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Question Bankfor
Re-AIPMT-2015
Corporate Office
ALLEN CAREER INSTITUTE
“SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005
+91-744-2436001 [email protected]
www.allen.ac.in
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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
(2) 0.70 m from mass of 0.7 kg
(3) 0.98 m from mass of 0.3 kg
(4) 0.98 m from mass of 0.7 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.
(2) 3 beats/s with intensity ratio between maxima
and minima equal to 9.
(3) 6 beats/s with intensity ratio between maxima
and minima equal to 4.
(4) 6 beats/s with intensity ratio between maxima
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
Electric linesof forces
(2)
Lower potential
Higher potential
x
y
(R) x
y
Electric linesof forces
(3)
Low
erpo
tent
ial H
igherpotential
x
y
(S) x
y
Electric linesof forces
(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
(2) Charge attained by inner sphere is –Q/4
(3) Charge attained by inner sphere is –Q/5
(4) Charge attained by inner sphere is –Q/3
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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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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 ˆE E cos t y xpæ ö= w -ç ÷è øl
r
(2) 0
2 ˆE E cos t y xpæ ö= - w +ç ÷è øl
r
(3) 0
2 ˆE E cos t y zpæ ö= - w +ç ÷è øl
r
(4) 0
2 ˆE 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
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
= 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
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Question Bank Physics for Re-AIPMT
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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