1. Acceleration on the smooth $45^\circ$ incline

For an object sliding down a smooth incline of angle $\theta=45^\circ$,

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

2. Acceleration on the rough $45^\circ$ incline

For the rough incline, friction opposes motion upward along the plane.

Normal reaction:

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

Kinetic friction:

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

So acceleration down the plane is

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

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

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

3. Use the time relation

If the object starts from rest and slides the same distance $s$ on both identical planes, then

$s = \frac{1}{2}at^2$

Hence for fixed $s$,

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

Given the rough plane takes $n$ times the time of the smooth plane,

$t_r = n t_s$

Therefore,

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

Squaring,

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

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

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

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

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

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

4. Check options

• A: $1-\frac{1}{n^2}$ ✅
• B: $\sqrt{1-\frac{1}{n^2}}$ ❌
• C: $1-n^2$ ❌
• D: $\sqrt{1-n^2}$ ❌

So the correct option is A.