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Question Bankfor

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Question Bank Physics for Re-AIPMT

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Basic Maths, Vectors1. Three coplanar vectors

r

A , r

B and r

C havemagnitudes 4, 3 and 2 respectively. If the anglebetween any two vectors is 120° then which of the

following vector may be equal to + +r r r

3A B C4 3 2

B60°

60°AC

(1) (2) (3) (4)

2. I f ˆˆ ˆ2 3A i j k= + +r

, ˆˆ ˆ 4B i j k= - + +r

and ˆˆ ˆ3 3 12C i j k= - -r

, then find the angle

between the vectors ( )A B C+ +r rr

and

( )A B´r r

in degrees.

(1) 30 (2) 60 (3) 90 (4) 03. For shown situation, what will be the magnitude of

minimum force in newton that can be applied in anydirection so that the resultant force is along eastdirection?

37°

5N3N

4NEastWest

South

North

(1) 3 N (2) 6 N (3) 9 N (4) 12 N4. Two balls are rolling on a flat smooth table. One

ball has velocity components ˆ3j and i while

the other has components ˆ2i and ˆ2 j . If both

start moving simultaneously from the same point,the angle between their paths is –(1) 15o (2) 30o

(3) 45o (4) 60o

5. Let ar

, br

, cr

are three unit vectors such that

ar

+ br

+ cr

is also a unit vector. If pairwise angles

between ar

, br

, cr

are q1, q2 and q3 respectively

then cosq1 + cosq2 + cosq3 equals(1) 3 (2) – 3 (3) 1 (4) – 1

6. The vecto r ( )a 3b+rr i s perpendicu lar to

( )7a 5b-rr

and ( )a 4b-rr

is perpendi cular

to ( )7a 2b-rr . The angle between a

r and b

r is :

(1) 30° (2) 45°

(3) 60° (4) None of these

7. The maximum value of function y = 4sinq – 3cosqis -

(1) 4 (2) 1 (3) 7 (4) 5

8. The radius of a circular plate increases at therate of 0.1 mm per second. At what rate doesthe area and perimeter increases when the radiusof plate is r = 1 m ?

(1) 0.2 p mm/s2, 0.2p mm/s

(2) 0.2 p mm/s, 0.2p mm/s2

(3) 0.4 p mm/s2, 0.2p mm/s

(4) 0.4 p mm/s2, 0.4p mm/s

Unit, Dimension & Measurement1. If A and B are two physical quantities having

different dimensions then which of the followingcannot denote a new physical quantity?

(1) +3A

AB

(2) expæ ö-ç ÷è ø

AB

(3) AB2 (4) 4

AB

2. Force on a particle in one-dimensional motion is

given by F = Av + Bt ++

CxAt D

, where F = force,

v= speed, t=time, x=position and A,B,C and Dare constants. Dimension of C will be-(1) M2L–2T0 (2) ML–1T0

(3) M2L0T–2 (4) None of these

3. Consider the equation ( ) ( )× = ×òrr r rd

F dS A F pdt

where ºr

F force , ºrs displacement , ºt time and

p momentumºr

. The dimensional formula of Awill be(1) M0L0T0 (2) ML0T0 (3) M–1L0T0 (4) M0L0T–1

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Directions : 4 to 6 –

A physical quantity X depends on another physical

quantities as ( )-b= + a2rX YFe ZWsin r where r,,

F and W represents distance, force and workrespectively & Y and Z are unknown physicalquantities and a, b are positive constants.

4. If Y represent displacement then YZ

dimF

æ öaç ÷è b ø

is

equal to

(1) M–1LT2 (2) M–1L2T–2

(3) M1L1T–2 (4) None of these

5. If Y represent velocity then dim (X) is equal to

(1) ML2T–3 (2) M–1L2T–3

(3) ML2T–2 (4) None of these

6. If Z represent frequency then choose the correctalternative

(1) The dimension of X is [ML1T–3]

(2) The dimension of Y is [M0LT–1]

(3) The dimension of b is [M0L–1T0]

(4) The dimension of a is [M0L1T0]

7. The density of a material in CGS system is 2g /cm3.In a system of units in which unit of length is 2 cmand unit of mass is 4 g, what is the numerical valueof the density of the material?

(1) 8 (2) 4 (3) 2 (4) 6

8. The van der waal’s equation of a gas is

2aTP

V

æ ö+ç ÷è ø Vc = (RT + b). Where a,b, c and R aree

constants. If the isotherm is represented by P = AVm

– BVn, where A and B depends on temperature :

(1) m = – c, n = – 1 (2) m = + c, n = 1(3) m = – c, n = +1 (4) m = c, n = –1

9. The dimensional formula of a physical quantity Xis [M

–1L

3T

–2]. The errors in measuring the quantities

M, L and T are 1%, 2% and 3% respectively. Themaximum percentage error in measurement of thequantity X is -

(1) 0% (2) 4% (3) 7% (4) 9%

Kinematics1. A particle is thrown vertically upwards from the

surface of the earth. Let TP be the time taken bythe particle to travel from a point P above the earthto its highest point and back to the point P. Similarly,let TQ be the time taken by the particle to travel fromanother point Q above the earth to its highest pointand back to the same point Q. If the distance betweenthe points P and Q is H, find the expression foracceleration due to gravity in terms of TP, TQ and H.

(1) 2 2p q

4H

T T-(2) 2 2

p q

8H

T T-

(3) 2 2p q

H

T T-(4) None

2. Two projectiles are projected at angles (q) and

pq

2-F

HGIKJ to the horizontal respectively with same

speed 20 m/sec. One of them rises 10 m heigherthan the other. Find the angles of projection.(Take g=10 m/s2)

(1) 45°, 45° (2) 30°, 60°

(3) 15°, 75° (4) All

3. A driver takes 0.20 s to apply the brakes after hesees a need for it. This is called the reaction time ofthe driver. If he is driving a car at a speed of 54 km/hand the brakes cause a deceleration of 6.0m/s2, findthe distance travelled by the car after he sees theneed to put the brakes on ?

(1) 20 m (2) 16.75 m

(3) 18.25 m (4) 21.75 m

4. In the figure shown, the two projectile are firedsimultaneously. Find the minimum distance betweenthem during their flight.

300600

20

20mA B

20ms-1

3 ms-1

(1) 20 m (2) 15 m

(3) 10 m (4) 5 m

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5. A ball is projected as shown in figure. The ball willreturn to point :

y(vertical)

x (horizontal)

gcotqwind

Oq

gravityu

(1) O (2) left to point O(3) right to point O (4) none of these

6. Two stones A and B are projected simultaneously asshown in figure. It has been observed that both thestones reach the ground at the same place after 7 secof their projection. Determine difference in their verticalcomponents of initial velocities in m/s. (g = 9.8 m/s2)

u1

B

q1

A

49m

u2

q 2

(1) 6 m/s (2) 7 m/s (3) 8 m/s (4) 9 m/s7. A bird moves from point (1, - 2, 3) to (4, 2, 3).

If the speed of the bird is 10 m/s, then thevelocity vector of the bird is :-

(1) 5 ( )ˆˆ ˆi 2 j 3k- + (2) 5 ( )ˆˆ ˆ4 i 2 j 3k+ +

(3) ˆ ˆ0.6 i 0.8 j+ (4) ˆ ˆ6 i 8 j+

8. The velocity of a particle moving along x–axis isgiven as v = x2 – 5x + 4 (in m/s) where x denotesthe x–coordinate of the particle in metres. Find themagnitude of acceleration of the particle when thevelocity of particle is zero ?

(1) 0 m/s2 (2) 2 m/s2

(3) 3 m/s2 (4) None of these

9. A, B, C and D are points in a vertical line such thatAB = BC = CD. If a body falls from rest from A,then the times of descend through AB, BC andCD are in the ratio :-

(1) 1 : 2 : 3

(2) 2 : 3 : 1

(3) 3 : 1 : 2

(4) 1 : ( 2 – 1) : ( 3 – 2 )

10. A particle is projected vertically upwards andit reaches the maximum height H in T seconds.The height of the particle at any time t will be :-

(1) H - g(t - T)2 (2) g(t - T)2

(3) H - 12

g(t - T)2 (4) g2

(t - T)2

11. A parachutist drops freely from an aeroplane for10s before the parachute opens out. Then hedescends with a net retardation of 2.5 m/s2. If hebails out of the plane at a height of 2495 m andg = 10 m/s2, his velocity on reaching the groundwill be:-

(1) 5 m/s (2) 10 m/s

(3) 15 m/s (4) 20 m/s

12. Initially car A is 10.5 m ahead of car B. Both start

moving at time t=0 in the same direction along

a straight line. The velocity time graph of two cars

is shown in figure. The time when the car B will

catch the car A, will be :-

v

t45°

car A

car B

10 m/s

(1) 21 sec (2) 2 5 sec

(3) 20 sec (4) None of these

13. The acceleration–time graph of a particle moving

along a straight line is as shown in figure. At what

time the particle acquires its initial velocity?

a(m/s )2

t(s)4

10

(1) 12 sec (2) 5 sec (3) 8 sec (4) 16 sec

14. A boat moving towards east with velocity 4 m/swith respect to still water and river is flowing towardsnorth with velocity 2 m/s and the wind is blowingtowards north with velocity 6 m/s. The directionof the flag blown over by the wind hoisted on theboat is :-

(1) North–west (2) South–east

(3) tan–1(1/2) with east (4) North

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15. Two trains, which are moving along different tracks

in opposite directions, are put on the same track

due to a mistake. Their drivers, on noticing the

mistake, start slowing down the trains when the

trains are 300 m apart. Given graphs show

their velocities as function of time as the trains

slow down. The separation between the trains

when both have stopped,, is :-

20

40

Train I 10t(s)

v(m/s)

Train II

8t(s)

v(m/s)

–20

(1) 120 m (2) 280 m (3) 60 m (4) 20 m

16. Two boats A and B are moving along perpendicular

paths in a still lake at night. Boat A move with a

speed of 3 m/s and boat B moves with a speed of

4 m/s in the direction such that they collide after

sometime. At t = 0, the boats are 300 m apart.

The ratio of distance travelled by boat A to the

distance travelled by boat B at the instant of collision

is :-

(1) 1 (2) 1/2 (3) 3/4 (4) 4/3

17. Pick the correct statements :-

(1) Average speed of a particle in a given time is

never less than the magnitude of the average

velocity.

(2) It is possible to have a situation in which

d udt

®¹ 0 but d

|u|dt

® = 0.

(3) The average velocity of a particle is zero in a time

interval. It is possible that the instantaneous velocity

is never zero in the interval.

(4) All of these

NLM & Friction1. An astronaut accidentally gets separated out of his

small spaceship accelerating in inter-stellar spaceat a constant acceleration of 10 m/s2. What is theacceleration of the astronaut at the instant he isoutside the spaceship?(1) 10 m/s2 (2) 9.8 m/s2

(3) » 0 m/s2 (4) could be anything2. Five situations are given in the figure (All surfaces

are smooth)

I ® F FA B

m 2m

II ® F F

ABm2m

III ® F 2FA B

m 2m

IV ® 2F F

A Bm2m

V ® F FA B

m 2m

Column–I Column–II(A) Accelerations of A & (P) I

B are same

(2) Accelerations of A & (Q) II

B are different

(3) Normal reaction (R) III

between A & B is

zero

(4) Normal reaction (S) IV

between A & B is

non zero

(T) V

(1) A®P, R, S, T; B®Q ; C®Q, R, S; D®P, T

(2) A®P, Q, S, T; B®Q ; C®Q, R, S; D®P, T

(3) A®Q, R, S, T; B®Q ; C®Q, R, S; D®P, T

(4) A®P, Q, R, S, T; B®Q ; C®Q, R, S; D®P, T

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3. A block of unknown mass is at rest on a roughhorizontal surface. A force F is applied to the block.The graph in the figure shows the acceleration ofthe block w.r.t. the applied force.

3

2

1

0

-2

2 4 6 8 10Applied forceF(N)

acceleration(m/s )2

The mass of the block and coefficient of friction are(g = 10 m/s2)

(1) 2 kg, 0.1 (2) 2 kg, 0.2

(3) 1 kg, 0.1 (4) can't be determined

4. A force F pushes a block weighing 10 kg against avertical wall as shown in the figure. The coefficient offriction between the block and wall is 0.5. Theminimum value of F to start the upward motion ofblock is [ g = 10 m/s2]

10kg

m=0.5

37°F

(1) 100 N (2) 500 N

(3) 500

3N (4) can't be determined

5. Two blocks A and B of masses m & 2m respectivelyare held at rest such that the spring is in naturallength. What is the acceleration of both the blocksjust after release?

A Bm 2m

(1) g ¯, g ¯ (2) g3

¯, g3

(3) 0, 0 (4) g ¯ , 0

6. A block is placed on an inclined plane movingtowards right horizontally with an accelerationa0 = g. The length of the plane AC = 1m. Frictionis absent everywhere. The time taken by the blockto reach from C to A is ( g = 10 m/s2)

A

30°B C

a =g0

(1) 1.2 s (2) 0.74 s (3) 2.56 s (4) 0.42 s7. A block is placed on a rough horizontal plane. A

time dependent horizontal force F = kt acts on theblock. Here k is a positive constant. Acceleration–time graph of the block is

(1)

a

t

(2)

a

t

(3)

a

t

(4)

a

t

8. A car is going at a speed of 6 m/s when it encountersa 15 m slope of angle 300. The friction coefficientbetween the road and tyre is 0.5. The driver appliesthe brakes. The minimum speed of car with whichit can reach the bottom is ( g= 10m/s2)

30°

(1) 4 m/s (2) 3 m/s(3) 7.49 m/s (4) 8.45 m/s

9. In the figure shown a ring of mass M and a blockof mass m are in equilibrium. The string is lightand pulley P does not offer any friction andcoefficient of friction between pole and M is µ. Thefrictional force offered by the pole on M is

m

PM

(1) Mg directed up(2) µ mg directed up(3) (M – m) g directed down(4) µ mg direction down

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10. If you want to pile up sand onto a circular areaof radius R. The greatest height of the sand pilethat can be erected without spilling the sand ontothe surrounding area, if µ is the coefficient offriction between sand particle is :-

(1) R (2) µ2R (3) µR (4) Rm

11. A sphere of mass m is kept in equilibrium with thehelp of several springs as shown in the figure.Measurement shows that one of the springs applies

a force Fr

on the sphere. With what acceleration

the sphere will move immediately after thisparticular spring is cut?

(1) zero

(2) F mr

(3) F m-r

(4) insufficient information12. Two forces are simultaneously applied on an object.

What third force would make the net force to pointto the left (–x direction)?

(1)

(2)

(3)

(4)

13. A light string fixed at one end to a clamp on groundpasses over a fixed pulley and hangs at the otherside. It makes an angle of 30° with the ground.A monkey of mass 5 kg climbs up the rope. Theclamp can tolerate a vertical force of 40 N only.The maximum acceleration in upward direction withwhich the monkey can climb safely is (neglectfriction and take g = 10 m/s2) :

300

a

(1) 2 m/s2 (2) 4 m/s2 (3) 6 m/s2 (4) 8 m/s2

14. A block A is placed over a long rough plank B ofsame mass as shown in figure. The plank is placedover a smooth horizontal surface. At time t=0,block A is given a velocity v0 in horizontal direction.Let v1 and v2 be the velocities of A and B at timet. Then choose the correct graph between v1 orv2 and t.

A v0

B

(1)

t

v or v1 2

v2

v1

(2)

tt

v or v1 2

v2

v1

(3)

t

v or v1 2

v1

v2

(4)

t

v or v1 2

v1

v2

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15. In figure shown, both blocks are released from rest.The time to cross each other is

2m

4 kg4m

1 kg

(1) 2 second (2) 3 second(3) 1 second (4) 4 second

16. A block A of mass m is placed over a plank B ofmass 2m. Plank B is placed over a smoothhorizontal surface. The coefficient of frictionbetween A and B is 0.5. Block A is given a velocityv0 towards right. Acceleration of B relative to A is

AB

v0

smooth

(1) g2

(2) g (3) 3g4

(4) zero

Work, Energy & Power1. Consider a roller coaster with a circular loop. A

roller coaster car starts from rest from the top of ahill which is 5 m higher than the top of the loop. Itrolls down the hill and through the loop. What mustthe radius of the loop be so that the passengers ofthe car will feel at highest point, as if they have theirnormal weight?

5m

(1) 5 m (2) 10 m

(3) 15 m (4) 20 m

2. A particle is projected along the inner surface of asmooth vertical circle of radius R, its velocity at the

lowest point being 1

95Rg5

. It will leave the circle

at an angular displacement.... from the highest point

(1) 37° (2) 53°

(3) 60° (4) 30°

3. A body of mass m is slowly halved up the rough hillby a force F at which each point is directed along atangent to the hill.

h

x

F

Work done by the force depends on(1) depends upon x.(2) depends upon h.(3) depends upon coefficient of friction (m)(4) All

Direction #4 to 5

A particle of mass m = 1 kg is moving along y-axisand a single conservative force F(y) acts on it. Thepotent ial energy of particle is given byU(y) = (y2–6y+14) J where y is in meters. At y = 3mthe particle has kinetic energy of 15 J.

4. The total mechanical energy of the particle is

(1) 15 J (2) 5 J

(3) 20 J (4) can't be determined

5. The maximum speed of the particle is

(1) 5 m/s (2) 30 m/s

(3) 40 m/s (4) 10 m/s

6. A person A of 50 kg rests on a swing oflength 1m making an angle 37O with the vertical.Another person B pushes him to swing on otherside at 53O with vertical. The work done byperson B is : [ g = 10 m/s2 ](1) 50 J (2) 9.8 J (3) 100 J (4) 10 J

7. The work done by the frictional force on a pencilin drawing a complete circle of radius r = 1/pmetre on the surface by a pencil of negligiblemass with a normal pressing force N = 5 newton(µ = 0.5) is :(1) + 4J (2) –3 J (3) – 2 J (4) – 5J

8. Work done in time t on a body of mass m which isaccelerated from rest to a speed v in time t1 as afunction of time t is given by :

(1) 2

1

1 vm t

2 t (2) 2

1

vm t

t

(3)

2

2

1

1 mvt t

2 t

æ öç ÷è ø

(4) 2

221

1 vm t

2 t

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9. Velocity–time graph of a particle of mass 2 kg

moving in a straight line is as shown in figure.

Work done by all the forces on the particle is :

2

20

v (m/s)

t (s)

(1) 400 J (2) –400 J

(3) –200 J (4) 200 J

10. A particle of mass m is moving in a circular pathof constant radius r such that its centripetalacceleration aC is varying with time t as aC = k2rt2,where k is a constant. The power delivered to theparticle by the force acting on it is :

(1) 2pmk2r2 (2) mk2r2t

(3) 4 2 5(mk r t )3

(4) zero

11. A weight is hung freely from the end of a spring.A boy then slowly pushes the weight upwardsuntil the spring becomes slack. The gain ingravitational potential energy of the weight duringthis process is equal to :

(1) The work done by the boy against thegravitational force acting on the weight.

(2) The loss of the stored energy by the springminus the work done by the tension in thespring.

(3) The work done on the weight by the boyplus the stored energy lost by the spring.

(4) The work done on the weight by the boyminus the workdone by the tension in thespring plus the stored energy lost by the spring.

12. In a simple pendulum, the breaking strength ofthe string is double the weight of the bob. The bobis released from rest when the string is horizontal.The string breaks when it makes an angle q withthe vertical–

(1) 1 1cos

3- æ öq = ç ÷è ø (2) q = 60°

(3) 1 2cos

3- æ öq = ç ÷è ø

(4) q = 0

13. A bob hangs from a rig id support by aninextensible string of length l. If it is displacedthrough a distance l (from the lowest position)keeping the string straight & released, the speedof the bob at the lowest position is :

(1) gl (2) 3gl

(3) 2gl (4) 5gl

14. A cube of mass M starts at rest from point 1 ata height 4R, where R is the radius of thecircular track. The cube s lides down thefrictionless track and around the loop. The forcewhich the track exerts on the cube at point 2 is:

4R

1

2

R

(1) 3 mg

(2) mg

(3) 2 mg

(4) cube will not reach the point 2

15. Figure shows the roller coaster track. Each car willstart from rest at point A and will roll with negligiblefriction. It is important that there should be at leastsome small positive normal force exerted by thetrack on the car at all points, otherwise the carwould leave the track. With the above fact, theminimum safe value for the radius of curvature atpoint B is (g = 10 m/s2) :

///

/

///

////

//

/////////

/

//////

/

25m

A

B 15m

(1) 20 m (2) 10 m

(3) 40 m (4) 25 m

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16. A particle 'A' of mass 107

kg is moving in the

positive x–direction. Its initial position is x = 0 andinitial velocity is 1 m/s. The velocity at x = 10mis : (use the graph given)

4

2

10

Power (in watts)

(in m)x

(1) 4 m/s (2) 2 m/s

(3) 3 2m/s (4) 100/3 m/s

17. A particle is projected vertically upwards with a speedof 16 m/s, after some time, when it again passesthrough the point of projection, its speed is found tobe 8 m/s. It is known that the work done by air resistanceis same during upward and downward motion. Thenthe maximum height attained by the particle is :(Take g = 10 m/s2)

(1) 8 m (2) 4.8 m

(3) 17.6 m (4) 12.8 m

18. A force ( )ˆ ˆF 3i 4j= +ur

N acts on a 2 kg movable

object that moves from an initial position

( )iˆ ˆd 3i 2j= - -

uur

m to final position ( )fˆ ˆd 5i 4 j= +

uur

in 6s. The average power delivered by the forceduring the interval is equal to :

(1) 8 watt (2) 506

watt

(3) 15 watt (4) 503

watt

19. A 1.0 kg block collides with a horizontal weightlessspring of force constant 2.75 Nm–1. The blockcompresses the spring 4.0 m from the rest position.If the coefficient of kinetic friction between theblock and horizontal surface is 0.25, the speedof the block at the instant of collision is :

(1) 0.4 ms–1 (2) 4 ms–1

(3) 0.8 ms–1 (4) 8 ms–1

Centre of mass & Collision

1. A ball of mass 2 kg dropped from a height H abovea horizontal surface rebounds to a height h after onebounce. The graph that relates H to h is shown infigure. If the ball was dropped from an initial heightof 81 m and made ten bounces, the kinetic energyof the ball immediately after the second impact withthe surface was

H(m)90

40

h(m)

O(1) 320 J (2) 480 J(3) 640 J (4) Can't be determined

2. An object is moving through air at a speed v. If thearea of the object normal to the direction of velocityis A and assuming elastic collision with the airmolecules, then the resistive force on the object isproportional to– (assume that molecules striking theobject were initially at rest)(1) 2Av (2) 2Av2

(3) 2Av1/2 (4) Can't be determined3. A small sphere of mass 1kg is moving with a

velocity (6i j)+$ $ m/s. It hits a fixed smooth wall and

rebounds with velocity (4i j)+$ $ m/s . The coefficient

of restitution between the sphere and the wall is-

(1) 32

(2) 23

(3) 9

16(4)

49

4. Two blocks A and B are joined together with acompressed spring. When the system is released,the two blocks appear to be moving with unequalspeeds in the opposite directions as shown in figure.Select correct statement :

A

10m/s

B

15m/sK=500Nm -1

(1) The centre of mass of the system will remainstationary.

(2) Mass of block A is equal to mass of block B.

(3) The centre of mass of the system will movetowards right.

(4) It is an impossible physical situation.

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5. A man of mass 80 kg stands on a plank of mass40kg. The plank is lying on a smooth horizontalfloor. Initially both are at rest. The man startswalking on the plank towards north and stops aftermoving a distance of 6 m on the plank. Then(1) The centre of mass of plank-man system remains

stationary.(2) The plank will slide to the north by a distance 4 m(3) The plank will slide to the south by a distance 2 m(4) The plank will slide to the south by a distance 12 m

6. A ball moving vertically downward with a speed of10 m/s collides with a platform. The platformmoves with a velocity of 5 m/s in downwarddirection. If e = 0.8, find the speed (in m/s) of theball just after collision.

(1) 1 m/s (2) 5 m/s (3) 4 m/s (4) 10 m/s

7. For shown situation, if collision between block A andB is perfectly elastic, then find the maximum energystored in spring in joules.

3kg 3kg 6kg2m/sCBA

smooth\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

(1) 2 J (2) 6 J (3) 8 J (4) 4 J

8. A thin rod of length 6 m is lying along the x-axis withits ends at x=0 and x = 6 m. Its linear density(mass/length) varies with x as kx4. Find the positionof centre of mass of rod in meters.

(1) 2 m (2) 4 m (3) 5 m (4) 6 m

9. A ball of mass 1 kg is projected horizontally asshown in figure. Assume that collision betweenthe ball and ground is totally inelastic. The kineticenergy of ball just after collision is -

5m

u=10ms–1

(1) 100 J (2) 50 J (3) 30 J (4) 20 J10. The velocity of centre of mass of the system as

shown in the figure

y

x’ x

y’

1kg 2m/s

3002 kg

2m/s

(1) 2 2 3 1ˆ ˆi j

3 3

æ ö--ç ÷è ø (2)

2 2 3 2ˆ ˆi j3 3

æ ö+-ç ÷è ø

(3) ˆ4i (4) None of these

11. A ball of mass 1 kg drops vertically on to the floorwith a speed of 25 m/s. It rebounds with an initialvelocity of 10 m/s. What impulse acts on the ballduring contact?

(1) 35kg m/s downwards

(2) 35 kg m/s upwards

(3) 30 kg m/s downwards

(4) 30kg m/s upwards

12. A particle of mass 4m which is at rest explodesinto masses m, m & 2m. Two of the fragmentsof masses m and 2m are found to move with equalspeeds v each in opposite directions. The totalmechanical energy released in the process ofexplosion is

(1) mv2 (2) 2mv2

(3) 1/2 mv2 (4) 4mv2

13. A cannon of mass 5m (including a shell of massm) is at rest on a smooth horizontal ground, firesthe shell with its barrel at an angle q with thehorizontal at a velocity u relative to cannon. Findthe horizontal distance of the point where shellstrikes the ground from the initial position of thecannon:

(1) 24u sin25g

q(2)

2u sin25g

q

(3) 23u sin25g

q(4)

28u sin25g

q

14. A particle moving horizontally collides with a fixedplane inclined at 60o to the horizontal. If it bouncesvertically, the coefficient of restitution is:

(1) 1

3(2)

2

3

(3) 13

(4) None of these

15. A ball of mass 2m impinges directly on a ball ofmass m, which is at rest. If the velocity with whichthe larger ball impinges be equal to the velocityof the smaller mass after impact then the coefficientof restitution

(1) 13

(2) 34

(3) 12

(4) 25

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16. If both the blocks as shown in the given arrangementare given together a horizontal velocity towardsright. If acm be the subsequent acceleration of thecentre of mass of the system of blocks then acm

equals

1kg

2 kg

µ=0.1

µ=0.2

(1) 0 m/s2 (2) 53

m/s2

(3) 73

m/s2 (4) 2 m/s2

17. Two balls of same mass are dropped from the sameheight onto the floor. The first ball bounces upwardsfrom the floor elastically. The second ball sticksto the floor. The first applies an impulse to the floorof I1 and the second applies an impulse I2. Theimpulses obey

(1) I2= 2I1 (2) I2= 1I2

(3) I2= 4I1 (4) I2= 1I4

Circular & Rotational Motion

1. A disc starts from rest and on the application of atorque, it gains an angular acceleration givenby a = 3t – t2 . Calculate the angular velocity after 2s.

(1) 5/3 rad/s (2) 10/3 rad/s

(3) 8/3 rad/s (4) 7/3 rad/s

2. A cyclist is riding with a speed of 18 km/h. As heapproaches a circular turn on the road of radius

25 2 m, he applies brakes and reduces his speed

at the constant rate of 0.5 m/s every second.Determine the magnitude of the net accelerationof the cyclist on the circular turn will be

(1) 0.86 m/s2 (2) 0.70 m/s2

(3) 0.5 m/s2 (4) 1.41 m/s2

3. A car starts from rest with a constant tangentialacceleration a0 in a circular path of radius r. At timet0, the car skids, find the value of coefficient of friction.

(1) +2 4

0 0 02

a a t1

g r(2) +

2 40 0 0

2

2a a t1

g r

(3) +2 4

0 0 02

2a a t1

g r(4) none

4. A particle of mass m is connected from a light

inextensible string of length l such that it behaves

as a simple pendulum. Now string is pulled to point

A making an angle q1 with the vertical and it is

released from the point A then the speed of particle

will be :

q2 q1 A

(1) q - ql 2 12g (cos cos )

(2) q - ql 2 1g (cos cos )

(3) q - ql 2 14g (cos cos )

(4) None of these

5. A light rod carries three equal masses A, B and C asshown in figure. What will be velocity of B in verticalposition of rod, if it is released from horizontalposition as shown in figure ?

