1. Use motion on an incline

For an object starting from rest and sliding a distance $s$ down an incline with constant acceleration $a$,

$s=\frac{1}{2}at^2 \implies t=\sqrt{\frac{2s}{a}}$

So, for the same distance $s$, time is inversely proportional to $\sqrt{a}$:

$t \propto \frac{1}{\sqrt{a}}$

2. Acceleration on a smooth $45^\circ$ incline

On a perfectly smooth incline,

$a_s=g\sin 45^\circ=\frac{g}{\sqrt{2}}$

Let the time taken on the smooth incline be $t_s$.

3. Acceleration on a rough $45^\circ$ incline

On the rough incline, friction acts up the plane.

Normal reaction:

$N=g\cos 45^\circ \cdot m$

Kinetic friction:

$f_k=\mu N=\mu mg\cos 45^\circ$

Net acceleration down the plane:

$a_r=g\sin 45^\circ-\mu g\cos 45^\circ$

Since $\sin 45^\circ=\cos 45^\circ=\frac{1}{\sqrt{2}}$,

$a_r=\frac{g}{\sqrt{2}}(1-\mu)$

Let the time taken on the rough incline be $t_r$.

4. Use the given time relation

The question says the object takes $n$ times more time on the rough incline than on the smooth incline:

$t_r=n t_s$

Using $t\propto \frac{1}{\sqrt{a}}$,

$\frac{t_r}{t_s}=\sqrt{\frac{a_s}{a_r}}=n$

So,

$n^2=\frac{a_s}{a_r}$

Substitute $a_s=\frac{g}{\sqrt{2}}$ and $a_r=\frac{g}{\sqrt{2}}(1-\mu)$:

$n^2=\frac{\frac{g}{\sqrt{2}}}{\frac{g}{\sqrt{2}}(1-\mu)}=\frac{1}{1-\mu}$

Hence,

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

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

5. Match with the options

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

This matches Option B.

6. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

They agree.