1. Motion on the smooth incline

Let the length of the incline be $s$ and angle be $\theta = 30^\circ$.

For the smooth plane, acceleration is $a_1 = g\sin 30^\circ = \frac{g}{2}$

Since the body starts from rest, $s = \frac{1}{2} a_1 T^2$ $s = \frac{1}{2}\cdot \frac{g}{2} \cdot T^2 = \frac{gT^2}{4}$

2. Motion on the rough incline

Let the acceleration on the rough plane be $a_2$.

Given that time taken is $\alpha T$, and distance is the same $s$. So, $s = \frac{1}{2} a_2 (\alpha T)^2$ $s = \frac{1}{2} a_2 \alpha^2 T^2$

But from the smooth case, $s = \frac{gT^2}{4}$

Hence, $\frac{1}{2} a_2 \alpha^2 T^2 = \frac{gT^2}{4}$

Cancelling $T^2$, $\frac{1}{2} a_2 \alpha^2 = \frac{g}{4}$ $a_2 \alpha^2 = \frac{g}{2}$ $a_2 = \frac{g}{2\alpha^2}$

3. Acceleration on rough incline using friction

Along the incline, $a_2 = g(\sin 30^\circ - \mu \cos 30^\circ)$ $a_2 = g\left(\frac{1}{2} - \mu \cdot \frac{\sqrt{3}}{2}\right)$

Equating with the value found above: $g\left(\frac{1}{2} - \frac{\mu \sqrt{3}}{2}\right) = \frac{g}{2\alpha^2}$

Cancel $g$ and multiply by 2: $1 - \mu \sqrt{3} = \frac{1}{\alpha^2}$

So, $\mu \sqrt{3} = 1 - \frac{1}{\alpha^2} = \frac{\alpha^2 - 1}{\alpha^2}$

Thus, $\mu = \frac{1}{\sqrt{3}}\left(\frac{\alpha^2 - 1}{\alpha^2}\right)$

4. Compare with the given form

Given, $\mu = \frac{1}{\sqrt{x}}\left(\frac{\alpha^2 - 1}{\alpha^2}\right)$

Comparing, $\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{3}}$

Therefore, $x = 3$

5. Comparison with stored answer

Derived answer is $3$, which matches the stored correct answer.