(1) 8g7

l(2)

4g7

l(3)

2g7

l(4)

10g7

l

6. A child's top is spun with angular accelerationa = 4t3 – 3t2 + 2t where t is in seconds and a is inradian per second-squared. At t =0, the top hasangular velocity w0 = 2 rad/s and a reference lineon it is at angular position q0 = 1 rad.

Statement I : Expression for angular velocity

( )2 3 42 t t tw = + - + rad/s

Statement II : Expression for angular position

( )2 31 2t 3t 4tq = + - + rad

(1) Only statement-I is true

(2) Only statement-II is true

(3) Both of them are true

(4) None of them are true

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7. Figure shows a uniform disk, with mass M = 2.4 kgand radius R = 20 cm, mounted on a fixedhorizontal axle. A block of mass m = 1.2 kg hangsfrom a massless cord that is wrapped around the rimof the disk. The tension in cord is

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

(1) 12 N (2) 20 N(3) 24 N (4) None of these

8. A thin uniform rod of mass m and length l is freeto rotate about its upper end. When it is at rest, itreceives an impulse J at its lowest point, normalto its length. Immediately after impact, which ofthe following statement (s) is true -(A) the angular momentum of the rod is Jl

(B) the angular velocity of the rod is 3Jml

(C) the kinetic energy of the rod is 23J

2m(D) the linear velocity of the midpoint of the rod

is 3J2m

(1) A & B (2) A, B & C(3) B, C & D (4) All

9. In the figure, the blocks have unequal masses m1

and m2 (m1 > m2). m1 has a downward accelerationa. The pulley P has a radius r, and some mass. Thestring does not slip on the pulley, which of thefollowing statement (s) is true -

P

m2

m1a

(A) The two sections of the string have unequaltensions.

(B) The two blocks have accelerations of equalmagnitude.

(C) The angular acceleration of P is ar

(D) 1 2

1 2

m ma g

m m

æ ö-< ç ÷+è ø

(1) A & B (2) A, B & C(3) B, C & D (4) All

10. Two gear wheels which are meshed together haveradii of 0.50 cm and 0.15 cm. The number ofrevolutions does the smaller turns when the largerturns through 3 revolution is(1) 5 revolution (2) 20 revolution(3) 1 revolution (4) 10 revolution

11. A rod of mass M and length L is placed in ahorizontal plane with one end hinged about the

vertical axis. A horizontal force of F=Mg2

is applied

at a distance 5L6

from the hinged end. The angular

acceleration of the rod will be :-

(1) 4g5L

(2) 54

gL

(3) 34

gL

(4) 43

gL

12. A person supports a book between finger andthumb as shown (the point of grip is assumed tobe at the corner of the book). If the book has aweight of W then the person is producing a torqueon the book of

b

a

(1) Wa2

anticlockwise

(2) Wb2

anticlockwise

(3) Wa anticlockwise(4) Wa clockwise

13. A particle starts from the point (0m, 8m) and moves

with uniform velocity of ˆ3i m/s . At t = 0, theangular velocity of the particle about the originwill be

3m/s

8m

O

y

x

(1) 8

289 rad/s (2)

38

rad/s

(3) 24

289rad/s (4)

817

rad/s

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14. A disc of mass M and radius R rolls on a horizontalsurface and then rolls up an inclined plane as shownin the figure. If the velocity of the disc is v, the heightto which the disc will rise will be :–

hv

(1) 23v

2g (2) 34

2vg

(3) vg

2

4 (4) vg

2

2

15. A particle of mass m is projected with a velocityv making an angle of 45° with the horizontal. Themagnitude of the angular momentum of theprojectile about the point of projection when theparticle is at its maximum height h is :-

(1) zero (2) 3mv

(4 2g)

(3) 3mv

2g(4) 3m 2gh

16. Two point masses of 0.3 kg and 0.7 kg are fixedat the ends of a rod of length 1.4 m and ofnegligible mass. The rod is set rotating about anaxis perpendicular to its length with a uniformangular speed. The point on the rod through whichthe axis should pass in order that the work requiredfor rotation of the rod is minimum, is located ata distance of :-

(1) 0.42 m from mass of 0.3 kg




17. A disc of mass M and radius R is rolling with angularspeed w on a horizontal

plane as shown. The magnitude of angular

momentum of the disc about the origin O is :–

O

wy

x

M

(1) w21MR

2(2) MR2w (3) w23

MR2

(4) 2MR2w

18. Four 2kg masses are connected by 14

m long spokes

to an axle as in shown figure. A force F of 24N

acts on a lever 12

m long to produce an angular

acceleration a. Determine the magnitude of a.

(1) 3 rad/s2 (2) 6 rad/s2

(3) 12 rad/s2 (4) 9 rad/s2

19. A uniform metre scale of mass m is suspended bytwo vertical string attached to its two ends as shownin figure. A body of mass m is placed on the 80cmmark. Calculate the ratio of tension is string.

(1) 12 : 7 (2) 7 : 13(3) 5 : 3 (4) 3 : 5

20. A thin uniform rod of length l and mass m isswinging freely about a horizontal axis passing

through its end. Its maximum angular speed is w.Its centre of mass rises to a maximum height of :

(1) 2 21

2 gwl

(2) 2 21

6 gwl

(3) 2 21

3 gwl

(4) 16 g

wl

21. A cubical block of side 'a' moving with velocity von a horizontal smooth plane as shown. It hits aridge at point O. The angular speed of the blockafter it hits O is :-

M vO

a

(1) 3v4a

(2) 3v2a

(3) 3

2a(4) zero

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22. One quarter section is cut from a uniform circular discof radius R. This section has a mass M. It is madeto rotate about a line perpendicular to its plane andpassing through the centre of the original disc. Itsmoment of inertia about the axis of rotation is :-

(1) 21MR

2(2) 21

MR4

(3) 21MR

8(4) 22 MR

23. A disc is rolling (without slipping) on a horizontalsurface. C is its centre & Q and P are two pointsequidistant from C. Let vP, vQ and vC be themagnitude of velocities of points P, Q & Crespectively, then

PC

Q

(1) vQ > vC > vP (2) vQ < vC < vP

(3) vQ = vP, vC = 12

vP (4) vQ < vC > vP

24. A small object of uniform density rolls up a curvedsurface with an initial velocity v. It reaches up to

a maximum height of 23v

4g with respect to the initial

position. The object is :-

v

(1) ring (2) solid sphere(3) hollow sphere (4) disc

25. A wheel of radius R rolls without slipping on theground with a uniform velocity v. The relativeacceleration of the topmost point of the wheel withrespect to the bottommost point is

A

B

(1) 2v

R(2)

22vR

(3) 2v

2R(4)

24vR

26. A tangential force F acts at the top of a thin spherical

shell of mass m and radius R. Find the acceleration

of the shell if it rolls without slipping.

(1) 3F5m

(2) 6F5m

(3) 7F5m

(4) None

27. A hollow cylinder with inner radius R, outer radius2R and mass M is rolling without slipping with speedof its centre v. Its kinetic energy is

R

2R

(1) 211Mv

16(2) 27

Mv4

(3) 213Mv

16(4) None of these

28. A uniform rod of mass M and length L is heldvertically on a smooth horizontal surface. When therod is released, choose the incorrect alternative(s)

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

L

M

(1) The centre of mass of the rod accelerates in thevertical direction.

(2) Initially, the magnitude of the normal reaction isMg.

(3) When the rod becomes just horizontal, themagnitude of the normal reaction becomes Mg/2.

(4) When the rod becomes just horizontal, themagnitude of the normal reaction becomes Mg/4.

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Gravitation1. Kepler's second law is a consequence of :

(1) conservation of kinetic energy(2) conservation of linear momentum(3) conservation of angular momentum(4) conservation of speed

2. One projectile after deviating from its path startsmoving round the earth in a cirular path of radiusequal to nine times the radius of earth R. Its timeperiod will be :

(1) R

2g

p (2) R

27 2g

´ p

(3) Rg

p (4) R

0.8 3g

´ p

3. A planet of mass m is moving in an elliptical orbitabout the sun (mass of sun = M). The maxium andminimum distances of the planet from the sun arer1 and r2 respectively. The period of revolution ofthe planet will be proportional to :

(1) 32

1r (2) 32

2r

(3) ( )3

2

1 2r r- (4) ( )3

2

1 2r r+

4. Assume that a tunnel is dug through earth fromNorth pole to south pole and that the earth is anon-rotating, uniform sphere of density r. Thegravitational force on a particle of mass m droppedinto the tunnel when it reaches a distance r fromthe centre of earth is

(1) 3

mG r4

æ örç ÷pè ø

(2) 4

mG r3pæ örç ÷

è ø

(3) 24mG r

3pæ örç ÷

è ø(4) 24

m G r3pæ örç ÷

è ø

5. Read the following statements :

S1 : An object shall weigh more at pole than atequator when weighed by using a physical balance.

S2 : It shall weigh the same at pole and equatorwhen weighed by using a physical balance.

S3 : It shall weigh the same at pole and equatorwhen weighed by using a spring balance.

S4 : It shall weigh more at the pole than at equatorwhen weighed using a spring balance.

Which of the above statements is /are correct ?

(1) S1 and S2 (2) S1 and S4

(3) S2 and S3 (4) S2 and S4

6. The radii of circular orbits of two satellites A andB of the earth, are 4R and R, respectively. If thespeed of satellite A is 3V, then the speed of satelliteB will be :-

(1) 3V/2 (2) 3V/4

(3) 6V (4) 12V

7. The figure shows a spherical hollow inside a solidsphere to radius R; the surface of the hollow passesthrough the centre of the sphere and “touches” theright side of the sphere. The mass of the spherebefore hollowing was M. With what gravitationalforce does the hollowed -out lead sphere attracta small sphere of mass m that lies at a distanced from the centre of the lead sphere, on the straightline connecting the centre of the spheres and thehollow?

d

R m

(1) 2 2

GMm 11

d R8 1

2d

é ù-ê úæ öê ú-ç ÷ê úè øë û

(2) 2 2

GMm 11

d R4 1

2d

é ù+ê úæ öê ú-ç ÷ê úè øë û

(3) 2 2

GMm 11

d R4 1

2d

é ù-ê úæ öê ú+ç ÷ê úè øë û

(4) 2 2

GMm 11

d R8 1

2d

é ù-ê úæ öê ú+ç ÷ê úè øë û

8. The gravitational field in a region is given by

ˆ ˆE (3i 4 j)= -r

N/kg. Find out the work done (in joule)

in displacing a particle by 1 m, along the line4y = 3x + 9 :

(1) 2 J (2) 0 J

(3) 4 J (4) 6 J

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9. The Earth may be regarded as a spherically shaped

uniform core of density r1 and radius R2

surrounded

by a uniform shell of thickness R2

and density r2.

Find the ratio of 1

2

rr if the value of acceleration

due to gravity is the same at surface as at depth R2

from the surface.

(1) 3/7 (2) 7/3 (3) 5/3 (4) 3/5

10. A particle of mass 1 kg is placed at a distance of 4 m fromthe centre and on the axis of a uniform ring of mass5 kg and radius 3 m. Calculate the work done to increase

the distance of the particle from 4 m to 3 3 m.

(1) 2.22 × 10–11 J (2) 1.11 × 10–11 J

(2) 3.33 × 10–11 J (4) 4.44 × 10–11 J

11. A double star system consists of two stars A andB which have time period TA and TB. Radius RA

and RB and mass MA and MB. Choose the correctoption :–

(1) If TA > TB then RA > RB

(2) If TA > TB then MA > MB

(3)

2 3

A A

B B

T RT R

æ ö æ ö=ç ÷ ç ÷è ø è ø

(4) TA = TB

12. Two bodies of masses m and 4m are placed at adistance r. The gravitational potential at a point onthe line joining them where the gravitational fieldis zero is :-

(1) –6Gm

r(2) –

9Gmr

(3) zero (4) –4Gm

r

13. The height at which the acceleration due to gravitybecomes g/9 (where g = the acceleration due togravity on the surface of the earth) in terms of R,the radius of the earth, is :-

(1) R2

(2) 2R

(3) 2R (4) R

2

14. A particle of mass 10 g is kept on the surface of

a uniform sphere of mass 100 kg and radius 10 cm.

Find the work to be done against the gravitational

force between them, to take the particle far away

from the sphere

(you may take G = 6.67 × 10–11 Nm2/kg2)

(1) 13.34 × 10–10 J (2) 3.33 × 10–10 J

(3) 6.67× 10–9 J (4) 6.67 × 10–10 J

15. A planet in a distance solar system is 10 times more

massive than the earth and its radius is 10 times

smaller. Given that the escape velocity from the

earth is 11 km s-1, the escape velocity from the

surface of the planet would be

(1) 1.1 km s-1 (2) 11 km s-1

(3) 110 km s-1 (4) 0.11 km s-1

16. Suppose the gravitational force varies inversely asthe nth power of distance. Then the time periodof a planet in circular orbit of radius R around thesun will be proportional to-

(1) n 1

2R+æ ö

ç ÷è ø (2)

n 12R-æ ö

ç ÷è ø

(3) Rn (4) n 2

2R-æ ö

ç ÷è ø

17. An object weighs 10 N at the north pole of the

Earth. In a geostationary satellite distance 7R from

the centre of the Earth (of radius R), the true weight

and the apparent weight are–

(1) 0 N, 0 N

(2) 0.2 N, 0

(3) 0.2 N, 9.8 N

(4) 0.2 N, 0.2 N

18. The rotation of the Earth having radius R about its

axis speeds upto a value such that a man at latitude

angle 600 feels weightless. The duration of the day

in such case will be

(1) R

8g

p (2) g

8R

p

(3) Rg

p (4) g

4R

p

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Properties of Matter & Fluid Mechanics1. Find the depth of lake at which density of water is

1% greater than at the surface. Given

compressibility K = 50 × 10–6 /atm.

(1) 1 km (2) 2 km

(3) 3 km (4) 4 km

2. A steel wire 1.5 m long and of radius 1 mm is

attached with a load 3 kg at one end the other end

of the wire is fixed it is whirled in a vertical circle

with a frequency 2Hz. Find the elongation of the

wire when the weight is at the lowest position–

(Y = 2 × 1011 N/m² and g = 10 m/s²)

(1) 1.77 × 10–3 m (2) 7.17 × 10–3 m

(3) 3.17 × 10–7 m (4) 1.37 × 10–7 m

3. A wire elongates by l mm when a load W is hanged

from it. If the wire goes over a pulley and two

weights W each are hung at the two ends, the

elongation of the wire will be (in mm)-

(1) l (2) 2l (3) zero (4) l/2

4. Two wires are made of the same material and have

the same volume. However wire 1 has cross-

sectional area A and wire 2 has cross-sectional area

3A. If the length of wire 1 increases by Dx on

applying force F, how much force is needed to

stretch wire 2 by the same amount ?

(1) 6F (2) 9F

(3) F (4) 4F

5. The adjacent graph shows the extension (Dl) of

a wire of length 1m suspended from the top of

a roof at one end and with a load W connected

to the other end. If the cross–sectional area of

the wire is 10–6 m2, calculate the Young's modulus

of the material of the wire :–

1

2

3

4

20 40 60 80W(N)

Dl(×10 m)–4

(1) 2 × 1011 N/m2 (2) 2 × 10–11 N/m2

(3) 3 × 1012 N/m2 (4) 2 × 1013 N/m2

6. Work done in increasing the size of a soap bubble

from a radius of 3 cm to 5cm is nearly (Surface

tension of soap solution = 0.03 Nm–1) :-

(1) 2p mJ (2) 0.4p mJ

(3) 4p mJ (4) 0.2p mJ

7. Two merucry drops (each of radius 'r') merge to form

a bigger drop. The surface energy of the bigger drop,

if T is the surface tension, is :

(1) 532 pr2T (2) 4pr2T

(3) 2pr2T (4) 8

32 pr2T

8. A thin liquid film formed between a U-shaped wire

and a light sl ider supports a weight of

1.5 × 10–2 N (see figure). The length of the slider

is 30 cm and its weight negligible. The surface

tension of the liquid film is :-

Film

w

(1) 0.025 Nm–1 (2) 0.0125 Nm–1

(3) 0.1 Nm–1 (4) 0.05 Nm–1

9. Two very wide parallel glass plates are heldvertically at a small separation d, and dipped inwater. Some water climbs up in the gap betweenthe plates. Let S be the surface tension of water,P0 = atmospheric pressure, P = pressure of waterjust below the water surface in the region betweenthe plates–

d

(1) P=P0 –2Sd

(2) P=P0 + 2Sd

(3) P=P0 –4Sd

(4) P=P0 + 4Sd

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10. A hemispherical portion of radius R is removedfrom the bottom of a cylinder of radius R. Thevolume of the remaining cylinder is V and massM. It is suspended by a string in a liquid of densityr, where it stays vertical. The upper surface of thecylinder is at a depth h below the liquid surface.The force on the bottom of the cylinder by theliquid is :–

h

r

(1) Mg (2) Mg – Vrg(3) Mg + pR2hrg (4) rg (V + pR2h)

11. A wooden block, with a coin placed on its top,floats in water as shown in figure. The distancel and h are shown there. After sometime the coinfalls into the water. Then :–

hl

coin

(1) l decreases and h increases

(2) l increases and h decreases

(3) both l and h increase

(4) both l and h decrease

12. A solid uniform ball having volume V and density

r floats at the interface of two unmixible liquids

as shown in figure. The densities of the upper and

the lower liquids are r1 andr2 respectively, such

that r r r1 2< < .What fraction of the volume of

the ball will be in the lower liquid :–

(1) 2

1 2

r - rr - r (2)

1

1 2

rr - r

(3) 1

1 2

r - rr - r (4)

1 2

2

r - rr

13. During blood transfusion the needle is inserted ina vein where the gauge pressure is 2000 Pa. Atwhat height must the blood container be placed sothat blood may just enter the vein ? [Density ofwhole blood = 1.06 × 103 kg m–3].

(1) 0.192 m (2) 0.182 m

(3) 0.172 m (4) 0.162 m

14. Water from a tap emerges vertically downwardswith an in it ia l speed of 1.0 m/s. Thecross–sectional area of tap is 10–4 m2. Assume thatthe pressure is constant throughout the stream ofwater and that the flow is steady, thecross–sectional area of stream 0.15 m below thetap is:–

(1) 5.0 × 10–4 m2

(2) 1.0 × 10–4 m2

(3) 5.0 × 10–5 m2

(4) 2.0 × 10–5 m2

15. A large open tank has two holes in the wall. Oneis a square hole of side L at a depth y from thetop and the other is a circular hole of radius Rat a depth 4y from the top. When the tank iscompletely filled with water, the quantities of waterflowing out per second from the holes are bothsame. Then, R is equal to :–

(1) L

2p(2) 2pL

(3) L (4) L2p

16. There are two identical small holes on the oppositesides of a tank containing a liquid. The tank is openat the top. The difference in height between thetwo holes is h. As the liquid comes out of the twoholes, the tank will experience a net horizontalforce proportional to–

h

(1) h (2) h

(3) h3/2 (4) h2

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17. The pressure of water in a water pipe when tap is

opened and closed is respectively 3 × 105 Nm–2 and

3.5 × 105 Nm–2. With open tap, the velocity of

water flowing is

(1) 10 m/s (2) 5 m/s

(3) 20 m/s (4) 15 m/s

18. A hole is made at the bottom of a large vessel open

at the top. If water is filled to a height h, it drains

out completely in time t. The time taken by the

water column of height 2h to drain completely is

(1) Ö2t (2) 2t (3) 2Ö2t (4) 4t

19. If the terminal speed of a sphere of gold

(density = 19.5 kg/m3) is 0.2 m/s in a viscous

liquid (density = 1.5 kg/m3), find the terminal speed

of a sphere of silver (density=10.5 kg/m3) of the

same size in the same liquid.

(1) 0.4 m/s (2) 0.133 m/s

(3) 0.1 m/s (4) 0.2 m/s

20. A spherical solid ball of volume V is made of a

material of density 1r . It is falling through a liquid

of density ( )2 2 1r r < r . Assume that the liquid applies

a viscous force on the ball that is propoertional to

the square of its speed v, i.e., Fviscous = –kv2 (k > 0).

Then terminal speed of the ball is

(1) ( )1 2Vg

k

r - r(2) 1Vg

kr

(3) 1Vgkr

(4) ( )1 2Vg

k

r - r

Thermal Physics(a) Thermal Expansion

1. The graph AB shown in figure is a plot of

temperature of a body in degree Celsius and degree

Fahrenheit. Then

32°F 212°F FahrenheitA

Cen

tigra

de

100°C B

(1) slope of line AB is 9/5(2) slope of line AB is 5/9(3) slope of line AB is 1/9(4) slope of line AB is 3/9

2. Suppose there is a hole in a copper plate. Onheating the plate, diameter of hole, would :(1) always increase(2) always decrease(3) always remain the same(4) none of these

3. If two rods of length L and 2L having coefficientsof linear expansion a and 2a respectively areconnected so that total length becomes 3L, theaverage coefficient of linear expansion of thecomposition rod equals:

(1) 32

a (2) 52

a

(3) 53

a (4) none of these

(b) Calorimetry4. A body of mass 5 kg fal ls from a height of

30 metre. If its all mechanical energy is changedinto heat, then heat produced will be:-(1) 350 cal (2) 150 cal (3) 60 cal (4) 6 cal

5. 2 kg ice at – 20°C is mixed with 5 kg water at20°C. Then final amount of water in the mixturewould be; Given specific heat of ice = 0.5cal/g°C,specific heat of water = 1 cal/g°C,Latent heat of fusion of ice = 80 cal/g.(1) 6 kg (2) 5 kg (3) 4 kg (4) 2 kg

6. 1 kg of ice at – 10°C is mixed with 4.4 kg of waterat 30°C. The final temperature of mixture is :(specific heat of ice is 2100 J/kg/k)(1) 2.3°C (2) 4.4°C (3) 5.3°C (4) 8.7°C

7. 10 gm of ice at 0°C is kept in a calorimeter ofwater equivalent 10 gm. How much heat shouldbe supplied to the apparatus to evaporate the waterthus formed? (Neglect loss of heat)(1) 6200 cal (2) 7200 cal(3) 13600 cal (4) 8200 cal

(c) Heat Transfer8. The ratio of cofficient of thermal conductivity of

two different materials is 5:3. If the thermalresistance of rods of same thickness of thesematerial is same, then what is ratio of lenght ofthese rods -(1) 3:5 (2) 5:3 (3) 25:9 (4) 9:25

9. Rate of heat flow through a cylindrical rod is Q1.Temperatures of ends of rod are T1 and T2. If all thelinear dimensions of the rod become double andtemperature difference remains same it's rate of

heat flow is Q2, then :–

(1) Q1 = 2Q2 (2) Q2 = 2Q1

(3) Q2 = 4Q1 (4) Q1 = 4Q2

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10. A spherical body of area A, and emissivity e = 0.6 iskept inside a black body. What is the rate at whichenergy is radiated per second at temperature T

(1) 0.6 s AT4 (2) 0.4 s AT4

(3) 0.8 s AT4 (4) 1.0 s AT4

11. A composite rod made of three rods of equal lengthand cross-section as shown in the fig. The thermalconductivities of the materials of the rods are K/2,5K and K respectively. The end A and end B are atconstant temperatures. All heat entering the faceA goes out of the end B there being no loss of heatfrom the sides of the bar. The effective thermalconductivity of the bar is

BA

K5KK/2

(1) 15K/16 (2) 6K/13(3) 5K/16 (4) 2K/13

12. The figure shows the face and interface temperatureof a composite slab containing of four layers of twomaterials having identical thickness. Under steady statecondition, find the value of temperature q.

(1) 5°C (2) 10°C (3) –15°C (4) 15°C

13. If a liquid takes 30 sec. in cooling of 95°C to 90°C

and 70 sec. in cooling of 55°C to 50°C then temp.

of room is -

(1) 16.5°C (2) 22.5°C

(3) 28.5°C (4) 32.5°C

14. The power radiated by a black body is P and it

radiates maximum energy around the wavelength

l0. If the temperature of the black body is now

changed so that it radiates maximum energy around

wavelength 3/4l0, the power radiated by it will

increase by a factor of

(1) 4/3 (2) 16/9

(3) 64/27 (4) 256/81

15. Two spherical bodies A(radius 6 cm) and B(radius

18 cm) are at temperature T1 and T2, respectively.

The maximum intensity in the emission spectrum

of A is at 500 nm and in that of B is at 1500 nm.

Considering them to be black bodies, what will be

the ratio of the rate of total energy radiated by A to

that of B ?

(1) 9 (2) 6 (3) 12 (4) 3

(d) KTG16. Find the approximate number of molecules

contained in a vessel of volume 7 litres at 0°C at

1.3 × 105 pascals

(1) 2.4 × 1023 (2) 3 × 1023

(3) 6 × 1023 (4) 4.8 × 1023

17. A real gas behaves like an ideal gas if its

(1) pressure and temperature are both high

(2) pressure and temperature are both low

(3) pressure is high and temperature is low

(4) pressure is low and temperature is high

18. An ideal gas follows a process PT = constant. The

correct graph between pressure & volume is

(1) (2)

(3) (4)

19. An ideal gas expands in such a way that PV2 =

constant throughout the process.

(1) The graph of the process of T-V diagram is a

parabola.

(2) The graph of the process of T-V diagram is a

straight line.

(3) Such an expansion is possible only with heating.

(4) Such an expansion is possible only with cooling

(e) Thermodynamics20. P-V plots for two gases during adiabatic processes

are shown in the figure. Plots 1 and 2 should

correspond respectively to

(1) He and O2 (2) O2 and He(3) He and Ar (4) O2 and N2

21. An ideal gas at 270C is compressed adiabatically to8/27 of its original volume. If g = 5/3, then the risein temperature is:-

(1) 450 K (2) 375 K

(3) 675 K (4) 405 K

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22. Three processes form a thermodynamic cycle as

shown on P-V diagram for an ideal gas. Process

1 ® 2 takes place at constant temperature (300K).

Process 2 ® 3 takes place at constant volume. During

this process 40J of heat leaves the system. Process

3 ® 1 is adiabatic and temperature T3 is 275K. Work

done by the gas during the process 3 ®1 is

(1) -40J (2) -20J (3) +40J (4) +20J

23. An ideal gas is taken through the cycle A ® B ® C ® A,as shown in the figure. If the net heat supplied tothe gas in the cycle is 5J, the work done by the gasin the process C ® A is

(1) –5J (2) –10 J (3) –15 J (4) –20 J

24. A shown in the figure, the amount of heat absorbedalong the path ABC is 90J and the amount of workdone by the system is 30 J. If the amount of workdone along the path ADC is 20 J the amount ofheat absorbed will be :-

B

A

C

D

P

V

(1) 80 J (2) 90 J (3) 110 J (4) 120 J

25. A gas for which g = 5/3 is heated at constant

pressure. The percentage of total heat given that

will be used for external work is :

(1) 40% (2) 30% (3) 60% (4) 20%

26. During an experiment an ideal gas obeys an addition

equation of state P2V = constant. The initial

temperature and pressure of gas are T and V

respectively. When it expands to volume 2V, then

its temperature will be :

(1) T (2) 2 T

(3) 2 T (4) 2 2 T

27. A Carnot engine takes 3 × 106 cal of heat fromreservoir at 627°C and gives it to a sink at 27°C.Then work done by the engine is

(1) 4.2 × 106 J (2) 8.4 × 106 J

(3) 16.8 × 106 J (4) zero

28. A carnot engine, having an efficiency of h = 1/10as heat engine is used as a refrigerator. If the workdone on the system is 10 J, then amount of energyabsorbed from the reservoir at lower temperatureis :-

(1) 99 J (2) 90 J

(3) 1 J (4) 100 J

29. A reversible refrigerator operates between a lowtemperature reservoir at TC and a high temperaturereservoir at TH. Its coefficient of performance isgiven by :

(1) (TH – TC)/TC (2) TC/(TH – TC)

(3) (TH – TC)/TH (4) TH/(TH – TC)

Oscillation (SHM)1. Infinite spring with force constants k, 2k, 4k, 8k, .....

respectively are connected in series. Calculate theeffective force constant of the spring.

(1) K (2) K2

(3) 2K (4) 4K

2. Frequency of oscillation of a body is 6 Hz whenforce F1 is applied and 8 Hz when F2 is applied. Ifboth forces F1 & F2 are applied together then findout the frequency of oscillation.

