A smooth block is released at rest on a $45°$ incline and then slides a distance ‘$d$’. The time taken to slide is ‘$n$’ times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

When surface is smooth

$$\begin{gathered} d=\frac{1}{2}\left(g \sin \theta \right)t_1^2, \\ {t}_1=\sqrt{\frac{2d}{g \sin \theta }} \end{gathered}$$

When surface is rough

$$\begin{gathered} d=\frac{1}{2}\left(g \sin \theta -\mu g \cos \theta \right)t_2^2, \\ {t}_2=\sqrt{\frac{2d}{g \sin \theta -\mu g \cos \theta }} \end{gathered}$$

According to question, $t_2-n{t}_1$

$n\sqrt{\frac{2d}{g \sin \theta }}=\sqrt{\frac{2d}{g \sin \theta -\mu g \cos \theta }}$

$\mu$, as applicable here, is coefficient of kinetic friction as the block moves over the inclined plane.

$n=\frac{1}{\sqrt{1-{\mu }_k}} \left(\because \cos 45°=\sin 45°=\frac{1}{\sqrt{2}}\right)$

$n^2=\frac{1}{1-{\mu }_k}$ or $1-{\mu }_k=\frac{1}{n^2}$ or ${\mu }_k=1-\frac{1}{n^2}$