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A block of mass m is placed at rest on an inclined plane of inclination $θ$ to the horizontal. If the coefficient of friction between the block and the plane is $θ$ , then the force, which the inclined plane exerts on the block is

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mgcos $θ \sqrt{1 + μ^{2}}$

mgsin $θ$

mgcos $θ$

answer is A.

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Detailed Solution

$\begin{array}{l} N = m g \cos θ; f = f_{s} = m g \sin θ \\ R = \sqrt{f^{2} + N^{2}} = m g \end{array}$

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