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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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The upper half of an inclined plane of inclination $\theta$ 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 if the coefficient of friction between the block and the lower half of the plane is given by

$\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 $\mu=2\tan\theta$
answered Aug 27, 2014 by sreemathi.v
 

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