Let the block have mass $m$ and the incline angle be $\theta$.

On the rough incline:

• Component of weight along plane = $mg\sin\theta$
• Normal reaction = $N = mg\cos\theta$
• Maximum static friction = $f_s^{\max} = \mu N = \mu mg\cos\theta$

Case 1: Maximum force down the plane for rest is $2\,\text{N}$.

If we apply a force downward, the block is about to move down the plane, so friction acts up the plane at limiting value.

Therefore, at limiting equilibrium: $mg\sin\theta + 2 = \mu mg\cos\theta \qquad (1)$

Case 2: Maximum force up the plane for rest is $10\,\text{N}$.

If we apply a force upward, the block is about to move up the plane, so friction acts down the plane at limiting value.

Therefore: $10 = mg\sin\theta + \mu mg\cos\theta \qquad (2)$

Solve equations (1) and (2).

Let $A = mg\sin\theta, \qquad B = \mu mg\cos\theta$

Then equations become: $A + 2 = B$ $10 = A + B$

Substitute $B = A+2$ into second equation: $10 = A + (A+2) = 2A+2$ $2A = 8$ $A = 4$

Hence, $B = A+2 = 6$

So, $mg\sin\theta = 4, \qquad \mu mg\cos\theta = 6$

Find $\mu$.

$\frac{\mu mg\cos\theta}{mg\sin\theta} = \frac{6}{4}$ $\mu\cot\theta = \frac{3}{2}$

Hence, $\mu = \frac{3}{2}\tan\theta$

Use the figure information.

From the given figure, the incline is at $30^\circ$.

Therefore, $\tan 30^\circ = \frac{1}{\sqrt{3}}$

So, $\mu = \frac{3}{2}\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{2}$

Match with options.

$\mu = \frac{\sqrt{3}}{2}$

Hence the correct option is C.