MULTIPLE CHOICE QUESTIONS1. If a vector X is represented as x
sin i y cos j then another vector Y which is normal to X can be
represented as a. x sin i y cos j b. y sin i x cos j c. y cos i x
sin j d. y cos i x sin j Sol: Correct option is (c)
As X an Y are normal to each other, therefore their scalar
product should be equal to zero If y= y cos i x sin j then X.Y = xy
cos sin xy sin cos 0 2. Two vectors are as follows V1 3 i 8 j and
V2 9 i 4 j . Find the magnitude of the scalar product of these two
vectors. a. 27 b. 32 c. 72 d. 59 Sol: Correct option is (d)
Scalar product of V1 and V2 = 9 3 ( i . i ) 8 4 ( j . j ) 27 32
59
3. If X=Y+Z and the magnitude of X, Y and Z are 13, 12 and 5
units, then the angle between X and Z is 13 a. sin 1 5 12 b. sin 1
5 5 c. sin 1 13 5 d. cos 1 13
Sol: As
Correct option is (d) y = 2 and z =5
(12)2 (5)2 2 12 5cos 169 cos 0
where is angle between Y an Z
90o
Therefore Y and Z are as shown
5 From above figure cos 1 13 4. If for vectors X and Y x+y a.
90o b. 0o c. 75o d. 15o = x-y then the angle between the vectors X
and Y is
Sol: x+y
Correct option is (a) = 2 x + y x-y 2 +2 x y cos = x 2 + -y 2 +2
x -y cos
where is the angle between X and Y
4
2 x x + y
2 +2 x
y cos =
2 x + y
2 -2 x
y cos
y cos
cos =0 900
Hence angle between X and Y is 900 5. A car traveling towards
east at 40 m/s and turns towards left and travels at same speed.
The change in velocity is a. 40 m/s due north. b. 20 m/s due east.
c. 40 2 m/s due north-west d. 20 2 m/s due north-east Sol: Correct
option is (c) Initial velocity is 40 m/s due east. Let this vector
be B. B 40 i
Final velocity is 40 m/s due north. Let this vector be A. A 40 j
Change in velocity = A B 40 j 40 i 40 2 m/s due north-east.
6.
A
force
F 2 i 6 j 4 k moved
a
body
from
position
vector
r1 3 i 2 j 4 k to the position r2 6 i 4 j 7 k . The work done
is
a. 10 units b. 6 units c. 8 units d. 4 units Sol: Correct option
is (b)
Work done = F .(r2 r1 )
(2 i 6 j 4 k )(6 i 4 j 7 k 3 i 2 j 4 k ) (2 i 6 j 4 k )(3 i 2 j
3 k ) (6 12 12) 6 units
7. Vector X of magnitude 10 units is inclined at 300 with the
horizontal and vector Y of magnitude 20 is inclined at 900 with the
horizontal. Find the resultant of X and Y. a. 500 b. 600 c. 700 d.
800 Sol: Correct option is (c)
Angle between X and Y is 600 Hence R =(10) 2 (20) 2 2 10 20 cos
600 500 200 700
8. At what angle should two forces P and 3 P act such that the
resultant force has the magnitude? 13P a. 00 b. 450 c. 900 d.
600
Sol:
Correct option is (d)( P ) 2 (3P) 2 2( P )(3P) cos
Resultant force =
13P 10 P 2 6 P 2 cos 13P 2 10 P 2 6 P 2 cos 3P 2 6 P 2 cos 3P 2
cos 6P 2 1 cos 2 0 60
9. The component of a vector is a. always equal to its magnitude
b. always less than its magnitude c. always greater than its
magnitude d. sometime equal and sometime less than its magnitude
Sol: Correct option is (d)
The component of a vector X can be either X sin or X cos and
value of sin and cos lies between 0 and 1, hence the component is
sometimes less or sometimes equal to the magnitude of the vector.
11. A force of 80 N and a force of 40 N acting simultaneously at a
point may produce a resultant force of a. 150 N b. 20 N c. 25 N d.
100 N Sol: Correct option is (d)
The pair of forces 80 N and 40 N can produce maximum of 120 N
and minimum of 40 N. So the resultant of both forces should lie
between 120 N and 40 N. Among above options only the option (d)
lies between the bounds.
12. If A 3 i 4 j 2 k and B 6 i 3 j 4 k the angle which the A+B
makes with x-axis is
a. cos1 b. sin 1 c. cos1 d. sin 1 Sol:
9 86 9 86
86 9 86 9
Correct option is (a)
A+B= 9 i j 2 k A+B 92 12 22 86 Let A+B makes an angle with
x-axis. Then scalar product of A+B with xaxis is (A+B). i = 9 A+B .
