$\begin{array}{1 1}\mu=2\tan\theta\\\mu=\tan\theta\\\mu=\large\frac{2}{\tan\theta}\\\mu=\large\frac{1}{\tan\theta}\end{array} $

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Answer : $\mu=2\tan\theta$

The acceleration of the block while it is sliding down the upper half of the inclined plane is $g\sin \theta$.

If $\mu$ is the coefficient of kinetic friction between the block and the lower half of the plane,the retardation of the block while it is sliding down the lower half = $-(g\sin \theta-\mu\cos\theta)$

For the block to come to rest at the bottom of the inclined plane,the acceleration in the first half must be equal to the retardation in the second half,i.e.

$g\sin \theta =-(g\sin \theta-\mu g\cos\theta)$

$\mu \cos \theta = 2\sin \theta$

$\mu =2\tan\theta$

Hence the correct answer is $\mu=2\tan\theta$

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