AIIMS2019Physics-Mechanics

AIIMS 2019 Physics Friction MCQ Question

Type: MCQ-conceptual-Medium-Class 11

A body of 5 kg weight kept on a rough inclined plane of angle 30° starts sliding with a constant velocity. Then the coefficient of friction is (assume g = 10 ms⁻²)

A

1/√3

B

2/√3

C

√3

D

2√3

Correct Answer

Option A

Detailed Explanation

When a body slides down a rough inclined plane at a constant velocity, the forces acting on it are balanced. The gravitational force component along the incline is given by mgsinθmg \sin \theta and the frictional force opposing this motion is μmgcosθ\mu mg \cos \theta, where μ\mu is the coefficient of friction. For equilibrium, these forces must be equal:

mgsinθ=μmgcosθmg \sin \theta = \mu mg \cos \theta

Dividing both sides by mgmg and rearranging gives μ=tanθ\mu = \tan \theta. For an incline of 3030^\circ, tan30=13\tan 30^\circ = \frac{1}{\sqrt{3}}, making option A correct. Other options are incorrect as they do not accurately represent the tangent of the angle, which is essential for calculating the coefficient of friction in this scenario.

Found an issue with this question?