1. Forces along the incline

For a block sliding down an incline of angle $\theta = 30^\circ$:

• Component of weight down the plane: $mg\sin\theta$
• Normal reaction: $N = mg\cos\theta$
• Kinetic friction opposes motion, so it acts up the plane: $f_k = \mu_k N = \mu_k mg\cos\theta$

2. Apply Newton’s second law along the plane

Taking downward along the plane as positive, $mg\sin\theta - \mu_k mg\cos\theta = ma$

Given: $a = \frac{g}{4}, \qquad \theta = 30^\circ$

So, $mg\sin 30^\circ - \mu_k mg\cos 30^\circ = m\frac{g}{4}$

Cancel $mg$: $\sin 30^\circ - \mu_k \cos 30^\circ = \frac{1}{4}$

3. Substitute trigonometric values

$\sin 30^\circ = \frac{1}{2}, \qquad \cos 30^\circ = \frac{\sqrt{3}}{2}$

Hence, $\frac{1}{2} - \mu_k \frac{\sqrt{3}}{2} = \frac{1}{4}$

Multiply by $4$: $2 - 2\sqrt{3}\,\mu_k = 1$

$2\sqrt{3}\,\mu_k = 1$

$\mu_k = \frac{1}{2\sqrt{3}}$

4. Match with options

$\mu_k = \frac{1}{2\sqrt{3}}$

This corresponds to Option C.

5. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So, the derived answer agrees with the stored correct answer.