100 MCQ of Physics

MULTIPLE CHOICE QUESTIONS 



1. 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:

Correct option is (d)

As

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:

Correct option is (a)

x+y 

=

x-y

2 x

+

y

2 +2 x

y cos  =

x

2 +

-y

2 +2

x

-y cos 

where  is the angle between X and Y 2



x

 4

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











F  2 i  6 j  4 k moved

force 



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:


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:


Resultant force =

( P ) 2  (3P) 2  2( P )(3P) cos 

 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:


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:


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. cos1    b. sin 1    c. cos1    d. sin 1  

9   86  9   86 

86   9  86   9 

Sol:

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     cos1    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 m

t=3

9 39  3u  a 2

and

144  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/s 14. 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:


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:


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:


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 second S10  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:


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:


As car gains 10 m in 2 s and let relative acceleration between car and bicycle is a therefore 1 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:


Let time taken by body to go to the water level be t. Then 1 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:


Time taken by a ball to go up =

1 second 2

Therefore height up to which ball rises is 2

H=

1 1  9.8     9.8 10 2 2

H = 98 

9.8  98  1.225  96.775m 8

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:


Let the time for which stone remained in the air is t Then distance traveled in last second 1 2 1 g gt  g (t  1) 2  tg  2 2 2

S=

Also distance traveled in first 5 seconds S=

1  g  25 2

Equating both distances we get tg 

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:


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=-g H 

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:


For 1st ball S1  19.6  t 

1 2 gt 2

For second ball S 2  19.6(t  1) 

1 g (t  1)2 2

For collision S1  S 2 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 2

19.6  t 

2

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 h 1

c. e h h d. eh

Sol:


V  H= 2

2g

Velocity after first rebound =ev e2v e 21v 2  h'   h'  2g 2g Height attained after first rebound h ' 

e21v 2 2g

Hence height attained after (t)th rebound ht  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 =

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:


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:


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 P1 , P2 and P3 are projected from same point with same initial speeds making angles 400 , 450 and 500 respectively with the horizontal. Which of the following statement is correct? a. Range P1 and P2 are equal but less than P3 . b. Range P2 is greater than P1 but less than P3 c. Range of P1 is maximum d. None of the above

Sol:


At 450 range is maximum and P1 and P2 will have same ranges as the angles 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:


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:


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 is a. 75 m/s b. 50 m/s c. 100 m/s d. 200 m/s Sol:


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 9.8

Hence velocity at W1 =

200  100 m/s 2

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  3V2 a. V1 

Sol:


Time taken by first particle to reach the highest point =

V1 sin 60 g

Time taken by second particle to reach the highest point =

V2 g

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:


At highest point let tension be T, then T  mg 

T 

mV 2 mV 2 T   mg r r

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:


Change in velocity = 6 m/s – (-6 m/s)=12 m/s Time taken =

r  2    v 6 3

Average acceleration =

v 12m / s 36   m / s2  t  s 3

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:


Centripetal acceleration =

 ac 

V2 r

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:


Velocity at mean position =

2gr

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:


Centripetal acceleration =

Therefore

v2 r

v2 3  v3 r r r2

 momentum  mv 

3m r

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:


De-acceleration provided to body =

7 m / sec  2m / s 2 3.5sec

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:


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:


Let original length = l When tension = 7 N 7 = kx, where x, is extension Therefore l 

7 x k

Also l 

7 y k

Therefore l 

15  (8 y  7 x) k

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:


Horizontal distance covered = 3 m/s x 2 s = 6 m Acceleration in vertical direction =

8N  4m / s 2 2kg

Therefore distance covered in vertical direction = Total distance =

1  4  (2) 2  8m 2

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:


Acceleration of rope =

F where M is mass of whole rope. M

Mass of rope of length (X-x) is M ' 

M ( X  x) X

Also F '  M ' a Hence F ' 

M F x  ( X  x) X  F 1   X M  X

x  Hence tension = F 1    X 45. 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:


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) x

Sol:


Applying conservation of momentum MV  ( M  m)V0  M   V0   V M m 2

2  M  V De-acceleration provided =     M  m  2x

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:


mv  mv1  mv2  v  v1  v2 Also

1 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. mM mV d. mM

a.