1. cos 9
86. cos 9 9 cos1 86 13. A body travels 39 m in first 3 seconds
and 105 m in next 5 seconds. What will be the velocity of the body
at the end of 10 seconds? a. 20 m/s b. 30 m/s c. 10 m/s d. 50
m/s
Sol: S= 39 m9 39 3u a 2
Correct option is (d) t=3 and144 8u 64 a 2
From the above two equations u = 10 m/s and a = 2 m / s 2
Velocity of the body at the end of 10 s V= u+at = 10 + 2 (10) =
30 m/s14. Two bodies traveling towards each other on a straight
road at velocity 6 m/s and 8 m/s respectively. When they are 110 m
apart both bodies start deaccelerating at 1 m / s 2 until they
stop. How far apart will they be when they have both come to stop?
a. 10 m b. 20 m c. 30 m d. 60 m Sol: Correct option is (d)
62 0 Distance traveled by first body before stopping is S1 18 m
2 Distance traveled by second body before stopping is S 2 82 0 32 m
2
Therefore distance between the two bodies after stopping is 110-
( S1 S2 )= 110- (18+32)=60 m 15. A car starts from rest with an
acceleration of 3 m / s 2 . While another car 300 m behind starts
from rest with an acceleration of 5 m / s 2 . How long will it take
for both cars to collide? a. 5 3 s b. 30 s c. 10 3 s d. 3 10 s
Sol:
Correct option is (c)
Let the first car travel the distance s, then distance traveled
by second car before colliding is S+300 m S=1 3 t 2 2
1 Also S 300 5 t 2 2 S 300 5 3S 900 5S S 3
S 450 m
1 450 3 t 2 2 t 300 10 3 Seconds
16. A truck starts from rest with an acceleration of 4 m / s 2 .
At the same time a bus traveling with a constant velocity of 60 m/s
overtakes and passes the truck. At what distance will the truck
overtakes the bus? a. 30 m b. 1500 m c. 60 m d. 1800 m Sol: Correct
option is (d)
1 Studying the motion of truck St 4 t 2 2t 2 2
Studying the motion of bus Sb 60t For truck to overtake bus
2t 2 60t t 30s
Hence distance = 60 x 30 = 1800 m
17. A body moving with constant acceleration covers distance of
70 m in 4th second and it also covers a distance of 100 m in 7th
second. What is the distance traveled in 10th second? a. 130 m b.
65 m c. 35 m d. 10 m Sol: Correct option is (a)a Distance traveled
in nth second is S n U (2n 1) 2 70 U 7a 2
and
10 U
13a 2
From above two equations U = 35 m/s and a = 10 m / s 2 Therefore
distance traveled in 10th secondS10 35 10 (20 1) 130 m 2
18. A particle starts from rest and moves with constant
acceleration. The ratio of distance covered by particle in t th
second to that covered in t seconds is 2t (2t 1) b. t 2 :1 2t 1 c.
2 t d. t:1 a. Sol: Correct option is (c)
a Distance covered in t th second = U (2t 1) and 2 1 Distance
covered in t seconds = Ut at 2 2
a St U 2 (2t 1) Ratio = 1 S Ut at 2 2 As particle starts from
rest U 0 Therefore Ratio =2t 1 t2
19. A car chases a bicycle 60 m ahead of it and gains 10 m in 2
s after the chase started. After 4 s the distance between the car
and bicycle is a. 10 m b. 20 m c. 30 m d. 15 m Sol: Correct option
is (b)
As car gains 10 m in 2 s and let relative acceleration between
car and bicycle is a therefore1 10 a 4 a 5m / s 2 2 1 After 4 s the
distance gained by car is S 5 16 40m 2
Hence distance between car and bicycle = 60 -40 = 20 m 20. A boy
sitting by the window of a train moving with velocity V1 30m / s
sees for 5 s a train moving with a velocity V1 5m / s in opposite
direction. The length of the second train is a. 125 m b. 100 m c.
175 m d. 75 m Sol: Correct option is (C)
Relative velocity of both train = 30+5 = 35 m/s Length of second
train = 35 m/s x 5 s = 175 m
21. A body is dropped into a well 19.6 m deep. After how much
time the sound will be heard by the person who had thrown the body
if velocity of sound is 300 m/s? a. 0.065 s b. 3.065 s c. 2.198 s
d. 2.065 s Sol: Correct option is (d)
Let time taken by body to go to the water level be t. Then1 19.6
9.8 t 2 t 2 s 2
Also time taken by sound to come up =
19.6 0.0653 300
Therefore total time = 2 + 0.065 = 2.065 s 22. A person is
throwing ball into the air, throwing one whenever the pervious one
is at its highest point. How much high does the ball rise if he
throws thrice a second, and initial velocity is 10 m/s? a. 96.775 m
b. 98 m c. 1.225 m d. 100 m Sol: Correct option is (a)1 second
2
Time taken by a ball to go up =
Therefore height up to which ball rises is H=1 1 9.8 9.8 10 2 2
9.8 98 1.225 96.775m 82
H = 98
23. A body falls freely from rest and the total distance covered
by it in the last second of its motion equals the distance covered
by it in the first 5 seconds of its motion. The time for which
stone remained in the air is a. 10 s b. 12 s c. 13 s d. 11 s Sol:
Correct option is (c)
Let the time for which stone remained in the air is t Then
distance traveled in last second S=1 2 1 g gt g (t 1) 2 tg 2 2
2
Also distance traveled in first 5 seconds S=1 g 25 2
Equating both distances we gettg g 1 g 25 t 13s 2 2
24. A ball is dropped on the ground from a height of 2 m. If the
coefficient of restitution is 0.6, the height to which the ball
will rebound is a. 1.72 m b. 0.72 m c. 2.86 m d. 1 m Sol: Correct
option is (b)
Let final velocity of ball on reaching ground is V, then V 2 2 g
2 V 2 g For upward motion U= 2 g e V=0
S=H a=-gH 2 2 g e2 2e 2 2 0.6 0.6 .72m 2g
25. Two balls are shot one after another at an interval of 1
second along the same vertical line with same initial velocity of
19.6 m/sec. Find the height at which they collide. a. 13.87 m b.
