JEE Physics 2023 Question with Solution

Given: A block of mass $m$ slides down a plane inclined at $30^\circ$ with acceleration $a = \frac{g}{4}$.

Find: The coefficient of kinetic friction $\mu$.

Along the plane, the component of weight is $mg\sin 30^\circ$ downward and kinetic friction $f_k = \mu mg\cos 30^\circ$ acts upward.

Using Newton's second law along the incline:

$$mg \sin 30^\circ - \mu mg \cos 30^\circ = ma$$

Substitute $a = \frac{g}{4}$, $\sin 30^\circ = \frac{1}{2}$, and $\cos 30^\circ = \frac{\sqrt{3}}{2}$:

$$mg \cdot \frac{1}{2} - \mu mg \cdot \frac{\sqrt{3}}{2} = m \cdot \frac{g}{4}$$

Cancel $mg$:

$$\frac{1}{2} - \mu \frac{\sqrt{3}}{2} = \frac{1}{4}$$

So,

$$\mu \frac{\sqrt{3}}{2} = \frac{1}{4}$$

Hence,

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

Therefore, the correct option is B.

Given: Inclination $30^\circ$, acceleration $\frac{g}{4}$, coefficient of kinetic friction $\mu$.

Find: $\mu$ using force balance along the incline.

Resolve the weight $mg$ into two components:

• Along the incline: $mg\sin 30^\circ$
• Perpendicular to the incline: $mg\cos 30^\circ$

Therefore, the normal reaction is $N = mg\cos 30^\circ$ and kinetic friction is:

$$f_k = \mu N = \mu mg\cos 30^\circ$$

Since the block moves downward, friction acts upward along the plane. Hence net force downward is:

$$mg\sin 30^\circ - \mu mg\cos 30^\circ$$

Equating to $ma$:

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

Now substitute trigonometric values:

$$mg\frac{1}{2} - \mu mg\frac{\sqrt{3}}{2} = m\frac{g}{4}$$

Divide throughout by $mg$:

$$\frac{1}{2} - \mu\frac{\sqrt{3}}{2} = \frac{1}{4}$$

Rearranging:

$$\mu\frac{\sqrt{3}}{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

Thus,

$$\mu = \frac{1}{4} \cdot \frac{2}{\sqrt{3}} = \frac{1}{2\sqrt{3}}$$

So the coefficient of kinetic friction is $\frac{1}{2\sqrt{3}}$.