Sol:


Applying conservation of momentum mV  ( 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:


Applying conservation of energy 1 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


Work done = FSCos F= 4 N S=6m Work done = 12 2 jules Hence cos  

12 2 1  24 2

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. 3

Sol:


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:


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:


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 is E 3 4E b. 3 E c. 2 3E d. 4

a.

Sol:


Initial momentum = mv Final momentum = 2 mv' 2mv '  mv  v ' 

v 2

Initial kinetic energy = E =

1 2 mv 2

2

1 mv 2 E v Final kinetic energy = (2m)     2 4 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 g a.

Sol:


According to conservation of momentum 2mv  mVo  V 

Also

V0 2

1 1 2 gh 4 gh (m)V 2  2mgh  V 2  V2  2 2 2 2 2

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:


Work done = Change in kinetic energy 1 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

Correct option is (c) Gm1m2 r2

If r '  2r

then

F' 

F 4

59. A simple pendulum has a time period T1 when on earth’s surface and T2 when taken to a height R above the earth’s 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:


The acceleration due to gravity at earth’s surface is g and at a distance R from g earth’s surface is . 4 Hence

T2 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:


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:


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:


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:


 FSCos60o 1 Work done  20  20  2  200 joules 65. 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:


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:


Let initial momentum = P Final momentum = 1.5 P Initial K.E=

Final K.E. =

p2 2m

(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:


Let initial K.E. = K Final K.E. = 4 K Initial momentum =

2mK

Final momentum =

2m(4 K )  2 2mK

Hence % increases =

2 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:


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:


As net force = 0 Hence work done = F.S. = 0

71. Work done by a simple pendulum in a complete vibration is l g

a. 2 b. 0 c. 2

g l

d. gl Sol:


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 is mV t b. mVt mV 2 c. t mV 3 d. t

a.

Sol:


Acceleration =

Force =

V t

mV t

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 that 80  480.2  mgh 100  384.16  2  9.8  h  h  19.6m

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:


As kinetic energy =

P2 2m

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:


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:


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 is 3V cos  2 b. V cos  2V c. cos  3 d. 3V cos 

a.

Sol:


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:


Kinetic energy =

p  mV  V 

1 mV 2 2

p m

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 be 4V 3 b. 2V c. V 2V d. 3

a.

Sol:


Applying conservation of momentum mV 

3m V ' 4

V' 

4V 3

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:


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:


Applying principle of conservation of momentum, we get 3m  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:


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:


We know that velocity of second ball V2 

2m1U1 , but m2  m1 m2  m1

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:


Initial momentum = 2 m x 3 V = 6 mV Final momentum = 3 m x V ' Applying conservation of momentum

6mV  3mV '  V '  2V 85. 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:


Velocity of system =

2 1  1m / s 11

Hence height obtained h 

h

V2 2g

1  .05m 2  9.8

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:


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 8g a.

Sol:


Velocity of system =

mVo V0  2m 2

 V0    2 Hence height reached =   2g

h

2

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:


Kinetic energy =

p2 6  6   6J 2m 2  3

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:


As kinetic energy

p2 2m

As p is same for both bodies But

K .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:


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:


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

b. 

Q 4

c. Q d. 

Q 3

Sol:


For system to be in equilibrium KQ 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:


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

Velocity attained in time t =

Hence K.E. =

 K.E. =

qEt m

1 q 2 E 2t 2  m 2 m

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 zero q 3 b.  q q c. 3 q d. 2

a.

Sol:


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 

Correct option is (c) Kq1q2 r2

1 1 but for medium K  40 40  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:


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:


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:


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:


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

100 MCQ of Physics

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