19.75 m c. 18.375 m d. 20 m Sol: Correct option is (c)1 2 gt 2 1 g
(t 1)2 2
For 1st ball S1 19.6 t
For second ball S 2 19.6(t 1) For collision S1 S 219.6 t
1 2 1 g gt 19.6t 19.6) gt 2 tg 2 2 2 g 0 19.6 tg 2 g 19.5 1 5 tg
19.6 t t 2 9.8 2 22
5 1 5 Therefore height S1 19.6 9.8 49 30.625 18.375m 2 2 2
26. A ball is dropped from height h on the ground. If the
coefficient of restitution is e, than the height to which the ball
goes up after t rebounds? a. e 2t h b. et h1
c. e h h d. eh
Sol:
Correct option is (a)2
V H=2g
Velocity after first rebound =ev e2v e 21v 2 h' h' 2g 2g Height
attained after first rebound h '
e21v 2 2g
Hence height attained after (t)th reboundht e 2t h
27. The range of the projectile when launched at angle of 300
with horizontal is 100 m. What is the range of the projectile when
it is launched at an angle of 450 with horizontal?3 200 200 b. 3 3
c. 200 200 d. 3
a.
Sol: 100 m =
Correct option is (d)
u 2 sin 2(30) 100 g u2 g sin 60
Therefore range when inclination is 450 is 100 g sin 90 R= sin
60 g R 100 2 200 3 3
28. A man aims a gun at a bird from a point at horizontal
distance of 300 m. IF the gun can impart a velocity of 300 m/s to
the bullet, at when height above the bird must he aims his gun in
order to hit it? a. 2 m b. 4.9 m c. 9.8 m d. 19.6 m Sol: Correct
option is (b)
Time taken to cover horizontal distance of 300 m at 300 m/s t =
1 s Now the distance traveled in vertical direction during 1 s is
h=1 9.8 (1)2 4.9 m 2
Hence the man should aim 4.9 m above the bird in order to hit
it. 29. A body is dropped vertically. Another identical body B is
projected horizontally, from the same point at same instant then a.
First body will reach first. b. Second body will reach first. c.
Second body will not reach the ground. d. Both bodies reach ground
at same instant. Sol: Correct option is (d)
The acceleration in vertical direction is same for both bodies
hence in vertical direction both bodies will take same time to
reach the ground. 30. Three particles P , P2 and P3 are projected
from same point with same 1 initial speeds making angles 400 , 450
and 500 respectively with the horizontal. Which of the following
statement is correct? a. Range P and P2 are equal but less than P3
. 1 b. Range P2 is greater than P but less than P3 1 c. Range of P
is maximum 1 d. None of the above
Sol:
Correct option is (a)
At 450 range is maximum and P and P2 will have same ranges as
the angles 1 of projection of both are complementary. 31. Two
bodies are projected with same speed but making different angles
and its maximum with horizontal. If the angle of projection of one
is 6 height is h , then the maximum height of other will be ranges
of both are equal.3h 2 b. 3h h c. 3 2h d. 3
a.
Sol:
Correct option is (b)
h ' u 2 sin 2 1 2g h ' sin 2 1 2 2 h 2g u sin h sin 2 h' h sin 2
1 sin 2
Now ' 90
as ranges are equal 2 6 3
3 sin 2 60 h h 2 h 4 3h 1 sin 30 4'
32. A projectile is projected with a linear momentum M, making
angle with the horizontal. The change in momentum of the projectile
on return to the ground will be a. M cos b. M sin c. 2 M cos d. 2 M
sin
Sol:
Correct option is (d)
Initial momentum in horizontal direction is P cos which is also
the final momentum in horizontal direction, hence no change of
momentum in horizontal direction. Initial momentum in vertical
direction = M sin Final momentum in vertical direction = M sin
Change = M sin -( M sin )= 2 M sin 33. Two walls W1 and W2 are
respected b a distance of 200 m. A bullet pierces W1 and then W2 .
