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. tan(theta)
B. sin(theta)
C. cos(theta)
D. cot(theta)
Answer: Option A
Solution (By JKExamLibrary)
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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