finite volume of e, because R is finite C : For current from F to C direction 2 0 6i e R

D : For current in 2Ω from B to A direction, 1 22 8 0 2 8 0 4

4 3 2R e Ri i R e e

R

Depending upon value of e can take any value from zero to infinity.

37 A sample of gas goes from state A to state B in four different manners, as shown by the grpahs. Let W be the work done by the gas and U be change in internal energy along the path AB . Correctly match the graphs with the statements provided.

Column I Column II

A.

A BV

P P. Both Q and U are positive

B.

A

B

T

P

Q. Both W and U are negative

C.

A

B

V

T

R. W is positive whereas U is negative

D.

V

P

AB

S. W is negative whereas U is positive

Answer: A-S; B-Q; C-R; D-Q;

Solution: ‘W’is +ve if volume of gas V increases in a process and –ve if V decreases. Similarly U is +ve if T increases and U is –ve If T decreases. Sign of Q is determined by Q U W .

38 An ideal monoatomic gas undergoes different types of process which are described in Column I. Match the corresponding effects in Column II. The letters have their usual meanings.

Column I Column II

A. 22P V P. If volume increases then temperature will also increase

B. 2 constantPV Q. If volume increases then temperature decreases

C. 2VC C R R. For expansion, heat will have to be supplied to the gas

D. 2VC C R S. If temperature increases then work done by gas is positive

Answer:A-P,R,S; B-Q; C-P,R,S; D-Q,R;

Solution: A) If 22P V , from ideal gas equation, we get 32V nRT

1) Hence as volume increases temperature will also increase u is +ve

2) positivedW is as V increases in expansion

Hence dQ dU dW is positive

B) If 2PV = constant, from ideal gas equation, we get 2

K V nRTV

VT K (constant)

Hence with increase in volume, temperature decreases.

Now 2v vPKdQ dU PdV nC dT nC dT dTT

2

KQdV dTT

2 ( )v vPTnC dT dT n C R dTT

Therefore, with increase in temperature dT = positive. And sice vC R for monoatomic gas, dQ positive as

C) vdQ nCdT nC PdV

( 2 )v vn C R dT nC dT pdV

2nRdT PdV

dV vedT

Hecne with increase in temperature, volume increase and vice versa. Therefore,

dQ u dW PdV

( 2 )v rn C R dT nC dT pdV

2nRT PdV

dV vedT

Therefore, with increase in volume, temperature decreases.

Also, ( 2 )vdQ n C R dT

With increase in temperature dT ve but 2vC R for monoatomic gas. Therefore dQ ve

with increase in temperature.

39 A monochromatic parallel beam of light of wavelength is incident normally on the plane containing slits 1S and 2S . The slits are of unequal width such that intensity only due to one slit is four times the other. The separation between the slits is d and that between screen and slit is D. Match the statement in column- I with results in column- II. ( )d D , ( )y D . Where y is distance of point of observation from central bright.

D

y

xO

1S

2S

1Sscreen

Column I Column II

A. The distance between two points on screen having equal intensities, such

that intensity at those points is 19

th of maximum intensity. P.

3D

d

B. The distance between two points on screen having equal intensities, such


th of maximum intensity. Q. D

d

C. The distance between two points on screen having equal intensities, such

that intensity at at those points is 59

th of maximum intensity. R. 2D

d

D. The distance between two points on screen having equal intensities, such


th of maximum intensity. S. 3D

d

Answer: A-Q,R,S; B-P,Q,R,S; C-Q,R,S; D-P,Q,R,S;

Solution: Initially at a distance x from central maxima on screen is

0 0 0 024 2 4 cos XI I I I I

, where Dd

max 09I I and min 0I I

At points where intensity is 19

th of maximum intensity, minima is formed

Distance between such points is , 2 ,3 ,4 ........

At points where intensity is 39

th of maximum intensity, 2 1cos2

X

or 3

x

Distance between such points is 2 2, , , , , 2 .........3 3 3 3

C) 2 1cos2

X

or 6

X .

Distance between such points is , , , 2 ,.......2 2

D) 2 1cos2

X

or 6

X .

Distance between such points is 2 2, , , , , 2 ,..............3 3 3 3

40 Three wires are carrying same constant current I in different direction. Four loops enclosing the wires in different manners are shown. The direction of dl is shown in the figure: (Loops should not touch)

Column I Column II

A. Along closed Loop-1 P. 0.B dl i

∮

B. Along closed Loop-2 Q. 0.B dl i

∮

C. A long closed Loop-3 R. . 0B dl

∮

D. A long closed Loop-4 S. net work done by the magnetic force to move a unit charge along the loop is zero

Answer:A-Q; B-P; C-Q; D-P;

Solution: Work done by magnetic force on a charge = 0 in any part of its motion.

‘S’ is matching for all parts (i),(ii),(iii),(iv)

For loop 1 inI i i i i 0. ( )B dl i

∮


∮


∮


∮

(Note: That current will be taken as positive which produces lines of magnetic field in the same sense in which dl

is taken)

QUICK REVISION TEST

PARAGRAPH TYPE QUESTIONS

Paragraph: A body is projected from a point ‘P’ on the ground. x,y are its position co-ordinates at time t. Assuming that it is projected from origin, the equations governing its motion are x=30t, 240 10y t t , x and y are in meters, t is in seconds. 1 Its time of flight is

A. 8s B. 4s C. 2s D. 1s Answer :B Solution : Conceptual

2 The maximum height reached by the body is A. 80m B. 40 m C. 20 m D. 10 m Answer :B Solution : Conceptual

3 Its horizontal range is A. 240 m B. 120 m C. 60 m D. 30 m Answer :B Solution : Conceptual

Paragraph: 16m

AB

Two persons B and A are separated by 16m as shown in the figure. A moves with an acceleration 21ms

starting from rest. B moves with a constant velocity 16 ms 4 A crosses B after time _____

A. 4s B. 6s C. 8s D. 10s Answer :C Solution : Let A move through ‘y’. B moves through x+y in that interval.

212

y at x y ut 212

x ut at

yx AB 2116 6

2t t

232 12t t 2 12 32 0t t

( 8)( 4) 0t t t=4 t=8 sec Time after which ‘B’ crosses ‘A’ is 4 seconds Time after which ‘A’ crosses ‘B’ is 8 seconds

5 Velocity of ‘A’ when ‘B’ crosses ‘A’ is _______ 1ms A. 6 B. 8 C. 16 D. 4

Sai charan

Typewritten text


Quick Revision Test COMPREHENSION Page 2

Answer :D Solution : 1 4 4 /V m s

6 Velocity of ‘A’ w.r.t. B when ‘A’ crosses ‘B’ is _______ 1ms A. 2 B. 8 C. 10 D. 16 Answer :A Solution : 16BV ms 18 1 8AV ms

r A Bv v v 18 6 2ms

Paragraph: Consider a system of three equal masses and 4 pulleys arranged as shown . The mass of each block is m

CB

Am

m m

7 The acceleration of C is

A. 223g down words B. 2

23g upwards C.

4g upwards D.

9g upwards

Answer :D Solution :

8 Tension in the single thread running through all the pulleys is

A. 59

mg B. 2523

mg C. 34

mg D. 79

mg

Answer :A Solution : Let acceleration of A is B and C are , &A B Ca a a in down ward direction Constrain relations gives.

2 2 0A C Ba a a …………………….(1) Also, mg –T = Ama for A ----------------(2) 2 for BBT mg ma ------------------(3) 2 for CCT mg ma -----------------(4)

Paragraph: In the figure shown, 1p and 2p are massless pullies. 1P is fixed and 2P can move. Masses of

A,B and C are 9 , 4 and 2 kg32

kg kg , respectively . All contacts are smooth and the string is massless.

0 0 2( 37 , tan 37 0.75& 10 )g ms


B

A

1P

2PC

9 Acceleration of block ‘c’ is _________ 2ms

A. 1 B. 2 C. 3 D. 4 Answer :C Solution :

10 Tension in the string is ____________ A. 10 N B. 13 N C. 24 N D. 36 N Answer :B Solution :

11 Normal reaction between the two blocks is ___________ A. 75 N B. 2.75 N C. 3.75 N D. 4.75 N Answer :C Solution : Let the acceleration of B downwards be 0a a From constraint ; acceleration of A and C are

4cot3Aaa a towards left

2Caa up wards

Free body diagram of A, B and C are

A

N sin

N sin

Ncos

+T

N1

2mg

B

C

2T

mg 9sin ( cot )64mN a ……………………………………..(1)

2 cos 2mg T N ma ………………………………….(2)

22aT mg m …………………………………….(3)

Solving we get 23 /

2caa m s

Ans: 23 /m s up wards Paragraph: We know that when a boat travels in water, its net velocity w.r.t ground is the vector sum of two velocities. First is the velocity of boat itself in river and other is the velocity of water w.r.t. ground. Mathematically:


,boat boat water waterv v v Now given that velocity of water w.r.t. ground in a river is u. Width of the river is d.

d

y

u

xO A A boat starting from rest aims perpendicular to the river with an acceleration of a = 5t, where t is time. The boat starts from rest from point A(1, 0) of the coordinate system as shown in figure. Assume SI units 12 Obtain the total time taken to cross the river.

A. 1/3(3 / 5)d B. 1/3(6 / 5)d C. 1/2(6 / 5)d D. 1/3(2 / 3)d Answer :B

Solution : 255 5

2y

y

dv ta t t vdt

2 35 52 6

dy t y tdt

Putting y = d, we get 1/3

35 62 5

dd t t

13 Find the equation of trajectory of the boat.

A. 1/331

5yx

B.

1/365yx u

C.

1/3615yx u

D.

1/2615yx u

Answer :C Solution : Let at any time the position coordinates of boat be (x, y)

(x, y)

(1, 0)

35 , 16

y t x ut

1/3615yx u

14 Find the drift of the boat when it is in the middle of the river

A. 1/33

5du

B. 1/33 1

5du

C.

1/365du

D. 1/23

5du

Answer :A Solution : y = d/2, drift 1/31 (3 / 5)x u d

Paragraph: A large heavy box is sliding without friction down a smooth, fixed plane of inclination . From a point P on the bottom of the box, a particle is projected inside the box. The initial speed of the particle with respect to the box is u and the direction of projection makes an angle with the bottom of the box as shown


in fig.

P

Qu

15 Magnitude of acceleration of the particle relative to box is ____

A. cosg B. sing C. g D. cos( )g Answer :A

Solution : sing

boxsing

cosg g

particle

clearly arelative = gcos only (along normal to the inclined plane) 16 Find the distance along the bottom of the box between the point of projection P and the point Q where

the particle lands in the box. (Assume that the particle does not hit any other surface of the box. Neglect air resistance).

A. sin 2cos

ug

B. 2 sin 2

cosug

C. 2 cos 2

sinu

g

D. 2 cossin

ug

Answer: B Solution: Consider the motion of the particle along the x and y - axis as shown in fig. With respect to box, we have

x

y

P

g cosg sing

sing

[ ] [ ]x particle x box xu u u

cos 0 cosu u

and [ ] [ ]x particle x box xa a a

or sin sin 0xa g g and sinyu u

cosya g Particle will hit the box after time t , then we have

212y yy u t a t

or 210 ( sin ) ( cos )2

u t g t 2 sincos

utg

Thus distance traveled in time t inside the box

cosPQ u t 2 sincoscos

uug


2 22sin cos sin 2cos cos

u ug g

17 If the horizontal displacement of the particle as seen by an observer on the ground is zero, the speed of

the box with respect to the ground at the instant when the particle was projected is

A. cos( )sin

u B.

2 cos( )cos

u C. cos( )

cosu

D.

2 cossin

u

Answer :C Solution : Horizontal component of initial velocity of particle w.r.t. box is cos( )u . Let xv is the initial velocity of box which is down the inclined plane and v is the velocity component of box along horizontal direction. The horizontal displacement as seen by the observer on the ground to be zero, we have

cos( ) 0v u or cos( )v u Then cosxv v

vxv

cos( )

cos cosxv uv

Paragraph: The figure shows a rod which starts rotating in horizontal plane with angular acceleration about vertical axis passing through one of its end (A). A bead of mass m just fit’s the rod and is situated at a distance ‘r’ from end A. Friction exists between rod and the bead with coefficient . As the angular velocity of rod increases, the bead starts sliding over the rod (say after time 0t ).

A

Based on above information, answer the following questions: 18 The normal force acting on bead at time 0( )t t is :

A. Mg B. 2( )m r t C. 2 2 4( )m g r t D. 2 2( )m g r Answer :D Solution : 1N mg

2N mr

2N1N

90(3 dfigure)

2 2

1 2N N N

19 Friction force acting on bead at time 0( )t t is given by :


A. mg B. 2( )m r t C. 2 2 4( )m g r t D. 2 2( )m g r Answer :B Solution : Only frictional force gives the required centripetal force as it is the only force acting along the surface or rod 2 2( )f mr mr t

20 If the bead start sliding at 0t t then value of 0t is given by :

A. 2 2

2

( )g rr

B. 2

mgr

C. 2 2

2 2

( )g r mgr r

D. None of these

Answer :A Solution : When 2

maxf mr That at this moment sliding just occurs 2 2 2

0( ) ( ) ( )mr t mg mr

Paragraph: A tool moving in x-y plane under a quality control process under the action of various forces. One force is 2 ˆF x y j

, a force in the negative y-direction whose magnitude depends on the position of the

tool. The constant is 2.50 N/m3. Consider the displacement of the tool from the origin to the point x = 3.00 m, y = 3.00 m. 21 Calculate the work done on the tool by the force F

if this displacement is along the straight line y = x

that connects these two points A. – 50.6 joule B. – 67.5 joule C. – 77.5 joule D. – 60.5 joule Answer :A Solution : ˆ( )ˆw F dxi dyj

3

3

0

w y dy ( x = y)

= 34

0

2.52.5 .(81)4 4y

50.6w J 22 Predict the nature of the force given in the paragraph.

A. Conservative B. Non-conservative C. Can’t be predicted D. Restoring force Answer :B

Solution : (3,0)

A

P

x

y

(3,3)

0 0Aw

3 3

2

0

(3 ).3. . 3 2.53A Pw y dy


= - 67.5 OP O A Pw w Hence the given force is non-conservative

Paragraph: A block of mass 1 kg is moving towards a movable wedge of mass 2 kg as shown in the figure. All surfaces are smooth. When the block leaves the wedge from top, its velocity is making an angle 30 with horizontal.

1 kg

v02 kg

600

1.45

m

23 The value of v0 in m/s is

A. 4 B. 7 C. 10 D. 9 Answer :B Solution : Let velocity of wedge is v1 (with respect to ground) and velocities of block is v2 (with respect to wedge when block reaches the top of the incline. From momentum conservation in horizontal direction.

00 2 1 11 1 ( cos60 ) 2v v v v

20 13

2vv v ……………………(1)

From given condition, 0

0 20

2 1

sin 60tan 30cos60v

v v

…………..(2)

From energy conservation,

2 20 0222 1 201

1 cos60 sin 60 12 1 1.452 2 2

v v v vv g …(3)

Solving above questions, we get 0 7 /v m s

1 2 2 /v v m s Maximum height attained by block is,

0 22( sin 60 )1.45 1.6

2vH m

g

600

1.45

m

v r

v1

Y

X

24 To what maximum height above the base of wedge will the block rise?

A. 1.9 m B. 2.7 m C. 1.6 m D. 1.45 m Answer :C Solution : Let velocity of wedge is v1 (with respect to ground) and velocities of block is v2 (with respect


to wedge when block reaches the top of the incline. From momentum conservation in horizontal direction.

00 2 1 11 1 cos 60 2v v v v

20 13

2vv v ……………………(1)


0 20

2 1

sin 60tan 30cos60v

v v

…………..(2)


2 20 0222 1 201

1 cos60 sin 60 12 1 1.452 2 2

v v v vv g …(3)



202 sin 60

1.45 1.62

vH m

g

600

1.45

m

v r

v1

Y

X

25 Let J be the magnitude of net impulse given to the block by the wedge. JH be its horizontal component

and JV be its vertical component.

A. 8 .3

J N s B. 4 .HJ N s C. 4 .3VJ N s

D. All of these

Answer :D Solution : Let velocity of wedge is v1 (with respect to ground) and velocities of block is v2 (with respect to wedge when block reaches the top of the incline. From momentum conservation in horizontal direction.

00 2 1 11 1 ( cos 60 ) 2v v v v

20 13

2vv v ……………………(1)


0 20

2 1

sin 60tan 30cos60v

v v

…………..(2)


2 20 0222 1 201

1 cos60 sin 60 12 1 1.452 2 2

v v v vv g …(3)




202 sin 60

1.45 1.62

vH m

g

600

1.45

m

v r

v1

Y

X

Paragraph: Two identical balls A and B, each of mass 2 kg and radius R, are suspended vertically from an inextensible strings as shown in fig. Third ball C of mass 1 kg and radius ( 2 1)r R falls and hits A and B symmetrically with 10 m/s. Speed of both A and B just after the collision is 3 m/s.

g

BA2kg

R

mc

1kg

2kg

26 Speed of C just after collision is A. 2 m/s B. 2 2 /m s C. 5 /m s D. ( 2 1) /m s Answer :A Solution : As the balls A and B are constrained to move horizontally (immediately after collision). If ‘I’ be the impulse imparted by ball ‘C’ to each of A and B, the impulse received by ball C from them would be 2I cos

A B

C

Now, each of ball B and C received impulse ‘I’ as shown in Fig. 2.413, but moves horizontally as its vertical comp. gets balanced by impulse imparted to ball B and C by the respective strings and hence, I cos M MA A B BV V For C Initial momentum + momentum imparted = final momentum 2 cosc c c cm u J m v -----------(1) For A

sin A AJ m v


6 2J From diagram 45 Substitute J equation (i) Vc = -2 m/sec. Coefficient of restitution:

A C

c A

V Veu u

0Au 10cos 45

322

12

C

A

C

u

v

V

e

27 Impulse provided by each string during collision is A. 6 2Ns B. 12 Ns C. 3 2Ns D. 6 Ns Answer :D Solution : As the balls A and B are constrained to move horizontally (immediately after collision). If ‘I’ be the impulse imparted by ball ‘C’ to each of A and B, the impulse received by ball C from them would be 2I cos

A B

C

Now, each of ball B and C received impulse ‘I’ as shown in Fig. 2.413, but moves horizontally as its vertical comp. gets balanced by impulse imparted to ball B and C by the respec tive strings and hence, I cos M MA A B BV V For C Initial momentum + momentum imparted = final momentum 2 cosc c c cm u J m v -----------(1) For A

sin A AJ m v


A C

c A

V Veu u

0Au


10cos 45322

12

C

A

C

u

v

V

e

28 The value of coefficient of restitution is

A. 14

B. 12

C. 2 1 D. 12

Answer :D Solution : As the balls A and B are constrained to move horizontally (immediately after collision). If ‘I’ be the impulse imparted by ball ‘C’ to each of A and B, the impulse received by ball C from them would be 2I cos

A B

C

Now, each of ball B and C received impulse ‘I’ as shown in Fig. 2.413, but moves horizontally as its vertical comp. gets balanced by impulse imparted to ball B and C by the respective strings and hence, I cos M MA A B BV V For C Initial momentum + momentum imparted = final momentum 2 cosc c c cm u J m v -----------(1) For A

sin A AJ m v


A C

c A

V Veu u

0Au 10cos 45

322

12

C

A

C

u

v

V

e


Paragraph: In the figure shown, frictional force between the bead and the light string is4

Mg . The system is

released from rest, with a bead of mass M at a distance from the free end of the string as shown. Assume string and pulley as mass less.