The hole in W2 is 19.6 m below the hole in W1 . If the bullet is
traveling horizontally at the time of hitting W1 , then the
velocity at bullet at
W1 isa. 75 m/s b. 50 m/s c. 100 m/s d. 200 m/s Sol: Correct
option is (c)
The bullet dropped 19.6 m between W1 and W2 . Hence we can
calculate time taken by bullet between W1 and W2
t
2 19.6 2s 9.8200 100 m/s 2
Hence velocity at W1 =
34. A particle is projected with velocity V1 at angle angle of
600 with the horizontal. Another particle is thrown vertically
upwards with velocity V2 from a point vertically below the highest
point of path of first particle. The necessary condition for the
two particles to collide at highest point is
3V2 2 2V2 b. V1 3 V c. V1 2 2 d. V1 3V2a. V1
Sol:
Correct option is (b) V1 sin 60 g V2 g
Time taken by first particle to reach the highest point =
Time taken by second particle to reach the highest point = For
collision V1 sin 60 V2 2V V1 2 g g 3
35. A body of mass 1 kg is rotated at the end of the string in a
vertical circle of radius 1 m at a constant speed of 6 m/s. The
tension in the string at highest point of its path is a. 26.2 N b.
19.6 N c. 9.8 N d. 36 N Sol: Correct option is (a)
At highest point let tension be T, then T mg mV 2 mV 2 T mg r
r
T
1 36 1 9.8 36 9.8 26.2 N 1
36. A particle is moving along a circular path of radius 2 m and
with uniform speed 6 m/s. What will be the average acceleration
when the particle completes half revolution? a. 12 m / s 2 12 b. m
/ s2 3 c. m / s 2 36 d. m / s2
Sol:
Correct option is (d)
Change in velocity = 6 m/s (-6 m/s)=12 m/s Time taken =
r 2 v 6 3v 12m / s 36 m / s2 t s 3
Average acceleration =
37. The speed of revolution of particle along a circle is halved
and its angular speed is doubled. What happens to the centripetal
acceleration? a. becomes one fourth b. halved c. remains same d.
doubled Sol: Correct option is (c) V2 r
Centripetal acceleration =
ac
V 2 (rw)2 rw2 (rw) w vw r r
Now here V is halved and w is double hence the centripetal
acceleration remains same. 38. The string of a pendulum of length l
is displaced through 900 from vertical and was released. Then the
minimum strength of the string in order to with stand the tension
as the pendulum passes through the mean position is a. 3 mg b. 7 mg
c. mg d. 2 mg
Sol:
Correct option is (a) 2gr
Velocity at mean position =
mv 2 Also at mean position T mg r T mv 2 mg r
T mg 2mg 3mg
39. A particle of mass M describes a circle of radius R. The
centripetal 3 acceleration of the particle is 2 . What will be the
momentum of the r particle? m r 3m b. r c. m r d. 3m r a. Sol:
Correct option is (b) v2 r
Centripetal acceleration =
Therefore
v2 3 v3 r r r2 3m r
momentum mv
40. A body of mass 5 kg moving on a horizontal surface with an
initial velocity of 7 m/s comes to rest after 3.5 sec. If one wants
to keep this body moving on the same surface with a velocity of 7
m/s the force required is a. 4 N b. 6 N c. 10 N d. 8 N
Sol:
Correct option is (c)7 m / sec 2m / s 2 3.5sec
De-acceleration provided to body =
Force which provided the de-acceleration = 5 x 2 = 10 N Now this
force is to be opposed hence a force of 10 N is required to keep
the body moving with a velocity of 7 m/s 41. Two bodies having
masses m1 3kg and m2 7 kg are attached to the ends of a string of
negligible mass and suspended from a light frictionless pulley. The
acceleration of the bodies is a. 2.92 m / s 2 b. 3.92 m / s 2 c.
1.96 m / s 2 d. 29.2 m / s 2 Sol: Correct option is (b)
Acceleration of bodies =
m2 m1 4 9.8 g 3.92m / s 2 m1 m2 10
42. An elastic string has a length x when tension is 7 N. Its
length is y when tension is 8 N. On subjecting the string to a
tension of 15 N, its length will be a. (3y-7x) b. (2y-3x) c.
(8y-7x) d. (7y-8x) Sol: Correct option is (c)
Let original length = l When tension = 7 N 7 = kx, where x, is
extension Therefore l 7 x k
Also l
7 y k 15 (8 y 7 x) k
Therefore l
Hence when tension is 15 N length is (8y-7x) 43. A block of mass
2 kg is moving horizontally at 3 m/s. A vertically upward force of
8 N acts on it for 2 s. What will be the distance of the block from
the point where the force started acting? a. 8 m b. 9 m c. 12 m d.
10 m Sol: Correct option is (d)
Horizontal distance covered = 3 m/s x 2 s = 6 m Acceleration in
vertical direction = 8N 4m / s 2 2kg1 4 (2) 2 8m 2
Therefore distance covered in vertical direction = Total
distance = 6 2 82 10m
44. A uniform rope of length X, resting on a frictionless
horizontal surface is pulled at one end by a force F. What is the
tensions of the rope at a distance x from the end where force is
applied?
x a. F 1 X 1 b. F X x 1 c. F x X d. F x X
Sol:
Correct option is (a)F where M is mass of whole rope. M M ( X x)
X
Acceleration of rope =
Mass of rope of length (X-x) is M ' Also F ' M ' a Hence F '
M F x ( X x) X F 1 X M X
x Hence tension = F 1 X45. A jet of water with a cross-sectional
area A is striking against a wall at an angle to the horizontal and
rebounds elastically. If the velocity of water jet is V and density
is d, the normal force acting on the wall is a. 2 AV 2 d sin b. 2
AV 2 d cos c. AV 2 d cos d. AV 2 d sin Sol: Correct option is
(b)
Mass of water striking = Avd Also Force = Rate of change of
momentum
( AVd )V cos ( AVd )(V cos ) 2 AV 2 d cos 46. A hammer of mass M
strikes a nail of mass m with velocity of V m/s and drives it x
metre into fixed block of wood. The average resistance of wood to
the penetration of nail is
( M m)V 2 2x mV 2 b. ( M m) x M 2V 2 c. ( M m)2 x a. d.