(1) 7Hz (2) 10 Hz

(3) 14Hz (4) 2 Hz

3. Periodic time of oscillation T1 is obtained when amass is suspended from a spring if another springis used with same mass then periodic time ofoscillation is T2. Now if this mass is suspended fromseries combination of above springs then calculatethe time period.

(1) T1 + T2 (2) +2 21 2T T

(3) +1 2T T2

(4) 1 2T T

4. If length of a simple pendulum is increased by 4%.Then determine percentage change in time period.

(1) 1% (2) 2%

(3) 3% (4) 4%

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5. A simple pendulum of length L and mass M is

suspended in a car. The car is moving on a circular

track of radius R with a uniform speed v. If the

pendulum makes oscillation in a radial direction

about its equilibrium position, then calculate its time

period .

(1) p

+4

22

L2

vg

R

(2) p

+4

22

2L2

vg

R

(3) p

+4

22

4L2

vg

R

(4) p

+4

22

8L2

vg

R

6. The amplitude of a damped oscillator becomes half

in one minute. The amplitude af ter

3 minutes will be 1x

times of the original. Determine

value of x.

(1) 4 (2) 8

(3) 6 (4) 10

7. On the superposition of two harmonic oscillationsrepresented by x1 = a sin (wt + f1) and x2 = a sin (wt + f2)

a resulting oscillation with the same time period and

amplitude is obtained. Find the value of f1 – f2.

(1) 90° (2) 60°

(3) 120° (4) 45°

8. The potential energy of a particle oscillating on x-

axis is U = 20 + (x – 2)2. Here U is in joules and x

in meters. Total mechanical energy of the particle is

36 J. Find the maximum kinetic energy of the

particle.

(1) 6 J (2) 36 J

(3) 20 J (4) 16 J

9. A bob of simple pendulum is suspended by a metalic

wire. If a is the coefficient of linear expansion and

dq is the change in temperature then calculate that

percentage change in time period.

(1) 20 a dq (2) 30 a dq

(3) 40 a dq (4) 50 a dq

10. The angle made by the string of a simple pendulum with

the vertical depends upon time as q = p

90 sin pt. Find

the length (in m) of the pendulum if g = p2 m/s2

(1) 1 (2) 2

(3) 3 (4) 4

11. Values of the acceleration &&x of a particle moving in

simple harmonic motion as a function of itsdisplacement x are given in the table below.

16 8 0 –8 –16

x (mm) –4 –2 0 2 4

( )&&2x mm/s

The period of the motion is

(1) p1

s (2) p2

s

(3) p

s2

(4) p s

12. The potential energy of a particle of mass 100 g,moving along the x-axis, is given by U = 5x(x–4) J,where x is in metre. Select correct alternative(s).

(1) The particle execute SHM with mean positionat x=4 m

(2) The particle execute SHM with mean positionat x=1 m

(3) The particle execute SHM with time periodp/5 second

(4) The particle execute SHM with time periodp/10 second

13. The time period of a particle executing SHM is T.After a time T/6 after it passes its mean position,its :

(1) velocity will be one-half of its maximum velocity

(2) kinetic energy = 1/3 (potential energy)

(3) acceleration will be 32

times of its maximum

acceleration

(4) All

14. A simple pendulum has time period 2s. The pointof suspension is now moved upward according torelation y=(6t – 3.75 t2)m where t is in second andy is the vertical displacement in upward direction.The new time period of simple pendulum will be

(1) 2s (2) 1s

(3) 4s (4) None of these

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15. A simple harmonic oscillator consists of a blockattached to a spring with k = 200 N/m. The blockslides on a frictionless horizontal surface, withequilibrium point x = 0. A graph of the block'svelocity v as a function of time t is shown (use p2 =10)

Column I Column II

(A) The block's mass (P) –0.20

in kg

(B) The block's (Q) –200

displacement at t = 0

in meters

(C) The block's acceleration (R) +0.20

at t = 0.10 s in m/s2

(D) The block's maximum (S) +4.0

kinetic energy in joule

(1) A®R, B®P, C®Q, D®S(2) A®P, B®R, C®Q, D®S(3) A®R, B®Q, C®R, D®S(4) A®Q, B®P, C®Q, D®S

16. The speed (v) of a particle moving along a straightline, when it is at a distance (x) from a fixed pointon the line, is given by : v2 = 144 – 9x2.

Column I Column II

(A) Motion is simple (P) p2

3units

harmonic of period(B) Maximum displacement (Q) 12 units

from the fixed point is(C) Maximum velocity of (R) 27 units

the particle(D) Magnitude of (S) 4 unit

acceleration at adistance 3 units fromthe fixed point is

(1) A®P, B®S, C®Q, D®R

(2) A®P, B®R, C®Q, D®S

(3) A®R, B®Q, C®R, D®S

(4) A®Q, B®P, C®Q, D®S

17. Time period of a spring mass system can bechanged by

(1) changing the mass

(2) cutting the spring (i.e. changing the length of thespring)

(3) immersing the mass in a liquid

(4) All

18. Spring of spring constant 1200 Nm–1 is mounted ona smooth frictionless surface and attached to a blockof mass 3 kg. Block is pulled 2 cm to the right andreleased. The angular frequency of oscillation is

(1) 5 rad/sec

(2) 30 rad/sec

(3) 10 rad/sec

(4) 20 rad/sec

19. A point particle of mass 0.1 kg is executing SHM of

amplitude 0.1 m. When the particle passes through

the mean position, its KE is 8 × 10–3 J. Find the

equation of motion of this particle if the initial phase

of oscillation is 45° .

(1) y = 0.1 cos (3t + p/4)

(2) y = 0.1 sin (6t + p/4)

(3) y = 0.1 sin (4t + p/4)

(4) y = 0.1 cos (4t + p/4)

Wave Motion1. A plane wave is described by the equation

y =3 cospæ ö- -ç ÷

è ø

x10t

4 2 . The maximum velocity of

the particles of the medium due to this wave is

(1) 30

(2) 32p

(3) 3/4

(4) 40

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2. Column I Column II

A Longitudinalwaves

P Particles of themedium vibrateperpendicular to thewave propagation.

B Transversewaves

Q Two progressive wavesof slightly differentfrequencies superposein the same direction

C Beats R Two progressive wavesof same frequencysuperpose in theopposite directions

D Stationarywaves

S Particles of themedium vibrate alongthe wave propagation.

(1) A-Q, B-R, C-Q, D-P(2) A-S, B-P, C-Q, D-R(3) A-Q, B-S, C-P, D-R(4) A-P, B-Q, C-S, D-R

3. The equation of a plane progressive wave is

y = 0.02 sin 8p é ù-ê úë û

xt

20 . When it is reflected at

a rarer medium (medium with higher velocity) atx = 0, its amplitude becomes 75% of its previousvalue. The equation of the reflected wave is

(1) é ù= p -ê úë û

xy 0.02sin8 t

20

(2) é ù= p +ê úë û

xy 0.02sin8 t

20

(3) é ù= + p +ê úë û

xy 0.015sin8 t

20

(4) é ù= - p +ê úë û

xy 0.015sin8 t

20

4. The tension in a stretched string fixed at both endsis changed by 2%, the fundamental frequency isfound to get changed by 15 Hz. Select the correctstatement(s)

(A) Wavelength of the string of fundamentalfrequency does not change

(B) Velocity of propagation of wave changes by 2%

(C) Velocity of propagation of wave changes by 1%

(D) Original frequency is 1500 Hz

(1) A, C & D (2) A, B & C

(3) A & B (4) Only D

5. Two waves traveling in a medium in the x–direction

are represented by y1 = A sin(at – bx) and

2y Acos x t4pæ ö= b + a -ç ÷è ø , where y1 and y2 are the

displacements of the particles of the medium, t is

time, and a and b are constants. The two waves

have different:–

(1) speeds

(2) directions of propagation

(3) wavelengths

(4) frequencies

6. Dependence of disturbances due to two waves on

time is shown in the figure. The ratio of their

intensities I1 / I2 will be:–

t

1

2

y

(1) 1 : 1 (2) 1 : 2 (3) 4 : 1 (4) 16 : 17. A uniform rope having some mass hinges vertically

from a rigid support. A transverse wave pulse isproduced at the lower end. The speed (v) of thewave pulse varies with height (h) from the lowerend as:–

(1)

V

h

(2)

V

h

(3)

V

h

(4)

V

h

8. The equation y = a sin 2p/l (vt – x) is expression

for:–

(1) Stationary wave of single frequency along x–axis.

(2) A simple harmonic motion.

(3) A progressive wave of single frequency along

x–axis.

(4) The resultant of two SHM's of slightly different

frequencies.

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9. A closed organ pipe of radius r1 and an open organpipe of radius r2 and having same length 'L'resonate when excited with a given tuning fork.Closed organ pipe resonates in its fundamentalmode where as open organ pipe resonates in itsfirst overtone, then:–

(1) r2 – r1 =L (2) r2 = r1 = L/2(3) r2 – 2r1 = 2.5 L (4) 2r2 – r1 = 2.5 L

10. Two open pipes of length 25 cm and 25.5 cmproduced 0.1 beat/second. The velocity of soundwill be:–

(1) 255 cm/s (2) 250 cm/s

(3) 350 cm/s (4) none of these

11. Two open pipes of length L are vibratedsimultaneously. If length of one of the pipes isreduced by y, then the number of beats heard persecond will be if the velocity of sound is v and y<< L:–

(1) 2

vy2L

(2) 2

vyL

(3) vy2L

(4) 22L

vy

12. Two tuning fork when sounded together produces5 beats per second. The first tuning fork is inresonance with 16.0 cm wire of a sonometer andsecond is in the resonance with 16.2 cm wire ofthe same sonometer then the frequencies of thetuning forks are:–

(1) 100 Hz, 105 Hz (2) 200 Hz, 205 Hz(3) 300 Hz, 305 Hz (4) 400 Hz, 405 Hz

13. The equation of a wave travelling along the positivex–axis, as shown in figure at t=0 is given by :–

0-0.5

-0.1

y1

x

(1) sin kx t6pæ ö

- w +ç ÷è ø

(2) sin kx t6pæ ö

- w -ç ÷è ø

(3) sin t kx6pæ ö

w - +ç ÷è ø

(4) sin t kx6pæ ö

w - -ç ÷è ø

14. A police car moving at 22 m/s chases amotocyclist. The police man sounds his horn at176 Hz, while both of them move towards astationary siren of frequency 165 Hz. Calculatethe speed of the motorcycle. If it is given that themotorcyclist does not observe any beats :–

Police car

22 m/s, 176 Hz v

Motorcycle

Stationary siren (165 Hz)

(1) 33 m/s (2) 22 m/s(3) zero (4) 11 m/s

15. A star moves away from earth at speed 0.8 c whileemitting light of frequency 6 × 1014 Hz. Whatfrequency will be observed on the earth in Hz ?(c = speed of light)(1) 0.24 × 1014 (2) 1.2 × 1014

(3) 2 × 1014 (4) 150 × 1014

16. A car has two horns having a difference in frequency

of 180 Hz. The car is approaching a stationary observer

with a speed of 60 ms–1. Calculate the difference in

frequencies of the notes as heard by the observer, if

velocity of sound in air is 330 ms–1.

(1) 210 Hz (2) 220 Hz

(3) 230 Hz (4) 240 Hz

17. Two vibrating tuning forks produce progressive

waves given by y1= 4 sin(500pt) and

y2= 2 sin(506pt). These tuning forks are held near

the ear of a person. The person will hear

(1) 3 beats/s with intensity ratio between maxima

and minima equal to 4.






and minima equal to 9.18. A 100 m long rod of density 10.0 × 104 kg/m3

and having young's modulus Y = 1011 Pa, isclamped at one end. It is hammered at the otherfree end as shown in the figure. The longitudinalpulse goes to right end, gets reflected and againreturns to the left end. How much time, the pulsetake to go back to initial point?

(1) 0.1 sec (2) 0.2 sec(3) 0.3 sec (4) 2 sec

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Electrostatics1. Consider a neutral conducting sphere. A positive

point charge is placed outside the sphere. The net

charge on the sphere is then :

(1) Negative and distributed uniformly over the

surface of the sphere.

(2) Negative and appears only at the point on the

sphere closest to the point charge

(3) Negative and distributed non-uniformly over

the entire surface of the sphere

(4) Zero

2. Four charges q1=1mC, q2=2mC, q3=3mC and

q4=4mC are placed at (0,0,0) , (1m,0,0), (0,1m,0),

(0,0,1m) respectively. Let iFr

be the net force acting

on ith charge of the given charges then S iFr

=.....

(1) 0.018 N (2) 0.02 N

(3) 0.036 N (4) zero

3. Under the influence of the Coulomb field of charge

+Q, a charge –q is moving around it in an elliptical

orbit. Find out the correct statement(s) :-

(1) The angular momentum of the charge –q is

constant

(2) The linear momentum of the charge –q is

constant

(3) The angular velocity of the charge –q is

constant

(4) The linear speed of the charge –q is constant

4. A metal sphere A of radius r1 charged to a potential

f1 is enveloped by a thin walled conducting

spherical shell B of radius r2. Then f2 of the sphere

A after it is connected by a thin conducting wire

to the shell B will be:

A

r1

B

(1) 1

1

2

rr

f (2) 2

1

1

rr

æ öf ç ÷è ø

(3)2

1

1

r1

ræ öf -ç ÷è ø

(4) 1 2

1

1 2

r rr r

æ öf ç ÷+è ø

5. A, B, C, D, P and Q are points in a uniform electricf ield. The potentia ls of these points areV (A) = 2 volt. V (P) = V (B) = V (D) = 5 volt.V (C) = 8 volt. The electric field at P is :-

(1) 10 Vm–1 along PQ

(2) 215 V m–1 along PAA

(3) 5 V m–1 along PC

(4) 5 V m–1 along PA

6. Charge Q coulombs is uniformly distributedthroughout the volume of a solid hemisphere ofradius R metres. Then the potential at centre Oof the hemisphere in volts is

(1) 0

1 3Q4 2Rpe

O

R

(2) 0

1 3Q4 4Rpe

(3) 0

1 Q4 4Rpe (4)

0

1 Q4 8Rpe

7. List I gives certain situations in which electric fieldis represented by electric lines of forces in x-y plane.List II gives corresponding representation ofequipotential lines in x-y plane. Match the figuresin List I with the figures in List II and indicate youranswer.

List - I List - II

(P) x

y

Electric linesof forces

(1)

Higher potential

Lower potential

x

y

(Q) x

y


(2)

Lower potential

Higher potential

x

y

(R) x

y


(3)

Low

erpo

tent

ial H

igherpotential

x

y

(S) x

y


(4)

Hig

herp

oten

tial Low

erpotential

x

y

Codes P Q R S(1) 1 2 3 4

(2) 4 3 2 1

(3) 3 4 2 1

(4) 2 1 3 4

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8. When the separation between two charges isincreased, the electric potential energy of the charges:

(1) Increases

(2) Decreases

(3) Remains the same

(4) May increase or decrease

9. A dipole having dipole moment p is placed in frontof a solid uncharged conducting sphere as shownin the diagram. The net potential at point A lyingon the surface of the sphere is

fp

A

R r90°

(1) f

2

kpcosr

(2) f2

2

kpcosr

(3) f2

2

2kpcosr

(4) zero

10. A big hollow metal sphere A is charged to 100

volt potential difference and another smaller hollow

sphere B is charged to 50 volt potential difference.

If B is put into A and joined with a metallic wire,

then the direction of charge flow is:

(1) From A to B

(2) From B to A

(3) No flow of charge

(4) The direction of charge flow depends on the

radii of spheres

11. An uncharged sphere of metal is placed in betweentwo charged plates as shown. The lines of forcelook like :-

(1)

+ + + + + + +

(2)

+ + + + + + +

(3)

+ + + + + + +

(4)

+ + + + + + +

12. The adjacent diagram shows a charge +Q held onan insulating support S and enclosed by a hollowspherical conductor. O represents the centre of thespherical conductor and P is a point such thatOP = x and SP = r. The electric field at point Pwill be :-

SO P

Charge +Q oninsulating support

rxSP = rOP = x

(1) pe 20

Q4 x (2) pe 2

0

Q4 r

(3) 0 (4) None of the above13. Consider a thin conducting spherical shell of radius

R with its centre at the origin, carrying uniformpositive surface charge density. The variation of themagnitude of the electric field E(r)

r and the electric

potential V(r) with the distance r from the centre,is best represented by which graph :-[Dotted line represents (V-r) curve and Bold linerepresents (E-r) curve ]

(1)

O R r

E(r)V(r)

(2)

E(r)V(r)

O R r

(3)

E(r)V(r)

O R r

(4)

E(r)V(r)

O R r14. A solid conducting sphere of radius a has a net

positive charge 2Q. A conducting spherical shellof inner radius b and outer radius c is concentricwith the solid sphere and has a net charge – Q.If the two spheres are connected by a wire for aninstant, The surface charge density on the inner andouter surfaces of the spherical shell will be

aBb

cA

(1) – 2Q/4pb2, Q/4pc2 (2) –Q/4pb2, Q/4pc2

(3) 0, Q/4pc2 (4) none of the above

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15. A dipole of dipole moment 'p' is placed innon-uniform electric field along x-axis. Electric fieldis increasing at the rate of 1 V/m2 then the forceon dipole is

(1) 0 (2) 2p (3) p/2 (4) p

16. In a region, the potential is represented byV(x, y, z) = 6x – 8xy – 8y + 6yz, where V is involts and x, y, z are in metres. The electric forceexperienced by a charge of 2 coulomb situated atpoint (1, 1, 1) is :-

(1) 6 5 N (2) 30 N

(3) 24 N (4) 4 35 N

17. The concentric conducting spheres are of radii r1and r2. The outer sphere is given a charge q. Thecharge q’ on the inner sphere will be (inner sphereis grounded):

q'r1

r2

(1) q (2) -q

(3) 1

2

rq

r- (4) Zero

Capacitor1. The net charge on capacitor is :-

(1) 2q (2) q/2

(3) 0 (4) infinity

2. The two parallel plates of a condenser have been

connected to a battery of 300 V and the charge

collected at each plate is 1 m C. The energy supplied

by battery is :

(1) 6 × 10–4J (2) 3 × 10–4J

(3) 1.5 × 10–4J (4) 4.5 × 10–4J

3. In the given figure, a capacitor of non–parallel

plates is shown. The plates of capacitor are

connected by a cell of emf V0. If s denotes surface

charge density and E denotes electric field. Then

B

FDAV0

(1) sA > sB (2) EF > ED

(3) EF =ED (4) sA = sB

4. An uncharged capacitor having capacitance C isconnected across a battery of emf V. Now thecapacitor is disconnected and then reconnectedacross the same battery but with reversed polarity.Then which of the statement is incorrect

(1) After reconnecting, heat energy produced in thecircuit will be equal to two–third of the totalenergy supplied by battery.

(2) After reconnecting, whole of the energy suppliedby the battery is converted into heat.

(3) After reconnecting, thermal energy producedin the circuit will be equal to 2CV2

(4) Final energy stored in capacitor is 12

CV2

5. Two thin conducting shells of radii R and 3R areshown in the figure. The outer shell carries a charge+Q and the inner shell is neutral. The switch S isopen then.

RS

3R

(1) Potential of inner sphere is greater than that

of outer sphere

(2) Potential of inner sphere is less than that of outer

sphere

(3) Potential of inner sphere is equal to that of outer

sphere

(4) Potential of both spheres will become zero

6. Two thin conducting shells of radii R and 3R are

shown in the figure. The outer shell carries a charge

+Q and the inner shell is neutral. The switch S is

closed then.

RS

3R

(1) Charge attained by inner sphere is –Q/2




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7. The charge q on a capacitor varies with voltageas shown in figure. The area of the triangle AOBrepresents :

B

A

V

qO

(1) electric field between the plates

(2) electric flux between the plates

(3) energy density

(4) energy stored by the capacitor

8. A parallel plate air capacitor

d

has a capacitance C. When itis half filled with a dielectric ofdielectric constant 5, thepercentage increase in thecapacitance will be :-

(1) 400% (2) 66.6%

(3) 33.3% (4) 200%

9. Figure shows two capacitors connected in seriesand joined to a battery. The graph shows thevariation in potential as one moves left to right onthe branch containing the capacitors :-

C1 C2

X

V

(1) C1 > C2

(2) C1 = C2

(3) C1 < C2

(4) The information is not sufficient to decide therelation between C1 and C2

10. Capacitance C1 = 2C2 = 2C3 and potentialdifference across C1, C2 & C3 are V1, V2 & V3respectively then :-

C1C2

C3

V+ –

(1) V1 = V2 = V3 (2) V1 = 2V2 = 2V3

(3) 2V1 = V2 = V3 (4) 2V1 = 2V2 = V3

11. In the circuit shown in figure, the battery is an idealone with emf V. The capacitor is initially uncharged.The switch S is closed at time t = 0.

R/2 A 5R/2

R

R/2

V

B

S

C

The final charge Q on the capacitor is -

(1) CV2

(2) CV3

(3) CV (4) CV6

12. Find the potential difference across the capacitorin volts.

R

R

R RR

10V

C

(1) 4V (2) 2V

(3) 1 V (4) 8V

13. A light bulb, a capacitor and a battery are connectedtogether as shown here, with switch S initially open.When the switch S is closed, which one of thefollowing is true :-

S

(1) The bulb will light up when the capacitor startscharging and its brightness decreases graduallyand becomes zero.

(2) The bulb will light up when the capacitor isfully charged

(3) The bulb will not light up at all

(4) The bulb will light up and go off at regular intervals

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14. Two thin dielectric slabs of dielectric constants K1

and K2 (K1 < K2) are inserted between plates of aparallel plate capacitor, as shown in the figure. Thevariation of electric field 'E' between the plates withdistance 'd' as measured from plate P is correctlyshown by :-

+++++++++

–––––––––

P Q

K1 K2

(1) 0 d ®

E

(2) 0 d ®

E

(3) 0 d ®

E

(4)

0 d ®

E

Current Electricity1. Find the current I & voltage V in the circuit shown.

V

8W

10W

4W

2W

0.4W

7W5W

41WI

60V

20V

7W

(1) I = 3.5A, V = 2.5 V

(2) I = 2.5A, V = 3.5V

(3) I = 3A, V = 4V

(4) I = 2A, V = 4A

2. In the circuit shown what are the potential differenceacross 1W, 2W and 3W respectively.

D

C

B

A

1W

2W

3W

12 V

6 V

0 V

(1) 3V, 2V, 1V (2) 2V, 1V, 3V

(3) 1V, 2V, 3V (4) 3V, 1V, 2V

3. The figure shows a network of resistor each heavingvalue 12W.Find the equivalent resistance betweenpoints A and B.

A B

(1) 6W (2) 9W (3) 12W (4) 3W4. Find the current through 25V cell & power supplied

by 20V cell in the figure shown.

10V

5W

5V

10

W

20V

5W

30V

11

W

25V

(1) 12A, –20W (2) 12A, –12W(3) 20A, +12W (4) 20A, –12W

5. For what value of R in circuit, current through 4Wresistance is zero ?

4V 6V

2W

4W

10V

R

(1) 1W (2) 2W(3) 3W (4) 4W

6. An electrical circuit is shown in figure. Calculatethe potential difference across the resistor of 400Was will be measured by the voltmeter V of resistance400W either by applying Kirchhoff's rules orotherwise.

I1

100W 100W 200W

400W

V

10V

I2

I

100W

(1) 5/3 V (2) 10/3 V

(3) 20/3 V (4) 8/3 V

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7. In the figure shown for gives values of R1 and R2

the balance point for Jockey is at 40 cm fromA. When R2 is shunted by a resistance of 10W,balance shifts to 50 cm. Find R1 and R2. (AB = 1m)

G

R1

A B

R2

(1) 1 2

10R ,R 5

3W

= = W

(2) 1 2

10R 5 ,R

3W

= W =

(3) R1 = 5W, R

2 = 10W

(4) R1 = 10W, R

2 = 5W

8. The resistance of the galvanometer G in the circuitis 25W. The meter deflects full scale for a currentof 10 mA. The meter behaves as an ammeter ofthree different ranges. The range is 0–10A, if theterminals O and P are taken; range is 0–1 Abetween O and Q; range is 0–0.1 A betweenO and R. Calculate the resistance R1, R2 and R3.

R1 R2 R3

G

O+

P Q R10A 1A 0.1A

(1) R1 = 0.0278 W, R

2 = 0.25 W, R

3 = 2.5W

(2) R2 = 0.0278 W, R

1 = 0.25 W, R

3 = 2.5W

(3) R2 = 0.0278 W, R

3 = 0.25 W, R

1 = 2.5W

(4) R3 = 0.0278 W, R

1 = 0.25 W, R

2 = 2.5W

9. An ammeter reads upto 1 A. Its internal resistanceis 0.81 W. To increase the range to 10 A, the valueof the required shunt is -(1) 0.03 W (2) 0.3 W(3) 0.9 W (4) 0.09 W

10. The total current supplied to the circuit by thebattery is-

1.5 W

6 W2 W

6V 3 W

(1) 1 A (2) 2 A(3) 4 A (4) 6 A

11. In the circuit, the galvanometer G shows zerodeflection. If the batteries A and B have negligibleinternal resistance, the value of the resistor R willbe-

G

A

12V

2V

B

R

500 W

(1) 200 W (2) 100 W(3) 500 W (4) 1000 W

12. Shown in the figure below is a meter - bridge set upwith null deflection in the galvanometer

G

20cm

55W R

The value of the unknown resistor R is

(1) 13.75W (2) 220 W (3) 110W (4) 55W13. A 5V battery with internal resistance 2W and a 2V

battery with internal resistance1W are connectedto a 10 W resistor as shown in the figure. Thecurrent in the 10W resistor is

2V1W

P1

10W

P2

5V2W

(1) 0.27 A P2 to P1 (2) 0.03 A P1 to P2

(3) 0.03 A P2 to P1 (4) 0.27 A P1 to P2

14. A 100 W bulb B1, and two 60 W bulbs B2 and B3,are connected to a 250 V source, as shown in thefigure. Now W1, W2 and W3 are the output powersof the bulbs B1, B2 and B3 respectively. Then

B1 B2

B3

250V

(1) W1 > W2 = W3 (2) W1 > W2 > W3

(3) W1 < W2 = W3 (4) W1 < W2 < W3

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15. A wire has a non–uniform cross–section as shownin figure. A steady current flows through it. Thedrift speed of electrons at points P and Q isvP and vQ

P Q

(1) vP = vQ (2) vP < vQ

(3) vP > vQ (4) data insufficient

16. The current density across a cylindrical conductorof radius R varies in magnitude according to the

equation J = æ ö-ç ÷è ø0

rJ 1

R where r is the distance

from the central axis. Thus, the current denisty isa maximum J0 at that axis (r = 0) and decreaseslinearly to zero at the surface (r = R). The currentin terms of J0 and conductor's cross-sectional areaA is :-

(1) 0J A3

(2) 0J A6

(3) 0J A2

(4) 0J A5

17. A conductor with rectangular cross section hasdimensions (a × 2a × 4a) as shown in figure.Resistance across AB is x, across CD is y and acrossEF is z. Then

(1) x = y = z (2) x > y > z

(3) y > x > z (4) x > z > y

18. A metal wire of resistance R is cut into three equalpieces that are then connected side by side to forma new wire, the length of which is equal to onethird of the original length. The resistance of thisnew wire is :-

(1) R (2) 3R

(3) R9

(4) R3

19. Find the current I1, I2, & I3 in the circuit.10W

3W 6W

3V

4.5V

I1

I2

I3

(1) 12

, 12

, 1 (2) 1, 1, 2

(3) 12

, 12

, 3 (4) 12

, 12

, 0

20. In the circuit shown in figure, cells of emf 2, 1, 3and 1 V, respectively having resistance 2W, 1W, 3Wand 1W are their internal resistances respectively.The potential difference between D and B. (in Volt)

2W

1W

3W

1W

1V

3V

1V

2V

2W

D

BA

C

(1) 5

13(2)

213

(3) 1013

(4) 7

13

21. In the shown circuit, when the voltage between Aand B is 16 V then voltage between C and D is8 V. When the voltage between C and D is 15 V,what is the voltage (in V) between A and B?

8W2W

2W 8WR

C

D

A

B

(1) 4 (2) 3(3) 2 (4) 0

22. The potential difference between points A and B

in a section of a circuit shown is :-

(1) 5 volts (2) –13 volt

(3) zero volts (4) +13 volts

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23. Resistors R1 and R2 have an equivalent resistanceof 6 ohms when connected in the circuit shownbelow. The resistance of R1 could be :-

ER1 R2

(1) 1 (2) 5 (3) 8 (4) 424. A battery of 10 V and internal resistance 0.5 W

is conneted across a variable resistance R. Thevalue of R for which the power delivered by batteryis maximum, is equal to :-(1) 0.5 W (2) 1 W (3) 1.5 W (4) 2 W

Magnetic field & Magnetism1. If the intensity of magnetic field at a point on the

axis of current coil is half of that at the centre ofthe coil, then the distance of that point from thecentre of the coil will be :–

(1) R2

(2) R

(3) 32R

(4) 0.766R

2. Current flows through uniform, square frames asshown. In which case is the magnetic field at thecentre of the frame not zero?