M

M bead

29 Acceleration of the block is

A. 2g downward B.

2g upward C.

4g upward D.

4g downward

Answer :A

Solution : M

M

t

bead

Tensions is string = friction between bob and string mg-2t=Ma

Mg-24

mg Ma

2blockga down

,344beed ground

mgmg ga downm

, beed stringa beed string a a 34g

+g

74g down

2, , ,

1 ( )2b s b s b ss u a t


87

ltg

30 Acceleration of the bead with respect to ground is

A. 74g downward B. 3

4g downward C. g downward D.

2g downward

Answer :B

Solution : M

M

t

bead

Tension is string = friction between beed and string mg-2t=Ma

Mg-2[ ]4

mg Ma

2blockga down

,344beed ground

mgmg ga downm


+g

74g down

2, , ,


87

ltg

31 Time taken by the bead to loose contact with the string is

A. 47g B. 8

7g C. 5

7g D. 3

7g

Answer :B


Solution : M

M

t

bead

Tension in string = friction between box and string mg-2t=Ma

Mg-24

mg Ma

2blockga down

,344beed ground

mgmg ga downm


+g

74g down

2, , ,


87

ltg

Paragraph: For a particle to move in a circular path, it must be provided centripetal acceleration with some force. The force could be gravitational or electromagnetic in nature. 32 A vehicle of mass m moves on a horizontal circular track of friction coefficient . Radius of the circular

track is R and the vehicle moves with constant speed V. Frictional force acting on the vehicle is: A. mg

B. 2mV

R C.

2mVmgR

D. 4

2 22

Vm gR

Answer: B Solution: A speed is constant, there is no tangential acceleration. Horizontal external force on the vehicle is friction only. This provides necessary centripetal force.

Friction 2mv

R

33 Let coefficient of static and kinetic frictions between the vehicle and the track mentioned in Q.No 32 be 0.4S and 0.3K respectively and the vehicle be moving with maximum speed without skidding .

The driver suddenly applies breaks hard enough to skid the wheels of the vehicle. Find the magnitude of change in frictional force between the wheels and the road just before breaks are applied and just after the wheels start skidding.


A. 0.5 mg B. 0.1 mg C. 0.3 mg D. data insufficient Answer: B Solution: Just before breaks are applied, the vehicle is moving with maxv without skidding. Hence the friction is limiting.

lim sF mg Just after breaks are applied, as the vehicle skids, sliding friction acts on it.

k kF mg Change in friction lim ( ) 0.1k s kF F F mg mg

Paragraph: A particle A’ strike a fixed inclined plane (making an angle with horizontal) at point A vertically with velocity 0V . Another particle B’ strikes the same inclined plane at point B horizontally. The collisions are elastic. If the times of flight for the projectile like paths above the inclined plane after collisions are same for both the particles, attempt the following questions.

α

A

B

B '

A '

34 Find the velocity of the particle B’ :

A. 0V tan B. 0 cotV C. 0 sinV D. 0 cosV Answer :B

Solution :

0V0V

sinV

VV

0 sin cosV V

0V Cos

'0V Cos V

090

yx

y

y

x

090

'For Particle A 'For particle B

ax = g sin and ay = -g cos

e = 1 = 0

1 0 '2 cos 0y y

Vu T aV

or 0' cosV V 212y y yS u T a T

20

10 ( cos ) ( cos )2

V T g T

T = 0 02 cos 2cos

V Vg g

Particle B’ strikes the plane horizontally sin ,xa g cosya g Velocity of particle B’ is V

Sy = uy T + 212 ya T 0 = V sin 21 cos

2T g T

T = 2 sin 2 tancos

V Vg g

From (1) and (2)

02 cos 2 sinV V = V = V0 cot .


35 Find the range of the particle A’ :

A. 2

04 sinVg

B. 2

02 sinVg

C. 2

02 cosVg

D. 2

0 cosVg

Answer :A

Solution : 21

12x xR u T a T

= 2

0 00

2 21sin sin2

V VV gg g

= 2

04 sinVg

36 Find the ratio of ranges of the particles A’ and B’ :

A. 22sin

cos 2

B. 22cos

sin 2

C. 2sin

cos 2

D. 2cos

sin 2

Answer :A

Solution : 22

12x xR u T a T

= 2

2 sin 1 2 sincos sincos 2 cos

V VV gg g

= 22

02 sin2 sin VVg g

= 2 22

0 02 2 sincossin

V Vg g

= 2

02

2 cos 2sinsin

Vg

R1 / R2 = 22sin

cos 2

.

Paragraph: A system of two blocks is at rest. A variable horizontal force is applied on the upper block. The coefficient of friction for both contacts is . Then

m

2m

F

37 When acceleration of the upper block is 2 g , net force on the lower block by the ground is


A. 3mg B. mg C. 2 mg D. 29mg

Answer: D Solution : Limiting friction on upper block (fL)1 = mg

Limiting friction on lower block(fL)2 = 3 mg

2mmg

1 3N mgmg

Net force = 29mg

38 When the acceleration of the upper block is 2 g , acceleration of lower block is

A. Zero B. g C. 2 g D. 2g

Answer :A Solution : Lower block does not move for any acceleration of upper block

39 The net horizontal force on lower block is

A. Always zero B. Always mg C. mg towards left D. 2 mg toward left

Answer :A Solution : Net force in horizontal direction on lower block is always zero as it does not move

Paragraph: A uniform rod of length ‘l’ and mass 2m rests on a smooth horizontal table. A point mass ‘m’ moving horizontally at right angle to the rod with velocity ‘v’ collides with one end of the rod and sticks to it, then :

40 The angular velocity of the system after collision is vl

A. 2vl

B. vl

C. 2vl

D. 3vl

Answer :B


Solution :

2m C

C1 Vm

Distance of common centre of mass from ‘c’ is

/ 23 6

mk l lxm

Conserve the angular momentum of system about ‘c’ before and after collision

2 2 212 2

2 12 36 9m l m ll mlmV

2 2 21

2 6 18 9l ml ml mlmV

2 13 1 23 18lmV ml

21

3 3l ml VmV

l

21.2iK E mV

2 2 2

2

1 1. (3 )2 9 2 3f

V ml VK E ml

2 21 1 1 1.3 2 3 2fK E mV mV

21 2.2 3fK E mV

41 The loss in kinetic energy of the system as a whole as a result of the collision is

A. 2

6mv B.

2724mv C.

2

3mv D.

223

mv

Answer :A


Solution :

Distance of common centre of mass from ‘c’ is

/ 23 6

mk l lxm

Conserve the angular momentum of system about ‘c’ before and after collision

2 2 21(2 ) (2 )

2 12 36 9l m l m l mlmV

2 2 21

2 6 18 9l ml ml mlmV

2 13 1 23 18lmV ml

21

3 3l ml VmV

l

21.2iK E mV

2 2 2

2

1 1. (3 )2 9 2 3f

V ml VK E ml

2 21 1 1 1.3 2 3 2fK E mV mV

21 2.2 3fK E mV

Paragraph: A uniform rod of mass m and length L is free to rotate about hinge O and is vertical initially. A slight disturbance causes the rod to rotate freely about O and it strikes the ground.

O


42 Angular velocity with the non-hinged end of rod strikes the surface is:

A. 2gL B. 3gL C. 6gL D. gL

Answer :B Solution : By conservation of energy

221 3

2 2 3L mLmg L gL

43 Horizontal force applied on rod by hinge just before the rod hits the surface is:

A. mg B. 4

mg C. 34mg D. 3

2mg

Answer :D

Solution : 22

3 32 2 2xL mL gL mgF m

L

44 Vertical force applied by hinge on rod at the moment it strikes the ground is:

A. mg B. 4

mg C. 34mg D. 3

2mg

Answer :B Solution : Taking torque about O.

2 32 3 2L mL gmg

L

3 32 4 4 4y yL mg mg mgmg f m f mg

Paragraph: The vessel in the figure is closed and contains water as shown. The space between water and top of tube is vacuum. The base area is 2

2 100A cm and the tube area 21 10A cm (take 31000 /w kg m ,

210 /g m s )

1m

5m2

1 10A cm

22 100A cm


For this arrangement, answer the following questions.

45 Find the total force acting on the bottom of the vessel due to liquid.

A. 100N B. 150N C. 600N D. 300N

Answer :C Solution : Force at the bottom 3 4(5 1) 10 10 1 10N

46 What is the weight of the water?

A. 100N B. 150N C. 300N D. 600N

Answer :B Solution : Wt of water = Vdg

Paragraph: A dumbbell is floating on water. It is observed that by attaching a point mass m (neglect its volume) to the rod, at 'l' distance from the center of sphere B, the dumbbell floats with the rod horizontal on the surface of the water and each sphere exactly half submerged, as shown. Take density of water as and volume of each sphere as V. Masses of the spheres are shown in the diagram.(Neglect the mass of rod)

A B

water surface2 ,M Vm,V M

l

d

47 The value of m is

A. – 3MV B. 2 – 3MV C. 2 –V M D. –V M

Answer :A Solution : (3 )V g M m g

48 The value of l is

A. ( )2( )d V M

V M

B. ( )(2 )d V M

M V

C. ( 2 )2( 3 )d V M

V M

D. 2 3d V M

V M

Answer :C


Solution : Torque about point ‘m’ is zero 22 2

V g V gMg Mg d

49 If the density of sphere A is o , the values of o in terms of which is not possible (allowed) is

A. 6o

B. 5o

C. 4o

D. 3o

Answer :D Solution : 3Mg V g

03V g V g

0 / 3

Paragraph: An infinitely long wire lying along z-axis carries a current I , flowing towards positive z- direction. There is no other current. Consider a circle in x-y plane with centre at (2 m, 0, 0) and radius 1 meter. Divide the circle in small segments and let d l

denote the length of a small segment in anticlockwise

direction, as shown.

1m

(2,0,0)

y

I x

dl

50 The path integral B∮ . dl

of the total magnetic field B

along the perimeter of the given circle is ,

A. 0

8I B. 0

2I C. 0I D. 0

Answer :D Solution : Acc. If amp. law B = 0

51 Consider two points A (3,0,0) and B (2,1,0) on the given circle. The path integral

B

A

B

. dl

of the total

magnetic field B

along the perimeter of the given circle from A to B is ,

A. 10 1tan2

I

B. 10 1tan2 2

I

C. 10 12 2

I sin

D. 0

Answer :B


Solution : Since . 0BDAB

B dl ∮

BDAB

. .B dl B dl

BD DB

A

B

C

D

I

1 1. (2 ) (1/ 2) (1/ 2)2

oI

DB

oIB dl Tan Tan

1 1.2 2AB

oIB dl Tan

52 The maximum value of path integral B

. dl

of the total magnetic field B

along the perimeter of the given circle between any two points on the circle is

A. 0

12I B. 0

8I C. 0

6I D. 0

Answer :C

Solution : Since .PQP

B dl o∮

CI0

P

Q

P Q P

. . (2 )2QP PQ

oIB dl B dl


For QP

Bdl to be max, be Max, it is possible for chord OP act as tangent to the dotted circle

6

The maximum value if 0.6PQ

IB dl

Paragraph: To convert a galvanometer into a voltmeter, we need to calculate its internal resistance and figure of merit. The electrical arrangement shown can be used for this purpose.

G

/\/\/\/\/\

/\/\/\/\/\

1000

2k

1k2V

Shunt resistance

When 1k is closed and 2k is open, galvanometer needle is deflected by 20 divisions which is also full scale deflection. When 2k is also closed and 100 is taken as shunt resistance, deflection shown by the galvanometer is halved.

53 Resistance of the galvanometer is approximately

A. 100 B. 99 C. 111 D. 121

Answer :C

Solution : 21000gI

G

1

and 100 21002 100 1000

100

gIGGG

2

1000 111.19

G

54 Figure of merit (inverse of current sensitivity of the galvanometer) is


A. 59 10 /A div B. 49 10 /A div C. 39 10 /A div D. 29 10 /A div

Answer :A

Solution : Figure of merit = 50.0018/ 9 10 /20gI A div

55 If the galvanometer is to be converted into a voltmeter of range 4.5 V, resistance required to be connected to the galvanometer is

A. 2500 B. 2389 C. 2000 D. 2486

Answer :B Solution : ( )gV I G R

g

g g

V GI VR GI I

2389R

Paragraph: A uniform solid cube of side a and mass m is suspended vertically from one of its edges as shown figure

aO

56 Find the period of small oscillations

A. 2 22 ag

B. 2 223

ag

C. 22 ag

D. / 22 ag

Answer :B

Solution : Moment of inertia of the cube about 223

O ma

Equation of motion

22 . . .sin ;3 2

ama mg angle from the vertical

32 2

ga


2 32 2

gT a

Or 2 223

aTg

57 Find the length of equivalent simple pendulum

A. 2

a B. 2a C. 2 23

a D. 2 2a

Answer :C

Solution : Length of equivalent simple pendulum 2 23

a

Paragraph: A uniform bar of mass M and length L is hanging from point S as shown in figure. The Young’s modulus of elasticity of the bar is Y and the area of cross-section of the bar is A

S

58 Find the stress at a distance x, (x<L) distance from the bottom end

A. /MgL Ax B. /Mgx AL C. 2Mgx

AL D.

2MgL

Ax

Answer :B Solution : The weight x length of the bar is

S

Lxdx

MgW xL

So stress at x distance from bottomW MgxA AL

59 Find the total elongation of bar, due to its own weight


A. MgLAY

B. 2MgLAY

C. 2MgL

AY D. 1

3MgLAY

Answer :C

Solution : Total elongation in wire 0 0 2

L LMg MgLL dL L x dxALY AY

Paragraph: A cylinder of radius r = 10 cm is placed between two planks as shown. Mass of the cylinder is 2kg.

r10 m/s

4 m/s

60 Assuming that there is no slipping at any point, find the angular velocity of the cylinder

A. 13 .rad s B. 130 .rad s C. 1300 .rad s D. 10.3 .rad s

Answer :B Solution : Let velocity of C.M. is CMv and angular velocity , then for no slipping Velocity of points of contact must be equal.

r

10 m/s

4 m/s

CMv r

CMv r

10CMv r

4CMv r 7CMv

7 10r 2

3 3 3010 10r

1rads

61 Find K.E. of the cylinder by assuming that there is no slipping at any point

A. 44.5 J B. 49 J C. 53.5 J D. 50 J

Answer :C

Solution : 2

2 2 2 21 1 1 1 2. . 2 72 2 2 2 2CM CM

rK E mv


2 21 0.1 30 30 0.149 (0.2) (30) 49 49 4.5 53.52 2

joule

Paragraph: Two uncharged capacitors A and B, each of capacitance C, and an ideal inductor of inductance L are connected to a battery of emf E as shown in the adjacent circuit diagram. The switches 1 2S and S are operated in two stages as follows:

State 1: At time t = 0, switch 1S is closed while switch 2S remains open.

Stage 2: At time 0 2t t LC

, switch 1S is opened and switch 2S is closed.

A B

E

S1 S2

L

62 The charge on capacitor A at time 0t t is

A. 2

CE B. CE C. 4

CE D. 2CE

Answer :B

Solution : netq LdI qE EC dt C

dI qEdt C

2

2

1 ( )d q q CEdt LC

( )q CE ASin t sin( )q CE A t and cos( )dqi A tdt

with t = 0, q = 0, i = 0, = 1LC

(1 cos )q CE t

At 0 0,2

t t q CE


63 The current flowing through the inductor at 0t t is

A. CEL

B. 2 CEL

C. 12

CEL

D. Zero

Answer :A

Solution : i ,2

CE Sin t t

01 Ci CE CE E

LLC

64 The maximum current through the inductor in stage 2 is

A. 32C EL

B. 54C EL

C. 25C EL

D. C EL

Answer :A Solution : At time 0t t , let the charge on B is q and that on A will be CE-q and I is the current in the circuit.

2 2 22 20

01 1 1 1 1 ( )2 2 2 2 2

q q CE qLi LiC C C

2 2 2 2 2 22 21 1 1 2

2 2 2 2 2 2 2C E C q C E q CEqL E Li

C L C C C C

2 22 21

2 2q CECE Li EqC

22 21 1

2 2qLi CE EqC

22 2 2 2C q Eqi E

L CL L

i is max, when 0didt

2 20 (2 ) 0Eqi iCL L

2CEq


222max

2 22 2

CE CE E CEiL CL L

2

max3 32 2

CE Ci EL L

Paragraph: A tank is filled with a liquid of density up to a height H. Water flows out from the tank through a nozzle of uniform cross-sectional area ‘a’. The exit of the nozzle makes angle 45 with the horizontal. At a given instant, the tank rests in equilibrium against a compressed spring of force constant k. The top of the tank is open to the atmosphere of pressure 0P . The cross-sectional area of the tank is very large

in comparison to that of the nozzle. The exit of the nozzle is at a height 4H from the base of the tank.

H

H4

45

A

k

65 The compression of the spring at the given instant is

A. 3 2

agHk

B. 3 28

agHk C. 4

3 2agH

k D. 3

2 2agH

k

Answer :D

Solution : 2 3 1 1cos 45 2 .4 2Hav kx a g x

k 3

2 2agHx

k

2 324Hv g

66 The pressure inside the nozzle just after the entry point (at A) is

A. 0P H g B. 0 4HP g C. 0P H g D. 0

34

P gH

Answer :B

Solution : 2 20 0

1 12 4 2 4A a

H Hp v p g v p P g

67 The maximum height, from the base of the tank, reached by the liquid coming out of the nozzle at the


given instant is

A. 34 2

H B. 38H C. 5

8H D. 3

4H

Answer :C

Solution : 2 2

maxsin 45 3 1 1 52

2 4 4 2 2 4 8v H H H Hh g

g g

Paragraph: A thin ring of radius R meters is placed in x-y plane such that its centre lies on origin. The half ring in region x < 0 carries uniform linear charge density C/m and the remaining half ring in region x > 0 carries uniform linear charge density /C m .(Take v = 0 at infinity)

x y

y

y

+λ -λ

++++++

+++++

++ + +

68 The electric potential (in volts) at point P whose coordinates are 0 ,

2Rm m

is

A. 0

14πε 2

B. 0 C. 0

14πε 4

D. cannot be determined

Answer :B Solution : Consider two small elements of ring having charges +dq and –dq symmetrically located about y-axis

The potential due to this pair at any point on y-axis is zero.