MV 2 ( M m) xCorrect option is (c)
Sol:
Applying conservation of momentumMV ( M m)V0 M V0 V M m2 M V
De-acceleration provided = M m 2x 2
Resistance = (M+m)x de-acceleration = M2 V2 M m 2x
47. A particle of mass m moving with a velocity V makes a head
on elastic collision with another particle of same mass and
initially at rest. The velocity of first particle after collision
is a. 2V b. V c. 0 d. 3 V Sol: Correct option is (b)
mv mv1 mv2 v v1 v2Also1 2 1 1 mv mv12 mv2 2 2 2 2
v 2 v12 v2 2
Therefore 2v1v2 0 Now v1 cannot be zero hence v2 0 . Therefore
v1 v Hence velocity of first particle is V. 48. A bullet of mass m
and velocity V is fired into a large block of mass M. The final
velocity of system is(m M ) V m (m M ) b. V M MV c. mM mV d. mM
a.
Sol:
Correct option is (d)
Applying conservation of momentummV ( M m)Vsystem
Vsystem
mV ( M m)
49. A block of mass m slides down along the surface of the bowl
having radius R from the rim to the bottom. The velocity of the
block at bottom will be a. 2gR b. 2gR c. 3gR d. 3gR
Sol:
Correct option is (a)
Applying conservation of energy1 mV 2 mgR 2 V 2 gR
50. A body moves a distance of 6 m along a straight line under
the action of a force of 4 Newtons. If the work done is 12 2 jules,
the angle which the force makes with the direction of motion of the
body is a. 00 b. 450 c. 600 d. 900 Sol Correct option is (b)
Work done = FSCos F= 4 N S=6m Work done = 12 2 jules Hence cos
Hence 450 51. A body of mass 9 kg moving with a velocity of 2 m/s
collides head on with a body of mass 3 kg moving with a velocity of
6 m/s. After collision the two bodies stick together and move with
a common velocity. Which in m/s is equal to? a. 1 b. 6 c. 2 d. 312
2 1 24 2
Sol:
Correct option is (d)
Initial momentum = 9 2 + 6 3 = 36 Final momentum = (9 + 3) V
Applying conservation of energy (9 + 3) V = 36 V = 3 m/s 52. Two
balls B1 and B2 having masses 2 kg and 4 kg respectively are moving
in opposite directions with velocity of B1 equal to 3 m/s. After
collision balls come to rest when velocity of B2 is a. 1 m/s b. 1.
5 m/s c. 3 m/s d. 4.5 m/s Sol: Correct option is (b)
For both of balls to come to rest after collision, final
momentum should be zero. According to conservation of momentum law,
initial momentum should also be zero. Hence 2 x 3 + 4 x (-v)= 0 v=
1.5 m/s 53. A bomb of mass 20 kg explodes into two pieces of 8 kg
and 12 kg. The velocity of 8 kg mass is 3 m/sec. The kinetic energy
of other mass is a. 0 b. 6 c. 12 d. 24 Sol: Correct option is
(d)
Applying conservation of momentum 0 = 8 x 3 + 12 (v)
v= -2 m/s
Hence kinetic energy =
1 12 (2) 2 24 2
54. An electric motor creates a tension of 4000 newtons in a
hosting cable and reels it in at the rate of 3 m/sec. What is the
power of electric motor? a. 9 kW b. 12 kW c. 18 kW d. 24 kW Sol:
Power = F.V = 4000 x 3 = 12000 W = 12 kW 55. A ball of mass m
moving with a certain velocity collides against a stationary ball
of mass m. The two balls stick together during collision. If E be
the initial kinetic energy, then loss of kinetic energy in the
collision isE 3 4E b. 3 E c. 2 3E d. 4
Correct option is (b)
a.
