A smooth block is released at rest on a $$45^\circ$$ 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

A block of mass $$m$$ is allowed to slide a distance $$d$$ down an incline making an angle $$\theta = 45^{\circ}$$ with the horizontal.

Case 1: Smooth incline (no friction).
The only component of gravity along the plane is $$mg\sin\theta$$, so the acceleration is
$$a_1 = g\sin\theta = g\frac{1}{\sqrt{2}}$$ because $$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$$.

Starting from rest, the time to cover distance $$d$$ is obtained from $$d = \tfrac12 a_1 t_1^{2}$$, giving
$$t_1 = \sqrt{\frac{2d}{a_1}}$$ $$= \sqrt{\frac{2d}{g/\sqrt{2}}}$$ $$= \sqrt{\frac{2\sqrt{2}d}{g}}$$.

Case 2: Rough incline (kinetic friction present).
Along the plane the net force is
$$mg\sin\theta - \mu_k mg\cos\theta,$$
so the acceleration is
$$a_2 = g\sin\theta - \mu_k g\cos\theta.$$

For $$\theta = 45^{\circ}$$ we have $$\sin\theta = \cos\theta = \tfrac1{\sqrt2}$$, hence
$$a_2 = g\frac{1}{\sqrt{2}}\bigl(1 - \mu_k\bigr)\;.$$

The time to slide the same distance $$d$$ now satisfies $$d = \tfrac12 a_2 t_2^{2}$$, so
$$t_2 = \sqrt{\frac{2d}{a_2}}$$ $$= \sqrt{\frac{2d}{g(1/\sqrt{2})(1-\mu_k)}}$$ $$= \sqrt{\frac{2\sqrt{2}d}{g}}\;\frac{1}{\sqrt{1-\mu_k}}.$$ Notice that $$\sqrt{\frac{2\sqrt{2}d}{g}}$$ is exactly $$t_1$$, hence

$$\frac{t_2}{t_1} = \frac{1}{\sqrt{1-\mu_k}}.$$

The problem states that the rough-incline time is $$n$$ times the smooth-incline time, i.e. $$t_2 = n\,t_1$$. Therefore

$$n = \frac{1}{\sqrt{1-\mu_k}}\; \Longrightarrow\; n^{2} = \frac{1}{1-\mu_k}\; \Longrightarrow\; 1-\mu_k = \frac{1}{n^{2}}.$$

Solving for $$\mu_k$$ gives
$$\mu_k = 1 - \frac{1}{n^{2}}.$$

Thus the coefficient of kinetic friction is $$\mu_k = 1 - \dfrac{1}{n^2}$$.

Option A which is: $$\mu_k = 1 - \frac{1}{n^2}$$