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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Starting from rest a body slides down a $45^{\circ}$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between body and plane is

\[(a)\;0.33 \quad (b)\;0.75 \quad (c)\;0.25 \quad (d)\;0.80 N \]
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on a rough surface acceleration of a body down a plane
$a=(g \;sin \theta-\mu g \cos \theta)$
'2t' is time taken to go down a rough inclined plane of length 's'
't' is time taken to go down a smooth inclined plane of length 's'
$s=\large\frac{1}{2}$$ [g \sin \theta -\mu g \cos \theta](2t)^2$-----(1)
On a smooth surface $a=g \sin \theta$
$s= \large\frac{1}{2} $$(g \sin \theta)t^2$-----(2)
Substituting (2) in (1) and solving
$\mu =1- \large\frac{1}{2^2}$
$\mu =\large \frac{3}{4}$
Hence b is the correct answer. 
answered Jul 11, 2013 by meena.p
edited May 27, 2014 by lmohan717

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