Q70 / 90 · JEE Main 2024 (31 Jan Shift 1) · Physics

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 $(\frac{2 \sqrt{2} + 1}{32}) \frac{G m^{2}}{L^{2}}$ , the length of the sides of the square is

Solution

Resultant force, $F_{n e t} = \sqrt{2} F + F'$ , where

$F = \frac{G m^{2}}{a^{2}}$ and $F' = \frac{G m^{2}}{{(\sqrt{2} a)}^{2}}$

Therefore, $F_{n e t} = \sqrt{2} \frac{G m^{2}}{a^{2}} + \frac{G m^{2}}{2 a^{2}}$

Comparing with the expression given in the question, we get

$(\frac{2 \sqrt{2} + 1}{32}) \frac{G m^{2}}{L^{2}} = \frac{G m^{2}}{a^{2}} (\frac{2 \sqrt{2} + 1}{2})$

$\Rightarrow a = 4 L$

Hence, the correct answer is option: 'B'.