A block of mass 'm' is placed on a rough inclined plane of inclination theta. If the block is just on the verge of sliding down, the coefficient of static friction (mu) is equal to: MCQ with Answer and Explanation

A block of mass 'm' is placed on a rough inclined plane of inclination theta. If the block is just on the verge of sliding down, the coefficient of static friction (mu) is equal to:
A. cos(theta)
B. cot(theta)
C. sin(theta)
D. tan(theta)
Answer: Option D
Solution (By JKSSB Mock Tests)
When the block is on the verge of sliding (limiting friction), the downward component of weight along the plane equals the maximum static friction force. mgsin(theta) = mu * N. The normal reaction N is balanced by the perpendicular component of weight: N = mgcos(theta). Dividing the two equations gives mgsin(theta) / mgcos(theta) = mu, so mu = tan(theta). This angle is called the angle of repose.

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Correct Answer: Option C


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