The sum of potential due to all such possible pairs is zero at all points on y-axis. Hence potential at

0,2RP

is zero

xxxxxx

xxxxxx

x x x x x

- - - - ------

- - - --

y

y

+λ -λ x

+dq -dq

dθ dθθ θ

'x


69 The direction of electric field at point P whose coordinates are 0 ,

2Rm m

is

A. along positive x-direction B. along negative x-direction

C. along negative y-direction D. none of these

Answer :A Solution : Since all charge lies in x-y plane, hence direction of electric field at point P should be in x-y plane

Also y-axis is an equipotential (zero potential) line. Hence direction of electric field at all point of y-axis is should be normal to y-axis.

The direction of electric field at P should be in x-y plane and normal to y-axis. Hence direction of electric field is along positive x direction

70 Dipole moment of the ring in C . m is

A. 2(2πR )i B. 2(2πR ˆ)i C. 2(4R )i D. 2(4R ˆ)i

Answer :C Solution : Consider two small elements of ring having charge +dq and –dq as shown in figure.

xxxxxx

xxxxxx

x x x x x

- - - - ------

- - - --

y

y

+λ-λ

x

+dq

-dq

dθθ'x

The pair constitutes a dipole of dipole moment.

dp dq 2R

(λRdθ)2R

The net dipole moment of system is vector sum of diplole

moments of all such pairs of elementary charges.

By symmetry the resultant dipole moment is along negative x-direction.

Net dipole moment


+π/2

-π/2

(dp cos ) i

+π/22

-π/2

(2λR cos dθ) i

24R i

Paragraph: Three point charges + 3Q, + 2Q and – Q respectively are located at distances each "a" from the

origin as shown in the fig. 0

1( )4

k

-Q + 2QOa a

a

+ 3Q

y -a x

i sx-axis

71 At the origin, the magnitudes of x and y-components of the net electric field E

are

A. 2

k3Q2Ex yEa

B. 2

k3Qx yE E

a C. 2

1 k3Q2x yE E

a D. 2

k3Q3x yE Ea

Answer :B

Solution : 2 20

1 2. ( )4

ˆQE ia

= 2

2 ( ˆ)QK ia

1 2

14 )ˆ(

QEw a i

1 2E E

aQ

12

2Qa

P

3E

3 3Q


= 2 ( ˆ)KQ ia

1 2E E E 4 3E E

3 20

1 3ˆ4 ( )

QEa i

Potential energy of system, u = (3 )( ) (2 )( )22

K Q Q K Q Qaa

13[ ]2

LKQUa

72 the electric potential energy 'U' for the configuration of the three charges is

A. 2kQ 11

2a

B. 2kQ 1 1

2a

C. 2kQ 3 1

2a

D. 2kQ 3 1

2a

Answer :D

Solution : 2 20

1 2. ( )4

ˆQE ia

= 2

2 ( ˆ)QK ia

1 2

14 )ˆ(

QEw a i

1 2E E

aQ

12

2Qa

P

3E

3 3Q

= 2 ( ˆ)KQ ia

1 2E E E

4 3E E

3 20

1 3ˆ4 ( )

QEa i


Potential energy of system, u = (3 )( ) (2 )( )22

K Q Q K Q Qaa

13

2

LKQUa

Paragraph: A thin cylindrical shell closed at both ends is subjected to a uniform excess internal pressure P over outside. The wall thickness is h and the inner radius is ( )r h r . Y = Young’s modulus of material of the cylinder.

73 The longitudinal normal stress existing in the walls due to P is

A. P B. Prh

C. Phr

D. 2Pr

h

Answer :D

Solution : The normal stress 2.

2 2p r pr

rh h

74 The circumferential normal stress existing in the walls due to P is

A. P B. Prh

C. Phr

D. 2Pr

h

Answer :B

Solution : Circum normal stress = 22

p RL pALh h

75 The increase in radius of the cylinder is (Poisson’s ratio is )

A. 2

12

PrYh

B.

2

1 2PhYh

C. 2

12

PrYh

D.

2

1 2PrYh

Answer :A Solution : Circum strain due to circum stress .

22

pr RYh r

2prrYh

Circum strain due to long. Stress rr

2

2prryh

2

( ) 12net

prryh

Paragraph:

A body of mass m falls from a height h onto a pan of negligible mass and lying on a light vertical spring of


force constant K. The mass sticks to the pan and executes simple harmonic motion in the vertical Line.

m

h

reference level (RL)(gravitational PE = 0)

K

u = 0

76 The mean position of oscillations is

A. at the original reference level of the pan B. above the RL by mgK

C. below the RL by mgK

D. below the RL by 2mgK

Answer :C

Solution : 0mgxk

77 The amplitude of oscillations is

A. 21mg khK mg

B. 1mg khK mg

C. 212mg kh

K mg D. 21 1mg hk

K mg

Answer :A

Solution : 2 21( ) 2 2 02

mg h x kx kx m gx mgh

21mg mg khxk k mg

=maximum compression of spring

Amplitude A mgxk

21mg hkk mg

78 The total energy of oscillations is

A. 2 2m gmghK

B. mgh C. 2 2

2m g

K D.

2 2

2m gmgh

K


Answer :D Solution : It is the energy that changes between PE and KE during oscillation. It is equal to maxKE

2 22

2

1 2. 12 2

k m g hkE kAk mg

2 2 22 2

m g hkmgk k

2 2

2m gmgh

k

Paragraph: A pulse is started at a time t = 0 along the +x direction on a long, taut string. The shape of the

pulse at t = 0 is given by function f(x) with 1 4 0

4( ) 1 0 1

0

x for xf x x for x

otherwise

Here f and x are in

centimeters. The linear mass density of the string is 50 g/m and it is under a tension of 5N.

79 The shape of the string is drawn at t = 0 and the area of the pulse enclosed by the string and the x-axis is measured. It will be equal to

A. 22 cm B. 22.5 cm C. 24 cm D. 25 cm

Answer :B Solution : Shape of the pulse is

14

1

area made by pulse with x-axis = 21 (5)1 2.52

cm

80 The vertical displacement of the particle of the string at x = 7 cm and t = 0.01 s will be

A. 0.75 cm B. 0.5 cm C. 0.25 cm D. Zero

Answer :C Solution : Shape of the pulse is


14

1

At t = 0.015 shape of pulse is

x = (1000)(0.01) = 10cm

14

1

at x = 7cm, 11 4y 0.25y cm

81 The transverse velocity of the particle at x = 13 cm and t = 0.015 s will be

A. -250 cm/s B. -500 cm/s C. 500 cm/s D. -1000 cm/s

Answer :A Solution : Shape of the pulse is

14

1

At t = 0.015 sec pulse is at and x = 13cm is

( )p wv slope v

11 161513


1 1000 250 /4

cm s

Paragraph: An Indian submarine is moving in “Arab Sager” with a constant velocity. To detect enemy it sends out sonar waves which travel with velocity 1050 m/s in water. Initially the waves are getting reflected from a fixed island and the reflected waves are coming back to submarine. The frequency of reflected waves are detected by the submarine and found to be 10% greater than the sent waves.

Sonar wavesInsian

submarine

FixedIsland

Now an enemy ship comes in front. Due to which the frequency of reflected waves detected by submarine becomes 21% greater than the sent waves.

82 If the wavelength received by enemy ship is 1 and wavelength of reflected waves received by

submarine is 2 then 1

2

equals

A. 1 B. 1.1 C. 1.2 D. 2

Answer :B Solution : 1v velocity of submarine

v velocity of sound

2v velocity of enemy ship

Case : submarine moves towards fixed island

Case (i) submarine approaches an island and receives a frequency

1 1

1

v vf fv v

Solving it 1 50 /v m s

1

1

1050110100 1050

vf fv

Case (ii) submarine approaches an enemyship

11 2 1

2 1

121100

v v v vf f fv v v v


2 50v

2

1 111

11 1

11

1050 501050 501050 50 1.11050 50

vff

v ff

83 Bulk modulus of sea water should be approximately 31000 /water kg m

A. 81.1 10 2/N m B. 91.1 10 2/N m C. 101.1 10 2/N m D. 111.1 10 2/N m

Answer :B Solution : 1v velocity of submarine

v velocity of sound

2v velocity of enemy ship

Case : submarine moves towards fixed island

Case (i) submarine approaches an island and receives a frequency

1 1

1

v vf fv v

Solving it 1 50 /v m s

1

1

1050110100 1050

vf fv

Case (ii) submarine approaches an enemy ship

11 2 1

2 1

121 ( )( )100

v v v vf f fv v v v

2 50v

kv

2 2 91000 (1050) 1.1 10k v

Paragraph: A system of men and trollies is shown in figure. To the left end of the string, a trolley of mass M is connected on which a man of mass m is standing. To the right end of the string, another trolley of mass m is connected on which a man of mass M is standing. Initially the system is at rest. All of a sudden, both the men leap upwards simultaneously with the same velocity u w.r.t. ground.


. .m MM m

84 Find the relative velocity of left man with respect to his trolley just after he leaps upwards.

A. mum M

B. Mum M

C. 2mum M

D. 2Mum M

Answer :C Solution : 1 2,I mu I Mu ,

1 0 2 0,I I Mu I I mu

. .m M

M m

u

0u

0u

u

I

1I 2I

2I1I I

Solve to get : 0( )m M uu

m M

Relative velocity of left man w.r.t his trolley :

0( ) 2m M u muu u u

m M m M

85 Find the impulse generated in the string connecting the trollies during this process.

A. Mmum M

B. 2 2( )M m u

m M

C. 2m u

m M D.

2M um M

Answer :B


Solution : 2 2

1 0 0( ) ( )M m M m M uI I Mu mu Mu mu um M M m

86 What is the correct statement among the following? Assume that trollies move with deceleration after jumping of men.

A. When the men are at the highest points of their motion, the trolleys will also be instantaneously at rest.

B. When the men are at the highest points of their motion, the left trolley will be moving downward.

C. Impulses acting on both the men will be same in the given process.

D. None of the above are true

Answer :A Solution : Time taken for men to come to rest :

utg

Acceleration of trolley (after the men have jumped):

( )m Ma gm M

Time taken by trolleys to come to rest :

01

( ) ( )( )

u m M u m M ut ta m M m M g g

Paragraph: A uniform cylinder of radius R and mass m is spines about its axis with angular velocity 0 and then placed at a corner, formed with rough horizontal floor and rough vertical wall as shown in figure. The coefficient of friction between the contact surfaces and the cylinder is k .

R0

87 The normal reaction imparted by the wall on the cylinder is

A. 2

21

k

k

mg

B. 21k

k

mg

C. 21 2k

k

mg

D. 22k

k

mg


Answer :B

Solution :

02N

1N

mg

2N

1N

As the centre of mass of the cylinder does not accelerate,

2 1 0kN N i

1 2 0kN N mg ii

On solving these equations we get 1 22 2,1 1

k

k k

mgmgN N

Torque on the cylinder about the axis of rotation

1 2 1 2 2

10 0 0

1k k

in k kk

mg N N N R N R mgR

Torque equation about its axis is 22

1 11 2

k k

k

I mgR mR

2

2 11k k

k

gR

Using equation 2 20 2 ,

20 2

2 12

1k k

k

gO

R

( and are in opposite sense)

2 201

4 1k

k k

Rg

Number of rotations accomplished is


2 2 2 20 01 1

2 2 4 1 8 1k k

k k k k

R RN

g g

88 Angular acceleration of the cylinder is

A. 2

11k k

k

kR

clockwise direction B.

2

11

k k

k

gR

anticlockwise direction

C. 2

2 11k k

k

gR

clockwise direction D.

2

2 11k k

k

gR

anticlockwise direction

Answer :C

Solution :

02N

1N

mg

2N

1N


2 1 0kN N i

1 2 0kN N mg ii


k

k k

mgmgN N


1 2 1 2 2

10 0 0

1k k

in k kk

mg N N N R N R mgR


1 11 2

k k

k

I mgR mR

2

2 11k k

k

gR



20 2

2 12

1k k

k

gO

R


2 201

4 1k

k k

Rg


2 2 201 1

2 2 4 1 8 1k k

k k k k

RN

g g

89 How many rotations will the cylinder accomplish before it stops?

A.

2 201

1k

k k

Rg

B.

2 201

8 1k

k k

Rg

C.

2 208 1

1k

k k

Rg

D.

2 203 1

1k

k k

Rg

Answer :B

Solution :

02N

1N

mg

2N

1N


2 1 0kN N i

1 2 0kN N mg ii


k

k k

mgmgN N


1 2 1 2 2

10 0 0

1k k

in k kk

mg N N N R N R mgR


1 11 2

k k

k

I mgR mR


2

2 11k k

k

gR


20 2

2 12

1k k

k

gO

R


2 20

2

14 1

k

k

Rg


2 2 2 20 01 1

2 2 4 1 8 1k k

k k k k

R RN

g g

Paragraph: Light having photon energy h v is incident on a metallic plate having work function to eject t he electrons. The most energetic electrons are then allowed to enter in a region of uniform magnetic field B as shown in Fig. The electrons are projected in X-Z plane making an angle with X-axis and magnetic field is 0

ˆB B i

along X-axis.

Y

Z

BX

e-1

Maximum pitch of the helix described by an electron is found to be p. Take mass of electron as m and charge as q. Based on above information, answer the following questions:

90 The correct relation between p and 0B is

A. 0 2 cos 2( )qpB hv m B. 0

2( )2 cos hvqpBm


C. 0 2 2( )pqB hv m D. 0

2 mp hvqB

Answer :A Solution : At any time t the location of electron is shown as P. In two dimensional view of electron in YZ- plane, the situation is more clear.

0

0

2 2( )

2cos

2( )2 cos

2 cos 2 ( )

KE hvvm m

mp vqB

hvpqB mm

m hv

P

X

Y

Z

tPO

Y

Z

X-coordinate, cosx v t

Y-coordinate, [ cos ]y R R t

Z-coordinate, sinz R t

So, 0

0

sin sin qBmvz tqB m

0

0

2 ( ) sinsin

m hv qB tqB m

From 2( )

cos coshv

x v t tm

As v increases, slope of x versus t graph (a straight line) increases.

91 Considering the instant of crossing origin at t = 0, the Z-coordinate of the location of electron as a function of time is


A. 0

0

2 ( )(sin ) 1 cos

m hv qB tqB m

B. 0

0

2 ( )(sin ) sin

m hv qB tqB m

C. 0

0

2 ( )(sin )sin

m hv qB tqB m

D. 0

0

2 ( )sin

m hv qB tqB m

Answer :B Solution : At any time t the location of electron is shown as P. In two dimensional view of electron in YZ- plane, the situation is more clear.

0

0

2 2( )

2cos

2( )2 cos

2 cos 2 ( )

KE hvvm m

mp vqB

hvpqB mm

m hv

P

X

Y

Z

tPO

Y

Z




So, 0

0

sin sin qBmvz tqB m

0

0

2 ( ) sinsin

m hv qB tqB m

From 2( )

cos coshv

x v t tm


92 The plot between X-coordinate of the location of electron as a function of time for different frequencies


v of the incident light, is

A.

x

t

1 2v v1v 2v

B.

x

t

1v 2v2 1v v

C.

x

t

1v

2v

1 2v v

D.

1 2v vx

t

1v2v

Answer :C Solution : At any time t the location of electron is shown as P. In two dimensional view of electron in YZ- plane, the situation is more clear.

0

0

2 2( )

2cos

2( )2 cos

2 cos 2 ( )

KE hvvm m

mp vqB

hvpqB mm

m hv

P

X

Y

Z

tP

O

Y

Z




So, 0

0

sin sin qBmvz tqB m

0

0

2 ( ) sinsin

m hv qB tqB m

From 2( )cos cos

hvx v t tm



Paragraph: A standing wave y = 2A sinkx cos t is setup in the wire AB fixed at both ends by two vertical walls (see the figure). The region between the wall contains a constant magnetic field B. Now answer the following questions.

xx

xx

xx

xxxx

xx

xx

xx

xx

xA

x

Bx

93 The wire is found to vibrate in the 3rd harmonic. The maximum emf induced is

A. 4ABk B. 3AB

k C. 2AB

k D. AB

k

Answer :A Solution : Induced emf is maximum when the elements are crossing the mean position.

2 2 2

max0 0 0

( )2 sin 2 sin .2 sin

4

e B dx A kx B dx A kx B dx A kx

BAk

94 In the above question, the time when the emf becomes maximum for the first time is

A. 2

B.

C. 2

D. 4

Answer :A Solution : 2 sin sinv A kx t emf is max when v is maximum

sin 1 sin2

2 2

t

t t

95 In which of the following modes the emf induced in AB is always zero?

A. Fundamental mode B. Second harmonic C. Second overtone D. Fourth overtone

Answer :A Solution : Second harmonic

Paragraph: Experiments show that the number of nucleous (protons and neutrons) per unit volume inside a nucleus is fairly constant near its center, and gradually decrease in the outer region. The near constant of nucleon density results from the fact that each nucleon in a nucleus interacts only with a small number of


nucleons in its surrounding through attractive nuclear force. The nucleon density ( )r as a function of the

distance r from the center of the nucleus is approximately given by 0( )1 exp

rr R

a

Where 0 0.17

nucleon/ 3fm , 1/3(1.1 )R fm A and 0.55a fm Here A = mass number.

96 The mass density in a nucleus near its center, in units of kg/ 3m , is in the range

A. 15 2010 10to B. 5 1010 10to C. 10 1510 10to D. 20 2510 10to

Answer :A Solution : Density at the centre

0 021R e

a

1+e

0.171.135

nucleons/ 3fm

2717 3

45

0.15 10 1.5 1010

kgm

97 For the nucleus 216 ,Te the value of r for which the nucleon density falls to half its value at the centre is in the range

A. 7 to 8 fm B. 5 to 6 fm C. 6 to 7 fm D. 8 to 9 fm

Answer :C

Solution : 0 01212 2(1 )

1ra

ee

12121 2(1 )

rae e

12 /[ 2] 1r ae e

/ 12( 2)r ae e 12ra

(neglecting of no comparison to 12e )

12 0.55 12 6.6r a fm fm

98 The plats of ( )r versus r for 28 Si and another nucleus X are shown in the figure. The nucleus X could


be

A. 42Ca B. 94 Zr C. 63Cu D. 142 Ba

Answer :D

Solution : 0 0( / 10) 101 2(1 )r ae e

( / 10) 101 2(1 )r ae e

10 /( 2) 1r ae e / 10( 2) 10 5.5r ae e r a fm

Nearest value of A is 142 Ba

Paragraph: A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let ‘N’ be the number density of free electrons, each of mass ‘m’, When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency ' 'p which is

called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency , where a part of the energy is absorbed and part of it is reflected. As approaches P , all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.

