Q 11-04-095JEE MainJEE Main 2024 (30 Jan, Shift 2)Medium
A block of mass $m$ is placed on a surface having vertical cross section given by $y = \dfrac{x^2}{4}$. If the coefficient of friction is $0.5$, the maximum height above the ground at which the block can be placed without slipping is:
Answer: (A) $\dfrac14\ \text{m}$
The block stays at rest while $\tan\theta \le \mu$, where $\tan\theta = \dfrac{dy}{dx} = \dfrac x2$.
$$\frac x2 = 0.5 \Rightarrow x = 1\ \text{m},\qquad y = \frac{1^2}{4} = \frac14\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics