JEE Physics 2025 Question with Solution

Given: mass of block $m$, angle of inclination $\theta = 60^\circ$, and acceleration down the plane $a = \frac{g}{2}$.

Find: coefficient of kinetic friction $\mu_k$.

Along the incline, the component of weight is $mg\sin\theta$ and the normal reaction is $N = mg\cos\theta$. Since the block is sliding down, kinetic friction acts up the plane with magnitude $F_f = \mu_k N = \mu_k mg\cos\theta$.

Applying Newton's second law along the incline:

$$mg\sin\theta - \mu_k mg\cos\theta = ma$$

Substitute $a = \frac{g}{2}$:

$$mg\sin\theta - \mu_k mg\cos\theta = m\frac{g}{2}$$

Cancel $m$ and $g$:

$$\sin\theta - \mu_k \cos\theta = \frac{1}{2}$$

Now use $\theta = 60^\circ$:

$$\sin 60^\circ - \mu_k \cos 60^\circ = \frac{1}{2}$$

With $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\cos 60^\circ = \frac{1}{2}$,

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

Multiply by $2$:

$$\sqrt{3} - \mu_k = 1$$

Hence,

$$\mu_k = \sqrt{3} - 1$$

Therefore, the coefficient of kinetic friction is $\sqrt{3} - 1$, so the correct option is A.

Given: $\theta = 60^\circ$ and $a = \frac{g}{2}$.

Find: $\mu_k$.

For a body sliding down a rough incline,

$$a = g(\sin\theta - \mu_k\cos\theta)$$

Substitute the given values:

$$\frac{g}{2} = g\left(\sin 60^\circ - \mu_k\cos 60^\circ\right)$$ $$\frac{1}{2} = \frac{\sqrt{3}}{2} - \mu_k\frac{1}{2}$$

So,

$$1 = \sqrt{3} - \mu_k$$ $$\mu_k = \sqrt{3} - 1$$

This works because the standard acceleration formula on a rough incline already combines the weight component and friction term. Therefore, the correct option is A.