99 Taking the electronic charge as ‘e’ and the permittivity as 0' ' , use dimensional analysis to determine

the correct expression for p .

A. 0

Nem

B. 0mNe

C. 2

0

Nem

D. 02

mNe

Answer :C Solution :

100 Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons 27 34 10N m . Take 11

0 10 and 3010m , where these quantities are in proper SI units. A. 800 nm B. 600 nm C. 300 nm D. 200 nm Answer :B Solution :

Paragraph: When subatomic particles undergo reactions, energy is conserved, but mass is not necessarily conserved. However, a particle’s mass ‘contributes’ to its total energy, in accordance with Einstein’s famous equation, 2E mc . In this equation, E denotes the energy a particle carries because of its mass. The particle


can also have additional energy due to its motion and its interactions with other particles. Consider a neutron at rest, and well separated from other particles. It decays into a proton, an electron, and an undetected third particle. 1 1 0

0 1 1 ?n H e

Particle 2( )Mass C MeV ( )Kinetic energy MeV

NeutronProton

Electron

940.97939.670.51

0.000.010.39

The table above summarizes some data from a single neutron decay. An MeV (mega electron volt) is a unit of energy. Column 2 shows the rest mass of the particle times the speed of light squared.

101 Assuming the table contains no major errors, what can we conclude about the (mass 2c ) of the undetected third particle ? A. It is 0.79MeV B. It is 0.39MeV C. It is less than or equal to 0.79MeV ; but we cannot be more precise D. It is less than or equal to 0.39MeV; but we cannot be more precise Answer :D Solution : D) According to the passage, subatomic reactions do not conserve mass. So, we cannot find the third particle’s mass by setting neutron proton electron third particlem equal to m m m . By constrast, the total

energy in this case, the sum of mass energy and kinetic energy, is conserved. If E denotes total energy, then neutron proton electron third particleE E E E

The neutron has energy 949.97MeV. The proton has energy 939.67 MeV+0.01MeV=939.68MeV. The electron has energy 0.51MeV + 0.39 MeV = 0.90 MeV. Therefore, the third particle has energy

third particle neutron protonE E E Electron = 940.97 – 939.68 – 0.90 = 0.39 MeV we just found the third

particle’s total energy, the sum of its mass energy and kinetic energy. Without more information, we cannot figure out how much of that energy is mass energy.

102 From the given table, which properties of the undetected third particle can we calculate?

A. Total energy, but not kinetic energy

B. Kinetic energy, but not total energy

C. Both total energy and kinetic energy

D. Neither total energy nor kinetic energy

Answer :A Solution : As just shown, energy conservation allows us to calculate the third particle’s total energy. But we don not know what percentage of that total is mass energy vs, kinetic energy.

Paragraph: In figure, light of wavelength 0

5000 A is incident on the slits (in a horizontally fixed place). Here, d = 1 mm and D = 1m. Take origin at O and XY plane as shown in the figure. The screen is released


from rest from the initial position as shown. (Take g = 10ms-2)

D Y

d

OS1 S2X

Screen

05000A

103 The velocity of central maxima at t = 5s is

A. 50 ms-1 along Y-axis B. 50 ms-1 along X-axis

C. 25 ms-1 along Y-axis D. 3 × 108 ms-1 along Y-axis

Answer :A Solution : At any time t, the situation is as shown in the figure below.

Y

S1 S2X

Screen

2

1gtD D2

0P P

xv = gt, a = g

Central maxima is always lying on Y-axis at 0P . Its velocity at any time t is given by v = gt along positive Y-axis. So, required velocity is 150m s .

104 Velocity of 2nd maxima w.r.t central maxima at t = 2s is

A. 1 1 ˆ(8 ) 20cm s i m s j B. 18 ˆcm s i

C. 12 ˆcm s i D. 186 ˆcm s i

Answer :C Solution : Path difference corresponding to point P.

sin tanx d d

1

dxxD

For 2nd maxima, 2x

11 22 Ddx D x

d


Location of central maxima is

210, (0, )

2gtD or D

Location of 2nd maxima is

112 ,D D

d

Velocity of 2nd maxima w.r.t central maxima is

2 12 2[0 ] 2ˆ ˆndvgt i gt i cm s i

D d d

105 Acceleration of a 3rd maxima w.r.t. 3rd maxima on other side of central maxima at t = 3s is

A. 2.02 ˆ0 m s i B. 2.03 ˆ0 m s i C. 210 ˆm s j D. 2.6 ˆ0 m s i

Answer :B

Solution : Location of 3rd maxima is 1

13 ,D Dd

Location of 3rd maxima on the other side is 1

13 ,D Dd

23 3 3 6( ˆ ˆ ˆ) 0.033

rda g gg i i ms ird d d d

Paragraph:

/\/\/\

/ \/\/\

/\/\ /\

/\ /\/\

/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ /\/\/\/\/\/\/\/\/\/\/\/\/\/\

AB

CD

ii

A plane divides space into two halves. One half is filled with a homogeneous conducting medium and a physicist works in this other. He marks the out line of a square of side ' 'a on the plane and let a current ' 'i in and out at two of its neighbouring corners A and B using fine electrodes. The potential difference between the


other two corners is found to be V as shown in figure. Using super position principle one can find resistivity of the conducting medium

106 The potential difference between points D and C is

A. 2 12 2i

a

B. ia

C. 2 12

ia

D. Zero

Answer :C Solution : To solve this problem we can apply principle of super position. Let us assume current ‘I’ enters at A and speeds out in the form of hemispherical region. So current density at a distance r from

22iAr

Now using ohm’s law, 22iE J

r

22

2

1 1 1 1 22 2 2 22 2

D

aVe a

C DaV a

i dr i i idV V Vr r a aa

2 12 2C DiV V

a

Net potential difference between D and C 2 2 1 2 12 2 2

i ia a

107 If r be the effective resistance of the battery and the connecting wires then the emf of the battery which is supplying current into the wires will be

A. 2 12 2i ir

a

B. i ira

C. 2 12

i ira

D. 2 12 2i ir

a

Answer :C

Solution : 2 1,2

imf E V ir ira

108 The resistivity of the material is

A. (2 2) aVi B. (2 2) aV

i C. ( 2 1) aV

i D. (2 2 4) aV

i

Answer :A


Solution : 2 (2 2) 2 (2 2) (2 2)(2 2)2 2(2 2) (2 2)i aV aV aVV

a l ii

Paragraph: Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photo electrons emitted from A is twice that of photoelectrons emitted from B.

109 Value of E (in eV) is

A. 20.8 B. 32.2 C. 24.6 D. 23.8

Answer :D Solution : AW ionization energy of electron in 2nd orbit of hydrogen atom 3.4eV

BW ionization energy of electron in the 2nd orbit of He ion 13.6eV

Now, given that 2 2( ) 2 23.8A B A B B AK K E W E W E W W eV

110 Radius of an electron moving in a circle in constant magnetic field is two times that of an - particle in the same field. Then de-Broglie wavelength of electron is x-times that of the -particle. Here x is

A. 2 B. 12

C. 1 D. 4

Answer :C

Solution : PrBq

Prq

. Given 2 . 22e e ePr r P P P

Now de-Broglie wavelength , 1, 1e

e

Ph xP P

111 An electron and a proton are separated by a large distance and the electron approaches the proton with a kinetic energy of 2eV. If the electron is captured by the proton to form a hydrogen atom in the ground state, the wave length of photon emitted in this process will be

A. 1262.2Å B. 793.Å C. 1204.6 Å D. 942.6 Å

Answer :B Solution : 2 , 13.6i fE eV E eV

2 ( 13.6) 15.6E eV

12375( ) 793.315.6

oor A


Paragraph: A body cools in a surrounding of constant temperature 030 C . Its heat capacity is 02 /J C . Initial temperature of the body is 040 C . Assume Newton’s law of cooling is valid. The body cools to 038 C in 10 minutes.

112 In further 10 min it will cool from 038 C to A. 035 C B. 036.4 C C. 033 C D. 037.5 C Answer :B Solution : We have 0( ) kt

s s e .

where 0 initial temperature of body = 040 C

= temperature of body after time t .

Since body cools from 40 to 38 in 10 min, we have

1038 30 (40 30) ke (i)

0 1030 (38 30) ke (ii)

( )( )iii

gives 8 10 , 30 6.430 8

036.4 C

113 The temperature of the body in 0 C denoted by . The variation of versus time t is best denoted as

A. t

040 C

030 C

B. t

040 C

030 C

C. t

040 C

030 C

D. t

040 C

Answer :A Solution : Self-explanatory

114 When the body temperature has reached 038 C , it is heated again so that it reaches 040 C in 10 min. The total heat required by the body from a heater is

A. 3.6 J B. 0.364 J C. 8 J D. 4 J

Answer :C Solution : During heating process from 038 C to 040 C in 10 min, the body will lose heat in the surrounding which will be exactly equal to the heat lost when it cooled from 040 C to 038 C in 10 min, which is equal to 2 2 4ms J . During heating process heat required by the body 4ms J . Therefore, total heat required = 8J


Paragraph: A process in which work performed by an ideal gas is proportional to the corresponding increment of its internal energy is described as a polytropic process. If we represent work done by a polytropic process by W and increase in internal energy as U , then W U , or

1W K U

For this process, it can be demonstrated that the relation between pressure and volume is given by the equation

2PV K (constant)

We know that a gas can have various values for molar specific heats. The molar specific heat ' 'C for an ideal gas in polytropic process can be calculated with the help of first law of thermodynamics. In polytropic process, the variation of molar specific heat ' 'C with for a monoatomic gas is plotted as in the graph shown.

A

B

C

D

EF

O

115 In the graph shown, the y-coordinate of point A is (for monoatomic gas)

A. 3 / 2R B. 5 / 2R C. 7 / 2R D. 4R

Answer :B Solution : For point , 0A

(0 ) 5(0 1)( 1) 1 2

R RC R

116 In the graph shown, the x-coordinate of point B is (for monoatomic gas)

A. 7/5 B. 5/3 C. 2/3 D. 8/3

Answer :B Solution : At point , 0B C

As ( )( 1)( 1)

RC

at point B

For monoatomic gas, 5 / 3

117 For a monoatomic gas, the values of polytropic constant for which molar specific heat is negative is


A. 2 53 3

B. 813

C. 513

D. 2 83 3

Answer :C

Solution : 1 1

R RC

As >1, 1

R

is +ve always

If 1,1

R

is –ve and ‘C’ will become +ve

For ‘C’ to become –ve, must be greater than 1.

Again 1 1

R R

for C to become -ve

1 1

' 'C is –ve when 1

QUICK REVISION TEST INTEGER TYPE QUESTIONS

1 Two balls are projected from a point in two mutually perpendicular vertical planes. Speed of projection of both the balls is 400 m/s. Angle of projection with horizontal is 053 for both the balls. After how much time (in seconds), their velocities will be at angle 060 from each other ? (Take g = 10 m/s2) Answer :8 Solution : Let x – y plane is horizontal plane and z-azis is vertical. If one particle moves in x – z plane, the other moves in y – z plane.

1 (400cos53) (400sin 53 ˆ ˆ) 240 320ˆ ˆu i k i k

2 (400cos53) (400sin 53 ˆ ˆ) 240 320ˆ ˆu j k j k

After time t, 1 1 240 320 10 2ˆ ˆ 40 ˆ(3ˆ 20 )ˆ 10v u gt i k tk i t k

2 2 240 320 10 2ˆ ˆ 40 ˆ(3ˆ 20 )ˆ 10v u gt j k tk j t k

2

1 22 2

1 2

. 1 (320 10 )cos 602 (240) (320 10 )

v v tv v t

2 22240 320 10 2 320 10t t

22240 320 10 320 10 240 10 80 8t t t t s

2 If the maximum and minimum speeds in the path of a projectile are 20 m/s and 10m/s respectively, what is the ratio of maximum and minimum radii of curvature at points in its path ? Answer :8 Solution : Initial speed is the maximum speed 20 /u m s

Speed at highest point is minimum 0cos 10 20cos 10 60u . Radius of curvature is minimum at highest point.

2 2 2

min( cos ) 10 10

10Nv ua RR g

R is maximum at the initial point.

2 2 2 2

maxmax

20cos 801cos 102

Nv u ua g RR R g

max

min

80 810

RR

.

3 Two men P & Q are standing at corners A & B of square ABCD of side 8m. They start moving along the track with constant speed 2m/s and 10m/s respectively. The time (in seconds) when they will meet for the first time, is equal to

Sai charan

Typewritten text


2m / s

A

B

P

Q10m/s C

D

Answer :3 Solution : They meet when Q displace 8 × 3m

2m /s

A

B

P

Q10m/s C

D

a = 8m

more than P displace relative displacement = relative velocity × time 8 × 3 = (10 – 2)t t = 3sec

4 A swimmer crosses a river with minimum possible time 10 seconds. When he reaches the other end, he starts swimming in the direction towards the point from where he started swimming. Keeping the direction fixed, the swimmer crosses the river in 15 s. The ratio of speed of swimmer with respect to water and the speed of river flow

is 2x

. Find x. (Assume constant speed of river water & swimmer).

Answer :5 Solution : V = velocity of man w.r.t river U = velocity of river

u

v

CA

B

resv

1resv

v

u

10 10d dA B d Vv v

(1)

15 15 cos (2)cos cos

d dB C d vv

(1) & (2) cos 2 / 3 sec 3 / 2

tan uv

2sec 1 uv

5 29 / 4 1

2 5u vv u

5 Two guns situated at a point on the top of a tall hill fire one shot from each gun with the same speed 5 3 /m s at

some interval of time. One gun fires horizontally and the other fires upwards at an angle of 060 with the horizontal. If the shots collide in air, what is the time interval between the firings in seconds? (Take g = 10 m/s2)

Answer :1

Solution : Horizontal velocity of oblique projectile is 0 55 3 cos 60 3 /2

m s .

For horizontal projectile, horizontal velocity is 15 3 ms . To get same horizontal displacement, time of journey for 1st projectile must be more as its horizontal velocity is less. Let t is the time o journey of 1st projectile and t is the time interval between firings.

u 5 3 m/ s

060u 2

1

y

x

1 25 3( ) 5 3( )2 2

tx x t t t t

2 2

221 2

1 1 15 3 sin 602 2 2 2 2 4

g t gty y t gt g t t t

215 1 3 22 2 4

t gt t s

12tt s

6 Figure shows the velocity and acceleration of a particle at the initial moment of its motion. Acceleration is constant

throughout the motion. If the time in seconds when the velocity reaches its minimum value is 16x

, find x.

2 006 , 24 / , 143a ms v m s

0v

a Answer :5 Solution : We can consider it like an oblique projectile from ground with angle of projection 053 . V is minimum at highest point i.e. after t = time of ascent ( )aT

vm

in

aa

900

53 0

v0

0 sin 24 4 / 5 16 56 5

vt xa

7 Each of the two blocks shown in the figure has a mass m . The coefficient of friction for all surfaces in contact is . A horizontal force P is applied to move the bottom block . The value of P , for which acceleration of block A is same in both cases is n times mg . Then ‘n’ is equal to _____

BAP

BAP

Answer :2

Solution :

1

2P mg mga

m

1fT

3P mgm

1fT

2f

1fT

2fT

1 2T f ma 2 1 2P T f f ma 2 1 22P f f ma

242

P mgam

For 2 6 4P mg P mg 2P mg

8 In the arrangement shown in figure , pulleys are mass less and frictionless and threads are inextensible. For the block of mass 1m to remain at rest, 1' 'm must be equal to______ kg 2 33 & 6m kg m kg

1m 2m 3m

Answer :8 Solution : cve When 1m remains at rest, the 1T m g

Aslo

2 31

2 3

42 om mT T g m g

m m

1 2 3

4 1 1m m m

9 Two identical spheres of radii 10 cm each and weight 3N each are placed between two rigid vertical walls as shown. The spacing between the walls is 36 cm. Then the force of contact between the two spheres is ________N.(All the surfaces are frictionless)

Answer :5

Solution : cv

12cos20

Also cosN W

20 5cos 12W WN N

10 A flexible chain of weight 3N hangs between two fixed points A and B at the same level. The inclination of the chain with the horizontal at the two points of support is 0( 37 ) . The tension at the mid point of the chain in

newtons is ________ 0sin 37 0.6

Answer :2 Solution : At mid point, tension is horizontal, Cosinder the FBD of half part.

sin2

mgT

cos oT T

2 tanOmgT

11 An inclined plane of angle of inclination 030 witha block of mass 2 kg kept on the inclined plane is given a

horizontal acceleration of 10 3 2ms towards left. Assuming frictionless contacts. The acceleration of mass 2

kg w.r.t ground is N times 22ms . Then N is equal to ____ (take g=10 2/m s )

2 kg

300

Answer :5 Solution : The force on the wedge is given to be

cotF mg

mM

F

In this case, m falls freely 210 /a m s

12 Two blocks of masses 4 kg and 6 kg are attached by a spring of spring constant k=200 N/m. Both the blocks are moving with same acceleration. Elongation of spring in cm is ______

Answer :4 Solution : For 4 kg block, Kx ma

Also 220 2 /4 6

a m s

200 4 2x 8

200x

4x cm

13 Two blocks of masses 4 kg and 6 kg are attached by massless springs. They are hanging in vertical position in equilibrium. If lower spring breaks due to excessive force, acceleration of 4 kg block just after breaking is 5 times

2x ms . Then x is equal to _______

Answer :3 Solution : Just before breaking 4 kg block is in equilibrium

4 6 10kx 100kx N

When the lower spring breaks,

40 4kx a 260 4 15 /a a m s

14 In the arrangement shown, neglect the masses of the ropes and pulley. What must be the value of m in kg to keep the system in equilibrium? There is no friction anywhere.(M=2kg)

030

M

Answer :1 Solution : By drawing F.B.D of both block it can be seen

0sin 30Mg mg

2Mm

15 Figure shows two blocks A and B connected to ideal pulley string system. In this system, when bodies are released, acceleration of block ‘A’ in 2ms is _____ (Neglect friction and take 210 /g m s )

B

A a40kg

10 kg

Answer :2 Solution : Appling NLM on 40kg block

40kg

a

4T

10kg

2T

T

2T

400 4 40T a For 10 kg block 10..(4 )T a Solving 22 /a m s

16 Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal forces F and 2F are applied on both the blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. If the normal reaction between the two blocks is N F, then ‘N’ is equal to __________

m 0302F

A BF m

Answer :3

Solution : Acceleration of two mass system is 2Fm

leftward

FBD of block A 0cos 60

2mFN F ma

m

Solving, 3N F 3N F

030

060

N

F

17 The surface of a smooth inclined plane of inclination 030 is ABCD. CD is in contact with horizontal ground. A particle is projected parallel to AB from A with a velocity of magnitude u such that it passes through diagonally opposite point C. If AB = BC = 10 m and 210g ms , its initial speed u in 1ms is ______ Answer :5

Solution : 21 110 10 22 2

t t s

10 2 5 /u u m s

18 A projectile is launched at time t = 0 from point A which is at height L above the floor with speed v m/s and at an angle 045 with horizontal. It passes through a hoop at B which is 1m above the level of A and B is the highest point of the trajectory. The horizontal distance between A and B is d meters. The projectile then falls into a basket, kept at ‘C’ on the floor at a horizontal distance 3d meters from A. Find L (in m).

