question_answer

A block of mass m is placed on a surface with a vertical cross section given by \[y=\frac{{{x}^{3}}}{6}.\]If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is :   JEE Main  Solved  Paper-2014

A) \[\frac{1}{3}m\]                                               

B) \[\frac{1}{2}m\]

C) \[\frac{1}{6}m\]               

D) \[\frac{2}{3}m\]

Correct Answer: C

Solution :

\[mg\sin \theta ={{\mu }_{s}}mg\cos \theta \] [when partied is just balanced] \[\Rightarrow \]\[\tan \theta ={{\mu }_{s}}\] \[\tan \theta =\frac{dy}{dx}=\left( \frac{{{x}^{2}}}{2} \right)\] \[\therefore \]\[\frac{{{x}^{2}}}{2}=0.5\Rightarrow x=1\]\[\Rightarrow \]\[y=\frac{1}{6}\]