(1) (2)

(3) (4)

3. A proton, a deuteron and an a–particle having thesame kinetic energy are moving in circulartrajectories in a constant magnetic field. If rp, rd

and ra denote respectively the radii of thetrajectories of these particles, then :

(1) ra = rp < rd (2) ra > rd < rp

(3) ra = rd > rp (4) rp = rd = ra

4. A particle of charge –16 × 10–18 C moving withvelocity 10 ms–1 along the x–axis enters regionwhere a magnetic field of induction B is along they–axis and an electric field of magnitude 104 V/mis along the negative z–axis If the charged particlecontinues moving along the x–axis, the magnitudeof B is–

(1) 103 Wb/m2 (2) 105 Wb/m2

(3) 1016 Wb/m2 (4) 10–3 Wb/m2

5. Two mutually perpendicular conductors carryingcurrents I1 and I2 lie in one plane. Locus of the pointat which the magnetic induction is zero, is a(1) circle with centre as the point of intersection

of the conductor(2) parabola with vertex as the point of intersection

of the conductors(3) straight line passing through the point of

intersection of the conductors(4) rectangular hyperbola

6. Equal current i is flowing in three infinitely long wiresalong positive x, y and z directions. The magnitudeof magnetic field at a point (0, 0, –a) would be

(1) ( )0i ˆ ˆj i2 am

-p

(2) ( )0i ˆ ˆj i2 am

+p

(3) ( )0i ˆ ˆi j2 am

-p

(4) ( )0i ˆˆ ˆi j k2 am

+ +p

7. An electron (mass = 9.1 × 10–31;charge = – 1.6 × 10–19 C) experiences nodeflection if subjected to an electric field of3.2 × 105 V/m and a magnet ic f ie ld of2.0 × 10–3 Wb/m2. Both the fields are normalto the path of electron and to each other. If theelectric field is removed, then the electron willrevolve in an orbit of radius(1) 45 m (2) 4.5 m(3) 0.45 m (4) 0.045 m

8. A long straight wire carries a current along thex–axis. Consider the points A(0, 1, 0), B(0, 1, 1),C(1, 0, 1) and D(1, 1, 1). Which of the followingpairs of points will have same magnetic fields-(1) A and B (2) A and C(3) B and C (4) B and D

9. In the loops shown, all curved sections are eithersemicircles or quarter circles. All the loops carrythe same current. The magnetic fields at thecentres have magnitudes B1, B2, B3 and B4

a

B2

b

a

B1

b

a

B4b

a

B2

ba

B3

b

(1) B4 > B3 > B2 > B1 (2) B1 > B2 > B3 > B4

(3) B4 > B1 > B2 > B3 (4) B1 > B4 > B3 > B2

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10. Two infinitely long linear conductors are arrangedperpendicular to each other and are in mutuallyperpendicular planes as shown in figure. If I1=2Aalong the y–axis and I2 = 3A along –ve z–axis andAP = AB = 1cm. The value of magnetic field

strength Br

at P is

I1I2

APy

xz

B

(1) (3 × 10–5 T) j + (–4 × 10–5 T) k

(2) (3 × 10–5 T) j + (4 × 10–5 T) k

(3) (4 × 10–5 T) j + (3 × 10–5 T) k

(4) (–3 × 10–5 T) j + (4 × 10–5 T) k

11. A conducting wire bent in the form of a parabolay2 = 2x carries a current i=2A as shown in figure.This wire is placed in a uniform magnetic field

ˆB 4k= -ur

Tesla. The magnetic force on the wiree

is (in newton)

x(m)2

y(m)

B

A

(1) ˆ16i- (2) ˆ32i

(3) ˆ32i- (4) ˆ16i

12. A charged particle enters a uniform magnetic fieldperpendicular to its initial direction travelling in air.The path of the particle is seen to follow the pathin figure. Which of statements 1–3 is /are correct?

(1) The magnetic field strength may have beenincreased while the particle was travelling inair

(2) The particle lost energy by ionising the air

(3) The particle lost charge by ionising the air

entry

(1) 1, 2, 3 are correct

(2) 1, 2 only are correct

(3) 2, 3 only are correct

(4) 1 only

13. A particle of mass m and charge q moves witha constant velocity v along the positive x–direction.It enters a region containing a uniform magneticfield B directed along the negative z–direction,extending from x = a to x = b. The minimum valueof v required so that the particle can just enterthe region x > b is

(1) qbBm

(2) q(b a)B

m-

(3) qaBm

(4) q(b a)B

2m+

14. A current carrying loop is placed in a uniformmagnetic field in four different orientations, I, II,III, IV, arrange them in the decreasing order ofpotential energy :

(I) n B

(II) n

B

(III) n

B

(IV)

n

B

(1) I > III > II > IV (2) I > II > III > IV

(3) I > IV > II > III (4) III > IV > I > II

15. A thin flexible wire of length L is connected to twoadjacent fixed points and carries a current I in theclockwise direction, as shown in the figure. Whenthe system is put in a uniform magnetic field ofstrength B going into the plane of the paper, thewire takes the shape of a circle. The tension in thewire is :

(1) IBL (2) IBL

p

(3) IBL2p

(4) IBL4p

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5

16. At a certain place a magnet makes 30 oscillations

per minute. At another place where the magnetic

field is double, its time period will be :-

(1) 4 s (2) 2 s

(3) 12

s (4) 2 s

17. Value of earth’s magnetic field at any point is

7 × 10–5 wb/m2. This field is neutralised by field

which is produced at the centre of a current carrying

loop of radius 5 cm. The current in the loop (approx)

(1) 0.56 A (2) 5.6 A

(3) 0.28 A (4) 28 A

18. Curie-Weiss law is obeyed by iron at a

temperature.......

(1) below Curie temperature

(2) above Curie temperature

(3) at Curie temperature only

(4) at all temperatures

19. The coercivity of a bar magnet is 100A/m. It is to

be demagnetised by placing it inside a solenoid of

length 100 cm and number of turns 50. The current

flowing the solenoid will be

(1) 4A (2) 2A

(3) 1A (4) zero

20. Soft iron is used to make the core of transformer,

because of its :

(1) low coercivity and low retentivity

(2) low coercivity and high retentivity

(3) high coercivity and high retentivity

(4) high coercivity and low retentivity

21. A proton is fired from origin with velocity

( )0ˆˆv v j k= +

r in a uniform magnetic field 0ˆB B j=

r

.

In the subsequent motion of the proton, wrong

statements is -

(1) the path of the proton is a helix

(2) the x-coordinate can never be zero

(3) the x and z-components will becomes zero

simultaneously after every pitch

(4) the y-coordinate will be proportional to its time

of flight

22. The figure shows three situations when an electron

with velocity vr travels through a uniform magnetic

field Br

. In each case, what is the direction ofmagnetic force on the electron?

y yy

B

BBx xxv

vv

z zz1 32

(1) positive z–axis, negative x–axis, positive y–axis(2) negative z–axis, negative x–axis and zero

(3) positive z–axis, positive y–axis and zero

(4) negative z–axis, positive x–axis and zero

23. Current i = 2.5 A flows along the circle x2 + y2 = 9 cm2

(here x & y in cm) as shown.

Magnetic field at point (0, 0, 4 cm) is

(1) ( )7 ˆ36 10 T k-p ´

(2) ( ) ( )7 ˆ36 10 T k-p ´ -

(3) 79 ˆ10 T k5

-pæ ö´ç ÷è ø

(4) ( )79 ˆ10 T k5

-pæ ö´ -ç ÷è ø

24. A proton is projected with a velocity 107 m/s, atright angles to a uniform magnetic field of induction100 mT. The time (in second) taken by the protonto traverse 90º arc is (mp = 1.65 × 10–27 kg andqp = 1.6 × 10–19 C) :-

(1) 0.81 × 10–7 (2) 1.62 × 10–7

(3) 2.43 × 10–7 (4) 3.24 × 10–7

25. If a charged particle goes unaccelerated in a region

containing electric and magnetic fields, then :-

(A) Er

must be perpendicular to Br

(B) vr

must be perpendicular to Er

(C) vr

must be perpendicular to Br

(D) E must be equal to vB

(1) A & B (2) B & C

(3) A & C (4) C & D

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26. In a cyclotron, if a deuteron can gain an energyof 40 MeV, then a proton can gain an energyof :-

(1) 40 MeV (2) 80 MeV

(3) 20 MeV (4) 160 MeV

27. A wire carrying a current i is placed in a magnetic

field in the form of the curve y = a sinx

Lpæ ö

ç ÷è ø

0 £ x £ 2L. Force acting on the wire is :-

O2L

× × × × × × × × ×××××××

××××××

××××××

××××××

××××××

××××××

××××××

××××××

××××××

(1) iBL

p(2) iBLp

(3) 2iBL (4) Zero

EMI & EMW1. When magnetic flux through a coil is changed, the

variation of induced current in the coil with timeis as shown in graph. If resistance of coil is 10W,then the total change in flux of coil will be–

4I(A)

0.1t(s)

(1) 4 (2) 8

(3) 2 (4) 6

2. A coil having 500 square loops each of side 10 cmis placed normal to a magnetic field which increasedat a rate of 1.0 T/sec. The induced e.m.f. (in volt)is :–

(1) 0.1 (2) 0.5

(3) 1.0 (4) 5.0

3. The magnetic field in a coil of 100 turns and40 cm2 an area is increased from 1 tesla to 6 teslain 2 second. The magnetic field is perpendicular tothe coil. The e.m.f. generated in it is :–

(1) 104 V (2) 1.2 V

(3) 1.0 V (4) 10–2 V

4. AB and CD are fixed conducting smooth railsplaced in a vertical plane and joined by a constantcurrent source at its upper end. PQ is a conductingrod which is free to slide on the rails. A horizontaluniform magnetic field exists in space as shown.If the rod PQ in released from rest then,

A C

B D

P Q

IA

(1) The rod PQ will move downward with constantacceleration

(2) The rod PQ will move upward with constantacceleration

(3) The rod will move downward with decreasingacceleration and finally acquire a constantvelocity

(4) Either A or B5. A rectangular loop with a sliding connector of

length 10 cm is situated in uniform magnetic fieldperpendicular to plane of loop. The magneticinduction is 0.1 tesla and resistance of connector(R) is 1 ohm. The sides AB and CD haveresistances 2 ohm and 3 ohm respectively. Findthe current in the connector during its motion withconstant velocity one metre / sec.

(1) 1

A110

R

A D

CB

2W 3W(2) 1

A220

(3) 1

A55

(4) 1

A440

6. Figure shows plane figure made of a conductor

located in a magnetic field along the inward normal

to the plane of the figure. The magnetic field starts

diminishing. Then select incorrect statement.P

Q

R

(1) the induced current at point P is anticlockwise

(2) the induced current at point Q is anticlockwise

(3) the induced current at point Q is zero

(4) the induced current at point R is zero

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7. A conducting wire frame is placed in a magneticfield which is directed into the paper. The magneticfield is increasing at a constant rate. The directionsof induced currents in wires AB and CD are

A

C

DB

(1) B to A and D to C

(2) A to B and C to D

(3) A to B and D to C

(4) B to A and C to D

8. Two different coils have self–inductances L1 = 8 mH

and L2 = 2mH. The current in one coil is increased

at a constant rate. The current in the second coil

is also increased at the same constant rate. At a certain

instant of time, the power given to the two coils is

the same. At that time, the current, the induced

voltage and the energy stored in the first coil are

i1, V1 and W1 respectively. Corresponding values for

the second coil at the same instant are i2, V2 and

W2 respectively. Then :

(A) 1

2

i 1i 4

= (B) 1

2

i4

i=

(C) 1

2

W 1W 4

= (D) 1

2

V4

V=

(1) A, C, D (2) B, C, D(3) Only A, C (4) None of these

9. When a 'J' shaped conducting rod is rotating in itsown plane with constant angular velocity w, about

one of its end P, in a uniform magnetic field r

Bdirected normally into the plane of paper) thenmagnitude of emf induced across it will be

Pw

Q

l L

(1) w + l2 2B L (2) w 21

B L2

(3) ( )w + l2 21

B L2

(4) wl21

B2

10. In the given figure if the magnetic field, which isperpendicular to the plane of the paper in theinward direction increases, then–

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

AB

(1) Plate B of the capacitor will become positivelycharged

(2) Plate A of the capacitor will become positivelycharged.

(3) The capacitor will not be charged.(4) Both plates will be charged alternately.

11. A semicircular wire of radius R is rotated withconstant angular velocity w about an axis passingthrough one end and perpendicular to the plane ofthe wire. There is a uniform magnetic field of strengthB. The induced e.m.f. between the ends is–

B

(1) B wR2/2 (2) 2B wR2

(3) is variable (4) none of these

12. The magnetic flux through a stationary loop withresistance R varies during interval of time T as f=at (T–t). The heat generated during this timeneglecting the inductance of loop will be

(1) 2 3a T3R

(2) 2 2a T3R

(3) 2a T

3R(4)

2 3a TR

13. As shown in figure, A and B are two coaxialconducting loops separated by some distance whenthe switch S is closed, a clockwise current I flowsin A (as seen by O) and an induced current I1 flowsin B. The switch remains closed for a long time.When S is opened, a current I2 flows in B. Then thedirection of I1 and I2 (as seen by O) are:

A B

O

S

(1) Respectively clockwise and anticlockwise(2) Both clockwise(3) Both anticlockwise(4) Respectively anticlockwise and clockwise

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14. A uniform magnetic field exists in the region given

by ˆ ˆ ˆB 3i 4 j 5k= + +r

. A rod of length 5 m placed

along y-axis is moved along x-axis with constantspeed 1 ms–1. Then induced e.m.f in the rodis :-

(1) Zero (2) 25 V (3) 5 V (4) 10 V

15. If given arrangement is moving towards left withspeed v, then potential difference between B andD and current in the loop are respectively :-

2R

R

B

A

C

D

Br

Ä

(1) BvR and non-zero (2) 2BvR and zero(3) 4BvR and non-zero (4) 4BvR and zero

16.

1 2 3

The figure shows three circuits with identical

batteries, inductors and resistances. Rank the

circuits according to the currents through the battery

just after the switch is closed, greatest first.

(1) I2 > I3 > I1 (2) I2 > I1 > I3(3) I1 > I2 > I3 (4) I1 > I3 > I2

17. If E®

and B®

are the electric and magnetic field

vectors of electromagnetic waves then the direction

of propagation of electromagnetic wave is along

the direction of –

(1) E®

(2) B®

(3) E®

× B®

(4) none of these

18. The rms value of the electric field of the light

coming from the sun is 720 N/C. The average

total energy density of the electromagnetic wave

is-

(1) 4.58 × 10–6 J/m3

(2) 6.37 × 10–9 J/m3

(3) 81.35 × 10–12 J/m3

(4) 3.3 × 10–3 J/m3

19. An electromagnetic wave with frequency w andwavelength l travels in the +y direction. Its magneticfield is along +x axis. The vector equation for theassociated electric field (of amplitude E0) is :-

(1) 0

2 Ê E cos t y xpæ ö= w -ç ÷è øl

r

(2) 0

2 Ê E cos t y xpæ ö= - w +ç ÷è øl

r

(3) 0

2 Ê E cos t y zpæ ö= - w +ç ÷è øl

r

(4) 0

2 Ê E cos t y zpæ ö= w -ç ÷è øl

r

Alternating Current1. The voltage supplied to a circuit is given by

V = V0t3/2, where t is time in second. Find the RMS

value of voltage for the period t = 0 to t = 1s :-

(1) 0V2

(2) V0 (3) 03V2

(4) 2V0

2. What is the approximate peak value of analternating current producing four times the heatproduced per second by a steady current of 2 Ain a resistor :-(1) 2.8 A (2) 4.0 A (3) 5.6 A (4) 8.0 A

3. An electric lamp marked 100 volt d.c. consumesa current of 10A. If it is connected to a 200 volt,50 Hz AC mains. Calculate the inductance of choke(inductor coil) required :-(1) 5.5 × 10–2 H (2) 5.5 × 10–1 H(3) 11.0 H (4) 11.0 × 10–2 H

4. A long solenoid connected to a 12 volt DC sourcepasses a steady current of 2A. When the solenoidis connected to an AC source of 12 volt at 50 Hz,the current flowing is 1A. Calculate inductance ofthe solenoid :-

(1) 33 mH (2) 66 mH

(3) 16.5 mH (4) 40 mH

5. When an electric device X is connected to a 220volt,50 hertz a.c. supply, the current is 0.5 A, and isin same phase as the applied voltage. When anotherdevice Y is connected to the same supply, theelectric current is again 0.5 A, but it leads the

potential difference by 2p

. When X and Y are

connected in series across the same source, whatwill be the current ?

(1) 0.35 A (2) 0.55 A (3) 1.35 A (4) 2.5 A

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6. In the circuit shown below, what will be the readingof the voltmeter and ammeter :-

220V, 50 Hz~

VC

A

100W

L

300V300V

(1) 200 V, 1 A (2) 800 V, 2 A

(3) 100 V, 2 A (4) 220 V, 2.2 A

7. In the circuit shown in figure neglecting sourceresistance, the voltmeter and ammeter reading willrespectively be :-

~

V4

A

240 V, 25 Hz

X =25L W

X =25C W 30W

(1) 0 V, 3 A

(2) 150 V, 3 A

(3) 150 V, 6 A

(4) 0 V, 8 A

8. An AC circuit containing 800 mH inductor and a60 µF capacitor is in series with 15W resistance.They are connected to 230 V, 50 Hz AC supply.Obtain total power consumed :-

(1) 20.07 watt (2) 5 watt

(3) 40 watt (4) 10 watt

9. In the following circuit the emf of source isE0 = 200 volt, R = 20W, L = 0.1 henry, C = 10.6farad and frequency is variable then the current atfrequency f = 0 and f = ¥ is :-

R

~

L C

(1) Zero, 10 A (2) 10 A, zero

(3) 10 A, 10 A (4) Zero, zero

10. Match the following items –Circuit component Phase differenceacross an ac source between current and(w = 200 rad/sec) source voltage

(A) 10W 500µF

(p) p2

(B) 5H

(q) p6

(C) 500µF

(r) p4

(D) 3µF4H

(s) p3

(E) 5H1kW (t) None of the above

(1) A-r, B-p, C-p, D-p, E-r

(2) A-q, B-p, C-p, D-p, E-s

(3) A-t, B-p, C-p, D-p, E-t

(4) A-s, B-p, C-p, D-p, E-s11. The switch in the circuit pictured is in position a for

a long time. At t = 0 the switch is moved from a tob. The current through the inductor will reach its firstmaximum after moving the switch in a time:-

(1) 2 LCp (2) 1

LC4

La bR

e C(3) LC

2p

(4) LCp

12. In a series CR circuit shown in figure, the appliedvoltage is 10V and the voltage voltage acrosscapacitor is found to be 8V. then the voltage acrossR, and the phase difference between current andthe applied voltage will respectively be

RC

8V VR

10V

(1) 6V, tan-1 43

æ öç ÷è ø (2) 3V, tan-1

34

æ öç ÷è ø

(3) 6V, tan-1 53

æ öç ÷è ø (4) None

13. In series LR circuit XL=3R and R is resistance. Nowa capacitor with XC = R is added in series. Ratio ofnew to old power factor is

(1) 1 (2) 2 (3) 1

2(4) 2

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Ray Optics1. A mirror is inclined at an angle of q with the horizontal.

If a ray of light is incident at an angle q as shown,then the angle made by reflected ray with the horizontalis

(1) q (2) 2q (3) 2q

(4) zero

2. The distance of an object from a spherical mirroris equal to the focal length of the mirror. Then theimage:

(1) must be at infinity (2) may be at infinity(3) may be at the focus (4) none

3. A ray of light travelling in a medium of refractiveindex m is incident at an angle q on a compositetransparent plate consisting of 50 plates of R.I.1.01 m , 1.02 m , 1.03 m , ........ , 1.50 m . The rayemerges from the composite plate into a mediumof refractive index 1.6 m at angle 'x' . Then :

(1) 50

1.01sin x sin

1.5æ ö= qç ÷è ø (2) sinx =

58

sin q

(3) sin x =85

sin q (4) sinx =50

1.51.01

æ öç ÷è ø sin q

4. A ray of light travels from an optical denser mediumto rarer medium . The critical angle for the two mediais C. The maximum possible deviation of therefracted light ray can be :–

(1) p – C (2) 2 C

(3) p – 2 C (4) 2p

– C

5. A ray of light from a denser medium strike a rarermedium. The angle of reflection is r and that ofrefraction is r¢. The reflected and refracted raysmake an angle of 90° with each other. The criticalangle will be:

(1) ( )1sin tan r- (2) ( )1tan sin r-

(3) ( )1sin tan r '- (4) ( )1tan sin r '-

6. A ray of light is incident upon an air/water interface(it passes from air into water) at an angle of 45°.Which of the following quantities change as the lightenters the water?(I) wavelength(II) frequency(III) speed of propagation(IV) direction of propagation(1) I, III only (2) III, IV only(3) I, II, IV only (4) I, III, IV only

7. A paraxial beam is incident on a glass (n = 1.5)

hemisphere of radius R=6 cm in air as shown. Thedistance of point of convergence F from the planesurface of hemisphere is :–

(1) 12 cm (2) 5.4 cm(3) 18 cm (4) 8 cm

8. A diverging lens of focal length 10 cm is placed10 cm in front of a plane mirror as shown in thefigure. Light from a very far away source falls onthe lens. The final image is at a distance :–

(1) 20 cm behind the mirror

(2) 7.5 cm in front of the mirror

(3) 7.5 cm behind the mirror

(4) 2.5 cm in front of the mirror

9. A man wishing to get a picture of a Zebra

photographed a white donkey after fitting a glass

with black streaks onto the objective of his camera.

(1) the image will look like a black donkey on the

photograph

(2) the image will look like a Zebra on the

photograph

(3) the image will be more intense compared to

the case in which no such glass is used

(4) the image will be less intense compared to the

case in which no such glass is used

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10. A beam of light consisting of red, green and blueand is incident on a right angled prism. Therefractive index of the material of the prism for theabove red, green and blue wavelengths are 1.39,1.44 and 1.47 respectively. The prism will–

45°

(1) separate part of the red color from the greenand blue colors

(2) separate part of the blue color from the red andgreen colors

(3) separate all the three colors from the other twocolors

(4) not separate even partially, any colors from theother two colors

11. The curve of angle of incidence versus angle ofdeviation shown has been plotted for prism. Thevalue of refractive index of the prism used is :

(1) 3

(2) 2

(3) 3

2

(4) 2

312. A beam of monochromatic light is incident at

i = 50° on one face of an equilateral prism, the angleof emergence is 40°, then the angle of minimumdeviation is :(1) 30° (2) < 30°(3) £ 30° (4) ³ 30°

13. Which of the following quantities related to a lens doesnot depend on the wavelength of the incident light ?(1) Refractive index (2) Focal length(3) Power (4) Radii of curvature

14. A diverging beam of light from a point source Shaving divergence angle a falls symmetrically ona glass slab as shown. The angles of incidence ofthe two extreme rays are equal. If the thicknessof the glass slab is t and its refractive index is n,then the divergence angle of the emergent betweenthem is :

(1) zero (2) a(3) sin–1 (1/n) (4) 2 sin–1 (1/n)

15. A ray of light passes through four transparentmedia with refractive indices µ1, µ2, µ3 and µ4 asshown in the figure. The surfaces of all media areparallel. If the emergent ray CD is parallel to theincident ray AB, we must have :–

(1) µ1 = µ2 (2) µ2 = µ3

(3) µ3 = µ4 (4) µ4 = µ1

16. A given ray of light suffers minimum deviation inan equilateral prism P. Additional prism Q and Rof identical shape and of the same material as Pare now added as shown in the figure

(1) greater deviation

(2) no deviation

(3) same deviation as before

(4) total internal reflection

17. Which one of the following spherical lenses doesnot exhibit dispersion ? The radii of curvature ofthe surfaces of the lenses are as given in thediagrams :–

(1) (2)

(3) (4)

18. The size of the image of an object, which is at

infinity, as formed by a convex lens of focal length

30cm is 2cm. If a concave lens of focal length 20cm

is placed between the convex lens and the image

at a distance of 26cm from the convex lens,

calculate the new size of the image :–

(1) 1.25 cm (2) 2.5 cm

(3) 1.05 cm (4) 2cm

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19. A ray of light is incident at the glass–water interfaceat an angle i, it emerges finally parallel to the surfaceof water, then the value of µg would be :–

(1) (4/3) sin i (2) 1/sin i(3) 4/3 (4) 1

20. White light is incident on the interface of glass andair as shown in the figure. If green light is just totallyinternally reflected then the emerging ray in aircontains.

(1) yellow, orange, red

(2) violet, indigo, blue

(3) all colours

(4) all colours except green

21. A point object is placed at distance of 20cm froma thin plano–convex lens of focal length 15cm. Theplane surface of the lens is now silvered. The imagecreated by the system is at :–

(1) 60cm to the left of the system

(2) 60cm to the right of the system

(3) 12cm to the left of the system

(4) 12cm to the right of the system

22. Figures shows the graph of angle of deviation dversus refractive index m of the material of constantthin angled prisms corresponding to light raysincident at a small angle of incidence. The prismangle and slope of the line are respectively :

543210

-1-2-3-4-5

Refractiveindex ( )m

Dev

iatio

n

(deg

ree)

d

(not to scale)

(1) 40 and 2 (2) 20 and 1/2(3) 20 and 4 (4) 40 and 4

23. A boy of height H is standing is front of mirror, whichhas been fixed on the ground as shown in figure. Whatlength of his body can the man see in the mirror ? Thelength of the mirror is (H/2) :-(1) H

(2) H2/(H2 + L2)1/2

(3) Zero

(4) 2H2/L

24. A square wire of side 3.0cm is placed 25 cm away

from a concave mirror of focal length 10cm. What

is the area enclosed by the image of the wire? (The

centre of the wire is on the axis of the mirror, with

its two sides normal to the axis)

(1) 4cm2 (2) 2cm2

(3) 8cm2 (4) 5cm2

25. A convex mirror of focal length f produces animage (1/n)th of the size of the object. Thedistance of the object from the mirror is

(1) nf (2) f/n

(3) (n + 1)f (4) (n - 1)f

26. If light travels a distance x in t1 sec in air and 10x

distance in t2 sec in a medium, find the critical angle

of the medium.

(1) sin–1 1

2

tt (2) cos–1 1

2

tt

(3) sin–1 1

2

10tt (4) sin–1 1

2

5tt

27. There is a small air bubble inside glass sphere

(µ = 1.5) of radius 10cm. The bubble is 4cm below

the surface and is viewed normally from the outsideas shown in figure. Find apparent depth of the bubble.

(1) 2cm (2) –3cm

(3) +3cm (4) – 4cm

28. An achromatic doublet of focal length 90cm is tobe made of two lens. The material of one having1.5 times the dispersive power of the other. The doubletis converging type find the focal length of each lens.

(1) 30 cm, – 45 cm (2) –20 cm, + 40 cm

(3) 15 cm, – 30 cm (4) 20 cm, – 40 cm

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29. The dispersive power of material of a lens of focallength 20 cm. is 0.08. Find the longitudinalchromatic abberation of the lens ?

(1) 1.6cm (2) 3.5cm

(3) 5.5cm (4) 10cm

30. A compound micro scope has a magnifying powerof 100 when the image is formed at infinity. Theobjective has a focal length of 0.5 cm and the tubelength is 6.5 cm. Find the focal length of theeyepiece.(1) 1.25cm (2) 2.0cm(3) 4.50cm (4) 5cm

31. The focal length of the objective and eye piece ofa microscope are respectively 1cm and 2cm. Thedistance between them is 12 cm. Where an objectshould be placed in order to view it at the leastdistant of distinct vision.(1) 4.05 cm (2) 0.05 cm(3) 2.05 cm (4) 1.1 cm

32. An astronomical telescope of length 50 cmproduces a magnification of 9 in normal adjustment.Calculate focal length of its objective and eye piece.(1) 4cm, 2cm (2) 30cm, 5cm(3) 55cm, 10cm (4) 45cm, 5 cm

Wave Optics1. In Young's double slit experiment, using light of

wavelength 400 nm, interference fringes of width

'X' are obtained. The wavelength of light is increased

to 600 nm and the separation between the slits is

halved. If one wants the observed fringe width on

the screen to be the same in the two cases, find the

ratio of the distance between the screen and the

plane of the slits in the two arrangements.

