JEE Physics 2024 Question with Solution

Given: Four identical masses $m$ are placed at the corners of a square of side $a$. The net gravitational force on one mass is given as

$$F_{\text{net}} = \left( \frac{2\sqrt{2}+1}{32} \right) \frac{Gm^2}{L^2}$$

Find: The side length of the square in terms of $L$.

For one corner mass, the two adjacent masses each exert force

$$F = \frac{Gm^2}{a^2}$$

These two equal perpendicular forces combine to give a resultant

$$\sqrt{2}F = \sqrt{2}\frac{Gm^2}{a^2}$$

The diagonally opposite mass is at distance $\sqrt{2}a$, so its force is

$$F' = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$$

This force acts along the same diagonal, so the net force is

$$F_{\text{net}} = \sqrt{2}\frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2}$$ $$F_{\text{net}} = \frac{Gm^2}{a^2}\left(\sqrt{2} + \frac{1}{2}\right) = \frac{Gm^2}{a^2}\left(\frac{2\sqrt{2}+1}{2}\right)$$

Equating with the given value,

$$\left( \frac{2\sqrt{2}+1}{32} \right) \frac{Gm^2}{L^2} = \frac{Gm^2}{a^2}\left(\frac{2\sqrt{2}+1}{2}\right)$$

Cancelling common factors,

$$\frac{1}{32L^2} = \frac{1}{2a^2}$$ $$a^2 = 16L^2$$ $$a = 4L$$

Therefore, the length of the side of the square is $4L$. The correct option is B.

Given: Two adjacent forces are equal and perpendicular, and the third force is along the diagonal.

Instead of resolving each adjacent force separately, combine the two adjacent forces directly:

$$F_{\text{adj}} = \sqrt{2}\frac{Gm^2}{a^2}$$

The diagonal force is

$$F_{\text{diag}} = \frac{Gm^2}{2a^2}$$

Since both act along the same diagonal,

$$F_{\text{net}} = \frac{Gm^2}{a^2}\left(\sqrt{2} + \frac{1}{2}\right)$$

Now compare this with the given expression

$$\left( \frac{2\sqrt{2}+1}{32} \right) \frac{Gm^2}{L^2}$$

Matching coefficients gives

$$a = 4L$$

Therefore, the correct option is B.