The upper half an inclined plane of inclination $θ$ is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom. The coefficient of friction between the block and the lower half of the plane is given by

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$μ = 2 \tan θ$

$μ = \frac{2}{\tan θ}$

$μ = \frac{1}{\tan θ}$

$μ = \tan θ$

answer is A.

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Detailed Solution

$v^{2} - u^{2} = 2 as$
For lower half of the incline
$\begin{array}{l} - 2 g \frac{l}{2} \sin θ = \frac{2 gl}{2} (\sin θ - μcos θ) \\ μcos θ = 2 \sin θ; μ = 2 \tan θ \end{array}$

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