Sol:
Correct option is (c)
Initial momentum = mv Final momentum = 2 mv'2mv ' mv v ' v 2 1 2
mv 2
Initial kinetic energy = E =
1 mv 2 E v Final kinetic energy = (2m) 2 4 2 2
2
Loss E
E E 2 2
56. A mass m moving horizontally with velocity V0 strikes a
pendulum of mass m. If two masses stick together after the
collision, then the maximum height reached by the pendulum is
Vo 2 8g V2 b. o 4g V2 c. o 2g V2 d. o ga. Sol: Correct option is
(d)
According to conservation of momentum2mv mVo V V0 2
Also
1 1 2 gh 4 gh (m)V 2 2mgh V 2 V2 2 2 2 22
4 gh Vo 2 4 gh Vo 2 2 4 2 Also V2 h o g 57. A sphere moving with
velocity V strikes a wall moving towards the sphere with a velocity
U. If the mass of the wall is infinitely large, the work done by
the wall during collision will be a. mUV b. mV (U V ) c. 2mU (U V )
d. mU (U V )
Sol:
Correct option is (c)
Work done = Change in kinetic energy1 1 m(V 2U )2 mV 2 2 2 1 m
4(U V ) 2mU (U V ) 2
58. If the distance between the two masses is doubled,
gravitational attraction between them a. is halved b. doubled c. is
reduced to a quarter d. none of the above Sol:F Gm1m2 r2
Correct option is (c)
If r ' 2r
then
F'
F 4
59. A simple pendulum has a time period T1 when on earths
surface and T2 when taken to a height R above the earths surface
where R is radius of T earth. The volume of 2 is T1 a. 2 1 2 c. 2 1
d. 2 b. Sol: Correct option is (a)
The acceleration due to gravity at earths surface is g and at a
distance R from g earths surface is . 4 HenceT2 2 T1
60. Let g e be the acceleration due to gravity at the equator
and g p be that at the poles. Assuming earth to be the sphere of
radius Re rotating about its own axis with angular speed W , then g
p g e is given by a. Re W 2 b. Re W 2 c. ReW 2 R d. e2 W Sol:
Correct option is (c)
We know that g ' g ReW 2 cos 2 At equator g e g ReW 2 At poles g
p g Therefore g p ge ReW 2 62. A man is riding a bicycle at
velocity 72 km/hr up a hill having a slope 1 m in 20. Total mass of
man and cycle is 90 kg. The power of man is a. 878 W b. 780 W c.
880 W d. 882 W Sol: Correct option is (d)
Power mg sin V 1 90 9.8 20 20 90 9.8 9 98 882W
63. A particle describes a horizontal circle of radius 2 m with
uniform speed. The centripetal force acting is 30 N. The work done
is describing a semi-circle is
a. 0 b. 30 N c. 80 N d. 70 N Sol: Correct option is (a)
As force is perpendicular to the displacement, hence no work is
being done. 64. A boy pushes a toy box 20 m along the floor of
means of a force of 20 N directed downward at an angle 60o to the
horizontal. The work done by the boy is a. 100 N b. 200 N c. 300 N
d. 400 N Sol: Correct option is (b)
FSCos60o 1 Work done 20 20 2 200 joules65. Two spheres of the
same diameter, one of mass 10 kg and other of 3 kg are dropped at
same time from top of a tower when they are 2 m above the ground
the two spheres have the same a. velocity b. kinetic energy c.
potential energy d. momentum Sol: Correct option is (a)
Both of the spheres will have same velocity as both started from
rest and are being acted upon by same acceleration, hence both the
spheres have same velocity. 66. A particle proves a point r1 2 i 3
j to another point r2 4 i 6 j during which a constant force F 4 i 3
j acts on it the work done by the force on the particle during the
displacement is
a. 10 J b. 9 J c. 8 J d. 17 J Sol: Correct option is (d)
Displacement = r2 r1 2 i 3 j Hence work done (4 i 6 j )(2 i 3 j
) 8 9 17 joules 67. The momentum of a body 50% then the percentage
increases in kinetic energy is a. 150% b. 125% c. 20 % d. 100 %
Sol: Correct option is (b)
Let initial momentum = P Final momentum = 1.5 P Initial K.E=p2
2m
Final K.E. =
(1.5 p ) 2 2.25 p 2 2m 2m
2.25 1 Hence % increases = 100 125% 1
68. The kinetic energy of body is increased by 300 %. The
momentum of the body would increase by a. 25% b. 70% c. 100 % d. 50
%
Sol:
Correct option is (c)
Let initial K.E. = K Final K.E. = 4 K Initial momentum = Final
momentum = Hence % increases =2mK
2m(4 K ) 2 2mK2 1 100 100% 1
69. A person raises 3 kg of weight to a height of 2 m and holds
it for 1 hour. How much work has he performed a. 9.8 J b. 3 x 9.8 J
c. 4 x 9.8 J d. 6 x 9.8 J Sol: Correct option is (d)
Work done = F x S =mxgxS = 3 x 9.8 x 2 = 6 x 9.8 J 70. A body is
under the action of two equal and opposite forces each of 8 N. The
body is displaced by 3 m. The work done is a. 0 b. 8 J c. 3 J d. 5
J Sol: Correct option is (a)
As net force = 0 Hence work done = F.S. = 0
71. Work done by a simple pendulum in a complete vibration is a.
2 b. 0 c. 2 d. gl Sol: Correct option is (b)l g
g l
As force is perpendicular to the displacement in case of simple
pendulum, hence the work done is zero. 72. A body of mass M
accelerates uniformly from rest to a speed V in t seconds. The
average power delivered ismV t b. mVt mV 2 c. t mV 3 d. t
a.
