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A block is released from the top of an inclined plane of inclination $θ$ . Its velocity at the bottom of plane is v. If it slides down a rough inclined plane of the same inclination, its velocity at bottom is v/2. The coefficient of friction is

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$\frac{3}{4} \cot θ$

$\frac{1}{4} \cot θ$

$\frac{1}{4} \tan θ$

$\frac{3}{4} \tan θ$

answer is D.

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Detailed Solution

$V = \sqrt{2 gLsin θ}$

$\begin{array}{l} \frac{V}{2} = \sqrt{2 Lg (\sin θ - μ_{k} \cos θ)} \\ \sin θ = 4 (\sin θ - μ_{k} \cos θ) \\ 3 \sin θ = 4 μ_{k} \cos θ \\ \Rightarrow μ_{k} = \frac{3}{4} \tan θ \end{array}$

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