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

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# Laws of Motion

#### Multiple Choice Questions

21.

Two weights w1 and w2 are suspended to the two strings on a frictionless pulley. When the pulley is pulled up with an acceleration g, then the tension in the string is :

A.

Equation of motion for first weight This is a frictionless and inextensible pulley.

for first weight

T - m1g = m1 ( a - g)

⇒ T - 2m1 g = m1a

For second weight

m2g -T = m2 ( a - g)

2 mg - T = m2 a

on solving equations (1) & (2)

22.

Figure shows two blocks A and B pushed against the smooth wall with the force F. The wall is smooth but the surfaces in contact of A and B arerough. Which of the following is true for the system of blocks ? • The system can not be in equilibrium

• F should be equal to weight of A and B

• F should be less than theweight of A and B

• F should be more than the weight of A and B

A.

The system can not be in equilibrium

Because the wall is smooth so whatever maybe the applied force, the system will move down due to its owriweight. Hence the system cannot be in equilibrium.

23.

A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is ) • • • • A. The moment of inertia of the uniform rod about an axis through the end and perpendicular to length is where m is mass of rod and l its length.
Torque acting on centre of gravity of rod is given by or  993 Views

24.

The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is

• • • • C. The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is where M is the mass of disc and R its radius. According to theorem of parallel axis, MI of circular disc about an axis touching the disc at it diameter and normal to the disc is  1958 Views

25.

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be • 30°

• 45°

• 60°

C.

45°

From force diagram shown in figure,

T = mg    .... (i)

and  2T cos θ =  $\sqrt{2}$ mg     .... (ii)

Combining Eqs. (i) and (ii), we have 26.

A block B is pushed momentarily along a horizontal surface with an initial velocity v. if μ is the coefficient of sliding friction between B  and the surface, block B will come to rest after a time: • v/gμ

• gμ/v

• g/v

• g/v

A.

v/gμ

Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,

force applied = frictional force
i.e., μmg = ma
μmg = m (v/t)
t =v/μg

738 Views

27.

A particle is placed at the origin and a force F = kx is acting on it (where k is positive constant). If u (O) = 0,  the graph of u(x) versus x will be (where u is potential energy function)

• • • • D. 28.

A small block is shot into each of the four tracks as shown below. Each of the  frictionless track rises to the same height. The speed, with which the block enters the tracks, is same in all cases. At the highest point of the track; normal reaction is maximum in

• • • • A. Since the block rises the same heights in all the four cases, from conservation of energy, speed of  the block at highest point will be same in all four cases. Say it is v0. Equation of motion will b R (the radius of curvature) in first case is minimum: Therefore, normal reaction N will be maximum in first case.

29.

A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is • 48 N

• 24 N

• 12 N

• 12 N

B.

24 N

The vector represents the momentum of the object before the collision, and the vector that after the collision. The vector represents the change in momentum of the object . As the magnitudes of are equal, the components of along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite. Thus, the change in momentum is due only to the change in direction of the perpendicular components.
Hence,  Its time rate will appear in the form of average force acting on the wall.  1088 Views

30.

A.

Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. When external force exceeds the maximum limit of static friction body begins to move. Once the body is in motion it is subjected to sliding or kinetic friction which opposes relative motion between the two surfaces in contact. At every instant, there is just one point of contact
between the body and the plane and this point has no motion relative to the plane. In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity.
In practice, this will not happen and some resistance to motion (rolling friction) does occur, i.e. to keep the body rolling. Static friction is always greater than kinetic friction and rolling friction is less than kinetic friction.