Q 11-04-154JEE MainJEE Main 2022 (27 Jul, Shift 2)Medium
A block $A$ takes $2$ s to slide down a frictionless incline of $30^\circ$ and length $l$, kept inside a lift going up with uniform velocity $v$. If the incline is changed to $45^\circ$, the time taken by the block, to slide down the incline, will be approximately:
Answer: (C) $1.68$ s
A lift moving with uniform velocity is an inertial frame, so the acceleration down the incline is just $g\sin\theta$.
$$l = \frac12 g\sin\theta\,t^2 \Rightarrow t \propto \frac{1}{\sqrt{\sin\theta}}$$
$$t_2 = 2\sqrt{\frac{\sin30^\circ}{\sin45^\circ}} = 2\sqrt{\frac{0.5}{0.707}} = 2\times0.841 \approx 1.68\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics