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A cylindrical vessel filled with water is released on an in inclined plane of angle $\theta$ as shown. The coefficient of friction of surfaces is $(\mu ( < \tan \theta)$. Then the contact angle made by the surface of water with the incline is

$ a)\;\tan^{-1}\;\mu \\ b)\;\theta - \tan^{-1}\mu \\ c)\;\theta + \tan^{-1}\mu \\ d)\;\cot^{-1} \mu $
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Force acting on the particle on the surface of water with respect to vessel
$mg\;\sin \theta$ and $\mu \;mg\;\cos \theta$ are pseudo force
$mg\; \sin \theta =\mu mg \cos \theta$ [Since no net force is acting along the surface]
$\tan \theta= \mu$
$\theta =\tan ^{-1} \mu$
Since free surface is $\perp$ to the resultant force acting on it .
$\theta$ is the angle between normal to inclined surfaces and resultant force
Hence a is the correct answer.


answered Oct 8, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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