A

L

C d

L+1

3d

B

C

450

Answer :3 Solution : The horizontal and vertical components of the velocity are the same, let it be 0cos 45u v .

0 (d,0) (3d,0) xC

1B (d, l + 1)

(0, l)

yv

l

From A to B: 2

21 22u u gg

At B: 2 1 /d ut t d u 2

21 1 21

2 2g d g dut t u

u u

2 2

21 12 4g d gdd d

u g

2 24 4 4 4 0d d d d 2d m

At C: 2 233 dd ut tu

2 22

2 2 2

1 3 9 932 4 2 4 4

d g d gdut gt u dg

29 93 3 2 4 6 9 34 4dd

3m

19 In a car race, car A takes 4 s less than car B at the finish and passes the finishing point with a velocity v m/s more than the car B. Assuming that the cars start from rest and travel with constant accelerations 2

1 4a ms and 2

2 1a ms respectively, find the velocity v in m/s Answer :8

Solution : 2 21 2 1 2 1 1 2 2

1 1, , ,2 2

t t t v v v S a t S a t

1 1 1 2 2 2 2 1 1,v a t v a t v v a t

12 2 1 1 1 2 2

1 2

v a ta t v a t a t ta a

2 1 2 1 2

1 2 2 1 1

( )1 1( )

a t a t a ata t t a v a t

2 21 1 2 1 2 1

11

a v a t a v a a t v a a a tv a ta

11 2 8v a a t ms

1 cosv u

20 The velocity of a projectile when it is at the greatest height is 2 / 5 times its velocity when it is at half of its

greatest height. Then its angle of projection is 010 n where n = Answer :6 Solution : At half of the greatest height y = h/2, , siny ya g u u

sin2y

uv

2 22 x yv v v

1

2

25

vv

tan 3 060

21 A particle is projected up from the bottom of a hollow wedge of inclination 030 , with a velocity 140 2 ms at 045 to horizontal. (Take point of projection as origin, horizontal direction as x-axis and vertical direction as y-

axis). There is a hole to the surface of wedge at (240 m, 60 m). This particle passes through that hole. If the base length of wedge is 280 m, the speed with which this particle hits the vertical face of the wedge is 110x ms . Then

x = ____ (Take g = 210 ms ) Answer :5 Solution : The vertical face is at a horizontal distance of 280m. 280 40 7t t s We have to find magnitude of velocity at 7s.

140 , 40 10 7 30 /x yv ms v m s 2 240 30 50 10 5 5v x

22 A body of mass m is slowly hauled up the hill by a force F which at each point is directed along a tangent to the trajectory. The work performed by this force is nmgh in moving the body from A to B if the height of the surface is h, the length of its base is ‘l’ and the coefficient of friction between the body and the surface is given by

tan where is the angle between the normal force applied by the surface and the vertical at every point. Find the value of n.

h

B

A

F

Rough ( )

Answer :2 Solution : cosfdw mg ds

dsdy

dx

tan cosmg ds ( sin )mg ds mg dy

f fw dw mg dy mgh

0F N g fw w w w

0Fw O mgh mgh 2Fw mgh mgh mgh

23 A plank of mass 4kg is placed on a smooth horizontal surface. A block of mass 2 kg is placed on the plank and is being acted upon a horizontal force F = 0.5 t where F is in newton and t is in s. If the coefficient of friction between the block and the plank is 0.10, the work done by friction on the system between t = 0 and t = 6s in joules is _________ (Take 210g ms )

F2kg

4kg

Answer :0 Solution : 2f N and they move together upto 6 sec. The frictional force between the block is static upto t = 6 sec. Hence 0fw

24 A body of mass 4 kg is moving at speed 1 m/s on circular path of radius 1m. Speed of particle is continuously increasing at the rate of 3 m/s2. Force acting on particle at the instant when speed 2 m/s is n x 10N. Find the value of n. Answer :2 Solution : 23 /ta m s

t ra a a

2

2 243 9 16 5 /1

a m s

F = 20 N

25 A particle of mass 2kg is tied to a light inextensible string of length 1m at one end. Other end of string is fixed at ‘O’. If the particle is released from rest with the string horizontal, the acceleration 2( )in ms of the particle when

the string becomes vertical is 10 n . find n. 210g ms .

O

V

1l m u=0(fixed)

Answer :2

Solution : 2 21 22

mg mv V g

At the bottom point 2

2normalVa a g

2n

26 A particle of mass 0.5 kg travels along x-axis with velocity of magnitude 3/2V a x where

1125 .a m s

and x is the position. If the work done by the net force during its displacement from x=0 to x=2 m is 10n joules, find n. Answer :5 Solution : at x=0, V = a(0)=0 at x=2, 3/ 2(5)2 10 2 /V m s

221 1 10 10 2 50 10 5

2 2 2W KE mV J

27 A gun is fired from a moving platform and the ranges of the shots are observed to be 02x and 0x when platform is

moving forward or backward respectively with velocity V. If the elevation of the gun is with horizontal then

tan = 022

gxkv

, where k = ____

Answer :6 Solution : Velocity components of body Platform Horizontal Vertical Range

Rest xu yu

Forward motion (v) 1x xu v u yu 10

22 x yu u

xg

------(1)

Backward motion (v) 2x xu v u yu 20

2( ) ( )x yu ux

g ------(2)

Angle of projection tan y

x

uu

-------(3)

On solving (1), (2) & (3)

02

( )tan 612g x k

v

28 Two blocks A and B connected by an ideal spring of spring constant 100 NK

m are moving on a smooth

horizontal plane due to the action of a horizontal force F. Mass of A is 5 kg, mass of B is 2 kg and F = 35 N. The extension of the spring at an instant when both A and B move with constant acceleration is _____ cm.

FABK

Answer :1 Solution : When A and B have constant acceleration (spring has maximum extension) And A Ba a a

25 /A B

Fa m sm m

1Bkx m a x cm

29 Two blocks of masses 1m and 2m are connected by massless threads. The pulleys are massless and smooth. If 1a

is the magnitude value of acceleration of 1m and 2a is the magnitude value of acceleration of 2m , find

the ratio 1

2

aa

.

m

m1a

2a

1

Answer :4

Solution :

1m

2m

1T

1T 1T

3T

3T

3T

2T

2T

Here 3 2 3 12 ; 2 4T T T T T Constraint equation:

1

2

. 0 4aT Va

30 In the figure shown, if all the surfaces are smooth, then the horizontal force, F required to keep the 2kg and 3kg

blocks stationary is 30g NK

. Then value of K is _______

Smooth

F

2 kg

3 kg5 kg

Answer :2

Solution : Smooth

F

2 kg

3 kg5 kg3a

a

N

2a

3g

T

T

T = 2a = 3g

32ga

3 30(5 3 2) 10 15 22g gF a g K

K

31 Coefficient of friction between two blocks shown in figure is 0.4 . The blocks are given velocities of 2 m/s

and 8 m/s in the directions shown in figure. The time when relative motion between them will stop is 5t/3 sec, then t is

1kg2kg

2m/s8m/ s

Answer :1 Solution : Relative motion between them stops when 1 2V V at an instant 0t t

Here 1a g ; 12

2

ma gm

1 2v v

0 0 01 52 8 sec2 3

gt gt t

32 Object A and B each of mass ‘m’ are connected by light inextensible cord. They are constrained to move on a friction less ring in a vertical plane as shown in figure. The objects are released from rest at the positions shown. The tension in the cord just after

T

T

mg

B

mg

A

relax will be mg

k then value of K is

Answer :2 Solution :

For A T cos 45 = ma ---------------- (1) For B mg – T sin 45° = ma

cos 45 1sin 45

Tmg T

2 2T Tmg

22T mg

2 T mg

2mgT

33 A train is moving along a straight track with a uniform acceleration. A boy standing in the train throws a ball

forward with a speed of 10 m/s relative to the train at angle 060 to the horizontal. The boy moves forward by 1.15 m inside the train to catch the ball at the initial height. Then the acceleration of the train is __________ m/s2. Answer :5

Solution : 21( cos )2

x u t gt , where2 sin 3ut s

g

1 1 3 31.15 10 3 3 5 1.73 8.652 2 2 2

x x ax x a a

23 7.5 5 /2

a or a m s

34 Two solid cylinders, each of mass m and radius r, are placed touching along their lengths on a rough horizontal surface of coefficient of friction . A third cylinder of same length, made of same material, but of radius 2r is placed lengthwise over them so that the system just remains at rest. There is no friction between the cylinders.

Find the value of 9 2 Answer :3 Solution : 2R = 6 mg 2 cos 4N mg

sinN R

Solving1

3 2

19 2 9 2 33 2

x

35 A ball of mass m = 0.5 kg is attached to the end of a string of length L = 0.5 m. The other end of the string is fixed. The ball is made to rotate on a horizontal circular path about the vertical axis through the fixed end of the string. The maximum tension that the string can bear is 324 N. The maximum possible angular velocity of the ball is n2 rad/s, where n = _________

Answer :6

m m

4m

R

R

Rr r

N

L

m

Solution : 2 2 324 324 41 12 2

TT m l xml x

18 2 36 n 6.

36 A mass less spring of force constant 1000 N/m is compressed through a distance of 20 cm between two discs of masses 2 kg and 8 kg on a smooth horizontal surface. The discs are not attached to the spring. The system is given an initial velocity of 3 m/s perpendicular to the length of the spring. Find the velocity (in m/s) of the 2 kg disc relative to ground when the spring regains its natural length.

Answer :5

Solution : 2 21 1(5 ) 1000 (0.2)2 2

x v x x

Where 1610

1 /v m s 2 24 3 5 /Av m s

37 A block of mass m = 2kg is moving with velocity v0 towards a massless unstretched spring of force constant k = 12 N/m. Coefficient of friction between the block and the floor is 0.2 . The block, after pressing the spring,

just stops there without returning. Find 20v in m2s-2.

Answer :6 Solution : mg kx

2 20

1 1( 1)2 2

mv mg x kx

A

2kg

B

8kg

y

x3m / s

Horizontal surface

fixed

3v

k0v

m

1m

2 22 2 20 0

1 3 3 2 2 4 62 2

mgmv kx mg v gk

38 A body is projected with velocity v0 to move along a vertical circular track of radius R as shown. It presses the

surface at B with a force of 65

mg , where m is the mass of the body. Neglecting friction, the value of initial

velocity v0 is 5

gRn where n = ___________

B

900Rv0

O

Answer :4

Solution : 2

2 20

6 6 25 5

mv gRmg v v gRR

20

6 1625 5gRv gR gR

0 45

gRv

39 Two blocks A and B of mass m and 2m are placed on a smooth horizontal surface. Two horizontal forces F and 2F are applied on blocks A and B respectively, where F = 3N. The block A does not slide on block B. Then the normal reaction acting between the blocks is _______ N.

300

F 2Fm 2mA B

Answer :8

Solution : 3Fam

4cos 60 83 2 3F R FR F ma R N

300

F 2Fm 2mA B

R

Rcos600600

600Rcos600

R

40 Two identical balls A and B are attached to the ends of a thread passed through a narrow hole on a smooth horizontal table. The distance of ball B from the hole is r = 20 cm. As ball B revolves with angular velocity about a vertical axis passing through the hole on the table, ball A neither rises nor falls. Then = ________ rad/sec.

Answer :7 Solution : 2T m r mg

9.8 49 7 /0.2

g rad sr

41 The acceleration of a particle vary with respect to time and is given by a = (2t - 6), where t is in seconds. Find the time (in seconds) at which velocity of particle in negative direction is maximum, if its initial speed is zero. Answer :3

Solution :

-63 t s

2a m / s

Acceleration – time graph of the particle is shown.

rB

A

T

T

mg

Maximum velocity in negative direction will be at t = 3 sec, as acceleration becomes positive after 3 sec.

42 Two small blocks A and B are released from rest on a fixed inclined plane of angle 300 and a circular track of radius R from different heights h1 and h2 respectively as shown in figure. The mass of each block is m. If F1 and F2 are the magnitudes of respective resultant forces experienced by two blocks at the bottom-most points of the tracks and F1 = F2, then find the value of h2 (in m) for R = 8 m.

B

A

h1

h2300

Answer :2 Solution : 0

1 1 sin 30 / 2F ma mg mg 2

2 2 ( / )F ma m v R

= 2 22 2gh hm mgR R

For 2 2 1/ 4, / 2h R F mg F So, 28 2

4 4Rh m

43 A massless string of 3 m length joins two small spheres A and B of mass 1 kg and 2 kg respectively. The spheres are placed on the horizontal surfaces at the same level. The string is horizontal and is rotated at 1 rad/s about a vertical axis passing through point O. The surface on which sphere B is placed is smooth and A is kept on rough surface. Find the value of frictional force between sphere A and the surface (in N) acting parallel to the string.

1m 2m

AO B

Answer :3 Solution : Centripetal force required for sphere B = mrw2 = (2)(2)(1)2 = 4N Centripetal force required for sphere A = (1)(1)(1)2 = 1N

Af

T T B

Since, there is no frictional force acting on B and T = 4N, is providing sphere A an extra force of 3N, which will be balanced by the friction, f = 3N

44 A sphere of mass 1

3

kg is placed on two smooth inclined planes of angles 300 and 600 with horizontal, as

shown. Find the normal reaction at point P (in N). (g = 10 ms-2)

300600 P

Answer :5

Solution :

06 0 030oN

pN

,O PN N and mg will pass through centre

00 0 0 0

(10 / 3)sin 90sin 90 30 sin 90 60

oP NN

0 10 310cos30 523PN N

45 A block A of mass 1 kg which lies on a rough horizontal surface has a velocity 0v directed towards a relaxed spring. After travelling a distance of 2.525m, it strikes the spring. After compressing the spring, the block comes to rest and remains there permanently. Find the maximum value of 0v in m/s. (g = 10 ms-2)

2.525 m

A

Answer :5 Solution : Spring is compressed by x. Block stops and will not return bock if kx = mg .

(0.4)(1)(10) 0.410

mgx mk

. Work done against friction 2 2

01 12.5252 2

mg x mv kx

(0.4)(1)(10)(0.4 +2.525) = 20

1 1(1) (10)(0.16)2 2

v

23.40 = 20 1.6v 5 /v m s

46 The coefficient of friction between a rough horizontal floor and a box of weight 1000 N kept on it, is

3k

. If a

minimum force of 600 N is required to start the box moving, then k = Answer :4

Solution : Fminimum = mg sin 600 3sin1000 5

3 3tan 44

kk

47 A pump motor delivers water at a certain rate. The power of motor is to be increased to obtain twice as much water from the same pipe and in same time. The power of motor has to be increased to how many times (in an integer)? Answer :8 Solution : Mass flowing out per per second, m = Avp Rate of increase of kinetic energy

= 2 31 12 2

mv Apv =' '3 '3

3 3

( )( )

P A v vP A v v

Now, ' ' 'm A v v

m A v v

As m’ = 2m, so 'v = 2v and thus 'P

P=(2)3 = 8

48 Blocks A and B each of mass 1 kg are moving with 4m/s and 2m/s respectively as shown. The coefficient of friction for all the surfaces is 0.10. Find the distance (in m) by which centre of mass will travel before the center of mass coming to rest. Assume that blocks do not collide before cmv becomes zero.

A B4 m/s 2 m/s

0.10 Answer :3 Solution : The block B will stop in 2 seconds. The block A will stop in 4 seconds. From 0 to 2 seconds equal force of friction are acting on the blocks in opposite directions and thus system will remain conserved as net force is zero. At t = 0.

1 1 2 2

1 2

(1)(4) (1)( 2)2CM

m V m VVm m

= 1m /sec

Up to t = 2 second, VCM is constant = d1 = VCM T = (1)(2) = 2m

After t > 2 second, retardation of centre of mass

a = 20.1(10) 1 /2 2 2g m s =

2 2

2(1) 1

2 2(1/ 2)CMVd ma

Total distance = 2 + 1 = 3 m

49 Two spheres A and B of masses 2 kg and 1 kg respectively are moving with 8 m/s and 4 m/s on a smooth horizontal surface. Let head on collision takes place between them. During collision, they exert impulse of magnitude J on each other. The minimum value of J (in N - s) for which sphere A will change its direction of velocity is 2I (in N-s) where I is an integer. Find the value of I.

A B4 m/s8 m/s

Answer :9 Solution : Initial momentum of A is 2 × 8 = 16 kg. m/s. To reverse the direction of velocity of A, impulse on it J must be greater than 16 N.s 2I > 16 I > 8 If I = 8, A will just stop only. Next allowed integer is 9.

50 Two particles are projected horizontally in opposite directions from a point on a smooth inclined plane of inclination 060 with the horizontal as shown in figure. Find the separation between the particles on the inclined plane when their velocity vectors become perpendicular to each other. 1 21 / , 3 / .v m s v m s Express your answer in the form of k/10 m. Then, find the value of k.

V1 V2

Answer :8

Solution : 1 1 siˆ ˆnv v i g t j

2 2 sinˆ ˆv v i g t j

1 2. 0v v

Or 2 2 21 2 sin 0v v g t

Or 2 2 22 2 sinv v g t

Or 1 2

sinv v

tg

The particles line will be parallel to x – axis. Separation between the particles will be

1 2 1 21 2

( )sin

v v v vx x

g

51 A system of uniform cylinders and plates is shown. All the cylinders are identical and there is no slipping at any contact. Velocity of lower and upper plates is ‘v’ and 2v respectively as shown. Then the ratio of angular speeds of the upper cylinders to lower cylinders is :

v

2v

Answer :3 Solution : In the absence of slipping, velocities of contact points of upper cylinders and lower cylinders are respectively.