Sol:
Correct option is (c)V t
Acceleration =mV t
Force =
Power = Force x Velocity mV mV 2 V t t
73. A machine which is 80 % efficient uses 480. 2 J of energy in
lifting up 2 kg of mass through a certain distance. The mass is
then allowed to fall through that distance. The velocity at the end
of its fall is
a. 19.6 m/s b. 9.8 m/s c. 3.92 m/s d. 20 m/s Sol: We know that80
480.2 mgh 100 384.16 2 9.8 h h 19.6m
Correct option is (a)
Now velocity =
2 gh 2 9.8 19.6 384.16 19.6m / s
74. A light and a heavy body have equal momentum which one has
greater kinetic energy? a. light body b. nothing can be said c.
heavy body d. none of the above Sol: Correct option is (a) P2
2m
As kinetic energy =
As P is same for both bodies hence light body has less m and
more kinetic energy. 75. A canon ball is fired with a velocity 200
m/s at an angle of 60o with horizontal. At highest point of its
flight it explodes into 3 equal fragments, one going vertically
upwards with a velocity 100 m/s, the second falling vertically
downwards with velocity 100 m/s. The third segment will be moving
with a velocity a. 20 m/s b. 100 m/s c. 300 m/s d. 400 m/s
Sol:
Correct option is (c)
Momentum is conserved in vertical direction. Applying
conservation of momentum in horizontal direction.m V 3 1 V 3 200
300m / s 2 m 200 cos 60
76. A body of mass m moving with a constant velocity V hits
another body of the same mass moving with same velocity V but in
opposite direction, and sticks to it. The velocity of the compound
body after collision is a. 2V b. 0 c. V d. 3V Sol: Correct option
is (b)
Initial momentum = mv + (-mv)=0 Hence final momentum = 0 Hence
velocity of compound body = 0 77. A shell if fired from a canon
with velocity v m/s at angle with horizontal direction. At the
highest point of its path it explodes into two pieces of equal
mass. One of the pieces retraces its path to the canon and the
speed of other piece immediately after collision is3V cos 2 b. V
cos 2V c. cos 3 d. 3V cos
a.
Sol:
Correct option is (d)
Applying conservation of momentum mV cos mV ' m V cos 2 2
mV ' 3mV cos 2 2 ' V 3V cos
78. If m , p and l denote respectively the mass, linear momentum
and angular momentum of a particle moving in a circle of radius r,
then the kinetic energy of particle can be express as l2 m p2 b. 2m
p2 c. 2l l d. m a. Sol: Correct option is (b)1 mV 2 2
Kinetic energy =p m
p mV V
Hence kinetic energy =
1 p2 p2 m 2 2 m 2m
79. A shell of mass m moving with velocity V suddenly breaks
into 2 pieces. m The part having mass remains stationary. The
velocity of other part will 4 be4V 3 b. 2V c. V 2V d. 3
a.
Sol:
Correct option is (a)
Applying conservation of momentummV 3m V ' 4 4V 3
V'
80. Two equal masses moving along same straight line with
velocities + 7 m/s and 8 m/s respectively collide elastically.
Their velocities after collision will be respectively. a. + 7 m/s,
-8 m/s b. + 8 m/s, -7 m/s c. -8 m/s, +7 m/s d. +8 m/s, +7 m/s Sol:
Correct option is (c)
As the mass of both bodies is same hence after collision their
velocities get exchanged hence the bodies have -8 m/s and + 7 m/s
velocity respectively. 81. A body of mass 5 m initially at rest
explodes into 3 fragments of mass ratio 3:1:1. Two of fragments of
mass m are found to move with a speed of 60 m/s in mutually
perpendicular directions. The velocity of third fragments is a. 20
b. 2 c. 10 2 d. 20 2 Sol: Correct option is (d)
Applying principle of conservation of momentum, we get3m V (m
60) 2 (m 60) 2 m 60 2
V 20 2
82. For a collision which is neither perfectly elastic nor
perfectly in elastic the coefficient of restitution is a. 0 b. 1 c.
0 < e < 1 d. 2 Sol: Correct option is (c)
For perfectly elastic collision e=1 For perfectly inelastic
collision e=0 But for collision to be neither perfectly elastic and
nor perfectly inelastic the value of e lies between 0 and 1 83. A
massive ball moving with speed V collides with a tiny ball of
negligible mass. The collision is perfectly elastic. The second
ball will move with a speed equal to a. 0 b. V c. 2V d. 3V Sol:
Correct option is (c)2m1U1 , but m2 m1 m2 m1
We know that velocity of second ball V2 Hence V2 2V
84. A block of mass 2 m, moving with a constant velocity 3 V
collides with another block of mass m, which is at rest and sticks
to it. The velocity of the compound block will be a. 2 V 2V b. 3 c.
3 V 4V d. 3
Sol:
Correct option is (a)
Initial momentum = 2 m x 3 V = 6 mV Final momentum = 3 m x V '
Applying conservation of momentum
6mV 3mV ' V ' 2V85. A mass of 1 kg moving with a velocity of 2
m/s strikes a pendulum bob of mass 1 kg. The two masses stick
together. The maximum height reaches by the system is a. 0.5 m b.
