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 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 :

[c]

\[mg\sin \theta ={{\mu }_{s}}mg\cos \theta \]

[when partied is just balanced]

\[\Rightarrow \]\[\tan \theta ={{\mu }_{s}}\]

\[\tan \theta =\frac{dy}{dx}=\left( \frac{{{x}^{2}}}{2} \right)\]

\[\therefore \]\[\frac{{{x}^{2}}}{2}=0.5\Rightarrow x=1\]\[\Rightarrow \]\[y=\frac{1}{6}\]