JEE Physics 2023 Question with Solution

Given: Mass of block is $10 \, \text{kg}$, coefficient of static friction is $\mu_s = 0.25$, angle of pull is $30^\circ$, and $g = 10 \, \text{m/s}^2$.

Find: The value of $F$ for which the block just starts to move.

Resolve the applied force into horizontal and vertical components:

$$F \cos 30^\circ \quad \text{and} \quad F \sin 30^\circ$$

The upward vertical component reduces the normal reaction. Therefore,

$$N = Mg - F \sin 30^\circ$$

Substituting the given values,

$$N = 10 \times 10 - F \times \frac{1}{2} = 100 - \frac{F}{2}$$

At the point of impending motion, the horizontal component of the pull equals the maximum static friction:

$$F \cos 30^\circ = \mu_s N$$

So,

$$F \frac{\sqrt{3}}{2} = 0.25 \left(100 - \frac{F}{2}\right)$$ $$\frac{\sqrt{3}F}{2} = 25 - \frac{F}{8}$$

Multiplying by $8$,

$$4\sqrt{3}F = 200 - F$$ $$F(4\sqrt{3} + 1) = 200$$ $$F = \frac{200}{4\sqrt{3} + 1} \approx 25.22 \, \text{N}$$

Therefore, the required force is approximately $25.2 \, \text{N}$. The numerical working matches option B, although the solution incorrectly labels it as D.

Given: The block is on a horizontal rough surface and is pulled upward at $30^\circ$.

Find: The pull needed to produce impending motion.

The key idea is that friction depends on the normal reaction, and the normal reaction is not equal to $mg$ here because the applied force has an upward component.

1. Vertical balance before motion:

$$N + F \sin 30^\circ = Mg$$

Hence,

$$N = Mg - F \sin 30^\circ$$

2. Limiting friction:

$$f_{\max} = \mu_s N$$

3. Horizontal balance at impending motion:

$$F \cos 30^\circ = f_{\max}$$

Combining these,

$$F \cos 30^\circ = \mu_s (Mg - F \sin 30^\circ)$$

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

$$F \frac{\sqrt{3}}{2} = 0.25\left(100 - \frac{F}{2}\right)$$

This gives the same result,

$$F \approx 25.2 \, \text{N}$$

So the correct choice from the given options is B.