0.05 m c. 1 m d. 0.08 m Sol: Correct option is (b)2 1 1m / s 11
Velocity of system =
Hence height obtained h 1 .05m 2 9.8
V2 2g
h
86. If a shell fired from a canon explodes in mid air then its
total a. momentum increases b. momentum decreases c. kinetic energy
increases d. kinetic energy decreases. Sol: Correct option is
(c)
As the chemical energy is converted into kinetic energy, hence
the total kinetic energy increases.
87. A mass m moving with velocity Vo strikes a pendulum of mass
m. If the two masses stick together after the collision, then
maximum height reached by the pendulum is
Vo 2 g 2V 2 b. o g 3V 2 c. o 8g V2 d. o 8ga. Sol: Correct option
is (d)mVo V0 2m 22
Velocity of system =
V0 2 Hence height reached = 2g
h
Vo 2 8g
88. The kinetic energy of a body of mass 3 kg and momentum 6 Ns
is a. 6 J b. 9 J c. 12 J d. 8 J Sol: Correct option is (a) p2 6 6
6J 2m 2 3
Kinetic energy =
89. Two bodies with kinetic energies in ratio of 4:1 are moving
with equal linear momentum. The ratio of their masses is a. 1:2 b.
4:1 c. 1:4 d. 1:1
Sol:
Correct option is (c) p2 2m
As kinetic energy
As p is same for both bodies ButK .E1 m2 m m 1 2 4 1 K .E2 m1 m1
m2 4
Hence ratio is 1:4 90. A billiard ball moving with a speed of 7
m/s collides with an identical ball, originally at rest. If the
first ball stops after collision then second ball will move forward
with a speed of a. 8m/s b. 9 m/s c. 2 m/s d. 7 m/s Sol: Correct
option is (d)
Applying conservation of momentum
m 7 m V ' V ' 7m / s
91. There are two charges of 2 C and 8 C. The ratio of
electrostatic force acting on them due to each other will be in
ratio a. 1:3 b. 3:1 c. 2:3 d. 1:1 Sol: Correct option is (d)
Electrostatic force on both the charges is same in magnitude but
opposite in direction. 92. A charge q is placed at the centre of
the joining of the two equal charges will be in equilibrium if q is
equal to
a.
Q 2 Q 4 Q 3
b. c. Q d. Sol:
Correct option is (b)
For system to be in equilibriumKQ 2 KqQ 0 2 r2 r 2 Q q 4
93. Four charges are arranged at the corners of a square WXYZ as
shown in figure. The force on the charge kept at the centre O
is
a. along OW b. along OY c. along OX d. along OZ Sol: Correct
option is (d)
Forces due to charges W and Y cancel each other and forces due
to charges X and Z act along OZ 94. A charged particle of mass m
and charge q is released from rest in an electric field of constant
magnitude E. The kinetic energy of the particle after a time t
is
q 2 E 2t 2 2m 2 2 qt b. mE q2m c. E qt d. mE a. Sol: Force = qE
Acceleration =qE m qEt m
Correct option is (a)
Velocity attained in time t = 1 q 2 E 2t 2 m 2 m
Hence K.E. =
K.E. =
q 2 E 2t 2 2m
95. Three charges +4q, Q, and q are placed in a straight line of
length l at l positions (o), and l respectively. What should be Q
in order to make 2 the net force on q to be zeroq 3 b. q q c. 3 q
d. 2
a.
Sol:
Correct option is (b)
Total force on q to be zero
K (4q )(q ) KQ (q ) 0 2 l2 l 2 Q q
96. Force between two charges when placed in free space is 20 N.
If they are in a medium of relative permittivity 4, the force
between them will be a. 3 N b. 6 N c. 5 N d. 2 N Sol:Fspace Kq1q2
r2
Correct option is (c)
1 1 but for medium K 40 40 r 20 Hence Fmedium 5N 4
Where K
97. Three charges each equal to +4 C are placed at the corners
of equilateral triangle. If the force between any two charges be F,
then the net force on either will be a. 2F b. 3 F F c. 2 d. 3F Sol:
Correct option is (d)
Resultant force =
F 2 F 2 2 F 2 cos 600 3F
98. Electric charges q, q, -2q are placed at the corners of an
equilateral triangle ABC of side l. The magnitude of electric
dipole moment of the system is
a. 2ql b. ql c.3ql ql d. 2
Sol:
Correct option is (c)
There will be two dipoles inclined to each other at an angle of
60o . The dipole moment of each dipole will be (ql). The resultant
dipole moment= (ql )2 (ql )2 2(ql ) 2 cos 60 3ql 99. What is the
angle between electric dipole moment and the electric field
strength due to it on the axial line?
a. 90o b. 0o c. 180o d. 270o Sol: Correct option is (b)
As we know that dipole moment is away from positive charge and
also the electric field is away from positive charge along same
line. Hence the angle between them is zero.
100. The electric flux through a hemisphere surface of radius R
placed in a uniform electric field of intensity E parallel to the
axis of its circular plane is a. R 2 E b. 2 RE c. 2 E d. RE
Sol:
Correct option is (a)
Electric flux through any surface is equal to the product of
electric field intensity at the surface and the component of the
surface perpendicular to electric field Flux = E R 2 R 2 E