• # 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) $\frac{2}{3}m$

 [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}$