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:

We consider the same object sliding the same distance $$s$$ along two different 45$$^\circ$$ planes – one perfectly smooth (no friction) and one rough (kinetic friction coefficient $$\mu_k=\mu$$). Because the initial speed is zero in both cases, the usual constant-acceleration kinematics apply.

First we recall the basic translational kinematic relation for motion with constant acceleration starting from rest:

$$s=\frac12\,a\,t^{2}\;.$$

Here $$a$$ is the magnitude of acceleration along the plane and $$t$$ is the time taken to cover the distance $$s$$.

Smooth 45$$^\circ$$ incline
The component of gravitational acceleration along a smooth plane is

$$a_{\text{smooth}}=g\sin45^\circ=\frac{g}{\sqrt2}\;.$$

Denote the corresponding time of slide by $$t_s$$. Substituting this acceleration into the displacement formula, we have

$$s=\frac12\,a_{\text{smooth}}\,t_s^{2} =\frac12\left(\frac{g}{\sqrt2}\right)t_s^{2}\;.$$

Rough 45$$^\circ$$ incline
With kinetic friction, the frictional force equals $$\mu N$$, where $$N$$ is the normal reaction. For a 45$$^\circ$$ plane, $$N=mg\cos45^\circ=\frac{mg}{\sqrt2}$$. Hence the frictional force magnitude is $$\mu\,\dfrac{mg}{\sqrt2}$$, acting upward along the plane. The net force along the plane therefore equals

$$mg\sin45^\circ-\mu mg\cos45^\circ =\frac{mg}{\sqrt2}-\mu\left(\frac{mg}{\sqrt2}\right) =\frac{mg}{\sqrt2}(1-\mu)\;.$$

Dividing by the mass $$m$$ gives the net acceleration down the plane:

$$a_{\text{rough}}=\frac{g}{\sqrt2}(1-\mu)\;.$$

Let $$t_r$$ be the time of slide on the rough plane. Using the same displacement formula for this case, we write

$$s=\frac12\,a_{\text{rough}}\,t_r^{2} =\frac12\left[\frac{g}{\sqrt2}(1-\mu)\right]t_r^{2}\;.$$

Given relation between the two times
The problem states that the rough slide takes $$n$$ times longer than the smooth slide, so

$$t_r=n\,t_s\;.$$

Equating the two expressions for the same distance $$s$$

$$\frac12\left(\frac{g}{\sqrt2}\right)t_s^{2} =\frac12\left[\frac{g}{\sqrt2}(1-\mu)\right]\!\!\left(n\,t_s\right)^{2}\;.$$

The common factors $$\dfrac12$$, $$\dfrac{g}{\sqrt2}$$, and $$t_s^{2}$$ cancel immediately, leaving

$$1=(1-\mu)\,n^{2}\;.$$

Solving this simple linear equation for $$\mu$$, we proceed step by step:

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

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

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

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

This matches Option 2 in the given list.

Hence, the correct answer is Option 2.