(1) 1 : 2 (2) 3 : 1

(3) 2 : 1 (4) 5 : 2

2. If the distance between the first maxima and fifth

minima of a double slit pattern is 7 mm and the

slits are separated by 0.15 mm with the screen

50 cm from the slits, then wavelength of the light

used is

(1) 600 nm (2) 525 nm

(3) 467 nm (4) 420 nm

3. If the ratio of the intensity of two coherent sources

is 4 then the visibility [(Imax – Imin)/(Imax + Imin)]

of the fringes is

(1) 4 (2) 4/5(3) 3/5 (4) 9

4. In the YDSE shown the two slits are covered withthin sheets having thickness t & 2t and refractiveindex 2m and m. Find the position (y) of centralmaxima

t,2m

m,2t

yd

D

(1) zero (2) tDd

(3) tDd

- (4) None of these

5. The resultant amplitude of a vibrating particle by

the superposition of the two waves

y1 = asin t3pé ùw +ê úë û

and y2 = a sin wt is :–

(1) a (2) 2 a

(3) 2a (4) 3 a

6. 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 : [assume d << D, l << d]

O

D

d 2d/3

(1) 0 (2) d/2 (3) d/3 (4) d/67. Wavefronts incident on an interface between the

media are shown in the figure. The refractedwavefront will be as shown in

45°

m1=1

m 2= 2

(1)

30°

(2)

30°

(3)

60°

(4)

60°

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8. If the first minima in a Young's slit experimentoccurs directly in front of one of the slits, (distancebetween slit & screen D = 12 cm and distancebetween slits d = 5 cm) then the wavelength of theradiation used can be :(1) 3 cm (2) 4 cm

(3) 23

cm (4) 43

cm

9. Plane wavefronts are incident on a spherical mirroras shown in the figure. The reflected wavefrontswill be

(1) (2)

(3) (4)

10. In Young's double slit experiment intensity at a pointis (1/4) of the maximum intensity. Angular positionof this point is :–

(1) sin–1

dlæ ö

ç ÷è ø

(2) sin–1

2dlæ ö

ç ÷è ø

(3) sin–1

3dlæ ö

ç ÷è ø

(4) sin–1

4dlæ ö

ç ÷è ø

11. Light of wavelength 6000Å is incident normallyon a slit of width 24 × 10–5 cm. Find out the angularposition of second minimum from centralmaximum ?

(1) 60° (2) 30°

(3) 45° (4) none

12. Light of wavelength 6328Å is incident normally on aslit of width 0.2 mm. Calculate the angular width ofcentral maximum on a screen distance 9 m ?

(1) 0.20° (2) 0.36°

(3) 0.48° (4) 0.24°

13. A Slit of width a is illuminated by monochromaticlight of wavelength 650nm at normal incidence.Calculate the value of a when the first minimumfalls at an angle of diffraction of 30° :

(1) 2.6 mm (2) 1.3 mm(3) 0.65 mm (4) 3.9 mm

14. A mixture of light, consisting of wavelength590 nmand an unknown wavelength, illuminates Young'sdouble slit and gives rise to two overlappinginterference patterns on the screen. The centralmaximum of both lights coincide. Further, it isobserved that the third bright fringe of known lightcoincides with the 4th bright fringe of the unknownlight. From this data, the wavelength of theunknown light is :-

(1) 442.5 nm (2) 776.8 nm

(3) 393.4 nm (4) 885.0 nm

15. At two points P and Q on screen in Young's doubleslit experiment, waves from slits S1 and S2 have a

path difference of 0 and 4l

respectively. the ratio

of intensities at P and Q will be :

(1) 3 : 2 (2) 2 : 1 (3) 2 : 1 (4) 4 : 1

16. In young's double slit experiment, slits are arrangedin such a way that besides central bright fringe,there is only one bright fringe on either side of it.Slit separation d for the given condition cannot be(if l is wavelength of the light used) :

(A) l (B) l/2

(C) 2l (D) 3l/2

(1) A & B (2) A & C

(3) B & C (4) A & D

17. White light is used to illuminate the two slits in aYoung's double slit experiment. The separationbetween the slits is b and the screen is at a distanced (>> b) from the slits. At a point on the screendirectly in front of one of the slits, certainwavelengths are missing. Some of these missingwavelengths are :

(A) l = b2/d (B) l = 2b2/d

(C) l = b2/3d (D) l = 2b2/3d

(1) A & B (2) A & D

(3) A & C (4) B & D

18. Two coherent sources of intensities I1 and I2 producean interference pattern. The maximum intensity inthe interference pattern will be :–

(1) I1 + I2 (2) 2 21 2I I+

(3) (I1 + I2)2 (4) 2

1 2( I I )+

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Modern Physics1. Figure shows the graph of stopping potential versus

the frequency of a photosensitive metal. The plank'sconstant and work function of the metal are

V0

V2

V1 n1 n2n

(1) 2 1 2 1 1 2

2 2 2 1

V V V Ve, e

æ ö æ ö+ n + nç ÷ ç ÷n + n n + nè ø è ø

(2) 2 1 2 1 1 2

2 1 2 1

V V V Ve, e

æ ö æ ö+ n + nç ÷ ç ÷n - n n - nè ø è ø

(3) 2 1 2 1 1 2

2 1 2 1

V V V Ve, e

æ ö æ ö- n - nç ÷ ç ÷n + n n + nè ø è ø

(4) 2 1 2 1 1 2

2 1 2 1

V V V Ve, e

æ ö æ ö- n - nç ÷ ç ÷n - n n - nè ø è ø

2. The positions of 2 41 2D, He and 7

3 Li are shown onthe binding energy curve as shown in figure.

42

He

73Li

21D

8

7

6

5

4

3

2

1Bin

din

g en

ergy

per

nuc

leon

(MeV

)

Mass Number (A)

2 4 6 8 10

The energy released in the fusion reaction.2 7 4 11 3 2 0D Li 2 He n+ ® +

(1) 20 MeV

(2) 16 MeV

(3) 8 MeV(4) 1.6 MeV

3. In a photoelectric experiment, electrons are ejectedfrom metals X and Y by light of intensity I andfrequency f. The potential difference V requiredto stop the electrons is measured for variousfrequencies. If Y has a greater work function thanX; which one of the following graphs best illustratesthe expected results?

(1) X

V

O f

Y(2) X

V

O f

Y

(3) X

V

O f

Y(4)

X

V

O f

Y

4. When a nucleus with atomic number Z and massnumber A undergoes a radioactive decayedprocess, which of the following statement isincorrect :(1) both Z and A will decrease, if the process is

a decay(2) Z will decrease but A will not change, if the

process is b+ decay(3) Z will decrease but A will not change, if the

process is b– decay(4) Z and A will remain unchanged, if the process

is g decay

5. Which of the following is correct for a nuclearreaction?

(1) A typical f ission represented by

92U235+0n

1®56Ba143+36Kr93+energy

(2) Heavy water is used as moderator in preferenceto ordinary water because H may captureneutrons, while D would not

(3) Cadmium rods increase the reactor power whenthey go in, decrease when they go outward

(4) Slower neutrons are more effective in causingfission than faster neutrons in case of U235

6. The half–life of 215At is 100 µs. The time takenfor the radioactivity of a sample of 215At to decayto 1/16th of its initial value is :

(1) 400µs (2) 63µs

(3) 40µs (4) 300µs

7. Which of the following process represent ag–decay ?

(1) A XZ + g ® A XZ – 1 + a + b

(2) A XZ + 1n0 ® A – 3 XZ – 2 + c

(3) A XZ ® A XZ + f

(4) A XZ + e–1 ® A XA – 1 + g

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8. A nucleus with mass number 220 initially at restemits an a–particle. If the Q value of the reactionis 5.5 MeV, calculate the kinetic energy of thea–particle.(1) 4.4 MeV(2) 5.4 MeV(3) 5.6 MeV(4) 6.5 MeV

9. If a star can convert all the He nuclei completelyinto oxygen nuclei. The energy released per oxygennuclei is : [Mass of the nucleus is 4.0026 amu andmass of oxygen nucleus is 15.9994](1) 7.6 MeV (2) 56.12 MeV(3) 10.24 MeV (4) 23.4 MeV

10. In the options given below, let E denote the restmass energy of a nucleus and n a neutron. Thecorrect option is :

(A) 236 137 9792 53 39E( U) > E( I) +E( Y)+2E(n)

(B) 236 137 9792 53 39E( U) < E( I) + E( Y) +2E(n)

(C) 236 140 9492 56 36E( U) < E( Ba)+E( Kr)+2E(n)

(D) 236 140 9492 56 36E( U) > E( Ba)+E( Kr)+2E(n)

(1) A & B (2) A & D(3) B & C (4) C & D

11. A radioactive sample S1 having an activity of 5mCihas twice the number of nuclei as another sampleS2 which has an activity of 10mCi. The half livesof S1 and S2 can be :–(1) 20 years and 5 years, respectively(2) 20 years and 10 years, respectively(3) 10 years each(4) 5 years each

12. Let mp be the mass of proton, mn the mass of

neutron. M1 the mass of 2010 Ne nucleus and M2 the

mass of 4020 Ca nucleus. Then :

(1) M2 = 2 M1

(2) M2 > 2M1

(3) M2 < 2 M1

(4) M1 > 10 (mn + mp)13. Assume that the nuclear binding energy per

nucleon (B/A) versus mass number (1) is as shownin the figure. Use this plot to choose the correctchoice(s) given below.

0100 200

A

2

4

6

8

B/A

(1) Fusion of two nuclei with mass numbers lyingin the range of 1 < A < 50 will release energy

(2) Fusion of two nuclei with mass numbers lyingin the range of 51 < A < 100 will releaseenergy

(3) Fission of a nucleus lying in the mass rangeof 100 < A < 200 will release energy whenbroken into two equal fragments

(4) Fission of a nucleus lying in the mass range of200 < A < 260 will release energy whenbroken into two equal fragments

(1) A & B (2) A & D(3) B & D (4) C & D

14. After 280 days, the activity of a radioactive sampleis 6000 dps. The activity reduces to 3000 dps afteranother 140 days. The initial activity of the samplein dps is :(1) 6000 (2) 9000(3) 3000 (4) 24000

15. The binding energy per nucleon for the parentnucleus is E1 an that for the daughter nuclei is E2.Then :-(1) E1 = 2E2 (2) E2 = 2E1

(3) E1 > E2 (4) E2 > E1

16. A radioactive nucleus (initial mass number A andatomic number Z) emits 3 a-particles and2 positrons. The ratio of number of neutrons to thatof protons in the final nucleus will be:-

(1) A Z 4

Z 2- -

-(2)

A Z 8Z 4- -

-

(3) A Z 4

Z 8- -

-(4)

A Z 12Z 4- -

-17. The half life of a radioactive substance is 20 minutes.

The approximate time interval (t2 – t1) between the

time t2 when 23

of it has decayed and time t1 when

13

of it had decayed is :-

(1) 20 min (2) 28 min(3) 7 min (4) 14 min

18. After absorbing a slowly moving neutron of mass mN

(momentum ~0) a nucleus of mass M breaks intotwo nuclei of masses m1 and 5m1(6m1 = M + mN),respectively. If the de Broglie wavelength of thenucleus with mass m1 is l, then de Brogliewavelength of the other nucleus will be:-(1) 25 l (2) 5l

(3) 5l

(4) l

19. The radioactivity of a sample is R1 at time T1 andR2 at time T2. If the half life of the specimen isT then number of atoms that have disintegratedin time (T2–T1) is proportional to(1) (R1T1–R2T2) (2) (R1–R2)T(3) (R1–R2)/T (4) (R1–R2) (T1–T2)

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20. The decay constant of the end product of aradioactive series is(1) zero(2) infinite(3) finite (non zero)(4) Depends on final product

21. At time t=0, N1 nuclei of decay constant l1 & N2

nuclei of decay constant l2 are mixed. The decayrate of the mixture is

(1) ( )1 2 t1 2N N e- l +l

(2) + ( )1 2 t1

2

Ne

N- l -læ ö

ç ÷è ø

(3) ( )1 2t t1 1 2 2N e N e-l -l+ l + l

(4) ( )1 2 t1 1 2 2N N e- l +l+ l l

22. The radioactive nuclide of an element X decays toa stable nuclide of element Y. Then, in a givensample of X, the rate of formation of Y is given bythe graph –

(1)

Rat

e(g /

s)m

time (s)

(2)

Rat

e(g /

s)m

time (s)

(3)

/R

ate(

gs)

m

time (s)

(4)

/R

ate(

gs)

m

time (s)23. The work functions of tungsten and sodium are

5.06 eV and 2.53 eV respectively. If the thresholdwavelength for sodium is 5896 Å, then thethreshold wavelength for the tungsten will be(1) 11792 Å (2) 5896 Å(3) 4312 Å (4) 2948 Å

24. The stopping potential necessary to reduce thephotoelectric current to zero–(1) is directly proportional to wavelength of incident

light.(2) uniformly increases with the wavelength of

incident light.(3) directly proportional to frequency of incident light.(4) Linearly increases with the frequency of incident

light.25. A photoelectric cell is illuminated by a point source

of light 1 m away. When the source is shifted to2m then–(1) each emitted electron carries one quarter of the

initial energy(2) number of electrons emitted is half the initial

number(3) each emitted electron carries half the initial

energy(4) number of electrons emitted is a quarter of the

initial number

26. 1.5 mW of 400 nm light is directed at aphotoelectric cell. If 0.10% of the incident photonsproduce photoelectrons, the current in the cell is:(1) 0.36 mA (2) 0.48 mA(3) 0.42 mA (4) 0.32 mA

27. Let K1 be the maximum kinetic energy ofphotoelectrons emitted by a light of wavelengthl1 and K2 corresponding to l2. If l1 = 2l2, then(1) 2K1 = K2 (2) K1 = 2K2

(3) K1 < 2K2

(4) K1 > 2K2

28. When a neutron is disintegrated, it gives :–(1) one proton, one electron and one anti neutrino(2) one positron, one electron and one neutrino(3) one proton, one positron and one neutrino

(4) one proton, g – rays and one neutrino

29. The half life of a radioactive element is 30 days, in90 days the percentage of disintegrated part is :–(1) 13.5 % (2) 46.5 %(3) 87.5% (4) 90.15%

30. Assuming that 200 MeV of energy is released perfission of 92U

235 atom. Find the number of fissionper second required to release 1 kW power :–(1) 3.125 × 1013 (2) 3.125 × 1014

(3) 3.125 × 1015 (4) 3.125 × 1016

Electronics & Communication systems1. Pure Si at 300 K has equal electron (ne) and hole

(nh) concentrations of 1.5 × 1016 m–3. Dopping byindium increases nh to 3 × 1022 m–3. Calculate ne inthe doped Si.(1) 7.5 × 109 m–3 (2) 7.5 × 1011 m–3

(3) 7.5 × 1010 m–3 (4) 7.5 × 108 m–3

2. An n-p-n transistor in a common emitter mode isused as a simple voltage amplifier with a collectorconnected to load resistance RL and to the basethrough a resistance RB. The collector-emittervoltage VCE = 4V, the base-emitter voltageVBE = 0.6V, current through collector is 4 mA andthe current amplification factor b = 100. Calculatethe values of RL and RB.

RBRL

8V

B E

C

(1) RL = 1 kW, RB = 185 kW(2) RL = 1 kW, RB = 200 kW(3) RL = 185 kW, RB = 1 kW(4) RL = 1 kW, RB = 200 kW

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3. Which logic gate is represented by the following

combination of logic gate :-

YB

A

(1) OR (2) NAND (3) AND (4) NOR

4. Which of the following pairs are universal gates

(1) NAND, NOT (2) NAND, AND(3) NOR, OR (4) NAND, NOR

5. The current flowing through the zener diode infigure is :–

5V 1kW10V

500WI1

(1) 20 mA 2) 25 mA(3) 15 mA (4) 5 mA

6. You are given two circuits as shown in followingfigure. The logic operation carried out by the twocircuit are respectively :–

A

BY

Y

A

B

(1) AND, OR (2) OR, AND

(3) NAND, OR (4) NOR, AND

7. The following configuration of gates is equivalentto :–

BA

Y

(1) NAND (2) OR(3) XOR (4) NOR

8. The logic symbols shown here are logicallyequivalent to :–

Y

(a)

AB

AB

Y

(b)

(1) 'a' AND and 'b' OR gate

(2) 'a' NOR and 'b' NAND gate

(3) 'a' OR and 'b' AND gate

(4) 'a' NAND and 'b' NOR gate

9. The real time variation of input signals A and Bare as shown below. If the inputs are fed intoNAND gate, then select the output signal fromthe following :-

A

B

AB

Y

(1)

0 2 4 6 8t(s)

Y

(2)

0 2 4 6 8t(s)

Y

(3)

0 2 4 6 8t(s)

Y

(4)

0 2 4 6 8t(s)

Y

10. A sinusoidal voltage of amplitude 25 volts andfrequency 50 Hz is applied to a half wave rectifierusing PN diode. No filter is used and the loadresistor is 1000 W. The forward resistance Rf idealdiode is 10 W. Calculate(i) Peak, average and rms values of load currrent.(ii) d.c. power output(iii) a.c. power input(iv) % Rectifier efficiency(v) Ripple factor

11. What is the value of current I in given circuits

(a)

(b)

(c)

SOLUTIONS

PHYSICS

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Basic, Maths and Vectors1. Ans (2)

As =r r

B C3 2

so + = -rr r

B C A3 2 4

therefore

+ + =r rr r

3A B C A4 3 2 2

2. Ans (3)

ˆˆ3 5P A B C i k= + + = -r rr r

and

ˆˆ ˆ

ˆˆ ˆ1 2 3 5 7 3

1 1 4

i j k

Q A B i j k= ´ = = - +-

r r r

Angle between & QPrr

is given by

15 15cos 0 90

P QPQ PQ

× -q = = = Þ q = °

rr

3. Ans (2)Let force be F so resultant is in east direction

( )ˆ ˆ ˆ ˆ ˆ4 3 5cos37 5sin 37i j i j F ki+ + ° + ° + =r

37°

5N3N

4NE( i )W

S

N( j )

Fmin

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ4 3 4 3 8 6i j i j F ki i j F kiÞ + + + + = Þ + + =r r

( ) ˆ ˆ8 6Þ = - -rF k i j

( ) ( )2 2

min8 6 6Þ = - + Þ =F k F N4. Ans (1)

Velocity of first ball 1

ˆ ˆv i 3j= +r

Veocity of second ball 2ˆ ˆv 2i 2j= +

r

Angle between their path :

cosq =( ) ( )

1 2

1 2

v .v 2 2 3 1 3v v 2 22 2 2

+ += =

r r

= 0.966

Þ q = 15°(See options)

5. Ans (4)

Q $ $|a b c| 1+ + =$

\ $ $ $ µ $ µ µ $2 2 2|a| |b| |c| 2(a b b c c a) 1+ + + × + × + × =$

Þ 1 +1 +1 +2(cos q1 + cos q2 + cosq3) = 1Þ cosq1 + cosq2 + cos q3 =–1

6. Ans (1)

( ) ( )a 3b . 7a 5b 0+ - =r rr r

Þ 7a2 – 15b2 + 16 a b×rr

=0 ...(i)

and ( ) ( )a 4b 7a 2b 0- × - =r rr r

Þ 7a2 + 8b2 – 30 a b 0× =rr

...(ii)By adding (i) and (ii)

Þ –23b2 + 46 a b 0× =rr

Þ 22a b b× =rr

So 7a2 – 15b2 + 8b2 = 0 Þ a2 = b2

Þ 2abcosq = b2 Þ 2cosq=1Þ q = cos–1(1/2) = 60°

7. Ans (4)

ymax = 2 2(4) ( 3) 5+ - =8. Ans (1)

Area, A = pr2

Þ 2

dA dr mm2 r 0.2

dt dt s= p = p

Perimeter, P = 2pr

Þ dP dr mm

2 0.2dt dt s

= p = p

Unit Dimensions & Measerment1. Ans (2)

For (1) : A and 3A

B may have same dimension.

For (2) : As A and B have different dimension so

expæ ö-ç ÷è ø

AB is meaningless.

For (3) : AB2 is meaningful.For (4) : AB–4 is meaningful.

2. Ans (3)

dim(A) = +æ ö é ùÞ =ç ÷ ê úë ûè ø

F F(At D)dim(C) dim

v x

[ ]-é ù é ùé ùæ ö= = = =ç ÷ ê ú ê úê úè ø ë û ë ûë û

2 22 2

2

F Ft F t Fdim dim dim M T

x v xv v

3. Ans (3)

( ) ( )é ù é ù× = ×ë ûê úë ûòrr r rd

F dS A F pdt

[ ] [ ] --

é ù é ùÞ = Þ = = =ê ú ê úë û ´ë û1

1

Fs s LAFp A M

t pt MLT T

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4. Ans (1)

Given

( )-b= + a2rX YFe ZWsin r

Here Dim(b)=L–2, Dim(a)=L–1

Also Dim(X) = Dim(YF) ...(i) & Dim(X) = Dim(ZW) ...(ii)Now If Dim(Y) = LSo Dim(X) = [LMLT–2] = [ML2T–2] from equation (i)

and Dim(Z) = Dim [ ]-

-

é ùæ ö = =ç ÷ ê úë ûè ø

2 20 0 0

2 2

X ML TM L T

W ML T

from equation (ii)

Thus -

- -

é ùæ öa=ç ÷ ê úb ë ûè ø

1

2 2

YZ L LDim

F L MLT= [M–1LTT2]

5. Ans (1)

If D(Y) = [LT–1] then D(X) = [LT–1MLT–2] = [ML2T–3]from equation (i)

6. Ans (2)

If D(Z) = T–1. Then D(Y) = ?

from equation (ii) : D(X) = [T–1ML2T–2] = [ML2T–3]

So from equation (i) :

D(Y) = ( )( )

--

-

é ùé ù= =ê ú ë ûë û

2 31

2

D X ML TLT

D F MLT

7. Ans (2)n1u1 = n2u2

-é ùé ù é ùÞ = ê úê ú ê úê úë û ë ûë û

1 3

1 12 1

2 2

M Ln n

M L

-é ùæ öæ ö= = ´ ´ =ê úç ÷ç ÷è øè øê úë û

31g 1cm 1

2 2 8 44g 2cm 4

OR

r = 2 3

2g 8(4g)2

(cm) 4(2cm)= = 3

(4g)4

(2cm)

Þ Numerical value = 48. Ans (1)

P + 2aT

V = (RT+b) V–c

Þ P = (RT +b) V–C – aT2 V–1 = AVm –BVn

Þm= – c and n = – 1

9. Ans (3)

As x µ m–1l3t–2 so x m t

3 2x m t

D D D Dæ ö= ± + +ç ÷è øl

l

% error = ±[1 + 3(2) + 2(3)] = ±7%Kinematics

1. Ans (2)Time taken from point P to point P

TP = 22(h H)

g+

H

h

Q

P

Highestpoint

Time taken from point Q to point Q TQ = 22hg

Þ Th H)

gT

hgP Q

2 28 8=

+=

(&

Þ T T8HgP

2Q2= +

Þ g = 8H

TP2 - TQ

2

2. Ans (2)

Maximum hight H = u

g

2 2

2sin q

Þ h1 = ( ) sin20

2

2 2 qg

= 20 sin2q

& h2 = ( ) sin20 2

2

2 2 p q-b gg

= 20 cos2q

h2 – h1 = 20 [cos2q – sin2q] = 10 Þ 20 cos 2q = 10

Þ cos2q = 12

Þ 2q = 60° Þ q = 30°

and q' = 90°-q = 60°3. Ans (4)

Distance covered by the car during the applicationof brakes by driver

s1 = ut = 554

18æ ö´ç ÷è ø

(0.2) = 15 x 0.2 = 3.0 m

After applying the brakes; v = 0,u = 15 m/s,a = 6 m/s2 s2 = ?Using v2 = u2 – 2as Þ 0 = (15)2 – 2 × 6 × s2 Þ12

s2 = 225 Þ 2

225s 18.75m

12= =

Distance travelled by the car after driver sees theneed for it s = s1 + s2 = 3 + 18.75 = 21.75 m.

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4. Ans (3)

Taking origin at A and x axis along AB

Velocity of A w.r.t. B

( ) ( )ˆ ˆ ˆ ˆ20 3 cos60i sin 60 j 20 cos150i sin150 j= + - +

1 3 3 1ˆ ˆ ˆ ˆ ˆ ˆ20 3 i j 20 i j 20 3i 20 j2 2 2 2

æ ö æ ö= + - - + = +ç ÷ ç ÷ç ÷ ç ÷

è ø è ø

300600

20m

A B

20

qq

x

900dmin

vAB®

20 3

020 1tan 30

20 3 3q = = Þ q =

so minmin

d 1sin sin 30 d 10m

20 2= q = ° = Þ =

5. Ans (1)

Here x x

y y

a ug cot 1a g tan u

q= = =

q Þ Initial velocity &

acceleration are opposite to each other.

Þ Ball will return to point O.

6. Ans (2)

In time of flight i.e. 7 s, the vertical displacement ofA is zero and that of B is 49 m so for relative motionof B w.r.t. A (u2sinq2–u1sinq1)× 7 = 49

Þ u2sinq2 – u1sinq1 = 7 m/s

7. Ans (4)

v =2 2 2

ˆˆ ˆ(4 1)i (2 2) j (3 3)k

3 4 0

- + + + -

+ +=

ˆ ˆ3i 4j5+

v v=r r

v = 10ˆ ˆ3i 4 j5

æ ö+ç ÷è ø = ˆ ˆ6i 8 j+

8. Ans (1)

For v = 0, x = 1,4 and a = vdvdx

so x 1a = = 0 × dvdx

=0 ; a|x=4 = 0 × dvdx

=0

9. Ans (4)

t1 = 2hg

t2 = 2 2h

g´

A

B

C

D

h2h

3h

t=0

t1

t2

t3t3 = 2 3h

g´

Required ratio t1 : (t2 – t1) : (t3 – t2)

= 1 : ( )2 1- : ( )3 2-

10. Ans (3)

H

t=T

t=t

h

t=0

h = H –12

g(t–T)2

11. Ans (1)Velocity after 10 sec is equal to

0 + (10) (10) = 100 m/sDistance covered in 10 sec is equal to

12

(10)(10)2 = 500m

Now from v2 = u2 + 2as.Þ v2=(100)2–2(2.5)(2495–500)=25 Þv=5ms–1

12. Ans (1)SB = SA + 10.5

2t10t 10.5

2= +

t2 = 20t + 21t2 – 20t – 21 = 0t = 21 sec

13. Ans (3)Area of a-t graph = change in velocityAccording to question A1 = A2

y

t – 4A1

A2

0

a10

time4t

Here y 10

t 4 4=

- Þ y = 2.5 (t – 4)

So [ ]1 1(4)(10) (t 4) 25(t 4)

2 2= - -

Þ (t – 4)2 = 16 Þ t – 4 = 4Þ t = 8 s

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14. Ans (1)

Flag blows in the direction of resultant of WVr

& BV-r

Vw = 6m/sVB

V =2 m/sR

V = 4m/sB/R

–VW

W BV V 6j (4i 2j)- = - +r r

$ $ $ = 4( i j)NW- +$ $

Þ N-W direction

15. Ans (4)

Distance covered by :

train I = (Area of D)train I = 200 m

train II = (Area of D)train II = 80 m

So the seperation = 300 – (200 + 80) = 20 m.

16. Ans (3)

Required ratio = ´

´

3 2 3

4 2 3 = 3 : 4

v =B

v = 3m/sA

4m/s300m

o

17. Ans (4)

For (1) Q Distance ³ Displacement

\ Average speed ³ Average velocity

For (2) ar

± 0 Þ D vr ± 0

velocity can change by changing its direction

For (3)Average velocity depends on displacement in

time interval e.g. circular motion ® after one

revolution displacement become zero hence

average velocity becomes zero but instantaneous

velocity never becomes zero during motion.

NLM & Friction1. Ans (3)

When the astronaut is outside the spaceship, thenet external force (except negligible gravitationalforce due to spaceship) is zero as he is isolated fromall interactions.

