Q 11-07-076JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
Four identical particles of mass $m$ are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is $\left(\dfrac{2\sqrt2 + 1}{32}\right)\dfrac{Gm^2}{L^2}$, the length of the sides of the square is
Answer: (B) $4L$
For side $a$: the two neighbours each pull with $\dfrac{Gm^2}{a^2}$ at right angles (resultant $\sqrt2\dfrac{Gm^2}{a^2}$ along the diagonal) and the opposite corner pulls with $\dfrac{Gm^2}{2a^2}$ along the same diagonal.
$$F = \frac{Gm^2}{a^2}\left(\sqrt2 + \frac12\right) = \frac{2\sqrt2 + 1}{2}\frac{Gm^2}{a^2}$$
Setting this equal to $\dfrac{2\sqrt2 + 1}{32}\dfrac{Gm^2}{L^2}$: $a^2 = 16L^2$, $a = 4L$.
Solution by Sreeraj P, M.Sc Physics