﻿ Engineering Entrance Exam Question and Answers | Laws of Motion - Zigya

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# Laws of Motion

#### Multiple Choice Questions

31.

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is

• • • • D.  By conservation of linear momentum 1144 Views

32.

A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release? • 2g/3

• g/2

• 5g/6

• 5g/6

B.

g/2

For the mass m,
mg-T = ma As we know, a = Rα    ... (i)
So, mg-T = mRα
T x R = mR2α ...... (ii)
From Eqs. (i) and (ii), we get
a = g/2
Hence, the acceleration of the mass of a body fall is g/2.

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33.

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

• 44%

• 50%

• 56%

• 56%

C.

56%

354 Views

34.

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :

• (0,1)

• (.89,.28)

• (.28,.89)

• (0,0)

B.

(.89,.28)

For collision of a neutron with deuterium: Applying conservation of momentum:

mv + 0 = mv1 + 2mv2 .....(i)
v2 -v1 = v ...... (ii)

Therefore, Collision is elastic, e = 1

From equ (i) and equ (ii) v1 = -v/3

Now, for the collision of neutron with carbon nucleus Applying conservation of momentum

mv + 0 = mv1 + 12mv2 ....; (iii)

v = v2-v1  ....(iv)

35.

Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is • 100N

• 80 N

• 120 N

• 120 N

C.

120 N

In the vertical direction, weight are balanced by frictional forces.
As the blocks are in equilibrium balance forces are in horizontal and vertical direction.
For the system of blocks (A+B)
F = N
For block A, fA = 20 N and for block B.
fB = fA +100 = 120 N

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