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

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# Laws of Motion

#### Multiple Choice Questions

11.

Assertion: The driver in a vehicle moving with a constant speed on a straight road is in a noninertial frame of reference.

Reason: A reference frame in which Newton's laws of motion are applicable is non-inertial.

• If both assertion and reason are true and reason is the correct explanation of assertion

• If both assertion and reason are true but reason is not the correct explanation of assertion

• If assertion is true but reason is false

• If both assertion and reason are false

C.

If assertion is true but reason is false

If one takes two frames with origin O1 and  O2 and if a body  P is  at rest, boith according to O1 as well as O2. The body is at rest. If the frame is moving with a constant velocity, the same observation will not be made. Therefore a frame moving with a constant velocity cannot be taken as  an inertial frame. Assertion is right.

But the  reason given is wrong because the frame on  which Newton's laws of motion are applicable is an inertial frame.

12.

A block of mass 1 kg is placed on the wall under a force F. If coefficient of friction ($\mathrm{\mu }$) is 0.5, then minimum value of force applied is :

• 9.8 N

• 49 N

• 19.6 N

• 1.96 N

C.

19.6 N

Weight of a block = mg

For block to be balanced,

mg = $\mu$F

$\therefore$ F = $\frac{\mathrm{mg}}{\mathrm{\mu }}$

=  = 19.6 N

13.

Two bodies of mass m and 4m have equal kinetic energy. What is the ratio of their momentum?

• 1 : 4

• 1 : 2

• 1 : 1

• 2 : 1

B.

1 : 2

Momentum p = $\sqrt{2{\mathrm{mE}}_{\mathrm{k}}}$

Both have equal kinetic energy

p1 : p2 = 1 : 2

14.

A particle of mass m moving with velocity v collides with a stationary particle of mass 2m. The speed of the system, will be

• 3v

• $\frac{\mathrm{v}}{2}$

• $\frac{\mathrm{v}}{3}$

• 2v

C.

$\frac{\mathrm{v}}{3}$

According to conservation of momentum

change in moment is eual to final momentum

m1v1+m2v2= (m1+m2)v'

mv+2m×0= (m+2m)v'

so,   v'=$\frac{\mathrm{v}}{3}$

15.

Assertion: Frictional forces are conservative forces.

Reason:  Potential energy can be associated with frictional forces.

• If both assertion and reason are true and reason is the correct explanation of assertion

• If both assertion and reason are true but reason is not the correct explanation of assertion

• If assertion is true but reason is false

• If both assertion and reason are false

D.

If both assertion and reason are false

Frictional force is resistive force to motion. When two bodies move against each other some of the kinetic energy is heat energy due to friction. This reduces the total kinetic energy in the system.

Frictional work is not completely recoverable. When the force of friction is the total mechanical energy is not conserved. The friction force is therefore called a non conservative or a dissipative force.

16.

A force of 49 newtons is just able to move a block of wood weighing 10 kg-on a rough horizontal surface. Its coefficient of friction is :

• 1

• 0.7

• 0.5

• zero

C.

0.5

Coefficient of friction

= 0.49 ≅ 0.5

17.

A cone filled with water is revolved in a vertical circle of radius 4 m and the water does not fall down. What must be the maximum period of revolution?

• 2 s

• 4 s

• 1 s

• 6 s

B.

4 s

When a cone filled with water is revolved in a vertical circle, then the velocity at the highest point is given by

18.

A car is moving with speed 30 m/s on a circular path of radius 500 m. If its speed is increasing at a rate of 5m/s2, then resultant acceleration at that instant is

• 6.83 m/s2

• 8 m/s2

• 5.31 m/s2

• 4 m/s2

C.

5.31 m/s2

The car will experience two types of accelerations :
(i) Radial acceleration due to circular path

ar$\frac{{\mathrm{\nu }}^{2}}{\mathrm{r}}$

=

(ii) Tangential acceleration due to tangential speed

a

$\therefore$ Resultant acceleration

a = $\sqrt{{\mathrm{a}}_{\mathrm{r}}^{2}+{\mathrm{a}}_{\mathrm{T}}^{2}}$

=

= $\sqrt{28.24}$

= 5.31 m/s2

19.

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking =10 m/s2, work done against friction is

• 200 J

• 100 J

• Zero

• Zero

B.

100 J

Net work done in sliding a body up to a height h on inclined plane
= Work done against gravitational force + Work done against frictional force

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20.

A particle is moving in a circle with uniform speed v. In moving from a point to another diametrically opposite point :

• the momentum changes by mv

• the momentum changes by 2mv

• the kinetic energy changes by $\frac{1}{2}{{\mathrm{mv}}}^{{2}}$

• the kinetic energy changes by mv2

B.

the momentum changes by 2mv

Initial velocity v1 = v

Final velocity v2 = -v

Initial momentum p1 = mv

Final momentum p= m (-v) = - mv

Change in momentum

Δp = p1 - p2

= mv - (-mv)

= 2mv