3 ;2

ABup

vvR AB

;2

CDlower

v vCD R

3up

lower

52 A solid sphere of mass ‘M’ and radius ‘R’ is initially at rest. Solid sphere is gradually lowered onto a truck moving with constant velocity V0.

The final speed of sphere’s centre of mass in ground frame when eventually pure rolling sets in (in multiples of

0

7V

?

Answer :2

Solution : 225

f t R MR ___________(1)

f0V

cmVw

cmf t M V _______(2)

From (1), (2) 25cmV R

0 cmV V Rw (Condition for pure rolling)

0CMV V Rw 052 cmV V

0512cmV V

027cmV V

53 In the figure shown, ends A and B of rod of length L slide on smooth horizontal ground and smooth inclined wall.

Instantaneous speed of end A of the rod is ‘v’ to the left. The angular velocity of the rod, in multiples of vL

is

A

B

060030

Answer :1 Solution : Draw normal at A and B to locate IC.

VL

54 A uniform solid sphere of radius ‘r’ is rolling on a smooth horizontal surface with velocity ‘v’ and angular velocity ( ).v r The sphere collides with a sharp edge on the wall as shown. The coefficient of friction between the

sphere and the edge is 15

. Just after the collision the angular velocity of the sphere becomes zero. The linear

velocity of the sphere just after the collision in multiples of 5v

is :

0edge

v

Answer :5

Solution : ' ( )Ndt mV mV ……….. (1)

225

VR Ndt mRR

……….. (2)

'V

0

Ndt

Ndt

From eqns. (1) and (2), we get

2Ndt mV

And 'V V

55 A hollow sphere is released from the top of a wedge, friction is sufficient for pure rolling of sphere on the wedge. There is no friction between the wedge and the ground. At the instant it leaves the wedge horizontally, velocity of

centre of mass of the sphere w.r.t ground is 7n gh . The value of ‘n’ is :

m

mh

Answer :3 Solution : 2 1 1 2mv mv v v (say v)

1v 2v

Constraint equation is 2 122 vv v R v RR

22 2 2 2 21 2 2

1 1 1 1 2 42 2 2 2 3

vmgh mv mv I mv mRR

3 37

v gh n

56 Uniform rod AB is hinged at end A in horizontal position as shown in the fig. The other end is connected to a block through a massless string as shown. The pulley is smooth and massless. Masses of block and rod are same

and equal to ‘m’. Then acceleration of block just after release from this position in multiples of 8g

is :

m, /m

AB

Answer :3 Solution : mg – T = ma

Rx

Ry mg

T T

a

Mg

2

2 3mgl mlTl

Also a l Acc. Of end B of rod is l

Thus3 5;8 8g mga T

38gl

57 A rod of length ‘l’ is traveling on a smooth horizontal surface with velocity uCM and rotating with an angular

velocity such that 2CMu

. The distance covered by the point B in multiples of ‘ ’ when the rod completes

one full rotation is :

CMU

A

B Answer :4

Solution : 2

2 2 cos2 2CM CMl lu u u

2 2cos( )2

ds ldt

2(1 cos )2

ds ld

2l

cmu

2

1/2

0

2 1 cos2lds d

2

20

0

cos22 sin 2 cos cos0 4

12 22

ls d l l l

58 Two blocks are connected by a massless string that passes over a frictionless peg as shown in Fig. One end of the string is attached to a mass m1 = 3 kg, i.e., a distance R = 1.2 m from the peg. The other end of the string is connected to a block of mass m2 = 6 kg resting on a table. From what angle (in multiples of 100) measured

from the vertical must the 3 kg block be released in order to just lift the 6 kg block off the table?

1m

2m

cosR R

R

Smooth peg

Answer :6

Solution : 2

12 1

m vm g m gR

2 (1 cos )v gR

59 A train has to negotiate a curve of radius 400 m. The distance between the rails is 1m. The height the outer rail should be raised with respect to inner rail for a speed of 36 km/hr is p/2 cm. Find the value of ‘p’ (take g = 10 m/s2) Answer :5

Solution : 2

tan h vl rg

60 A block of mass m1 = 150 kg is at rest on a very long frictionless table, one end which is terminated in a wall. Another block of mass m2 is placed between the first block and the wall, and set in motion towards m1 with constant speed u2. Assume that all collisions are perfectly elastic. Find the value of m2 (in multiples of 10 kg)for which both the blocks move with the same velocity after m2 collides once with m1 and once with the wall. The wall has effectively infinite mass.

1m2m2u

wall

Answer :5

Solution : 2 1 21 2 2 2

1 2 1 2

2 ( ),m m mv u v um m m m

1 2 2 1 22v v m m m

2 1 23 50m m m kg

61 A ball is released from position A and travels 5m before striking the smooth fixed inclined plane as shown. If the

coefficient of restitution in the impact is 12

e , the time taken by the ball to strike the plane again is (in sec) (g =

10 m/s2)

A5m

030 Answer :1

Solution : 2 cos , 2 10 /cos

euT u gh m sg

1T s

h

u

coseu

cosu

62 A uniform solid cylinder of density 0.8 g/cc floats in equilibrium in a combination of two non- mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7 g/cc and 1.2 g/cc respectively. The height of liquid A is 1.2Ah cm . The length of the part of the cylinder immersed in liquid B is 0.8Bh cm . The total force exerted by liquid A on the cylinder is (in N)

Answer :0 Solution :

The total force exerted by liquid A is Zero. This can be easily explained by pressure profile

63 A syringe of diameter D = 8mm and having a nozzle of diameter d = 2mm is placed horizontally at a height of 1.25 m as shown in the figure. An incompressible and non-viscous liquid is filled in syringe and the piston is moved at a speed of v = 0.25 m/s. Find the range of liquid jet on the ground (in meters). (g = 10m/ 2s ).

Answer :2

Solution : 1 1 2 2 22, .hA v A v R vg

64 A square gate of size 4 4m m is hinged at topmost point. A liquid of density fills the space left of it. The

force which acting at a height 1m from lowest point can hold the gate stationary is 29

N

g . The value of ‘N’ is

Answer :8 Solution : Force acting on any elementary strip at a distance y from O

dF gy ady Torque about O

( )d y dF ( )d y dF

2( )d gy a dy

Net torque 2

2

0 03

aa ygy dy ga

65 A fixed cylindrical tank having large cross-section area is filled with two liquids of densities a and 2 and in

equal volumes as shown in the figure. A small hole of area of cross-section 26a cm is made at height h/2 from

the bottom. Find the area of cross- section of stream of liquid in 2cm just before it hits the ground

h

hh/2

area = a

2

Answer :2 Solution : Applying Bernoulli’s equation at cross-section 1 & 2

2 20 0 0 ..... 1atm atmP gh P P P gh Again applying Bernoulli’s equation at section 2 & 3

22

10 2 2 ..... 22 2atmhP g P V

2V gh This is required velocity of efflux Applying continuity equation between 3 & 4 cross-section.

1 1aV a V This is required velocity of efflux Applying Bernoulli’s equation between 3 & 4

2 21

1 12 2 2 02 2 2atm atm

hP V g P V

2 2 21 1 3V gh V V gh

2

11

22

3ghaVa cm

V gh

66 An open tank having dimensions 1 m ´ 2 m ´ 3m completely filled with water is kept on a horizontal surface. The mass of water that spills out is 100x kg, when the tank is slowly accelerated horizontally at the rate of 2 m/ 2s . Then find the value of x. ao = 2 m/s2

2 m 1 m

3 m

Answer :4

Solution : tan a / tan 2 /10 tan 0.2o g tan y / 2 \ y = 0.4 m Þ volume of shaded region Þ V = (1/2) ´ 0.4 ´ 2 ´ 1 = 0.4 m3 mass = V ´ d = 0.4 ´ 1000 = 400 kg Þ x = 4

67 Water is filled in a uniform container of area of cross section A. A hole of cross section area a (<< A) is made to the wall of container at a height of 20 m above the base. Water streams out and hits a small block placed at some distance from container. With what speed (in ms-1) should the block be moved such that water stream always hits

the block before the level of water in the vessel reaches 20 m. (Given120

aA ). (Take g = 10 –2ms )

a

20m

A

Answer :1

Solution :

V

A

x

ay

h

Velocity of efflux 2v gy

Range 22 hx gyg

The velocity of the block must be ( )dxdt

.

2 122b

dx h dyV gdt g dty

.bh dyV

dty ………(i)

Using equation of continuity

2Ady a gydt

………..(ii)

equation (i) and (ii)

2bh aV gyy A

2baV ghA

12020

1 ms–1.

68 A long glass capillary tube of radius r is placed horizontally and filled with water (angle of contact for water –

glass = 0o ). If the tube is made vertical, then the length of water column that remains in the capillary is T

r g

where ‘T’ is the surface tension of water, is density of water. Find Answer :4

Solution :

0 02 2T TP gh PR R

Þ4Th

R g

69 A ball A moving with momentum 2ˆ 6 ˆi j collides with another identical moving ball B with momentum ˆ4 j

and momentum of ball B after collision is 2 j . The coefficient of restitution in the collision is15x

. Find the value

of x. Answer :3

Solution : Since the momentum exchange in j - direction line of impulse is in j -direction

ˆ ˆ 2ˆ2 6 ˆ4 2ˆA Ai j j P j P i

2 0 2 16 4 10 5me

m m

70 A moving sphere A of mass ‘m’ experience a perfectly elastic collision with a stationary sphere B of same mass ‘m’ as shown in the fig. At the instant of collision the velocity vector of A makes an angle of 300 with the line joining the centers of A and B. After collision the spheres fly apart then the angle between their velocity vectors is

010K . Then the value of ‘K’ is

u300

A

B Answer :9 Solution : After oblique elastic collision the balls are moving mutually perpendicular directions hence K = 9

71 The number of possible overtones of air column in a closed pipe of length 83.2 cms and diameter 6 cms whose frequencies lie below 1000 Hz will be, given velocity of sound in air =340 m/sec.(Consider end correction ; end correction equal to 0.3 times of diameter) Answer :4

Solution : 0.3 83.2 0.3 6 854

l d

340cm

100Vn Hz

(possible frequencies are 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz. No. of over times “4”).

72 The metal plate on the left in the figure carries a charge q . The metal plate on the right has a charge of 2q . The magnitude of that charge will flow through S when it is closed ( if the central plate is initially neutral) is xq where x = _____

q -2q

S

Answer :1 Solution : In steady state the following charges will appear on different faces of the plates.

2q2qqq

Net charge on the central plate is q . Thus, q charge will flow through the switch.

73 An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the each. If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find the speed (in km/sec) with which it hits the surface of the earth ( 210 /g m s and R = 6400 km).

Answer :8 Solution : Given 0 / 2ev v

1/2 1/21 22

GM GMR h R

On solving, .h R From law of conservation of energy,

21

2GMm GMmmvR h R

or 212 2 2

GMm GMm GMmmvR R R

or GMv gR

R

1/2610 6.4 10 8 /km s

74 A rope is held horizontally as shown in the figure. The rope is under a uniform tension 196T newton. The mass

per unit length of the rope is 49 /2025

kg m . At the free end of the rope, a vertical jerk is given with a frequency of

2.0 Hz , generating a transverse wave in the rope. Neglecting the fact that the weight of the rope may cause the

rope to have a curved shape rather than a straight one, The number of wave cycles in the rope’s length is 16 ,x

where x = _____

T

80.0m

Answer :9

Solution : 1196 2025 9049

V ms

90 452

V mf

No. of wave cycles 80 16 1.7745 9

, An: x = 9

75 Six capacitors are arranged as shown in the figure. The circuit is in steady state The current through the battery is steady state is 2x amp where x = _____

1 ; 1 ; 10C F R E Volt

/\/\/\

/\/\/\

/\/\/\

/\/\/\

R Ac

4c

8EB 4R

2c3c

2R3R

2cc

Answer :5 Solution : 8 8 \ 10R I E E R Amp ; An: x = 2

( ) 8 0A BV V E IR

^^^

^^^

^^^^^^

cR

8E 4RB

I

I

I

4c2c 3c

2ccI

I I>

>

>

>

>>

76 A cylinder of height 1h m has a narrow vertical slit along its wall running up to a length 40l cm from the bottom. The width of the slit 1b mm. The cylinder is filled with water with the silt closed. The force experienced by the vessel immediately after the silt is opened, is nearly ( 0.4)x N where x = ______

2( 10 / )g m s

h

l

Answer :6 Solution : Consider an element of the slit of length dx at a depth x from the top.

Velocity of the efflux at 2x gx Force on elementary area = rate of change of momentum ( ) (2 )bdx v v b dx gx

F ( total force experienced ) 2 (2 )h

h l

b d xdx b gl h l

Or , 6.27F N An; x = 6

77 A cubical block of volume v and density3 is placed inside a liquid of density and attached to a spring of spring constant k as shown in the figure. Assuming ideal spring and pulley and spring is attached at A which is at

/ 2R from centre. The compression in the spring at equilibrium is xpvg

k, where x = _____

k cA/\/\/\/\/\/\/\/\

Answer :4 Solution : Tension in the string 3 2BT mg F vg vg vg Balancing torque

\ 2kxR TR 2 / 4 /x T k Vg k An; x = 4

78 A soap film is created in a small wire frame as shown in the figure. The sliding wire of mass m is given a velocity

u to the right and assume that u is small enough so that film doesn’t break. Plane of the film is horizontal and the

surface tension is T. Then time to regain the original position of wire is xumT

, where x = ___

l uSoap film

Answer :1

Solution : 2Tlam

0v u at u at

2u umta Tl

2 umTotal timeT tTl

; here x = 1

79 A non-viscous incompressible liquid of mass m is filled fully inside a thin uniform spherical shell of mass m and radius R performing pure rolling on a rough horizontal surface. There are two points A and B inside the liquid on the vertical diameter separated by 2R as shown in the figure. Pressure difference between B and A is B Ap p . At

the given instant velocity of centre of mass of the given system is 0v and kinetic energy of this system is K. There

is no slipping of sphere on surface. If the value of B A

Kp p

is of the form 2 20 ,

9X v R

g

then find the value of X .

A

B

g

No slipping Answer :8 Solution : Liquid will be in pure translation.

80 A heavy uniform rope of length L is suspended from a ceiling. The rope is given a sudden sideways jerk at the bottom. A particle of mass 3kg is dropped from the ceiling at the instant at which jerk is given at the bottom of

rope. The kinetic energy of the particle when the particle and wave are at the same level is ( )x gL . Find the value of x. Find the value of x. Answer :4

Solution : T dxv gxdt

2

4gtx

2 21 32 4

L x gt L gt

43

Ltg

Velocity of the particle 43

Lgt gg

KE of the particle 21 2 2(3) 22 3 3

mg gm g

dxL

x

m

81 A small ball of radius r is falling in a viscous liquid under gravity. Find the power of radius r on which the rate of heat produce depends after the drop attains terminal velocity. Answer :5 Solution : From Stoke’s law,

6F Rate of heat produced 2. 6Q F v r

Terminal velocity 2

0( )29

r g

2

20( )26 ( )9

r gQ r

5 2 20( )4

3r g

5Q r The power is 5.

82 A block of mass m produces an extension of 9 cm in an elastic spring of length 60 cm when it is hung by it, and the system is in equilibrium. The spring is cut in two parts of 40 cm and 20 cm lengths. The same block hangs in equilibrium with the help of these two parts. Find the extension (in cm) in this case. Answer :2 Solution : ( )(9)mg K or K=mg/9 The spring constant is inversely proportional to length. So the spring constant for the spring of 40 cm

length is 32

K . The spring constant for the spring of 20 cm length is 3K. The spring constant for the combination

is 3 932 2

K K K

Now 092

mg Kx or 0992

K Kx

0 2x cm

83 Two blocks with masses 1 1m kg and 2 2m kg are connected by a spring of force constant K=24 N/m. The left block is imparted an initial velocity of 12 cm/s. The amplitude of the oscillations is _______cm

1m2m

smooth

Answer :2 Solution : Initial velocity of the centre of mass

1 0

1 2cm

mm m

21 2

1 ( )2translattional cmE m m

Energy imparted to 21 1 0

1' ',2

m E m

vibrational translationalE E E

1 2

1 2 1 2

E m E mEm m m m

2

2 1 2 0

1 2

1 12 2

m mkam m

Amplitude 1 20

1 2( )m ma

m m k

2(12 / )3 24

cm s

= 2 cm

1m2m

smooth

84 A wooden plank of length 1 m and uniform cross section is hinged at one end to the bottom of a tank as shown. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. The angle that the

plank makes with the vertical in the equilibrium position (exclude cos 0 ) is x . Find the value of x.

ABC

O

l

Answer :4 Solution : The forces acting on the plank are shown in the figure. The height of water level is l=0.5 m. The length of the plank is 1.0 m=21. The weight of the plank acts through the centre B of the plank. We have OB=l. The buoyant force F acts through the point A which is the middle point of the dipped part OC of the plank.

ABC

O

F

mgl

(i) We have 2 2cos

OCOA

Let the mass per unit length of the plank be . Its weight 2mg g

The mass of the part OC of the plank cos

The mass of water displaced 1 2

0.5 cos cos

Therefore, the buoyant force 2cos

gF

.

Now, for equilibrium the torque of mg about O should balance the torque of F about O. So, ( ) sin ( ) sinmg OB F OA

22cos 2cos

2 1cos2

01cos 4542

or

85 A uniform rod mass m= 5.0 kg, length L=90 cm rests on a smooth horizontal surface. One of the ends of the rod is struck with the impulse J=3.0 N.s in a horizontal direction perpendicular to the rod. As a result the rod obtains the momentum p=3.0 N.s. Find the force with which one half of the rod will exert on the other in the process of motion. Answer :9

Solution : 2 6

2 12L mL JJ

mL

Rod will rotate about its c.m., one half exerts centrifugal force on the other half, therefore 2 29 9

2 4 2m L JF N

mL

L

JL/4

2m/ 2 L/ 4

86 The bob is released from 04 , so that it collides with the inclined wall of inclination 02 . If the length of

the string is 10cm . The time after which the bob collides with the wall is 4

10s

X

, then value of X is

Answer :3 Solution : The total time period motion is

A

BC

( )AB BCT t t ……………. (i)

Where ABtg

…………….. (ii)

And BCt can given as sin( )BCt or 1 1 11 1sin sin sinBCtgg

…… (iii)

From equations (i), (ii) and (iii), 0

1 10

10 2[ 2sin ( )] [ 2sin ( )]100 10 4

Tg

1 2 4[ ]10 6 30

s .

