Q 11-04-196JEE MainJEE Main 2018 (15 Apr, Shift 1)Medium
A given object takes $n$ times more time to slide down a $45^\circ$ rough inclined plane as it takes to slide down a perfectly smooth $45^\circ$ incline. The coefficient of kinetic friction between the object and the incline is:
Answer: (B) $1 - \dfrac{1}{n^2}$
On the smooth incline: $a_1 = g\sin 45^\circ = \dfrac{g}{\sqrt2}$.
On the rough incline: $a_2 = g\sin45^\circ - \mu g\cos 45^\circ = \dfrac{g}{\sqrt2}(1-\mu)$.
For the same length $L$ starting from rest, $L = \frac12 a t^2$, so $t \propto \dfrac{1}{\sqrt a}$:
$$\frac{t_2}{t_1} = n = \sqrt{\frac{a_1}{a_2}} = \frac{1}{\sqrt{1-\mu}}$$
Hence $1 - \mu = \dfrac{1}{n^2}$, i.e.
$$\mu = 1 - \frac{1}{n^2}$$
Solution by Sreeraj P, M.Sc Physics