Q 11-04-089JEE MainJEE Main 2024 (8 Apr, Shift 2)Medium
A given object takes $n$ times the time to slide down a $45^\circ$ rough inclined plane as it takes to slide down an identical perfectly smooth $45^\circ$ inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is:
Answer: (C) $1 - \dfrac{1}{n^2}$
For the same distance from rest, $s = \frac12at^2$, so $t \propto \dfrac{1}{\sqrt a}$.
Smooth: $a_1 = g\sin45^\circ$. Rough: $a_2 = g(\sin45^\circ - \mu\cos45^\circ) = g\sin45^\circ(1-\mu)$.
$$n^2 = \frac{t_2^2}{t_1^2} = \frac{a_1}{a_2} = \frac{1}{1-\mu} \;\Rightarrow\; \mu = 1 - \frac{1}{n^2}$$
Solution by Sreeraj P, M.Sc Physics