87 For two satellites at distances R and 7R above the earth’s surface, the ratio of their total energies are ________ Answer :4 Solution : Distance of the two satellites from the centre of the earth are 1 2r R and 2 8r R respectively R=earth’s radius, Their potential energies are

11

GmMVr

and 22

GmMVr

Their ratio is 1 2

2 1

8 42

V r RV r R

The kinetic energy of a satellite can be obtained from relation 2

2

mV GmMr r

or 212 2

GmMK mvr

Thus 112

GmMKr

and 222

GmMKr

The ratio of their kinetic energies is 1 2

2 1

8 42

K r RK r R

Their total energies are

11 1 12 2

GmM GmM GmMEr r r

and 22 1 22 2

GmM GmM GmMEr r r

Their ratio is 1 2

2 1

8 42

E r RE r R

88 Two balls A and B are thrown vertically upwards from the same location on the surface of the earth with velocities

23

gR and

23gR

respectively, where R is the radius of the earth and g is the acceleration due to gravity on the

surface of the earth. The ratio of the maximum height attained by A to that attained by B above the earth’s surface is Answer :4 Solution : I h is maximum height attained, then we have

212 ( )

GMm GMmmvR R h

Which gives 2 2( )

ghRVR h

2

GMgR

For ball A, we have 24 43 ( )

AA

A

gh RgR h RR h

For ball B, we have 22

3 ( ) 2B

BB

gh RgR RhR h

8B

Hh

.

89 The electric potential V (in volt) varies with x (in metre) according to the relation V = 25 4x . The magnitude of force experienced by a negative charge of –62 10 C located at x = 0.5 m is –6z 10 N, then the value of z is ________ Answer :8

Solution : Electric field 25 4 8dV dE x x xdx dx

Force on charge ( ) 8q qE qx At x=0, 5m, force 6 68 2 10 1.5 8 10 N

8z

90 Two point charge 1 2q C and 2 1q C are placed at distances b = 1cm and a = 2cm from the origin on y and

x–axes as shown in figure. The electric field vector at point P(a,b) will subtend an angle with the x–axis given by tan ________

q1 P(a,b)

ba

O x

y

q2 Answer :2

Solution : The electric field, 1E at ( , )a b due to 1q has a magnitude 11 2

0

1 .4

qEa

And is directed along +x- axis. The electric field 2E at ( , )a b due to 2q

Has a magnitude 22 2

0

1 .4

qEb

and is directed along +y-axis. The angle subtended by the resultant field E

with the x-axis is given by 22

2 22

1 1

1 2tan . 22 1

E q a xE q b

91 A charge Q is distributed uniformly on a ring of radius r. A sphere of equal radius r is constructed with its centre at

the periphery of the ring. The flux of the electric field through the surface of the sphere is 0

Qn

. The value of n is

_______ Answer :3 Solution : Due to a charged non-conducting sphere, (i) Intensity inside sphere = 1E

1 30

14

QE rR

1 ,E r for r < R, as (OA)in figure.

Lotion (A) is correct. (ii) Intensity outside sphere = 2E

2 20

14

QEr

when R < r <

2 2

1 ,Er

as (AB) in the figure.

2E decreases as r increases, Option (C) is correct.

y

q1

q2xO a

b

P

E

E1

E2

(iii) Option (B) is incorrect in view of (A) and (C) above. (iv) Intensity at surface = 0E

0 20

14

QER

, at point A in the figure.

0E is continuous at r = R. Option (D) is incorrect.

92 A mass of 246 10 kg is to be compressed in a sphere in such a way that the escape velocity from its surface is 83 10 m/s. The radius of the sphere is _______ mm(nearly)

Answer :9

Solution : Kinetic energy 212 2

GmMmvr

or 1KEr

. Thus choice (A) is correct.

Angular momentum = mvr = GMmx xr m GMr

r , which is proportional to r . Hence choice (B) is wrong.

Linear momentum mv = ,GMmr

which is proportional to 1r

. Hence choice (C) is correct.

The freuqeuncy of revolution is 3

1 1 . . .2

GMv i eT r

3/2

1vr

.

Hence the correct choices are (A), (C) and (D)

93 A particle of mass 1kg is placed at a distance 4m from the centre and on the axis of a uniform ring of mass of 5kg and radius 3m. The work required against mutual gravitational attraction to increase the distance of the particle

from 4m to 3 3 m is 6G Jx

. The value of x is _____ (G = universal gravitational constant)

Answer :1 Solution : The potential at x = 0 due to set of infinite number of charges placed on the x- axis as shown in figure, is

x=0 x=1 x=2 x=4 x=8to infinity

0

1 [ ..... ]4 1 2 4 8

q q q qV

0

1 1 11 ....4 2 4 8

q to

=

0

11214 12

q

0 0

1 014 22

q qx

The charges are placed along the same straight line, the electric filed at x = 0 will be directed along the x – axis and its magnitude is given by

2 2 2 20

1 .....4 1 2 4 8

q q q qE to

=

0

1 1 11 .....4 4 16 64

q to

= 0 0 0

11 1 041 34 4 314 4

q q qx

94 Two identical small charged spheres are suspended by strings of equal lengths. The strings make an angle of 030 with each other in air. When suspended in a liquid of density 0.8 g/cc, the angle remains same. If the density of the material of the sphere is 1.6 g/cc, the dielectric constant of the liquid is _________ Answer :2 Solution : If the consecutive charges have opposite sign, the potential at x = 0 is given by

0

1 .....4 1 2 4 8 16 32

q q q q q qV to

0

1 1 1 1 11 ... .....4 4 16 2 8 32

q to to

0

1 1 11 14 21 14 4

q

0 0

4 1 44 3 2 3 6

q qx

2 2 2 2 2 20

1 .....4 1 2 4 8 16 32

q q q q q qE to

0

1 1 1 1 11 ..... ...4 16 256 4 64 1024

q to to

0 0 0

1 1 1 16 1 161 14 4 4 15 4 15 41 1

16 16

q q qx

95 A solid sphere of radius R has a charge Q distributed in its volume with a charge density akr , where 'k' and

'a' are constants and r is the distance from its centre. If the electric field at 2Rr is

18

times that at r = R, find the

value of a. Answer :2 Solution : Electric flux through a plane surface of area S is given by where is the angle which the normal to the surface makes with the direction of the electric filed. Now, S = 10cm x 10cm = 100 2cm =100 x 410 2m and 33 10E x 1NC along the positive x- direction. 00 Hence 3 4 0 1 23 10 100 10 cos 0 30x x x x NC m .

96 In a uniform electric field, the potential is 10V at the origin of coordinates, and 8V at each of the points (1,0,0), (0,1,0) and (0,0,1). The potential at the point (1,1,1) will be ________(in volts)

Answer :4 Solution : In this case 0 0 090 30 60 , hence 1 215NC m .

97 A child of mass 4 kg jumps from cart B to cart A and then immediately back to cart B. The mass of each cart is 20 kg and they are initially at rest. In both the cases the child jumps at 6 m/s relative to the cart. If the cart moves along the same line with negligible friction with final velocities of BV and AV , respectively, find the ratio of 6 Bv

and 5 .Av

A B

Answer :1 Solution : All the velocities shown in diagrams are w.r.t. ground. After first jump :

A 20 kg B 20 kg

4kgv2

v1

1 220 4v v and 1 2 6( )v v given

Solve to get 1 2/ , 5 /b m s v m s When child arrives on A:

v3 v1A B

3 2 3(20 4) 4 5 / 6 /v v v m s

After the second jump : v4

v1A BvA

4 3 46, 24 20 4A Av v v v v Solve to get 411 25/ , /6 6Av m s v m s

When child arrives on B :

vBA BvA

4 124 4 20Bv v v 25 5524 4 20 1 /6 36B Bv v m s

Now 6 6 55 6 15 36 5 11

B

A

vv

98 A uniform rod of length 1 m and mass 2 kg is suspended. Calculate tension ( )T in N in the string at the instant

when the right string snaps 2( 10 / ).g m s

M

l

Answer :5 Solution : yMg T ma

2

( )2 12L MLT and

2yLa on solving, we get

54

MgT N

T

mg

99 A stone of mass m, tied to the end of a string, is whirled around in a horizontal circle (neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then the tension in the string is given by / nT A r , where A is a constant , r is the instantaneous radius of the circle and n is _____ Answer :3

Solution : For circular motion of the stone . 2

[ 0]( )mv T as g ir

And as A.M is constant, , ( )Kmvr K v iimr

i.e.,

Eliminating v between Eqs. (i) and (ii) , we get 2 2

3 3,m K KT T r T Arr mr m

with

2

( )KA iiim

Comparing Eqs. (iii) with / ,nT A r we find n = 3.

100 A ball of mass 1 kg moving with a velocity of 5 m/s collides elastically with rough ground at an angle with the vertical as shown in Fig. What can be the minimum coefficient of friction if ball rebounds vertically after collision? (given tan 2)

Answer :1

Solution : From impulse momentum theorem ( 5cos )N dt m v

5sinf dt m

5sinN dt m

( 5cos ) 5sinm v m

f

5sin

5cos

v

According to Newton’s law of restitution, 5cosv e Solve to get 1

101 A blocks of mass 0.18 kg is attached to a spring of force constant 12Nm . The coefficient of friction between the block and the floor is 0.1 Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in 1ms is /10.V N Then N is

Answer :4

Solution :

v = 0u

v=0.06 2 21 10.06

2 2mu mg kx

2 21 10.18 0.1 1.18 10 0.06 2 (0.06)2 2

u

10.4u ms

0.4 410N N

102 A block is placed on an inclined plane moving towards right horizontally with an acceleration 0 .a g The length of the plane AC = 1m. Friction is absent everywhere. Find the time taken (in seconds) by the block to reach from C to A.

oa g

B C

A

37o

Answer :1 Solution : (1) Drawing free – body diagram of block with respect to plane. Acceleration of the block up the plane is

Pseudo - force

mg

N

B C

37o

ma0 = mg

2cos37 sin 37 4 3 25 5

o omg mga g msm

Applying, 212

s at we get

2 2 1 1sec2

sta

103 In the figure shown all the surface are frictionless, and mass of the block is m = 100 g. The block and the wedge are held initially at rest. Now the wedge is given a horizontal acceleration of 210 ms by applying a force on the wedge, so that the block does not slip on the wedge. Then find the work done in joules by the normal force in ground frame on the block in 1 s.

B C

m

M

10 ms-2

Answer :5 Solution : (5) If block does not slip then , tana g 10 10 tan 45o

m

mg

N

sinN

1sin 0.1 10 22

N ma N N N

2 21 1 10(1) 52 2

S at m

cos 2 5 1/ 2 5NW NS J

Alternatively : 2 21 1 (0.1)( ) 52 2NW mv at J

104 A binary star consists of two stars A (mass 2.2Ms) and B (mass 11 Ms). Where Ms is the mass of the sun. They are separated by distance d and are rotating about their centre of mass,which is stationary. The ratio of the total

angular momentum of the binary star to the angular momentum of star B about the centre of mass is. Answer :6

Solution : (6) 2

1 12

2 2

1total

B

L m rL m r

105 A steel wire of length 1 metre, mass 0.1 kg and uniform cross sectional area 10-6 m2 is rigidly fixed at both ends. The temperature of the wire is lowered by 200C. If transverse waves are setup by plucking the string in the middle, the frequency of the fundamental mode of vibration. [Young’s modulus of steel = 2×1011 N/m2,

coefficient of linear expansion of steel =1.21×10-5/0C] is 22 Hzx

find “ x ”

Answer :2

Solution : TlyA l

and al l

48.4T yA N

1 1 48.4 222 2 1 0.1 2

Tfl

2x

106 A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure. Where should a mass of 4.8 kg be placed on the rod so that the same tuning fork may excite the wire on left into its fundamental vibrations and that on right into its first overtone? from left in cm (Take g = 10 2/m s )

40 cm Answer :5 Solution : 1 2 6T T g

1 2f f

V V

1T 2T

l xx 1.2g4.8g

1 21 22 2

T Tl l

1 24T T

4.8 1.2 / 22Tg x g l l

5x cm

107 Figure shows a string of linear mass density 1.0 g/cm on which a wave pulse is travelling. The time taken by the pulse in travelling through a distance of 50 cm on the string is 210x s find x Take g = 10 2/m s

1Kg

Answer :5

Solution : 10 10 /0.1

Tv m s

Time 21/ 2 1 5 10 sec10 20

stv

108 Two identical piano wires have a fundamental frequency of 600 vib/sec. When kept under the same tension. What percentage (nearly) increase in the tension of one of wire will lead to the occurrence of six beats per second. Answer :2

Solution : 11

1 6002

Tfl

; 22

1 60.62

Tfl

Solving 2

2

1

101100

TT

2

2 12

1

101100 1 100

100T T

T

= 2%

109 Three sources of sound 1 2,S S and 3S producing equal intensity at P are placed in a straight line with

1 2 2 3S S S S (figure). At a point P, far away from the sources, the wave coming from 2S is 0120 ahead in

phase of that from 1S . Also, the wave coming from 3S is 0120 ahead of that from 2S . What would be the resultant intensity of sound at P?

1S 2S 3S p

Answer :0 Solution : 1A A

0 02

ˆcos 60 sin 60A A A j 0 0

3ˆcos60 sin 60A A A j

1 2 3 0A A A A Resultant intensity = 0

110 Consider the situation shown in figure. The wire which has a mass of 4.00 g oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound

in air is 340 m/s. The tension in the wire is 1110x

Find x

40 cm

1.0m

Answer :6

Solution : 22 4

T vl l

2

1 340 11.6 110.4 10 4 10

T xT

111 The equation of a longitudinal stationary wave in a metal rod is given by, y=0.002 sin

3x sin 1000 t where

x & y are in cm and t is in seconds. Maximum change in pressure ( the maximum tensile stress) at the point

x=2 cm, [if young’s modulus of the material is 3

8dynes/ 2Cm ] is 31 10

p dy/ 2Cm . Find p

Answer :8 Solution : Maximum change in pressure Maximum tensile stress = YAK

33 108 3

= 3 21 10 /8

dyne cm

112 A road passes at some distance from a standing man. A truck is coming on the road with some acceleration. The truck driver blows a whistle of frequency 500 Hz when the line joining the truck and the man makes an angle with the road. The man hears a note having a frequency of 600 Hz when the truck is closest to him.

Also the speed of truck has got doubled during this time. The value of ‘q’ is x find x.

Answer :3 Solution : Let Velocity of truck be SV

Then 600 500cosS

VV V

C

l

M

( )T x

As 2 S S SV V at at V

As ;sin

lVt V lVelocity of sound

21cot2Sl V t at

2cos 3

sin 2 Sl at V at

2cos 3 1sin 2 sin

p aV

2

2 sin cos3

alV

from (1) 65 cosV

VV V

65 cos

VV at

6 cos5

sin

ValV

V

6 2 sin coscos5 3sin

V VV

0cos 60

113 A source of sonic oscillations with frequency 0 1700v Hz and a receiver are located on the same normal to a wall. Both the source and the receiver are stationary, and the wall recedes from the source with velocity u = 6.0

cm/s. The beat frequency registered by the receiver is 10K HZ find K. The velocity of sound v = 340 m/s

Answer :6 Solution : Let cl be The velocity of wall

OV velocity of sound then

beat frequency 2

0.589

0.6

2 O

O H

ffV

114 A man standing in front of a mountain beats a drum at regular intervals. The drumming rate is gradually increased and be finds that the echo is not heard distinctly when the rate becomes 40 per minute. He then moves near to the mountain by 90m and finds that the echo is again not heard when the drumming rate becomes 60 per minute. The distance (in metre) between the mountain and the initial position of the man is 90x m find ' 'x value ? Answer :3

Solution : Initially time interval beats 60/ ] 1.5sec40

b n

S Initial distance and

V speed of the sound ; 251.5V tV

When man approaches 90m towards the wall. 2( 90) 1S

V

[ min 60 / min]dru g rate 2 (2)

Solve (1) & (2) 270 90 3; 3S m N

115 A ball is held at rest in position A by two light cords. The horizontal cord is now cut and the ball swings to the

position B. Then the ratio of the tension in the cord in position B to that in position A originally is 3x

, find x

value ……….

030 030

AB

Answer :3 Solution : . Tension in string at two position is same

3x

116 Find the current in amperes in the branch AB in the circuit shown in the figure.

10

10

5

5

20VA

B

Answer :5 Solution : 5 is in parallel to cell. Current through it should be 4A.

i-4

i-I

I

i 20V

10

105

54

4-I

10( 4) 5(4 ) 20

510( ) 5(4 ) 0

i II A

i I I

117 A series R-C combination is connected to an AC voltage of angular frequency 1. .600 rad s The

impedance

of the circuit is 1.25 .R Then the time constant of the circuit is10n

millisecond, where n = ________

Answer :3

Solution : 2

2 2 2 2 254 4C C

RZ X R R X

1 2 102 3R CR ms

C

118 In the circuit shown, switch 2S remains closed for a long time, with 1S open. It is given that 10 ,R

1L mH and 3V . Now switch 1S is also closed. Immediately afterwards, if the magnitude of rate of

change of current (in 1As ) in the inductor is 1000 x, find x.

L

R 2R

1S 2S

Answer :2

Solution : Steady current L (with 2S closed): 0 2i

R

V1i1 i2

L

i0

O Current is same in L just before and just after closing 1S .

1 11 2 0 2 2

V Vi i iR R R

132

V

0 01 3

2 1 2 3 2000 23 3 10

Ldi diV xdt dt L

119 Two spheres of relative densities 0.8 and 1.2 respectively, each having equal volume 3250 cm , are connected by

a string and the combination is immersed in a liquid of relative density 1.0. As the system freely floats in the

liquid, the tension in the string is 10n N , where n = _____ 2( 10 / )g m s .

Answer :5

Solution : 6 6(250 10 800 ) (250 10 1200 )g T g T

62 250 10 400 0.5T g T N

120 Each resistor in the finite network shown in the figure is1 . A current of 1 A flows through the last branch. The potential difference across the terminals A and B is (10 )V a b volt with a,b as integers , where a + b = _____

V

A

B

1A

Answer :7

Solution :

21 A 3A

13A

8A

1AV=34 volt

5A 2A

a = 3, b = 4 a + b = 7

121 From the top of a tower of height 80m, a body is projected up with velocity 50m/s at an angle of inclination of 037 with horizontal. If mass of the body is m=0.01 kg and acceleration due to gravity is 10 2ms , then find

power (in watts) supplied by gravitational force on the body at an instant seventh second after the projection . Answer :4

Solution : 40 30 10 7 sec 40 (30 70) 40ˆ ˆ ˆ ˆ 4 ˆ0ˆ ˆu i j and a j velocity after i j i j

0.01 10ˆ ˆ ˆ0.1mgj j j . 4power F v W

122 An object of mass 0.2kg executes SHM along x-axis about origin with a frequency of

25 .Hz

At the position x

= 0.04m, the object has K.E of 0.5J and P.E 0.4J. If the amplitude of osillations is A cm, find A. Answer :6

Solution : 2

1 25 1 500 \2 0.2

k kv k N mm

21 2 0.90.5 0.4 62 500

KA A cm

123 A block of mass 25kg rests on a horizontal floor ( 0.2) .It is attached by a 5m long light, straight horizontal rope to a peg fixed on floor. The block is pushed along the ground with an initial velocity of 10 ms-1 so that it moves in a circle around the peg. Find the time (in seconds) when tension in the rope just becomes zero.

2( 10 )g ms Answer :5

Solution : 110 0

2 10 2v tdv a g dv dt v t

dt

The tension in the rope will become zero, when centripetal acceleration becomes zero I,e., when speed becomes zero i.e., 10 2 0 5sect t

124 A mass M is in static equilibrium on a massless vertical spring as shown in fig. A ball of mass 'm' dropped from

certain height sticks to the mass 'M' after colliding with it. Combined mass executes SHM. During oscillations, it rises to a height ‘a’ above the equilibrium level of M and a depth b below it. What is the height (in meters) above the initial level of M from which the mass 'm' was dropped?(given M=m,b=3m, m=1kg, k=10N/m, g=10m/s2)

Answer :3

Solution : 2v gh After collision 1

2vv

m

2m

v

1v

m

2 2

'21 12 0 22 2s g

mg mgw w K k b mb mvk k

21 (15) 6 32 4

mvk mg h m

125 A square metal wire loop of side 10 cm and resistance 1 ohm is moved with a constant velocity 0v in a uniform magnetic field of induction B = 2 weber/ 2m as shown in the figure. The magnetic field lines are perpendicular to the plane of the loop (directed into

M

h

b

the paper). The loop is connected to a network of resistors each of value 3 ohm. The resistances of the lead wires OS and PQ are negligible. What should be the speed of the loop (in cm/s) so as to have a steady current of 1 mA in the loop?

Answer :2 Solution : Given network forms a balanced Whetstones bridge. The net resistance of the circuit is therefore 3 1 4 . Emf of the circuit is 0Bv l . Therefore, current in the

circuit would be 0Bv liR

or 3

0(1 10 )(4)

2 0.1iRvBl

= 0.02 m/s = 2 cm/s

126 An LCR series circuit with 10 resistance is connected to an AC source of 200 V and angular

frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60 .

When only the inductance is removed, the current leads the voltage by 60 . Calculate the power (in kW)

dissipated in the LCR circuit. Answer :4 Solution : When the capacitance is removed, the circuit becomes LR with

tan , . . tan 10 3LL

X i e X RR

and when inductance is removed, the circuit becomes CR with

tan , . . tan 10 3CC

X i e X RR

as here L CX X , so the circuit is series resonance and hence as

2000, . . , 2010

rms rmsL C rms

V VX X X i e Z R so I AZ R

and / cos 200 20 1 4av rms msP V kW

127 Two circular coils X and Y have equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as YB and that due to smaller coil X at O as

XB , then find x

y

BB

Answer :2 Solution : Conceptual

128 A current I, is flowing through a circular loop of radius ‘R’ made up of thin copper wire. When a uniform magnetic field ‘B’ is produced perpendicular to the plane of the loop, the tension developed in

the loop is found to be BIRx

. Find the value of ' 'x .

Answer :1 Solution : Conceptual

129 Two uniform rods of equal lengths 0( )l and equal masses have coefficient of linear expansion 2and are placed in contact on a smooth horizontal surface as shown. The temperature of system

is 0 .C Now the temperature is increased by 0 .C The junction of the rods will shift from its initial

position by 0

x . The value of ‘x’ is

Answer :4 Solution : The extension in left rod 0l l

The extension in right rod 02 2l l

Since the centre of mass of two rod system will not shift, the junction will shift by x as shown. Taking moment of mass about original centre of mass C,

0

2m x

= 0 22

m x

Solving we get x = 0

4 4

130 To determine the half life of a radioactive element, a student plots a graph of ( )ln dN t

dt versus t. Here

( )dN tdt

is the rate of radioactive decay at time ‘t’. If the number of radioactive nuclei of this element

decreases by a factor of P after 4.16 years, the value of ‘P’ is

65

4

3

2

12 3 4 5 6 7 8

years

Answer :8

Solution : , ln ln lndN dNN Ndt dt

1/ 2 put 0tN N e

0tN N e

1 4.160 2

0N N eP

2.08 8P e

131 A diatomic ideal gas is compressed adiabatically to 132

of its initial volume. If the initial temperature

of the gas is iT (in Kelvin) and the final temperature is iaT , the value of a is Answer :4 Solution : For adiabatic process

1 11 1 2 2TV T V ( 7 / 5)

Hence, we have

11

1 1 1. .32VT V T

Solving 4

132 Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is

observed to move from 253

m to 507

m in 30 seconds. What is the speed of the object in km per hour?

Answer :3

Solution : 1 1 1v u f

1

3 1 225 20u

1 50u m

Now, 2

7 1 250 20u

2 25u m

Speed of object 1 2u ut

50 ( 25) 5 /3 6

m s

5 18 36 5

kmph

Hence 25 1830 5

v

3kmph

133 A large glass slab ( 5 / 3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R ? Answer :6 Solution : As shown, at ‘P’ TIR will occur

R P

8cm

Thus, 0sin 3 / 5 . ., 37i e

Now R= 08 tan 37 = 6cm

134 Water (with refractive index = 43

) in a tank is 18 cm deep. Oil of refractive index 74

lies on water

making a convex surface of radius of curvature R = 6 cm as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. Then ‘x’ is

S

Answer :2 Solution : Refraction at curved surface

6R cm

4 / 3

S

1

18cm

L

For refraction at air-oil interface,

2 1 2 1

v u R

we have,

1

7 / 4 1 (7 / 4) 124 6v

1 21v cm

Now refraction at plane surface (oil-water surface)

2

4 / 3 7 / 4 4 / 3 7 / 421v

Hence 2 16v cm

Therefore,

18 16 2x

Answer is 2 cm

135 A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is 10zA (where 1 < A < 10). The value of ‘Z’ is Answer :7

Solution : 1242 6.21200photon

hc eV nmE eVnm

0

1 .4

neVR

Emission will stop when potential reached by sphere is capable to restrict fastest electron to escape. Hence

( ) 6.21 4.7 1.51hceV eV

19 2 919

2

(1.6 10 ) 9 10 1.51 (1.6 10 )10

n

715.1 10 101.6 9

Zn A

Z= 7

136 The activity of a freshly prepared radioactive sample is 1010 disintegrations per second. Whose mean life is 910 s. The mass of an atom of this radioisotope is 2510 kg. The mass (in mg) of the radioactive sample is Answer :1 Solution : A N

1010 9 1910 10 10 10N

19 25 610 10 10 1M Nm kg mg

137 The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25 cm in

front of it to 50cm, the magnification of its image changes from 25m to 50m . The ratio 25

50

mm

is

Answer :6 Solution : In first case 25; ?; ?u v m

1 1 1v u f

Solving 100v and 25 4m

In second case

50u

50 2 / 3m

Thus, 25

50

6mm

138 Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures 1T and 2T , respectively. The maximum intensity in the emission spectrum of A is at 500nm and in that of B is at 1500nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B? Answer :9 Solution : By Wien’s, displacement law,

max .T constant.

max max 2 2. .T T

Thus, 1 2(500 ) (1500)( )T T

Hence, 1 23T T

Also 2 2 41(4 )(6 10 ) ( )AE T

2 2 42(4 )(18 10 ) ( )BE T

Thus, 2

41 3 93

A

B

EE

139 A piece of ice of specific (heat capacity = 01 12100Jkg C and latent heat = 5 13.36 10 /J kg ) of mass

‘m’ grams is at 05 C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice water mixture is in equilibrium, it is found that 1gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is Answer :8 Solution : Heat lost= Heat gained

Mass=m grams, Thus, we have,

5 3[ (2100)(5) 1 (336 10 )] 10 420m

Solving, we get m = 8 gram

140 Monochromatic lights of wavelength 400 nm and 560 nm are incident simultaneously and normally on a double-slit apparatus whose slit separation is 0.1 mm and screen distance is 1m. Distance between areas of total darkness will be 4z mm. Find the value of z. Answer: 7 Solution : At the area of total darkness, minima will occur for both the wavelengths.

1 2(2 1) (2 1)

2 2n m

1 2(2 1) (2 1)n m

Or (2 1) 560 7(2 1) 400 5

nm

Or 10n = 14m + 2

By inspection: For m = 2, n = 3. For m = 7, n= 10. The distance between them will be the distance between such points, i.e.,

1 2 1(2 1) (2 1)2

D n nsd

Putting 2 10n and 1 3n and on solving, we get 28 .s mm

141 In Young’s double-slit experiment, the two slits act as coherent sources of equal amplitude A and of

wavelength . In another experiment with the same setup, the two slits are sources of equal amplitude

A and wavelength , but are incoherent. The ratio of intensity of light at the mid-point of the screen in

the first case to that in the second case is

Answer :2

Solution : In the first case, 00 0 0cos 0 4I I I or I I

In the second case, 00 0 0

0

4 2' 2' 2 1

III I I II I

142 The wave front of a light beam is given by the equation 2 3x y z c (where c is arbitrary constant),

then the angle made by the direction of motion of light with the y axis is 1cos14

find .

Answer :2 Solution : Here, direction of light is given by normal vector ˆˆ2 3n t j k

Angle made by the with y-axis is given by 2 2 2

2 2cos141 2 3

143 A screen is at a distance D = 80 cm from a diaphragam having two narrow slits 1S and 2S which are d

= 2 mm apart. Slit 1S is covered by a transparent sheet of thickness 1 2.5t m and 2S by another

sheet of thickness 2 1.25t m as shown in the figure. Both sheets are made of same material having refractive index 1.40 . Water is filled in space between diaphragm and screen. A monochromatic

light beam of wavelength 0

5000 A is incident normally on the diaphragm. Assuming intensity of beam to be uniform, calculate ratio of intensity of centre of screen ‘C’ to intensity of individual slit. ( 4 / 3)w .

1S

2S

1t

2t

C

Water

80 cm

Answer :3 Solution : Path difference at C,

1 2( 1) ( 1)x t t 1 2 1 2( ) ( )t t t t

1 214 3( )( 1) (2.5 1.25) 14 10

t t

2 2.5 11.2540 40 16

x m

610

2 2 4 1. . 105000 3 10 16 3W

x

I at C, 0 02 1 cos 33cI I I

Required ratio 0

3CII

144 A monochromatic beam of light of wavelength 0

5000 A is used in Young’s double slit experiment. If one of the slits is covered by a transparent sheet of thickness 51.4 10 m , having refractive index of its medium 1.25, then the number of fringes shifted is Answer :7 Solution : Number of fringes is

( 1) / ( 1) 7/D d tt

D d

145 Assuming that about 200MeV of energy is released per fission of 23592U nuclei, the mass of 235U

consumed per day in a fission reactor of power 1 megawatt will be approximately in grams is Answer :1 Solution : Power P of fission reactor,

6 6 110 10P W Js

Time = 21 24 36 10t day s

Energy produced, 6 2 810 24 36 10 24 36 10U Pt or U J

Energy released per fission of 235U

12200 32 10MeV J

Number of 235U atoms used

820

12

24 36 10 27 1032 10

mass of 236 10 atoms of 235 235U g

mass of 2027 10 atoms of 235 2023

235 27 10 1.058 16 10

U g g

146 There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially, both have equal number of nuclei. After n half-lives of A, rates of disintegration of both are equal. The minimum value of n is

Answer: 1 Solution: Let 2A Band . Half life of A is 2A BT T

Initially rate of disintegration of A is 0N and that of B is 02 N . After one half-life of A, rate of

disintegration of A will become 0

2N . (Half-life of B = one – half the half-life of A). So, after one

half-life of A or two half – lives of B B

dN dNdt dt

1n

147 The ratio of molecular mass of two radioactive substances is 32

and the ratio of their decay constants is

43

. Then, the ratio of their initial activity per mole will be3K . Find K

Answer :4 Solution : Activity , R N . Number of nuclei (N) per mole are equal for both the substances.

R

or 1 1

2 2

43

RR

148 At any instant, the ratio of the amounts of two radioactive substances is 2 :1. If their half-lives be respectively 12h and16h , then after two days, what will be the ratio of the substances? Answer :1

Solution : Let 1

2

2( )1

M mass ratioM

2 2 24 48days h h

For first substance, 4 half-life periods and for second substance 3 half-life periods are passed; the masses are reduced to

4

1 11'2

M M

3

2 21'2

M M

'1 1'2 2

1 2 1 12 1 2 1

M MM M

149 A steel rod of length l m is heated from 025 C to 075 C keeping its length constant. The longitudinal

strain developed in the rod is (Given: Coefficient of linear expansion of steel 6 012 10 / C ) 410n then n is Answer :6 Solution : Strain developed :

6 4(12 10 )(50) 6 10T

Strain will be negative, as the rod is in a compressed state.

150 Two light waves are given by, 01 2 sin(100 30 )E t kx and 0

2 3 cos(200 ' 60 )E t k x . The

ratio intensity of first wave to that of second wave is 4k

then k is

Answer :9

Solution : 2I A 2

12

2

2 4 / 93

II

151 In a hydrogen atom, the electron is in thn excited state. It comes down to first excited state by emitting 10 different wavelengths. The value of n is. Answer :6

Solution : ( 1)( 2)102

n n , 6n

152 The wavelengths of K X rays of two metal ‘A’ and ‘B’ are 41875R

and 1675R

respectively, where

‘R’ is rydebrg constant. The number of elements lying between ‘A’ and ‘B’ according to their atomic numbers is Answer :4

Solution : Using 22 22 1

1 1 1( 1)R zn n

For particle ; 1 22, 1n n

For metal A; 21

1875 3( 1)4 4

R R Z

1 26Z

For metal B; 675R 22

3( 1)4

R Z

2 31z

Therefore, 4 elements lie between A and B.

153 An electron in a hydrogen atom makes a transition from first excited state to ground state. The equivalent current due to circulating electron increases k times then k is Answer :8

Solution : qiT

Now 2 3 63

1T r n in

154 In the figure shown if a parallel beam of white light is incident on the plane of the slits then the

distance of the nearest white spot on the screen from O is: [assumed ,d D d ] dm

than ‘m’ is

D

Od 2d/3

Answer :6 Solution : The nearest white spot will be at P, the central maxima.

Od/2 P2d/3

y

23 2 6d d dy

155 An interference is observed due to two coherent sources 1S placed at origin and 2S placed at (0,3 ,0) . Here is the wavelength of the sources. A detector D is moved along the positive x-axis. Find the integral value of the x-coordinate (excluding x = 0 and x ) where maximum intensity is observed. Answer :4 Solution : At x = 0, path difference is 3 . Hence, third order maxima will be obtained. At x , path difference is zero. Hence, zero order maxima is obtained. In between, first and second order maximas will be obtained. First order maxima :

2 1S P S P or 2 29x x or 2 29x x . Squaring both sides, we get 2 2 2 2 2x p x x . Solving this, we get 4x

Second order maxima : 2 1 2S P S P

Or 2 29 2x x or 2 29 ( 2 )x x

Squaring both sides, we get 2 2 2 29 4 4x x x

Solving, we get 5 1.254

x

Hence, the desired x coordinates are,

1.25x and 4x .

156 Two kilograms of ice at 020 C is mixed with 5kg of water at 020 C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that specific heats of water and ice are 01 / /kcal kg C and 00.5 / /kcal kg C while the latent heat of fusion of ice is 80 /kcal kg Answer :6 Solution : Heat required to convert 5 kg of water at 020 C to 5 kg of water at 00 C

5 1 20 100umC T kcal

Heat released by 2kg ice at 020 C to convert 2 kg of ice at 00 C is

2 0.5 20 20icemC T kcal

Amount of ice that will convert into water at 00 C for giving another 80 Kcal of heat can be found as follows;

80 80 1Q mL m m kg

Therefore, the amount of water at 00 C is

5kg+1kg=6kg

Thus, at equilibrium we have, 6 kg water at 00 C +1kg ice at 00 C

157 A metal sphere of radius ' 'a is surrounded by a concentric metal spherical shell of radius b ( )b a . The space between the spheres is filled with material whose electrical conductivity varies with the electric field strength E according to the rotation KE , where K is a constant. A potential difference V is maintained between inner sphere and outer shell. If the current (in ampere) between the inner

sphere and outer shell is x, find 4x

.

[Here 2 1 11 10 , 20

4K V V volt

, b = ea ( e = exponential)

Answer :4

Solution : For the element,

2 2

2 2

.41 1. .4 4

dV Edxi i KE xdx dxx KE x

.4 4

b V

a o

i dv i dxEx x dVK dx K x

x

a

v

b

dx

22

2

14 10 4004 16ln

ln

KVi Aeb

a

158 Two identical parallel plate capacitors A and B are connected in series through a battery of potential difference V. Area of each plate is ' 'a and initially plates of capacitors are separated by a distance ' 'd. Now separation between plates of capacitor B starts increasing at constant rate v. The rate by which work is done on the battery when separation between

A B

+ -V

plates of capacitor B is 2d is

2

2oa vV

nd

. Find the value of n. Answer :9 Solution : Let at any instant separation between plates of capacitor be x , then

, ( )( ) ( )e

a aa xd ad xC a a xd d x d x

d x

and ( )e

aQ C V V

d x

2 2.( ) ( )

oa V a VdQ dx

dt d x dt d x

Rate of work done on the battery

2

29oa VdQ V

dt d

9n

159 For the circuit shown in the figure, find the peak current (in ampere) through the source

~

L = 0.1 H

E = 50 sin(100 t)

1 10R

2 10R 310C F

Answer :5

Solution : Peak current through 1 1

50 5,10 2 2

R i A

Peak current through 2 2

50 5,10 2 2

R i A

1 2 L CR R X X

place difference between 1 2i and i is 2

peak current through the source is ,

2 21 2

25 25 52 2

i i i A

160 A particle of mass m oscillates about C between A,B inside a smooth spherical shell of radius R and center O.

R

LX

CX

4545

At any instant kinetic energy of the particle is K. If the magnitude of force applied by particle on the shell at

this instant is nKR the value of n is ______

A BO

C

Answer :3

Solution :

O

R

R21K mv

2

0K

R

AB

(1 cos )mgR K mgR

2

cosmvN mgR

cosmgR K 2 2cosK K KmgR R R

cos KmgR

3KNR

quick revision test single correct choice type …

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