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(B) Let the side length of the square be $a$. Consider one of the masses at a corner. It experiences gravitational forces from the other three masses.

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$4 L$

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(B) Let the side length of the square be $a$. Consider one of the masses at a corner. It experiences gravitational forces from the other three masses.Two masses are at a distance $a$ (adjacent corners),and one mass is at a distance $\sqrt{2}a$ (opposite corner).The forces from the two adjacent masses are $F = \frac{Gm^2}{a^2}$ each,acting at an angle of $90^\circ$. Their resultant is $\sqrt{F^2 + F^2} = \sqrt{2}F = \sqrt{2} \frac{Gm^2}{a^2}$.The force from the opposite mass is $F' = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$.Both these forces act along the diagonal of the square,so the net force is:$F_{	ext{net}} = \sqrt{2} \frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2} \left( \sqrt{2} + \frac{1}{2} ight) = \frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} ight)$.Given that $F_{	ext{net}} = \left( \frac{2\sqrt{2} + 1}{32} ight) \frac{Gm^2}{L^2}$,we equate the two expressions:$\frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} ight) = \left( \frac{2\sqrt{2} + 1}{32} ight) \frac{Gm^2}{L^2}$.$\frac{1}{2a^2} = \frac{1}{32L^2} \implies a^2 = 16L^2 \implies a = 4L$.

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

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