JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Laws of Motion

A block rests without sliding as long as the surface is gentle enough for static friction to hold it, the limiting condition being $\tan \theta =\mu$, where $\theta$ is the angle the surface makes with the horizontal. For the profile $y=\frac{x^2}{4}$ the slope is $\frac{dy}{dx}=\frac{x}{2}=\tan \theta$, so the steepest point at which the block can still stay put satisfies $\frac{x}{2}=\mu =0.5$, that is $x=1$ m. The corresponding height is $y=\frac{x^2}{4}=\frac{1}{4}$ m $=25$ cm, the maximum height at which a stationary block will not slip down.