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 Multiple Choice QuestionsMultiple Choice Questions

11.

A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to

  • 300 N

  • 150 N

  • 30

  • 30


C.

30

(mv-0)
⇒ 0.15 x 20
F = 3/0.1 = 30 N

250 Views

12.

A block of mass ‘m’ is connected to another block of mass ‘M’ by a spring (mass less) of spring constant ‘k’. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is stretched. Then a constant force ‘F’ starts acting on the block of mass ‘M’ to pull it. Find the force on the block of mass ‘m’.

  • mF/M

  • (M+m)F/m

  • mF/(m+ M)

  • mF/(m+ M)


C.

mF/(m+ M)

195 Views

13.

Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is [µ =k 0.5]

  • 800 m

  • 1000 m

  • 600 m

  • 600 m


B.

1000 m

straight mu subscript straight k mgs space equals space 1 half mu squared
straight s equals space fraction numerator straight u squared over denominator 2 straight mu subscript straight k straight g end fraction space equals space 1000 space straight m
317 Views

14.

Consider a two-particle system with particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

  • straight m subscript 1 over straight m subscript 2 straight d
  • straight m subscript 2 over straight m subscript 1 straight d
  • d

  • d


A.

straight m subscript 1 over straight m subscript 2 straight d
straight m subscript 1 straight d space plus straight m subscript 2 straight x space equals space 0
straight x space equals space fraction numerator straight m subscript 1 straight d over denominator straight m subscript 2 end fraction
457 Views

15.

The block of mass M moving on the frictionless horizontal surface collides with a spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

  • square root of MK L
  • KL2/2M

  • zero

  • zero


A.

square root of MK L
1 half KL squared space equals space fraction numerator straight P squared over denominator 2 straight m end fraction
therefore space straight P space equals space square root of MK straight L
512 Views

16.

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m which applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2

  • 22 N

  • 4 N

  • 20 N

  • 20 N


C.

20 N

mgh = Fs
F = 20 N

395 Views

17.

A mass ‘m’ moves with a velocity v and collides inelastically with another identical mass. After collision, the 1st mass moves with velocity v/√3 in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision

  • v

  • √3 v

  • 2v/√3

  • 2v/√3


C.

2v/√3

mv= mv1 cos = θ



0 space equals space fraction numerator mv over denominator square root of 3 end fraction space minus mv subscript 1 sinθ
therefore space straight v subscript 1 space equals space fraction numerator 2 over denominator square root of 3 end fraction straight v

738 Views

18.

A particle of mass 0.3 kg is subjected to a force F=−kx with k=15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin?

  • 3 m/s2

  • 15 m/s2

  • 5 m/s2

  • 5 m/s2


D.

5 m/s2

a = kx/m =10m /s2

171 Views

19.

A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms−1 . The kinetic energy of the other mass is

  • 144 J

  • 96 J

  • 288 J

  • 288 J


C.

288 J

m1v1 = m2v2

KE space equals space 1 half straight m subscript 2 straight v subscript 2 superscript 2 space
space equals space 1 half space straight x space 4 space straight x space 144 space equals space 288 space straight J

506 Views

20.

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is

  • Mg left parenthesis square root of 2 minus 1 right parenthesis
  • Mg left parenthesis square root of 2 plus 1 right parenthesis
  • Mg square root of 2
  • Mg square root of 2

A.

Mg left parenthesis square root of 2 minus 1 right parenthesis
straight F calligraphic l space sin space 45 space equals space M g left parenthesis calligraphic l minus calligraphic l space cos space 45 right parenthesis
F space equals space M g left parenthesis square root of 2 minus 1 right parenthesis
285 Views