Engineering Physics Questions and Answers – Centripetal Force

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Centripetal Force”.

1. In stable equilibrium, a body has maximum potential.
a) True
b) False

Explanation: A body is said to be in stable equilibrium if it tends to regain its equilibrium position after being slightly displaced and released. In stable equilibrium, a body has minimum potential energy.

2. On being displaced, the centre of mass of unstable equilibrium goes ___________
a) Higher
b) Neither higher nor lower
c) First higher and then lower
d) Lower

Explanation: A body is said to be in unstable equilibrium if it gets further displaced from its equilibrium position after being slightly displaced and released. In unstable equilibrium, a body possesses maximum potential energy and its centre of mass goes lower on being slightly displaced.

3. Which of the following stays in equilibrium even after being slightly displaced?
a) Stable equilibrium
b) Unstable equilibrium
c) Neutral equilibrium
d) Rigid body

Explanation: If a body stays in equilibrium position even after being slightly displaced and released, it is said to be in neutral equilibrium. When a body is slightly displaced, its centre of mass is neither raised nor lowered and its potential energy remains constant.

4. Moment of inertia can be regarded as the measure of the rotational inertia of the body.
a) True
b) False

Explanation: The mass of a body resists change in its state of linear motion, it is a measure of its inertia in linear motion, Similarly, the moment of inertia of a body about an axis of rotation resists a change in its rotational motion. The greater the moment of inertia of body, the greater is the torque required to change its state of rotation. Thus moment of inertia of a body can be regarded as the measure of the rotational inertia of the body.

5. The flying wheel attached to the shaft of steam engine works on the principle of ___________
a) Centripetal action
b) Moment of inertia
c) Newton’s third law of motion
d) Conservation of momentum

Explanation: A flywheel is attached to the shaft of an engine. Because of its large moment of inertia, the flywheel opposes the sudden increase or decrease in the speed of the vehicle. It allows a gradual change in the speed and prevents jerky motions and hence ensures a smooth ride for the passengers.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. A wheel of mass 8kg and radius of gyration 25cm is rotating at 300rpm. What is its moment of inertia?
a) 0.5 kgm2
b) 10 kgm2
c) 5 kgm2
d) 0.25 kgm2

Explanation: M = 8kg, K = 25cm = 0.25m
Therefore, l = MK2 = 8×0.252 = 0.5 kgm2.

7. The moment of inertia of a uniform circular disc about its diameter is 100gcm2. What is its moment of inertia about its tangent?
a) 200 gcm2
b) 100 gcm2
c) 900 gcm2
d) 500 gcm2

Explanation: By the theorem of parallel axes, moment of inertia about a tangent parallel to the diameter,
I = Id+MR2 = 1/4 MR2+MR2=5/4 MR2
I = 5×100 = 500 gcm2.

8. The moment of inertia of a uniform circular disc about its diameter is 100 gcm2. What is its moment of inertia about an axis perpendicular to its plane.
a) 500 gcm2
b) 100 gcm2
c) 200 gcm2
d) 700 gcm2

Explanation: By theorem of perpendicular axes, moment of inertia of the disc about an axis perpendicular to its plane,
I = Sum of the moments of inertia about two perpendicular diameters
I = Id+Id=2×1/4×MR2=2×1000= 200 gcm2.

9. Calculate the moment of inertia of the earth about its diameter, taking it to be a sphere of 1025kg and diameter 12800km.
a) 1.64 kgm2
b) 16.4×1038 kgm2
c) 1.64×1038 kgm2
d) 0

Explanation: M = 1025kg, R = 6400km = 6.4×106 m
Moment of inertia of the earth about its diameter I = 2/5 MR2 = 2/5×1025×(6.4×106)2
I = 1.64×1038 kgm2.

10. A torque of 2×10-4 Nm is applied to produce an angular acceleration of 4rad/s2 in a rotating body. What is the moment of inertia of the body?
a) 0.5 kgm2
b) 5×104 kgm2
c) 0.5×10-4 kgm2
d) 0.5×104 kgm2

Explanation: Torque = Iα
I = Torque/α = (2×10-4)/4=0.5×10-4 kgm2.

Sanfoundry Global Education & Learning Series – Engineering Physics.

To practice all areas of Engineering Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]