2. Ans (2)(1) ® (P,R,S,T) ; (2) ® (Q) ; (3) ® (Q,R,S) ; (4) ® (P, T)I : aA = aB = 0 & N ¹ 0

II : A

Fa

2m= , B

Fa

m= & N = 0

III : A

Fa

m= ,

B

2F Fa

2m m= = & N = 0

IV : A

2F Fa

2m m= = , B

Fa

m= & N = 0

V : A B

2Fa a

3m= = & N ¹ 0

3. Ans (2)Acceleration of block,

F mg 1a a F g

m m- m æ ö= Þ = - mç ÷

è ø

From graph ; slope = 1 1

m 2kgm 2

= Þ =

and y-intercept; – mg = –2 Þ m = 0.24. Ans (2)

N = Fcos37° = 4

F5

and to start upward motion

Fsin37° = 10 g + mN

37°FFsin37° 10g mN

Fcos37°N

( )3 4F 100 0.5 F

5 5æ ö= + ç ÷è ø

ÞF = 500 N

5. Ans (1)Just after release T = 0 due to non–impulsive natureof spring. So acceleration of both blocks will beg¯

6. Ans (2)

FBD of block w.r.t. wedge 30°

N

mg

mg 30°

Acceleration of block w.r.t wedge

=

3 1mg mg

2 2m

æ ö- ç ÷è ø =

3 1g

2

æ ö-ç ÷è ø

Now from S = ut + 21at

2, 1 = 21 3 1

gt2 2

æ ö-ç ÷è ø

Þ t = ( )4

3 1 g- = 0.74 s

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7. Ans (3)Block starts sliding when kt0 = µmg

mµ µs, k F = kt

so for t £ t0, a = 0

and for t > t0, a = kF µ mgm

- = k

kt– µ

m8. Ans (3)

Acceleration of car along slope= g sinq – µgcosq

= 10 × 12

– (0.5)(10)3

2

æ öç ÷è ø = 5 – 4.33

= 0.67 ms–2

Now from v2 = u2 + 2as

v = 26 2(0.67)(15)+

= 36 20.1+ = 56.1 = 7.49ms–1

9. Ans (1)

FBD of ringf

Mg

Here 'M' is in equilibrium.So net force on 'M' must be zero.\ f = Mg (upwards)

10. Ans (3)

tanq = hR

= m Þ h = µR hqR

11. Ans (3)Let forces acting on mass m in equilibrium arer r r r r

1 2 3 4F,F ,F ,F ,F

+ + + + =rr r r r r

1 2 3 4F F F F F 0 [equilibrium condition]

Þ + + + = -r r r r r

1 2 3 4F F F F F ...(i)

After cutting the string with force r

F , the net forceon mass m

= + + +r r r r r

net 1 2 3 4F F F F F Þ = = -

r r

r netF Fa

m m12. Ans (2)

For (1) :

( ) ( )= + + - + =r

netˆ ˆ ˆ ˆ ˆF Fi Fj Fi Fj 2Fj

For (2) :

( ) ( ) ( )= + + - - = - -r

netˆ ˆ ˆ ˆ ˆF Fi Fj 3Fi Fj 3 1 Fi

For (3) :

( ) ( )= + + - - =rr

netˆ ˆ ˆ ˆF Fi Fj Fi Fj 0

For (4) :

( ) ( )= + + - = - +r

netˆ ˆ ˆ ˆ ˆF Fi Fj 2Fi Fi Fj

13. Ans (3)Maximum tension in string Tmaxsin 30° = 40

Þ Tmax = 401

2

= 80 N

For monkey Tmax – mg = ma Þ a = 805

– 10

= 6 ms–2

14. Ans (2)Velocity of Block 'A' at any timev1 = v0 – mgt

and velocity of 'B' is v2 = mg

tM

m

here v1–t graph is a straight line of negative slopeand v2–t graph is also a straight line of +ve slope.

15. Ans (3)Acceleration of system

= 4 14 1

-æ öç ÷è ø+ g =

35

× 10 = 6 ms–2

Relative acceleration of blocks = 12 ms–2

Now 2 + 4 = 1/2 (12) t2 Þ t = 1 sec16. Ans (3)

Acceleration of B,

aB = ( ) ( )

A

B

1 m gm g g2m 2m 4

m= =

aAB

m aA B

µm gAP & D of block A w.r.t. B

aAB = A A B

A

m g m am

m + = µg +aB = g g 3g2 4 4

+ =

Work, Power & Energy1. Ans (1)

According to mechanical energy conservation

between A and B

( ) 2 21mg 5 O mv v 10g

2= + Þ = ...(i)

5mB

MgN

V GP =0e

A

According to centripetal force equation

2mvN mg

r+ = for N = mg;

2mv2mg

r=

Þ 2v 10g

r 5m2g 2g

= = =

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2. Ans (2)

By conservation of mechanical energy [ betweenpoint A and B]

( )2 21 1mu mgR 1 cos mv

2 2= + q +

N=0Rq

R

uA

B

mg

v

q

( )2

1 1 1m 95Rg mgR 1 cos mgRcos

2 5 2æ ö = + q + qç ÷è ø

95 452 2cos cos 3cos

25 25Þ = + q + q Þ q =

15 3cos 53

25 5Þ q = = Þ q = °

3. Ans (4)

Work done by the force = Work done against gravity(Wg) + work done against friction (Wf)

F

dh

dx

q

ds

( )= q = q = =ò ò ògW mgsin ds mg dssin mg dh mgh

and

( )= m q = m q = m = mò ò òfW mgcos ds mg dscos mg dx mgx

4. Ans (3)Total mechanical energy = kinetic energy + potential

energy = 15+ [32–6(3)+14] = 15 +5 = 20 J

5. Ans (2)

At maximum speed ( i .e. maximum kineticenergy ), pot ent ia l energy i s min imum

U = y 2 – 6y + 14 = 5 + (y–3) 2

which is minimum at y=3 m so Umin = 5J

Therefore Kmax = 20 – 5 = 15 J

2max max

1mv 15 v 30 m/s

2Þ = Þ =

6. Ans (3)

By applying work energy theoram change inkinetic energy = W

g + W

ext.P

0 = mg(l cos 37° – l cos 53°) + Wext P

Þ 0 = 50 × 10 × 13 45 5

é ù-ê úë û+ W

ext P

Þ Wext

= 100 joule

7. Ans (4)

Work F.dr=rr

, Work =–0

(0.5)(5)Rdq

qò \F=mN

Þ [work] = (2.5) (R) (2p) = –5 J

8. Ans (4)

By applying work energy theorem

æ ö æ öD = = =ç ÷ ç ÷

è ø è ø

22 2

21 1 1

v 1 v mvKE fd m t t

t 2 t 2t

9. Ans (2)

Slope of v–t graph Þ Acceleration

Þ –10m/s2

Area under v–t graph ® displacement Þ 20 m

work = f.sr r = 2 (10) (20) Þ –400 J

OR

Work done = DKE = 2 2m(0 20 ) –400J21

- =

10. Ans (2)

ac = k2 rt2 Þ

22 2v

k rtr

=

Þ v2 = k2r2t2 Þ v = krt Þ aT =

dvdt

= kr

P = = =òr r 2 2

Tma .v m(kr).(krt) mk r t

11. Ans (3)

The gain in PE = Wboy

= stored energy lostby spring.

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12. Ans (3)

Tension at any point 2mv

T mgcos= q +l

As v2=2glcosq. So T = 3mgcosq

Given 3mg cosq = 2mg

Þ cosq =23

Þ q = cos–1 23

æ öç ÷è ø

13. Ans (1)

COME Þ K1 + U

1 = K

2 + U

2

0 + mgl (1–cos60°) = 12

mv2 + 0 Þ v = gl

14. Ans (1)

COME : K1 + U

1 = K

2 + U

2

0 + mg (4R) =12

mv2 + mg (2R) Þ mv2 = 4mgR

Forces at position 2 :

N=2mv

R–mg = 4mg – mg = 3 mg

15. Ans (1)

COME : KA + U

A = K

B + U

B

0 + mg × 25 = + ´2B

1mv mg 15

2Þ 2

Bmv = 20mg

Forces at B : N = mg –2Bmv

R=0 ÞR = 20 m

16. Ans (1)

Area of graph

=vdv

P.dx mv.a.dx mv. dxdx

æ ö= = ç ÷è øò ò ò

= ( ) ( )- -

= =´ò

v 3 3 32

u

m v u 10. v 1mv dv

3 7 3

= 12

(4+2) × 10 Þ v = 4 m/s

17. Ans (1)

For upward motion : mgh + fh =12

m × 162

downward motion : mgh – fh =12

m× 82 Þ h = 8m

18. Ans (1)

( ) ( )ˆ ˆ ˆ ˆ3i 4 j 8i 6jW F. SP 8W

t t 6

+ × +D D= = = =

D D

rr

19. Ans (4)

( ) ( )2 21 12 20 kx mgx 0 mv- + -m = - Þ v=8 m/s

Centre of mass & Collision1. Ans (1)

From graph e = h 40 2H 90 3

= =

Kinetic energy of the ball just after second bounce

= ( ) ( ) ( )= =22 4 2 41 1

m e u me u e mgH2 2

= ( ) ( ) ( )æ ö =ç ÷è ø

42

2 10 81 320J3

2. Ans (2)Velocity of air molecule after collision = 2v . Thenumber of air– molecules accelerated to a velocity2v in time Dt is proportional to AvDt. Therefore F

= pt

DD

µ (AvvDt) vt

2æ öç ÷è øD Þ F µ 2Avv2

3. Ans (2)Impulse = Change in momentum

1(4i j) 1(6i j) 2i= + - + = -$ $ $ $ $

Which is perpendicular to the wall.

Component of initial velocity along i 6i=$ $

Þ Speed of approach = 6 m/sSimilarly speed of separation = 4ms–1

4 2e

6 3Þ = =

4. Ans (1)As net force on system = 0 (after released)So centre of mass of the system remains stationary.

5. Ans (1)Let x be the displacement of the plank.Since CM of the system remains stationary

north

6m

south x

Then 80(6 – x) = 40xÞ x = 4 mÞ Plank moves 4m towards south.

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6. Ans (1)

5 m/s 5 m/s

10 m/s v

Just before collision Just after collision

By definition of e : 2 1

1 2

v ve

u u-

=-

;

we have0.8 =v 5

10 5+-

Þ v = 1 m/s

7. Ans (4)At maximum compression of spring, velocities ofblock B and C are same (say v0)then by conservation of linear momentum 3(2)

= (3+6)v0 Þ v0 = 23

m/s

At this instant energy stored in spring

= ( ) ( ) ( )2

21 1 23 2 3 6

2 2 3æ ö- + ç ÷è ø = 6 – 2 = 4 J

8. Ans (3)Centre of mass is given by -

( )

( )

666 64 5

0 0 0cm 6 6 654 4

0 0

0

xx kx dx x dxxdm 6x 5m

dm xkx dx x dx5

æ öç ÷è ø

= = = = =æ öç ÷è ø

ò òòò ò ò

9. Ans (2)Vertical velocity just before collision

vy = 2gh 2 10 5= ´ ´ = 10 m/s

10ms–1 10ms–1

10ms–1

Þ Kinetic energy of ball just after collision

= 12

× 1 × 102 = 50 J

10. Ans (2)

+ ´ + ´ -= =

+

r r $ $ $r 1 1 2 2

cm1 2

m v m v 1 2i 2 (2cos30i 2sin30j)v

m m 3

2 2 3 2i j

3 3

æ ö+= -ç ÷è ø

$ $

11. Ans (2)

Impulse = change in momentum

= pf – pi = 1 × 10 –1× (–25) = 35kg m/s ()

12. Ans (2)

v vm m 2m

v'

0= 2mv – mv + mv' Þ v' = –v

Total mechanical energy released

= 12

(m+m+2m)v2 = 2mv

2

13. Ans (1)

COLM :

Along horizontal

usinq

ucosq

u

qv

0 = m(u cos q – v) – 4mv Þ v = ucos

5q

\ velocity of shell along horizontal w.r.t ground

= u cos q –ucos

5q

=45

(u cos q)

Time of flight T=2u sin

gq

\ x = horizontal displacement

24 2usin 4u sin2ucos

5 g 5gæ öq qæ ö= q =ç ÷ ç ÷è ø è ø

14. Ans (3)

Along tangent

u cos q = v sin q...(i)

q

qu

qv

Along normal

2v cose cot

usinq

= = qq

= cot260° =

13

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15. Ans (3)

2m

u

mÞ

2m

v (Let)

m

u

COLM Þ 2mu + 0= 2mv + mu

Þ v = u2

2 1

2 1

v v u u/2 1e

u u 0 u 2æ ö- -æ ö= - = - =ç ÷ç ÷ è ø- -è ø

16. Ans (4)

2netcm

F (0.2)(3)(10)a 2ms

Totalmass 1 2-= = =

+

ORAcceleration of 1kg w.r.t. ground

=(0.1)(10)=1ms–2

Accceleration of 2 kg w.r.t. ground

= 2(0.2)(3)(10) (0.1)(10) 5ms

2 2--

=

21 1 2 2cm

1 2

m a m a (1)(1) (2)(5/ 2)a 2ms

m m 1 2-+ +

= = =+ +

17. Ans (2)mu – I1 = –mu \ I1 = 2mu &

mu – I2 = 0 Þ I2 = mu (for IInd ball) \ I2 = I1/2

Circular & Rotational Motion1. Ans (2)

ddtw

a = = 3t–t2 Þ t

2

0 0

d (3t t )dtw

w = -ò ò

Þ w = 2 33t t

2 3- Þ at t =2 s,

w = 103

rad/s

2. Ans (1)

v = 18 × 5

18=5 ms–1 and

aCP= vR

ms2

225

25 2

1

2= = - ,

aT = - = - -dvdt

ms12

2 aT

aCP®

®

v®

anet®

anet = 1

2

12

32

0862 2

2FHG

IKJ + F

HGIKJ = = -. ms

3. Ans (1)The tangential and centripetal acceleration isprovided only by the frictional force.

q

vR

2

f

a0

Thus, f sinq = ma0 and f cosq = 2mv

r =

( )2

0 0m a t

r

Þ f = m ( )

+4

0 020 2

a ta

r= ma0 +

2 40 02

a t1

r

Þ m = +2 40 0

0 2

a tmg ma 1

r

Þ m = +2 4

0 0 02

a a t1

g r

4. Ans (1)

h = l(cosq2 – cosq1)

q2

q1

l

h

mgcosq2mg

T

lcosq1

lcosq2

q2

Applying conservation of mechanical energy at

point A & B

21mv

2= mgh Þ v= 2gh = 2 12g (cos cos )q - ql

5. Ans (1)Loss in P.E. = Gain in K.E.

mg3l

+ mg 2 2

2 22 1 2mg m m m

3 2 3 3

æ öæ ö æ ö æ ö+ = + + wç ÷ç ÷ ç ÷ ç ÷è ø è ø è øè ø

l l ll l

B B

36g 2 36g 8gv

14 3 14 7Þ w = Þ = w = =

l ll

l l

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6. Ans (1)

( )w

w = a Þ w - = - +ò òt

t4 3 20

2 0

d dt 2 t t t

Þ w = + - +2 3 42 t t t

tt 3 4 5

1 0 0

t t td dt 1 2t

3 4 5

q æ öq = w Þ q - = + - +ç ÷è øò ò

3 4 5t t t1 2t

3 4 5Þ q = + + - +

7. Ans (4)

For block : mg – T = ma .......(i)

For disk (pulley) 2MR

TR I2

= a = a

R

T

T

am

mg

M

a

But aR

a = so Ma

T2

= ...(ii)

Therefore -= Þ = +

mg T 2m mg 2m1

T M T M

( )´Þ = = =

´æ ö æ ö+ +ç ÷ ç ÷è ø è ø

mg 1.2 10T 6N

2m 2 1.21 1

M 2.4

8. Ans (4)

By impulse momentum theorem

2m 3JJ

3 m= w Þ w =

ll

l

m

J

KE of rod = 22 2

21 1 m 3J 3JI

2 2 3 m 2m

æ ö æ öw = =ç ÷ç ÷ è øè øl

l

Linear velocity of midpoints = 3J

2 2mæ ö

w =ç ÷è øl

9. Ans (4)

In this situation

( )1 21 2

1 2 2

m m gaT T a

Ir m mr

-¹ Þ a = Þ =

+ +

m1m2

T1T2

P

a a

a

10. Ans (4)

r1 = 0.5 × 10–2 ; r2 = 0.15 × 10–2

Þ 2p r1 × 3 = 2p × r2 × n

\ 0.5 × 10–2 × 3 = 0.15 × 10–2 × n Þn = 10

11 Ans (2)

Rod rotates about its one end in a horizontal plane

\ t = Ia ÞMg 5L2 6

´ =2ML

3 × a Þ

5g4L

a =

12. Ans (2)

\ Book does not rotate so for rotationalequilibrium the net torque becomes zero.

weighttr + t =

rrman 0

\ mantr = – weightt

r = – b

W anticlockwise2

é ù´ê úë û13. Ans (2)

v = wr^ = wr sinq v 3

r sin 8Þ w = =

qrad/s

14. Ans (2)

12

Mv2 +12

2MR2

× 2

2

vR

= mgh Þ h =23v

4g

15. Ans (2)

Max height H = 2 2(v sin45) v

2g 4g=

45°

v

Hv cos45°

Angular momentum

= mv cos45° × H = mv

2×

2v4g

= 3mv

4 2g

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16. Ans (3)Work required is minimum for rotation when theMI is minimum. It is minimum about that axis whichpass through CM & perpendicular to the length.

0.3kg 0.7kg

1.4m

C.M from 0.3kg mass = 0.7 1.40.3 0.7

´+

= 0.98

17. Ans (3)Angular momentum of this disc about origin

= 2MR

2w + M(wR) × R =

32

MR2w

R

18. Ans (3)Moment of inertia about an axis passing throughintersection point

= mr2 × 4 = 2 × 2

14

æ öç ÷è ø

× 4 =12

2kg

2kg

2kg

2kg

1/2m 3 F2

=12 3

F 1 2N2

=

Torque of force about this point = 12 × 12

=6

\ Angular acceleration

a=It

=6× 2 = 12rad/s2

19. Ans (2)For translational equilibrium T1 + T2 = 2mg

T2T1

mgmg0.5m

0.8m

For rotatory equilibriumTake torque about extreme left point

= mg[0.5 + 0.8] = T2 × 1T2 = mg × 1.3 \ T1 = 2mg –T2 = 0.7 mg

\ Ratio 1

2

T 0.7mg 7T 1.3mg 13

= =

20. Ans (2)

w =21I mgh

2 where I=

l2m

3

Þ w =l

221 m

mgh2 3

Þ h = wl

2 2

6g

21. Ans (1)a

r 22

= Þ 2

2 ar

2=

M

O

a

M

O

Cw

ar

Net torque about O is zero. Therefore, angularmomentum (L) about O will be conserved

æ ö = wç ÷è ø 0a

Mv I2 = (ICM + Mr2) w

2 22Ma a 2

M Ma6 2 3

ì üæ ö æ öï ï= + w = wí ýç ÷ ç ÷è ø è øï ïî þÞ

3v4a

w =

22. Ans (1)

Mass of the whole disc = 4 M

Moment of inertia of the disc about the given axis

12

= (4M)R2 = 2MR2

\ Moment of inertia of quarter section of the disc

14

= (2MR2) = 12

MR2

23. Ans (1)

The velocity of all the points of pure rolling bodywhich lie above the centre is more than VCM andbelow the centre is less than VCM.

24. Ans (4)

12

mv2 + 12

I 2 2v 3v

mgR 4g

æ öæ ö =ç ÷ ç ÷è ø è ø \ I =

12

mR2

\ Body is disc.

25. Ans (2)

Acceleration w.r.t. centre of wheel are shown, the

relative acceleration = 22v

R

A

B

v /R2

v /R2

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26. Ans (2)Let ƒ be the force of friction between the shell andthe horizontal surface.For translational motion F + ƒ = ma

For rotational motion F R – ƒR = Ia = IaR

[Q

a= Ra for pure rolling] Þ F – ƒ = Ia

R2

By solving above equations. 2F = mI

R+

FHG

IKJ2

a =

m m+FHG

IKJ

23

a = 53

ma Q I mRshell

=LNM

OQP

23

2

Þ F = 56

ma Þ a=65

Fm

27. Ans (3)

v2R

w = , I = 25MR

2 Þ KE

= 2 21 1Mv I

2 2+ w =

1316

Mv2

28. Ans (3)For (1) : Only two vertical forces are acting on the

rod (Mg & N)

For (2) : Initially N = Mg

For (3/4) : FBD of rod (just before striking)

A

N

L/2

C

Mg

B

By taking torque about A and C

= a = a2

A

L MLMg I

2 3....(i)

æ ö= a = aç ÷è ø

2

C

L MLN I

2 12...(ii)

By dividing (i) by (ii) we get N = Mg/4

Gravitation1. Ans. (3)

2. Ans. (2)3. Ans. (3)

T2 µ a3

a = + +

=max min 1 2r r r r2 2

T = 3/2

1 2r r2+æ ö

ç ÷è ø

T µ (r1 + r2)3/2

4. Ans (2)

The gravitational field at P is due to the massenclosed. (Gauss's theorem)

\ The force acting = ¢Gm m

r m is the mass of the

particle released.

\ F = – G. r.p 3

2

4 r m3 r

m'O

Mr

P

= –æ öp rç ÷è ø4

G m r3

* In the option, the negative sign is omitted. Thissign is important because this is the restoring forceon the mass which will make S.H.M.

5. Ans (4)

S2 is correct because whatever be the g, the sameforce is acting on both the pans. Using a springbalance, the value of g is greater at the pole.Therefore mg at the pole is greater. S4 is correct.S2 and S4 are correct.

6. Ans (3)

Speed of satellite V = GM

r

Þ B A

A B

V r 4R2

V r R= = = Þ VB = (3V)(2) = 6V

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7. Ans (1)

d

R m

The gravitational force between masses m and M is

1 2

GMmF

d= ....(i)

Mass of the lead sphere before hollowing,

M = 4

R3

3p r ....(ii)

If M1 is the mass of the removed partis of sphere,

then 1

4 RM

3 2

3æ ö= p rç ÷è ø

1 4 MR

8 3 83æ ö= p r =ç ÷

è ø(Using (ii))

Force between m and hollow sphere M1 is

2 2 2

GMmGMm8F

R Rd 8 d

2 2

= =æ ö æ ö- -ç ÷ ç ÷è ø è ø

....(iii)

So the magnitude of gravitational force on the small

sphere of mass m by the hollowed out lead sphere

is, F = F1 –F2

2 2

GMm GMm

d R8 d

2

= -æ ö-ç ÷è ø

(Using (i) and (iii))

= GMm2 2

1 1d R

8 d2

é ù-ê úæ öê ú-ç ÷ê úè øë û

2 2

GMm 11

d R8 1

2d

é ù= -ê úæ öê ú-ç ÷ê úè øë û

8. Ans (2)

For the line 4y = 3 x + 9

4dy = 3dx; 4dy – 3 dx = 0....(i)

For work in the region,

dW = ( )ˆ ˆE. dxi dyj+r

= ( ) ( )ˆ ˆ ˆ ˆ3i 4 j . dxi dyj- +

= 3dx–4dy (from equation (i)) = 0

9. Ans (2)Total mass of earth

3 33

1 2

4 R 4 RM R

3 2 3 8

æ öæ ö= p r + p - rç ÷ ç ÷è ø è ø

( )3

1 2

4 RM 7

24p

= r + r

Acceleration due to gravity at earth's

surface = ( )1 22

GM 4 GR7

24Rp

= r + r

Acceleration due to gravity at depth R2

from the

surface

2

11

2

4 RG

4 G R3 26R

2

é ùæ öp rê úç ÷è ø p rë û= =æ öç ÷è ø

Now according to question

( )1 2 1 1

2

4 GR 7 4 GR 724 6 3

p r + r p r r= Þ =

r

10. Ans (2)

V1 = 2 2 1/2 1/ 2

GM G(5)G

(R x ) (16 9)= =

+ +

2 2 2 1/2

GMV

(R x )=

+4

3 3Ö

1/2

G(5) G56(9 27)

Þ Þ+

work done = m[V2 –V1] = G6

= 1.11 × 10–11 Joule

11. Ans (4)

Figure shows a binary star system.

MB RB

RAMA

The gravitational force of attraction between thestars will provide the necessary centripetal forces.In this case angular velocity of both stars is thesame. Therefore time period remains the same.

2Tpæ öw =ç ÷è ø

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12. Ans (2)

m 4m

r

x

2

G m

x

´= 2

G 4m

(r – x)

´ Þ x =

r3

Potential at point the gravitational field is zerobetween the masses.

V = – 3Gm

r –

3 G 4m2r

´ ´

= 3Gm

–r

[1 + 2] = 9GM

–r

13. Ans (3)

2

h 2

g g hg 1 9

9 Rh1

R

æ ö= = Þ + =ç ÷è øæ ö+ç ÷è ø

Þ 1+hR

=3 Þ h = 2R

14. Ans (4)Work done = Uf – Ui = Ui

= 1 2i

GM mU

R

- -

-

´ ´ ´ ´=

´

11 3

2

6.67 10 100 10 1010 10

-= ´ 106.67 10

15. Ans (3)

e

GMv

R= e

Mv

Rµ

pe e e e1

e p ee

Rv M R / 10

R M R 10Mev

M= ´ Þ ´

1e

11 110v

= ; 1ev 110= km/s

16. Ans (1)

2n

GMmF m r

r= = w

Þ n 1

GMr +

w = Þ ( )n 12

2T r

+p= =

w

17. Ans (2)

2

GM gg'

49(R h)= =

+ ; w' = mg49

=1049

= 0.20 N

Apparent weight of the rotating satellite is zerobecause satellite is in free fall state.

18. Ans (3)g' = g – w2r cos 60g' = g – w2R cos2 60g' = 0, g = w2R cos2 60

60°

m rw 2

r = R cos 60°

60°

4gR

= w , 2 R R

t 24g g

p= = p = p

w

Properties of Matter & Fluid Mechanics1. Ans (2)

B = P

V / VD

-D Þ VV

D=–

PB

D

m

VPatm

m

VP + Patm D

D rP=h g h

We know P = Patm + h gr and m = Vr = constant

d V dV 0r + r = Þ drr = –

dVV

i.e. Drr =

PB

D

Þ Drr =

1100

=h gBr

[assuming r = constant]; B

h g100

r = = 1

100K

Þ5

6

1 1 10h g

100 50 10-

´ ´r =

´ ´

Þ h = 5

6

105000 10 1000 10-´ ´ ´

= 3100 10

50´

m = 2km

2. Ans (1)Tension in wire at lowest positionT = mg+mw2r

So elongation (foLrkj) Dl = 2

2

FL (mg m L)LAY r Y

+ w=

p

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3. Ans. (1)4. Ans. (2)

F / AY\ =

Dll

2YA

FA

D\ =

l

l

D=

l2YA

FV

here V = volume of wire

F µ A2 Þ

2 2

2 2

1 1

F A 3A9

F A A

æ ö æ ö= = =ç ÷ç ÷ è øè ø

F2 = 9F5. Ans (1)

11 26 4

F 20 1Y 2 10 N / m

A 10 10- -

D ´= = = ´

´l

l

6. Ans (2)W = 8pT [(r2

2) – (r12)]

= 8 × p × 0.03 [25 – 9] × 10–4

= p × 0.24 × 16 × 10–4

= 3.8 × 10–4 p= 0.384 p mJ » 0.4 p mJ

7. Ans (4)By volume conservation

34R

3p = 2 34

r3

æ öpç ÷è ø

Þ R = 1

32 rSurface energy E = T (A)

= T (4p R2) = T (4p 22/3 r2) = 28/3 p r2 T

8. Ans (1)weight = mg = 1.5 × 10–2 N (given)length = l = 30 cm (given)

= 0.3 m2Tl = mg

2mg 1.5 10T

2 2 0.3

-´= =

´l= 0.025 N/m.

9. Ans (1)The free liquid surface between the plates is cylindricaland curved along one axis only so radius of curvature

r = d2

and P0 – P = s 2sr d

= Þ P = P0–2sd

10. Ans (4)

According to archemedes principle

h

PV

P1Upthrust = Wt. of fluid displaced.\ Fbottom = Ftop + V r g

= P1 × A + V r g= (h r g) × (pR2) + V r g = r g [pR2 h + V]

11. Ans (4)

l decreases as the block moves up. h will alsodecreases because when the coin is in the waterit will displace euqal volume of water, whereas whenit is on the block an equal weight of water isdisplaced.

12. Ans (3)

Let V1 volume of the ball in the lower liquid then

V r g = V1r2g + (V – V1) r1g

ÞVg(r–r1)=V1g(r2 – r1) Þ1 1 1

2 1 1 2

VV

r - r r - r= =

r - r r - r

13. Ans (1)

Pressure P = hrg

Þ h = ´ ´3

20001.06 10 9.8

= 0.192m

14. Ans (3)

From equation of continuity

v1A1= v2A2

and

2 22v u 2gs- = ; 2

2v 1 2 10 0.15- = ´ ´ Þ v2 = 2 m/s

Hence A2 = 4

1 1

2

v A 1 10v 2

-´= = 5 × 10–5m2

15. Ans (1)

Equating the rate of flow

2 2(2gy) L (2g 4y) R´ = ´ p

Þ L2 = 2pR2 Þ R = L

2p

16. Ans (2)

1v 2gx= and v2 = 2g(x h)+ .

Let cross section area of hole is a then rate of

flow = av

force = v(avr) =arv2

h

xv1

v2

\ F1 = arv12 and F2 = arv2

2

Net force

= (F2 – F1) = ar (v22 – v1

2) = ar(2g (x+h)–2gx)

= 2argh

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17. Ans (1)

+ r =2open closed

1P v P

2

( ) ( )- ´ - ´Þ = =

r

5closed open

3

2 P P 2 3.5 3 10v

10

10 m/s

18. Ans (1)

Here æ öç ÷è ø

dh–A

dt = (av) = a 2gh

Þ =ò òh t

–1 2

0 0

ah dh 2g dt

A Þ t = A 2ha g Þ t µ Öh

Therefore for height 2h, t' µ Ö2h Þ t' = Ö2t

19. Ans (3)

( )( )

sss

g g

vv 0.1 m/s

v

r - r= Þ =

r - rl

l

20. Ans (1)

( )r - rr - r = Þ = 1 22

1 2 T T

VgVg Vg kv v

k

Thermal Physics

(a) Thermal Expansion1. Ans (2)

Slope of line AB

= C 100 0 100 5

212 32 180 9FD -

= ==D -

2. Ans. (1)3. Ans (3)

1 2D = D + Dl l l

mi x (3 ) T T 2 (2 ) Ta D = aD + a Dl l l

m ix

53

a = a

(b) Calorimetry4. Ans (1)

Heat = mgh

J´ ´

= =5 9.8 30

Cal 350Cal4.2

5. Ans (1)2kg ice – 20°C + 5 kg water 20°Qgain = Qlost

12 20 M 80 5 1 20

2´ ´ + ´ = ´ ´

M = 1 kgWater = 5 + 1 = 6 kg

6. Ans (4)

10 C 30 C1kg ice 4.4kg

- ° °+ of water

Heat gain = Heat loss

11000 10 1000 80 1000

2´ ´ + ´ + (T – O)

= 4.4 × 1000 × 1 × (30–T)5.4 T = 4.4 × 30 – 85T = 8.7°C

7. Ans (4)DQ = mL + mSDT + mL + WDT= 10 × 80 + 10 × 1 × 100 + 10 × 540 + 10 × 100= 8200 Cal

(c) Heat Transfer8. Ans (2)

=L

RKA

Þ =1 2

1 2

L LK A K A

r = =1 1

2 2

L K 5L K 3

9. Ans (2)

2rQ µ

l Þ ( )

21

22

Q r 2LQ L 2r

= ´ Þ Q2 = 2Q

1

10. Ans. (1)11. Ans (1)

Req = R1 + R2 + R3

R1

K/2

R2 R3

5K K

eq

3 L L LKAAK 5KA KA2

= + +l

eq

15K

16= K

12. Ans (1)

Qt

DD

= same

So( )KA 20 10-

l=

( )2KA 10 - ql

Þ 2 q =10 Þ q = 5°C

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13. Ans (2)

q - q q + qæ ö= - - qç ÷è ø

1 2 1 20K

t 2

Þ - +æ ö= - - qç ÷è ø

0

95 90 95 90K

30 2.......(i)

Þ - +æ ö= - - qç ÷

è ø0

55 50 55 50K

70 2.......(ii)

By dividing equation (i) by (ii),

- q

=- q

0

0

185 273 105 2 Þ 735 – 14q0 = 555 – 6q0

8q0 = 180 Þ q0 = 180

8 = 22.5 °C

14. Ans (4)lT = constant

1 2

2 1

TT

l=

l

2 02 1

10

T 4 4T T

3T 3 34

l= = Þ =

l

4 442 1

4 4 256E ' AT A T E E

3 3 81æ ö æ ö= s = s = ´ =ç ÷ ç ÷è ø è ø

15. Ans (1)

1T µ

l4 4

1 1 1 1 2

2 2 2 2 1

Q A T AQ A T A

æ ö æ öl= =ç ÷ ç ÷lè ø è ø

2 44

2 4 2

6 15 13 9

18 5 3= ´ = ´ =

(d) KTG16. Ans (1)

5 3

23

PV 1.3 10 7 10N

KT 1.38 10 273

-

-

´ ´ ´= =

´ ´= 2.41 × 1023

17. Ans. (4)18. Ans (3)

PVT

R=

m

PVP C

Ræ ö

=ç ÷mè ø

P2V = C Þ 2 1P

Vµ

P

V

19. Ans (4)PV2 = constant

2TV

V = constant

TV = constant

1T

Vµ on cooling exp.

(e) Thermodynamics20. Ans (2)

Slopeadiabatic

µ gSlope of 1 < Slope of 2

g1 < g2

So, 2 is monoatomic & 1 is diatomic21. Ans (2)

TVg–1 = constant

1

2 1

1 2

T VT V

g-æ ö

= ç ÷è ø

51 2/3 23

22

T 27 27 3 9300 8 8 42

-æ ö æ ö= = = =ç ÷ ç ÷è ø è ø

2

9T 300 675K

4= ´ =

DT = T2 – T1 = 675 – 300 = 375 K22. Ans (1)

2 ® 3 volume constant

2

V

P

3

1DQ = – 40DW = 0 (V constant)DU = DQ = –40U3 – U2 = –40 ...(1)I ® 2 temperature constantT1 = T2

U1 = U2 ...(2)from (1) & (2)U1 – U3 = 40DU3®1 = 403 ® 1 Adiabatic processDW = –DUDW = –40

23. Ans (1)For cyclic process

DU = 0

DQ = DW

WAB + WBC + WCA = DQ

10 + 0 + WCA = 5

WCA = –5 J

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24. Ans (1)DQABC = 90 J and DWABC = 30 J

\ dU = 90 – 30 = 60 J

if DWADC = 20 J

DQADC = dUADC + DWADC = 60 J + 20 J = 80 J25. Ans (1)

P p

W R T R R 2100% 40%

5Q C T C 5R2

m D= = = = ´ =

m D

26. Ans (2)

RTP

Vm

=

P2V = C

2T

V CV

æ ö ´ =ç ÷è ø

2TC

V=

22 22

11

T VVT

= (Q V2 = 2V1 )

2T 2T=

27. Ans (2)

2

1

T W1

T Qh = - =

6

300 W1

900 3 10- =

´

6

2 W3 3 10

=´

W = 2 × 106 Cal = 2 × 4.2 × 106 J= 8.4 × 106 J

28. Ans (2)

2QW

b =

1

WQ

h =

1

10Q 100J

110

= =

Q2 = Q1 – W = 100 – 10 = 90J29. Ans (2)

b =-C

H C

TT T

Oscillation (SHM)1. Ans (2)

= + + + + ¥eff

1 1 1 1 1.............

k k 2k 4k 8k (For infinite

G.P. S¥ = -a

1 r where a = First term,

r = common ratio)

eff

1 1 1 1 1 1 1 21 ...........

1k k 2 4 8 k k12

é ùê úé ù

= + + + + = =ê úê úë û ê ú-ë û

so keff = k/22. Ans (2)

According to question F1 = – K1 x & F2 = – K2x

so n1 = 1

2pKm

1 = 6Hz ; n2 = 1

2pKm

2 =

8 HzNow F = F1 + F2 = – (K1 + K2)x Therefore n =

12p

K Km

1 2+

Þ n = 1

2p 4 42

12 2

22p pn m n m

m+

= n n12

22+ = 8 62 2+ = 10 Hz

3. Ans (2)

p= p Þ = p Þ =

22 2

1 1 1 21 1 1

m m 4 mT 2 T 4 k

k k T a n d

p= p Þ = p Þ =

22 2

2 2 2 22 2 2

m m 4 mT 2 T 4 k

k k T

m

k1

m

k2 k1

m

k2

Now = pm

T 2k '

where

æ ö æ öp pç ÷ ç ÷è ø è ø

= + Þ = =+ p p

+

2 2

2 21 21 22 2

1 2 1 22 21 2

4 m 4 mT Tk k1 1 1

k 'k ' k k k k 4 m 4 m

T T

;

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é ùpp ê ú pë û= =

+é ùp +ê ú

ë û

22

2 2 21 2

2 21 22

2 21 2

4 m4 m

T T 4 mk '

T T1 14 m

T T

2

2 21 2

m mT 2 2

k ' 4 mT T

\ = p = pp+

= +2 21 2T T

4. Ans (2)Percentage change in time period

T100%

TD

´ = 1

1002

D´

l

l [Q Dg = 0]

According to question 100D

´l

l = 4%

\T

100%T

D´ =

12

× 4% = 2%

5. Ans (1)

Centripetal acceleration ac = 2v

R & Acceleration

due to gravity = g

g geff

Rva

2

c =R

So

222

eff

vg g

R

æ ö= + ç ÷è ø

Þ T ime period

4eff 2

2

L LT 2 2

g vg

R

= p = p

+

6. Ans (2)Amplitude of damped oscillation is A = A0e

–gt

[from x = xme–g t]

at t = 1 min 0AA

2= so

00

AA e

2-g= or eg = 2

After 3 minutes A = 0Ax

so

300

AA e

x-g´= or x = e3g = (eg)3 = 23 = 8

7. Ans (3)

Resultant amplitude = a a a a12

22

1 22+ + cos Df

Þ a2 = 2a2 + 2a2cos(f1 – f2)

Þ cos(f1 – f2) = -12

Þ f1 – f2 = 120° or 23p

rad

8. Ans (4)

Maximum kinetic energy of the particle is,

Kmax = E – Umin = 36 – 20 = 16 J

Note : Umin is 20 J at mean position or at x = 2m.9. Ans (4)

With change in temperature dq, the effective lengthof wire becomes l' = l (1 + a dq)

'T ' 2

g= p

l and T 2g

= pl

HenceT ' 'T

=l

l = (1 + a dq)1/2 =

11 d

2+ a q

\ Percentage increase in time period

= T ' T

100T-é ù ´ê úë û

= T '

1 100T

é ù- ´ê úë û

= d

1 1 1002

a qé ù+ - ´ê úë û= 50 a dq

10. Ans (1)

Here p p

= = = p Þ =w p p

l l

2

2 2 1T 2

g g But g = p2

therefore l = 1 m11. Ans (4)

In SHM a = –w2x. So 16 = –w2(–4) Þ w = 2

Þ Time period 2 2

T s2

p p= = = p

w12. Ans (3)

( )= - Þ = - = - + = - -2 dUU 5x 20x F 10x 20 10 x 2

dx

Acceleration ( )= = - -F

a 100 x 2m

so w = Þ w =2 100 10

Time period p p p

= = =w2 2

T10 5

sec

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13. Ans (4)

Let x=Asinwt where p

w =2T

For (A) : At t =

p pæ ö æ ö æ ö= = =ç ÷ ç ÷ ç ÷è ø è ø è ø2 T 3

x A sin A sin AT 6 3 2

,

p pæ ö æ ö æ ö= = =ç ÷ ç ÷ ç ÷è ø è ø è ø2 T 3

x A sin A sin AT 6 3 2

For (B) : = = w wdx

v A cos tdt

At =T

t6

,

p p wæ öæ ö= w = w =ç ÷ç ÷è øè ø2 T A

v A cos A cosT 6 3 2

For (C) : KE =

( ) ( )æ ö= = =ç ÷è ø

2

2 maxmax

v1 1 1 1mv m KE TE

2 2 2 4 4

& PE = TE – KE = 34

TE ( )Þ =1

KE PE3

For (D) : Acceleration a =

= - w w = -w2 2dvA sin t x

dt

14. Ans (3)

T1 = 2s = 2pl

g , T2 = 2pl

g '

where g' = g+2

2

d ydt

= 10–7.5 = 2.5 = g4

Þ = pl

2T 2g / 4 = 2 × 2 = 4s

15. Ans (1)

vm = Aw Þ p

= =w pmv 2

A2

× (0.2) = 0.20 m; T =

2pmk

Þ m =p

2

2

T k4

= 0.2 kg

At t = 0.1, acceleration is maximum Þ –w2 A =–200 m/s2

Maximum energy =12

mvm2 = 4J or

12

kA2 = Emax

=12

× 200 × 0.04 = 4 J

16. Ans (1)For (A) :

= - Þ = -2 2 dvv 144 9x 2v 0 18x

dx

Þ = - Þ w = Þ w =2a 9x 9 3

Time period p p

= =w2 2

T3

units

For (B) : Q ³2v 0

\ - ³ Þ £ Þ £2 2144 9x 0 x 16 x 4ÞAmplitude = 4 unitsFor (C) : Maximum velocity = Aw = (4) (3) = 12 unitsFor (D) : At x = 3 units, a = – 9x = – 27 units

17. Ans (4)

Time period of spring mass system = pM

T 2K

18. Ans (4)Angular frequency

w = = =K 1200

20 rad / secm 3

19. Ans (3)At mean position KE =

-w = ´ Þ w =2 2 31m a 8 10 4

2rad/sec

Now let equation of SHM be y = 0.1 sin(4t+f). At

t=0, f = 45° =p4

. Therefore pæ ö

= +ç ÷è øy 0.1sin 4t4

Wave Motion1. Ans (1)

Vmax = aw = 3 × 10 = 302. Ans. (1)3. Ans (3)

Amplitude = 0.02 × 75

100 = 0.015

\ Reflected wave = 0.015 sin 8 px

t20

æ ö+ç ÷è ø

4. Ans (1)

f = D D

Þ µ Þ =ml

1/ 21 T f 1 Tf T

2 f 2 T

Þ æ ö= ç ÷è ø

15 1 2f 2 100

Þ f = 1500Hz

As v µ T so velocity of propagation have changes

by 1%As length = costant so l = constant.

5. Ans (2)y

1propagartes in +x-axis and y

2 along –ve x-axis.

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6. Ans (1)

I = 2 2

2

2 A vT

p r2

AI

Tæ öÞ µ ç ÷è ø

2 2

1 1 2

2 2 1

I A T 1 21 :1

I A T 2 1

æ ö æ öÞ = = ´ =ç ÷ç ÷ è øè ø

7. Ans (3)

T xgv gx

m= = =

m m Þ v2 = gx

(symmetrical about x)8. Ans (3)

The equation represents a progressive wave moving

along x-axis of single frequecy.

9. Ans (3)

4l

=L + 0.6 r1

(closed organ pipe)

l = L + 1.2 r1

(open organ pipe)

Þ 4 (L + 0.6 r1) = (L + 1.2r

2) Þ r

2 – 2r

1 = 2.5 L

10. Ans (1)

v = f1 × 50 = f

2 × 51

Þ f1 – f

2 =

v v0.1

50 51- = Þ v = 255 m/s

11. Ans (1)l

1 = 2L ; l

2 = 2(L–y)

Df = f2–f

1 =

2 1

v v-

l l =v 1 12 L y L

é ù-ê ú-ë û

2

vy vy2(L y)L 2L

Þ-

;

12. Ans (4)

1 Tn

2=

ml, n

1µ

l

n 16n 5 16.2

=+

Þ n = 400 Hz

n + 5 = 405 Hz

13. Ans (4)y = sin (wt – kx + f)

v = w cos (wt – kx + f)

at t =0, x = 0, y =–0.5, v >0 Þ f = –6p

y sin t kx6pæ ö\ = w - -ç ÷è ø

14. Ans (2)

The motorcyclist observes no beats. So, theapparent frequency observed by him from the twosources must be equal.

f1 = f2 \ 176330 v

330 22

-æ öç ÷è ø-

= 165330 v

330

+æ öç ÷è ø

Solving this equation, we get v = 22m/s

15. Ans (3)

n n'..

=-

+

F

H

GGGG

I

K

JJJJ=

-+

FHG

IKJ ´

1

1

1 081 08

6 1014

vcvc

e j

= 13

× 6 × 1014 = 2 × 1014 Hz

16. Ans (2)

By using Doppler Effect Dn' = v

v v s–

FHG

IKJ

Dn = 330

330 60–FHG

IKJ (180) = 220 Hz

17. Ans (2)

y1 = 4sin(500 pt) y

2 = 2 sin(506 pt)

Number of beats 1 2n n 506 5002 2

- -= = = 3 beat/

sec.

As I1 µ (16) and I

2 µ 4

Þ ( )( )

22 2

1 2max2

min1 2

I II 4 2 69

I 4 2 2I I

+ +æ ö æ ö= Þ = =ç ÷ ç ÷è ø è ø--

18. Ans (2)

Velocity of longitudinal

-= = =´l

113 1

4

Y 10u 10 ms

10 10

Required time

3

2 2 1000.2 s

v 10´

= =l

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Electrostatics1. Ans. (4)

net charge = 0

+q

2. Ans. (4)All forces make action-reaction pairs, and summa-tion of these pairs must be zero.

3. Ans. (1)

Q tnet = 0 (Q Central force)Angular momentum is constant

4. Ans. (1)

f1 = 1

KQr Þ KQ = r1f1

All charge moves to outer sphere after connection

f2 = 2

KQr =

1

2

rr

æ öç ÷è ø

f1

5. Ans. (2)

Er

= –dVdx

Þ Er

= ( )-

´

8 5

0.1 2 = 15 2 V/m along PAA

6. Ans. (1)Total charge on complete sphere = 2Q,

potential at centre =C

3 k(2Q)V

2 R

But due to hemisphere

V'C = 12

VC = 3 KQ2 R

7. Ans. (2)Electric field is directed from high potential to lowpotential.

8. Ans. (4)Depends upon nature of charges that's why mayincrease or may decrease.

9. Ans. (2)

90°

A

R

O x

r

fP

x= rcosf

Potential at A is same as centre O

then 2

2 2

kP kP cosx r

f=

10. Ans. (2)Direction of charge flow from inside to outside inconducting materials.

11. Ans. (3)electric field lines perpendicular to the conductingmaterial.

12. Ans. (1)

Apply Gauss theoremElectric field here due is +Q outer surface charge

Þ 2

KQx

13. Ans. (4)

E

E µ 1r2

r=Rr

V

V µ

r=Rr

R

O

1r

KQR

KQ

R2

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14. Ans. (3)

Here all charge of inner sphere comes on the outersurface of outer sphereHence qinner = 0 and qouter = Qsinner = 0

souter = 2

Q4 cp

15. Ans. (4)As U = –PE

then F = –dU PdEdx dx

=

But dEdx

= 1

Hence F = P16. Ans. (4)

¶ ¶ ¶= - - -

¶ ¶ ¶

r V V V ˆˆ ˆE i j kx y z

= –[(6 – 8y) i – (–8x – 8 + 6z) j – (6y) k ]

At (1, 1, 1), = + -r ˆˆ ˆE 2i 10j 6k

Þ (r

E ) = + +2 2 22 10 6 = 140 = 2 35

Force = qE = 2 × 2 35 = 4 35 N17. Ans. (3)

q'r1

r2

q

Let charge an inner sphere = q'\ Potantial of inner sphere = 0 (earthed)

\1 2

kq ' kq0

r r+ =

q' = 1

2

qrr

-

Capacitor1. Ans. (3)

Charge on capacitor = Charge of positve plate andnegative plate net charge on capacitor = 0

2. Ans. (2)

Energy supplied by the battery is

U = CV2 = Q2/C = QV

= (10–6)(300)

= 3 × 10–4J

3. Ans. (1)

0F D

VE E E

d= Þ < Also sA > sB

4. Ans. (1)

Uinitial = 12

CV2; Ufinal = 12

CV2

\ Heat = work done by battery

= [CV–(–CV)]V = 2CV2

5. Ans. (3)

S–open ; Vinner = Vouter

6. Ans. (4)

S –closed ; Vinner = 0

ÞKQ3R

+ KqR

=0 Þ q = – Q/3

7. Ans. (4)

q B

A

v

Area = 12

qV = Energy

8. Ans. (4)

C = 0 Ad

Î ; C1 = 0 A C2d 2

Î=

C2 = 05 A2dÎ

= 52

C

Ceq = C1 + C2 = 3C

C%

CD

= 2C

100C

´ = 200%

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9. Ans. (3)

C1 C2

V

X

V

Ef is more in C1 Þ =Q

VC

is more in C1

Þ C1 < C2

10. Ans. (1)C2 & C3 are in parallel so V2 = V3Let C1 = C

C2 = C3 = 1C2

= C2

Equivalent of C2 & C3 = C C

C2 2

+ =

So V1 = V2 = V311. Ans. (1)

In steady state VC = VAB = capacitor voltage= V/2Calculation of time constant (tC)effective resistance across C = 3R

q = q0c

t

1 e-

tæ ö

-ç ÷è ø

, q0 = C V2

Þ q = t

3RCCV1 e

2

æ ö-ç ÷è ø

Final charge Q = CV/2R/2 A 5R/2

R

R/2

V

B

C

12. Ans. (4)

R

R

R RR

10V

Ci = 0

10

10

10 0

0

0

i1i2

6R A

1

2

i R 1i 2R 2

= =

i1 = 6 2

1R 3 R

´ =´

VC = 8V13. Ans. (1)

Bulb light up during charging

14. Ans. (3)

Electric field, E µ 1K

As K1 < K2 so E1 > E2

Current Electricity1. Ans. (2)

40I

16= =

102.5A

4=

21

1 2

RI I

R Ræ ö

= ç ÷+è ø

9.6W 60V

0.4W

6W

20V

1

48I 2.5

60æ ö= ç ÷è ø

= 2A Þ I2 = 0.5

Þ V = 0.5(7) = 3.5 volt

2. Ans. (3)

D=6V

C =9V

B=11V

A=12V1A

6 V

0 V

12

12V

6V

1W

2W

3W

DV = IR

DVAB

= 1V

DVBC

= 2V

DVCD

= 3V

3. Ans. (2)By applying perpendicular

A B

P

D

Q

C

axis symmetry. Points lyingon the line 'PD' have samepotential thereforeResistance between PQand CD can be removed.So R

AB= 9W.

4. Ans. (2)

10V

5W

5V

10

W

20V

5W

30V

11

W

25V

25 25 25 25 25I

15V 30V 5V 55V

3A 3A 1A 5A

A 0

Taking point 'A' as reference potential and itspotential to be '0' : I = 12 A

Power supplied by 20 V cell = –20 × 1 = –20 W

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5. Ans. (1)

By taking 'O' as a reference potential as currentthrough '4W' is zero there should be no potentialdrop across it

2W

4W

10

R4 6 6

2A 2A

0 0 0

6V

10V

4V

Value of 'R' for this condition = 1W

6. Ans. (3)

Potential difference across voltmeter is same asthat of 200W

æ ö= =ç ÷è ø1

200 20V 10 V

300 3 10

100W 200W

7. Ans. (1)

1

2

R 40 4R 60 6

= = ...(i); 1 2

2

R (R 10)1

R 10+

=´

R1R

2 + 10R

1 = R

2× 10...(ii)

By solving (i) and (ii) 1

10R ;

3= W R

2 = 5W

8. Ans. (1)

(0.01) G = 0.1(R1+ R

2 + R

3)

G = 10 (R1 + R

2+R

3)...(i)

G

R1 R2 R3

0.01 25W

0.1

(0.01)G = 1(R1 + R

2)...(ii)

25

R1 R2

R30.01

1A

(0.01)G = 10R1...(iii)

25

R1

R20.01 R3

By solving Equation (i), (ii) and (iii)

R1 = 0.0278W ; R

2 = 0.25W; R

3 = 2.5W

9. Ans. (4)

Ig=1A; G=0.81W; I=10A

10A 1A 0.81

9A

S

g

g

IS G

I I

æ ö= ç ÷-è ø

; 1

S 0.81 0.099

= ´ = W

10. Ans. (3)

DA

2W

6W

1.5W

C6V

3W

3W

1.5W2W

6W

BA

6V

C Þ

3W

1.5W1.5W 1.5W

6V

6V

Þ

On redrawing the diagram, we get I = 6

1.5=4A

11. Ans. (2)Voltage across R=2VHence, voltage across 500W=10V

G500W

R 2V12V

Current through 500W=10 1500 50

= A

As 500W and RW are in series value of

R

R

V 2R 100

I 1 / 50= = = W

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12. Ans. (2)

55 R 55 8R 220

20 80 2´

= Þ = = W

13. Ans. (3)

Here Eeffective =

5 2– 12 1 V

1 1 32 1

=+

reffective = 23

W

So current through 10W

I =

13

210

3+

= 0.03A from P2 to P1

14. Ans. (4)

P=2V

Rso, R=

2VP

\ R1 = 2V

100& R2=R3 =

2V60

Now, W1 = ( )

( )

2

121 2

250R

R R×

+ and

W2 = ( )

( )

2

221 2

250R

R R×

+ and

( )23

3

250W

R=

W1:W2:W3 = 15:25 :64 Þ W1<W2<W3

15. Ans. (3)

d

iv

Ane= As A so v

d¯ Þ v

P > v

Q

16. Ans. (2)

0

rJ J 1

Ræ ö= -ç ÷è ø

( )di JdA J 2 rdr= = p

R

00

ri J 1 2 rdr

Ræ ö= - pç ÷è øò

i = 2pJ0

2 3R R2 3R

é ù-ê ú

ë û = 0J A

6

17. Ans. (4)

A

C

4aD

B

x > z > y

a

2a

18. Ans. (3)

RR/3R/3R/3

Parallel combination

R = eqR9

19. Ans. (4)

by KCL

x 4.5 x 0 x 30

3 6 10- - -

+ + =

10x – 45 + 5x + 3x – 9 = 0

x = 3V

i1 = 0.5 i2 = 0.5 i3 = 0

20. Ans. (3)

22

2W

2W

1V

1W 1W

1V

1

2V 2V

3V3

113

1613

B x

D3

2026 1V

4 2V 3

1

O

4

by KCL

x 2 x 3 x 10

3 2 4- - -

+ + =

4x – 8 + 6x – 18 + 3x – 3 = 0

13x = 29

x = 2913

; BD

10V V

13=

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21. Ans. (2)

16VR

8W2W

2W 8W

8V

D

C

16R 8

4 Ræ ö =ç ÷+è ø

R = 4W

8W

8WR=4 15V

A

B

2W

2W

i = 0.75V = 0.75 × R = 3 V

22. Ans. (4)

3A 2W 2W 1A 2A

A 3V 1W 1W2VB

VA – 6 – 3– 4 + 2 – 2 = VB

VA – VB = 13V23. Ans. (3)

1 2

1 2

R R6

R R= W

+

R1 could be = 8W24. Ans (1)

For maximum power Rinternal = Rext.

Magnetic Field & Magnetism1. Ans. (4)

Baxis = 12

(Bcentre)

Þ ( )0 0

3/22 2

2 R I 2 I 14 4 R 2R x

2m p m pæ ö= ´ç ÷è øp p+Þ x = 0.766 R

2. Ans. (3)3. Ans. (3)

r = 2mK.E.

qB

r µ mq

4. Ans. (1)

( )F q v B E 0= ´ + =r r rr ( )4 ˆˆ ˆq 10i Bj 10 k 0Þ ´ - =

Þ B = 103 Wb/m

2

5. Ans. (3)6. Ans. (1)7. Ans. (3)

qE = qvB Þ v = EB

r = 2

mv mEqB qB

=

8. Ans. (4)Z

B,DC

A Y

A and C have same M.F. and B and D have same M.F.9. Ans. (3)

In B1 and B4 M.F. add up.In B2 and B3, M.F. oppose each other.

10. Ans. (2)

$0 1 0 2I IB k j

2 (AP) 2 (PB)m m

= +p p

r$

$7 7

2 2

2 10 2 2 10 3k j

10 2 10

- -

- -

é ù é ù´ ´ ´ ´= +ê ú ê ú´ë û ë û

$

$5 5(3 10 T)j (4 10 T)k- -= ´ + ´$

11 Ans. (3)

Leffective = AB = 4j$

\ |F||IL B|= ´r r r

= ILeff B =2. 4j$ × 4 (– $k ) = –32 i$ N

12. Ans. (2)

R = mvqB

The radius may be decease if v decreases or Bincreases.

13. Ans. (2)If (b–a) > r (r= radius of circular path of particle)

The particle cannot enter the region x > bSo, to enter in the region x > b r> (b – a)

Þ mv

(b a)Bq

> - Þ q(b a)B

vm

->

14. Ans. (3)

U MB= -r r

= – MB cos q

Here, Mr

= magnetic moment of the loop

q = angle between Mr

and Br

U is maximum when q = 180° and minimum whenq = 0°. So, as q decrease from 180° to 0° its PE alsodecreases.

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15. Ans. (3)Tension in wire T = RBI

Here 2pR = L Þ R = L2p

T = IBL2p

16. Ans. (4)17. Ans. (2)

BC = 50I7 10

2R-m

= ´

I = 5

0

2R 7 10-´ ´m

18. Ans. (2)19. Ans. (2)

Coercivity = nI20. Ans. (1)21. Ans. (2)

As velocity of particle have component non-parallelto magnetic field so path is helical. The x and z-components will becomes zero. Simultaneously afterevery pitch. y-coordinate = v0t

22. Ans. (2)

Force on electron = ( ) ( )q v B e B v´ = ´r rr r

23. Ans. (1)Magnetic field on the axis of a circular loop

( )

20

3/22 2

2 iRB

4 R z

m pæ ö= ´ç ÷è øp +

--

-

p ´ ´ ´= ´

´

2 47

6

2 2.5 3 10 ˆ10 k125 10

( )- -pæ ö= ´ = p ´ç ÷è ø

5 79 ˆ ˆ10 T k 36 10 T k25

24. Ans. (2)

t = 2 m m

T2 2 qB qBq q p q

= =p p

so t = 27

19 3

1.65 102 1.6 10 100 10

-

- -

p ´´

´ ´ ´ = 1.62 × 10-7

Hence option (2)

25. Ans. (1)

Here ( )netF 0 qE q v B 0= Þ + ´ =rr r rr

Þ E B v= ´r r r

Therefore E B^r r

and E v^r r . Hence option (1,2)

26. Ans. (2)

Energy in cyclotron E = 2 2 2q B r2m

so E µ 22

1 1 2

2 2 1

q E q m2m E q m

æ ö æ öÞ = ç ÷ ç ÷è ø è øFor proton & deutron q1 = q2 = e, m1 = 1 amu, m2=2 amu

so 1

2

E 2E 1

= Þ E1 = 2E2 = 80 MeV

Hence option (2)27. Ans. (3)

xeff = 2Lso F = IB Xeff

F = 2IBL

××××

××××

××××

××

×x

y

x

Hence option (3)

EMI & EMW1. Ans. (3)

Total change in flux = Total charge flown throughthe coil × resistance

= 1

4 0.12

æ ö´ ´ç ÷è ø × Resistance= 0.2 × 10 = 2 Webers

2. Ans. (4)

2d dBe Na 5volt

dt dtf

= - = =

3. Ans. (3)

de

dtf

= - 4dB 6 1

NA (100)(40 10 )dt 2

- -æ ö= = ´ ç ÷è ø

= 1 volt4. Ans. (4)

IlB > mg Þ IlB < mg5. Ans. (2)

B v (0.1)(0.1)(1) 5 1I A

1 6/ 5 11/ 5 1100 220= = = =

+l

2WB vl

1W3W º 5/6W

B vl

1W

6. Ans. (4)P

Q

R

IP & IQ ® clockwise IR = 0

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7. Ans. (3)

8. Ans. (1)

1

2

i 1i 4

= Þ 1

2

W 1W 4

= ; 1

2

V4

VÞ =

9. Ans. (3)2B

e2w

=l

effective length for the given diagram is

leff2 = l2 + L2 Þ

2 2B ( L )e

2w +

=l

10. Ans. (1)

According to Lenzs law × × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

AB

Þ Plate B will become positively charged.11. Ans. (2)

e=2B L

2w

effective length of the wire frame is 2R Þ e = 2BwR2

12. Ans. (1)

f = at (T – t) Þd

aT 2atdtf

= -

Þf -

= =d (aT 2at)

iRdt R

Heat = -

òT 2

20

(aT 2at)Rdt

R

é ù

= + -ê úë û

T3 22 2 2 2

0

t t 1a T t 4a 4a T

3 2 R

2 32 3 2 34a T

a T 2a T R3

é ù= + -ê ú

ë ûHeat =

2 3a T3R

13. Ans. (4)As the switch is closed, Flux linked with ‘B’increases away from eye so the current will be‘ANTICLOCKWISE’. As the switch is opened,flux linked with ‘B’ decreases, So the current willbe ‘CLOCKWISE’.

Hence option (4) is correct.

14. Ans. (2)Induced emf = v × length × component ofmagnetic field ^r to the plane of n & l

= 1 × 5 × 5

= 25 V

Hence option (2) is correct.

15. Ans. (4)

Induced emf = B × v × effective length ^r to velocity

= Bv (lBD) = Bv(4R) = 4BvR

Since B is uniform & Area is fixed, then' changein flux is zero in the loop so current flow will bezero.

Hence option (4) is correct16. Ans. (1)

At t = 0 Inductor behaves as open circuit hence

I1 = 0, I2 = ER

, I3 = E

2R

So I2 > I3 > I1

Hence option (1) is correct.17. Ans. (3)18. Ans. (1)

20 RMSU E= e

19. Ans. (4)Alternating Current

1. Ans. (1)

Vrms = 2V Þ

2

1

t 2

t2 2rms

2 1

V dtV V

t t= =

-

ò

\( )21 3/2

102 2 30rms 0 0

V t dtV V t dt

1 0= =

-ò

ò

=

1 242 2 00 0

0

Vt 1 0V V

4 4 4 4

é ù é ù= - =ê ú ê úë û ë û

therefore, Vrms = 0V2

2. Ans. (3)Heat produced by ac = 4 (heat produced by steady

current )

( )2 2rms dci R 4 i R=

irms = 2idc = 2 × 2A = 4A

\ i0 = 2 irms = 4 2 A = 5.656 A

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3. Ans. (1)

For the lamp R = dc

dc

V 100i 10

=

Þ R = 10 W

with ac, Z = V 200

20i 10

= = W

Q Z = 2 2LR X+

Þ XL = 2 2Z R-

wL = 400 100- = 10 3 W

L = ( )210 3 17.32

10 H2 50 3.14

-= ´p

= 5.5 × 10-2 H

4. Ans. (1)With dc source

R = V 12V

6I 2A

= = W

with ac source

Z = V 12Vi 1A

= = 12W

Q Z2 = R2 + 2LX

Þ XL = 2 212 6 6 3- = W

use 2pfL = XL Þ L = 33 mH5. Ans. (1)

X is resistance (Q phase difference = 0 )

\ R = V 220

440i 0.5

= = W

Y is capacitor (Q phase difference = – 2p

)

\ XC = Vi

= 2200.5

= 440 W

on connecting X and Y in series

Z = 2 2cR X 440 2+ = W

\ i = V 220 0.5

A 0.35AZ 440 2 2

= = =

6. Ans. (4) V = 2 2R L CV (V V )+ -

Here VL = VC

V = VR = 220 V

I = RVV 2202.2A

R R 100= = =

7. Ans. (4)8. Ans. (1)

Total power consumed

P = vrms irms cos f = 2

2

VR

Z

æ öç ÷è ø

\ XL = wL = 100p × 800 × 10-3

= 25 W

XC = 1Cw

= 6

1

100 60 10-p´ ´

= 53W

so, Z = ( )22L CR X X+ -

= ( ) ( )2 215 251 53+ - = 200W

\ P = 2

23015 20W

200æ ö ´ »ç ÷è ø

9. Ans. (4)10. Ans. (1)11. Ans. (3)

t = T 2 LC4 4

p= Þ t = LC

2p

12. Ans. (1)2 2

RV 10 8= - = 6volt

1 1 1 1L

R

L V 8 4tan tan tan tan

R V 6 3- - - -æ öwæ ö æ ö æ öf = = = =ç ÷ ç ÷ ç ÷ç ÷è ø è ø è øè ø

13. Ans. (4)

Old power factor = 2 2

R 1

10R (3R)=

+

New power factor = 2 2

R 1

5R (2R)=

+

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Ray Optics

1. Ans (4)

2. Ans (2)Image formation by concave mirror when objectis real & placed at focus f, forms image at infinitybut not in other cases.

3. Ans (2)

q

m 1.014 1.02... 1.50m

For composite transparent IIr slabes

=msinq= constant msinq = 1.6 m sin xsin x = 5/8 sinq.

4. Ans (4)

Maximum possible deviation = d = p/2 –C

d

cc

Reflected ray)

Denser Rarer

5. Ans (1)As given : r + r' = 90°

u2 sinr = u1sin r' 90°

r

r'rr

Denser(u ) 2

Rarer (u )1

Þ u2 sin r = u1 sin (90–r)u2 sin C = u1 sin 90°

1

2tan r

m=

m ; sin C = 1

2

mm

(C: Critical angle)\ sin C = tan r; C = sin

–1 (tan r)

6. Ans (4)m = Speed of light in medium

m = cv

Þ velocity is different in different medium

From v = nlBut frequency is the fundamental property. It neverchange by changing the medium hence l is alsochanged as v is change.

7. Ans (4)

(i) For (i) m2

1v – 1¥

=m -2 1

6

v1 = 32

× 6 × 2 = 18 cm

For plane surface : I1 ® object

dapperent= 2

1

nn × dactual =

2

1

nn × (18–R)

= 1

3 /2 × ( )-18 6 = 8 cm

8. Ans (4)

I1 I2

Final image (by lens) object I2u = – 30, f = – 10

v = +uf

u f=

- -300

30 10 =

-30040

= – 7.5

\ 2.5 cm in front of mirror.9. Ans. (4)

10. Ans (1)

mred sin i = 1 sin r 45°

45°1.39 ×

11.414

= sin r

sin r < 1 Þ possible(refraction is possible)

mgreen × sin i Þ 1.44 × 1

1.41 >1 Þ not possible

Same for blue 1.47 × 1

1.41 > 1

Þ It also not possible.11. Ans (1)

dmin= 2i – A; 60 = 2 × 60 – Å; A = 60°

m =

min.Asin

2A

sin2

+ dæ öç ÷è ø

=

+æ öç ÷è ø60 60

sin260

sin2

m = 3 /2

1/2 = 3

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12. Ans (2)i = 50° , equilateral prism A = 60°; e = 40°Total angle of deviationd = i + e – A = 50 + 40 – 60; d = 30°Hence dmin < d Þdmin < 30°

13. Ans. (4)14. Ans (2)

Divergence angle will remain unchanged becausein case of a glass slab every emergent ray is parallelto the incident ray. However, the rays are displacedslightly towards outer side.

O

i iA D

r r

a

rEB

C Fi i

a

(In the figure OA|| BC and OD|| EF)15. Ans (4)

Applying Snell's law at B and C,

iB iC

m sin i = constant or m1 sin i

B = m

4 sin i

C

But AB||CD\ iB = i

C or m

1 = m

4

16. Ans (3)Figure (a) is part of an equilateral prism of figure(b) as shown in figure which is a magnified imageof figure (c).

Therefore, the ray will suffer the same deviationin figure (a) and figure (c).

17. Ans (3)

1 2

1 1 1( 1)

f R R

æ ö= m - -ç ÷è ø

For no dispersion

1d

f

ì üí ýî þ

=0 Þ 1 2

1 1d 0

R R

ì üm - =í ý

î þ Þ R

1 = R

2

18. Ans (2)Image formed by convex lens at I

1 will act as a

virtual object for concave lens. For concave lens

I2

26 cm 4cm

I1

1 1 1

v u f- = Þ

1 1 1

v 4 20- =

-Þ v = 5 cm

Magnification for concave lens v 5

m 1.25u 4

= = =

As size of the image at I1 is 2cm. Therefore, size

of image at I2 will be 2 × 1.25 = 2.5 cm.

19. Ans (2)

Applying Snell's law (m sin i = constant) at 1 and2, we have m

1 sin i

1 = m

2 sin i

2

GlassWater

Airr

r

i1

90°2

Here, m1 = m

glass, i

1 = i; m

2 =m

air = 1 and i

2 = 90°

\ mg sin i = (1) (sin 90°) or g

1

sin im =

20. Ans (1)

Critical angle - æ öq = ç ÷è m ø

1C

1sin

Wavelength increases in the sequence ofVIBGYOR.

According to Cauchy's formula refractive index (m)decreases as the wavelength increases. Hence therefractive index will increase in the sequence ofROYGBIV. The critical angle q

C will thus increase

in the same order VIBGYOR. For green light theincidence angle is just equal to the critical angle.For yellow, orange and red the critical angle willbe greater than the incidence angle. So,thesecolours will emerge from the glass air interface.

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21. Ans (3)

Refraction from lens : - =-1

1 1 1v 20 15

\ v = 60cm + ve directioni.e. first image is formed at 60 cm to the right oflens system.Reflection from mirror : After reflection fromthe mirror, the second image will be formed at adistance of 60 cm to the left of lens system.

Refraction from lens : - =3

1 1 1v 60 15

¬ +ve

direction or v3 = 12cm

Therefore, the final image is formed at 12 cm tothe left of the lens system.

22. Ans (4)For small angled prism A

Þ d = m -A A from graph A = 40,slope = A =4

23. Ans. (3)24. Ans (1)

Magnification of side m = f

f u-

= 10

10 ( 25)-

- - - = 2

3-

2

1

hh =

23

- Þ h2 =

23

- × 3 = – 2 cm

\ area of image = 2 × 2 = 4 cm2

25. Ans (4)

m = f

f u-;

1 fn f U

=-

f – U = nff – nf = Uf(1 – n) = U

26. Ans (3)

Refractive index m = air

medium

vv

= 1

2

xt

10 xt

= 2

1

t10t

also sinqc = 1m =

1

2

10 tt Þ qc = sin–1

1

2

10tt

æ öç ÷è ø

27. Ans (2)

Let v be the apparent depth of bubble then by using

m2 1 2 1

v u R-

m=

m - m

\1 1.5 1 1.5v 4 10

-- =

- -Þ v = –3 cm

28. Ans (1)

F= 90cm, w2 = 1.5w1 \ 1

2

ƒƒ

= – 1

2

ww

= –23

Þ ƒ1 = –23

ƒ2

By using 1 2

1 1ƒ ƒ

+ =1F

we get ƒ1 = 30 cm and

ƒ2 = –45 cm29. Ans (1)

Longitudinal chromatic abberation= w f = 20 × 0.08 = 1.6 cm.

30. Ans (2)Here ve = –¥, ue = –feL = |v0| + |ue| Þ v0 = L – |ue|v0 = 6.5 – fe

MP = 0 0

0e 0 e

f vD Dm

f f f

-æ ö= ç ÷è ø

–100 = e

e

(0.5 – 6.5 f )25(0.5)f

+

fe = 2 cm31. Ans (4)

f0 = 1 cm & fe = 2 cm

L = 12 cm & ve = D = –25 cm (given)

Q e e e

1 1 1–

v u f=

e

1 1 1(–25) u 2

- =

e

1 1 1–

u 25 2= - or

e e e

e e

v f vu f

-=

e

1 27–

u 50= = ue = –

5027

\ L = v0 + |ue| = 12 cm

12 = v0 +50

–27

= v0 + 5027

v0 = 12 – 5027

= 27427

Now, 0 0 0

1 1 1–

v u f= or

0 0

0 0 0

v fu f u

=+

0

1 1 1–

274 u 124

=æ öç ÷è ø

0

1 24–1

u 274=

0

1 24 274u 274

-=

u0 = – 274250

= – 1.1 cm

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32. Ans (4)Q Length of telescope(L) = f0 + feMeans 50 = f0 + fe ...(1)Magnifying power

(M) = 0

e

ff

- = –9 or f0 = 9 fe ...(2)

from (1) & (2) f0 = 45 cm & fe = 5 cm

Wave Optics1. Ans (2)

Conditions for sustained interference of light(i) Sources should be coherent.

(ii) There should be point sources

Q fringe width bl

=Dd

Here, bl

11 1

1=

Dd

and

bl

22 2

2=

Dd

As b1 = b2

Þ l l1 1

1

2 2

2

Dd

Dd

=

Þ DD

dd

1

2

2 1

1 2

600400

11 2

62

31

= = ´ = =ll

2. Ans (1)

For minima yn = (2n–1)

D2dl

where n is the no. of

minimum. For maxima n

n Dy

dl

=

th5 dark

9 Dy

2dl

= ; st1 max ima

Dy

dl

=

By Equation th st5 min 1 maxy y- = 7 × 10–3

3 53

2

9 D D 7 10 15 10 27 10

2d d 7 50 10

- --

-

l l ´ ´ ´ ´- = ´ Þ l =

´ ´\ l = 600 nm

3. Ans (2)

Given 1

2

II =4 \ I

1 = 4I (let the I

2 = I)

2max 1 2I ( I I )= + 2(2 I I )= + = 9I

Imin

= ( )2

1 2I I- = ( )22 I I- = I

\ - -

= = =+ +

max min

max min

I I 9I I 8I 4I I 9I I 10I 5

4. Ans (2)

At the central Maxima Dx=0

But upward shift m -

=(2 1)tD

d and

Downward shift m -

=( 1)2tD

d

So net shift y =tDd

[2m–1 – 2m + 2] Þ tD

yd

=

5. Ans (4)

y1 = a sin t

3pæ öw +ç ÷è ø and y

2 = a sin wt

2 21 2 1 2A a a 2a a cos= + + f where

pf =

3

2 2a a 2aacos3p

= + + 3a=

6. Ans (4)

Let the distance 'y'from 'O' where nearest whitespot occurs. Then the path difference = 0

y

D

2d/3d

æ ö æ öÞ + - - + + =ç ÷ ç ÷è ø è ø

2 22 22d d

D y D y 03 3

\ y =d6

7. Ans (2)

m1sin 45° = m

2sinq Þ q=30°

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8. Ans (3)

By Geometry, path difference at '0' for minima

should be (2n–1)2l

\ S2O

– S

1O = (2n – 1)

2l

S1

S2D=12

l/2

d=5c

m

0

Þ 2 2D d D (2n 1)2l

+ - = -

Þ (13–12)cm = (2n–1)2l

For n=1, 2,3 Þ l = 2cm, 23

cm

9. Ans (1)

10. Ans (3)

Given IR =

4I4

= I. So I = I + I + 2I cos f

cos f = 12

- Þ f = 23p

Corresponding path difference Dx = 3l

So d sin q = 3l

Þ q = sin–1

3dlæ ö

ç ÷è ø

11. Ans (2)a sinq = 2lgiven l = 6 × 10–7 m, a = 24 × 10–5 × 10–2 m

sinq = 2al

= 7

7

2 6 10 1224 10

-

-

´ ´=

´

\ q = 30°12. Ans (2)

given l = 6.328 × 10–7 m, a = 0.2 × 10–3 m

wq = 2al

= 7

4

2 6.328 10

2 10

-

-

´ ´´

radian

= 36.328 10 180

3.14

-´ ´ = 0.36

13. Ans (2)

For first minimum sin q1 = la

\ a = l

qsin1

= 650 10

30´

°

–9

sin

= 650 10

05´ –9

. = 1.3 × 10–6m

14. Ans (1)Third bright of known light

X3 = l13 Dd

..... (1)

4th bright of unknown light

X4 = l24 Dd

..... (2)

Given X3 = X43l1 = 4l2

l2 = 34

l1 = 34

× 590 = 442.5 nm

15. Ans (2)

At P : Dx = 0; p

Df = ´ =l

20 0

( )= + =2

PI I I 4I

At Q : x4l

D = , Df = 2

4 2p l p

´ =l

p= + + =QI I I 2 II cos 2I

2; = =P

Q

I 4I 2I 2I 1

16. Ans (1)

As path difference l

q = l Þ =q

ndsin n d

sinAs q <sin 1 so d > nl, where n is an integer

Therefore d ¹l and l

¹d2

So A and B both are correct17. Ans (3)

Seperation between slits =b, screen distance d (>>b)

Path difference at 'O' must be odd multiple of 2l

for missing wavelengths S2O–S

1O=

n2l

S1

S2

d

yO

b

Þ 2 2 nd b d

2l

+ - =

Þ d2 + b2 = d2+2 2n n

2 d4 2l l

+ ´ ´

Þ 2b

ndl = (Here n = 1,3,5)

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18. Ans (4)I1 µ a

12, I

2 µ a

22

Iresultant

= I1 + I

2 + 1 22 I I cos f

Imax

= ( )2

1 2I I+ , When cos f = 1, f = 0, 2p,....

Modern Physics1. Ans (4)

Equation of straight line

- -=

n - n n - n1 2 1

1 2 1

V V V V

æ ö æ ö- n - n

Þ = n -ç ÷ ç ÷n - n n - nè ø è ø2 1 2 1 1 2

2 1 2 1

V V V VV

But æ öf -æ ö= n - Þ = ç ÷ç ÷ n - nè ø è ø

0 2 1

2 1

V VhV h e

e e

æ ön - n

f = ç ÷n - nè ø2 1 1 2

02 1

V Vand e

2. Ans (2)Released energy = 2 × 4 × 7 – 2 × 1 – 7 × 5.4= 16 MeV

3. Ans (1)Greater work function greater intercept

4. Ans (3)a decay :

2He4, so both Z & A decreases.

b+ decay : +1

e0 ,so A will not change but Z will change (decreases)b- decay :

–1e0 ,

so A will not change but Z will change (increases)g decay : no change in A & Z.

5. Ans (1)A is balanced both in mass number & atomic no.

6. Ans (1)

n

0R R2

1æ ö= ç ÷è ø ...(i)

Here R = activity of radioactive substance after n

half–lives 0R

6= (given)

Substituting in equation (i), we get n =4

\ t = (n)t1/2 = (4) (100 ms) = 400 ms

7. Ans (3)During g–decay atomic number (Z) and massnumber (A) does not change. So, the correct optionis (c) because in all other options either Z,A or bothis/are changing.

8. Ans (2)Given that K1 + K2 = 5.5 MeV ...(i)From conservation of linear momentum, p1 = p2

Þ 1 22K (216m) 2K (4m)= as P 2Km=\ K2 = 54 K1 ...(ii)Solving equation (i) and (ii).We get K2 = KE of a–particle = 5.4 MeV

9. Ans (3)4(2He4) = 8O

16

Mass defect Dm = {4(4.0026) – 15.9994} = 0.011 amu\ Energy released per oxygen nuclei = (0.011) (931. 48) MeV = 10.24 MeV

10. Ans (2)Rest mass of parent nucleus should be greater thanthe rest mass of daughter nuclei.

11. Ans (1)

Activity = lN & T1/2 = l

ln2

So 5 = l

01

n2(2N )

T & 10 =

l

02

n2(N )

T Þ T1 = 4T2

12. Ans (3)Due to mass defect (which is finally responsible forthe binding energy of the nucleus), mass of anucleus is always less than the sum of masses of

its constituent particles. 201.0Ne is made up of 10

protons plus 10 neutrons.

Therefore, mass of 2010Ne nucleus,

M1 < 10 (mp + mn)Also, heavier the nucleus, more is the mass defect.Thus, 20 (mn + mp) – M2 > 10 (mp + mn) –M1

Þ 10 (mp + mn)>M2–M1ÞM2< M1 + 10(mp + mn)Now since M1 < 10 (mp + mn) \ M2 < 2M1

13. Ans (3)(A) For 1 < A < 50, on fusion mass number forcompound nucleus is less than 100

Þ Binding energy per nucleon remains sameÞ No energy is released

(B) For 51 < A < 100, on fusion mass numberfor compound nucleus is between 100 &200Þ Binding energy per nucleon increasesÞ Energy is released.

(C) For 100 < A < 200, on fission, the massnumber of product nuclei will be between50 & 100Þ Binding energy per nucleon decreasesÞ No energy is released

(D) For 200 < A < 260, on fission, the massnumber of product nuclei will be between100 & 130 Þ Binding energy per nucleon increases Þ Energy is released.

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14. Ans (4)Activity reduces from 6000 dps to 3000 dps in140 days. It implies that half–life of the radioactivesample is 140 days. In 280 days (or two half–lives)

activity will remain 1

4th of the initial activity. Hence,

the initial activity of the sample is 4 × 6000 dps= 24000dps

15. Ans (4)Because energy is releasingÞ Binding energy pernucleon of product > that of parent Þ E2 > E1.

16. Ans (3)

ZXA 3( )2a4 Z–6( )A–12 2( )+1b

0 Z–8( )A–12

( ) ( )A 12 Z 8No. of neutrons A Z 4No. of protons Z 8 Z 8

- - - - -\ = =

- -

17. Ans (1)

Q

t/T

0

N 1N 2

é ù= ê úë û

2tT1 1

3 2é ù\ = ê úë û

&

1tT2 1

3 2é ù= ê úë û

2 11

(t t )T

2 1(t t )1 11

2 2 T

- -é ùÞ = Þ =ê úë ûÞT = t2 – t1

Þ t2 – t1 = 20 min.18. Ans (4)

M+ mN m5ml 'p'

lp

According to Conservation of linear momentum

P' = P. Therefore l' = l19. Ans (2)

1T1 0N N e-l= ; 2T

2 0N N e-l=

R1 = lN

1 ; R

2 = lN

2

(N1 – N

2) = ( ) 1 2

1 2

(R R )N N

-l- =

l l

T = e elog 2 log 2;

Tl =

l

(N1 – N

2) =

1 2

e

(R R )T(log 2)

-

(N1 – N

2) µ (R

1 – R

2)T

20. Ans. (1)21. Ans (3)

1 1 2 2

dNN N

dt= l + l Þ -l -ll + l1 2t t

1 1 2 2N e N e

22. Ans. (4)23. Ans (4)

1 2

2 1

2.532948

5.06 5896f l l

= Þ = Þ l =f l

Å

24. Ans (4)eV

S = hn -f

25. Ans (4)

Q I µ 2

1r

\ intensity becomes 14

th

26. Ans (1)

(a) P = Nhc·

l (b)

0.1n N

100

· ·

= (c) i n e· ·

=

27. Ans (3)

22 2

1

1 2

hchc 2K

hcK hc2

æ ö- f- f ç ÷lè øl

= =æ ö- f - fç ÷l lè ø

Þ K1 < 2K

2

28. Ans (1)1 1 0

0 1 1n p e-® + + n

29. Ans (3)

90 days ® 3 half lives, left 18

i.e. 12.5%

Disintegrated = 100 – 12.5 = 87.5%30. Ans (1)

N x·

(200 × 106 × 1.6 × 10–19) = 1000

Electronics & Communication Systems1. Ans (1)

For a doped semi-conductor in thermal equilebrium

nenh = n i2 (Law of mass action)

2 16 2i

e 22h

n (1.5 10 )n

h 3 10´

= =´ = 7.5 × 109 m–3

2. Ans (1)Given, ic = 4 mA applying Kirchhoff's secondlaw in loop 1,

VCE = 8 – iC RL

Þ RL = 8 - V

iCE

C

=8 4

4 10 3

-

´ -

= 1× 103 W = 1 kW

II mA

BC= =b

4100 = 4 × 10–5 A

Q VBE = 8 – IBRB

Þ RB =8 8 06

4 10 5

-=

-´ -

VI

BE

B

.=185 kW

E


86

\\N

OD

E2\D

ATA

\201

5\RE

-AIP

MT-

2015

\PH

Y\EN

G\S

OLU

TIO

NS.

P65

3. Ans (3)

Y A B A.B A B AND gate= + = = × =

4. Ans. (4)5. Ans (4)

L

5VI 5mA

1k= =

W

10V 5V

I1 1kW

500WI IL

10 5I 10mA

500-

= =

I1 = I – IL =10 mA – 5mA = 5mA6. Ans (1)

For first circuit

Y = A.B A.B= Þ AND gate

For second circuit

Y = A.B A B= + Þ OR gate

7. Ans (3)

Y = (A + B).( A.B ) =(A+B).( A B+ )

= A. A + B. A + A. B + B. B

= A. B + B. A = XOR gate8. Ans (4)

For (a) : Y = A B A.B+ = Þ NAND gate

For (b) : Y = A.B A B= + Þ NOR gate

9. Ans. (2)

10. (i) mm

f L

V 2524.75 mA

R R (10 1000)I = = =

+ +;

mdc

24.757.88mA

3.14I

I = = =p

;

mrms

24.7512.38mA

2 2I

I = = =

(ii) Pdc = Idc2 × RL = (7.88 × 10–3)2 × 103 » 62 mW

(iii) Pac= Irms2(Rf + RL) = (12.38 × 10–3)2 × (10 +

1000) »155 mW(iv) Rectifier efficiency

dc

ac

P 62100 100 40%

P 155h = ´ = ´ =

(v) Ripple factor

II

1/22 2

rms

ac

12.381 1 1.21

7.88

é ù é ùæ ö æ öê ú= - = - =ê úç ÷ ç ÷ê ú è øê úè ø ë ûë û

11. (a) I = 27 07

1 103

. .-

´ = 2mA

(b) I =20

10 10+ =1 A (c) I =18 6500

-=24 mA

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