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(C) Let the side length of the square be $a$. The net gravitational force on any one mass (say mass $4$) is the vector sum of the forces exerted by th

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$4L$

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(C) Let the side length of the square be $a$. The net gravitational force on any one mass (say mass $4$) is the vector sum of the forces exerted by the other three masses $(1, 2, 3)$.Let $\vec{F}_{14}$,$\vec{F}_{34}$,and $\vec{F}_{24}$ be the forces exerted by masses $1, 3,$ and $2$ on mass $4$ respectively.The magnitudes are $F_{14} = \frac{Gm^2}{a^2}$,$F_{34} = \frac{Gm^2}{a^2}$,and $F_{24} = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$.The resultant of $\vec{F}_{14}$ and $\vec{F}_{34}$ is $F_{13} = \sqrt{F_{14}^2 + F_{34}^2} = \sqrt{\left(\frac{Gm^2}{a^2}\right)^2 + \left(\frac{Gm^2}{a^2}\right)^2} = \sqrt{2} \frac{Gm^2}{a^2}$.This resultant force $F_{13}$ acts along the diagonal,in the same direction as $\vec{F}_{24}$.Thus,the net force is $F_{net} = F_{13} + F_{24} = \sqrt{2} \frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2} \left(\sqrt{2} + \frac{1}{2}\right) = \frac{Gm^2}{a^2} \left(\frac{2\sqrt{2} + 1}{2}\right)$.Given that $F_{net} = \left(\frac{2\sqrt{2} + 1}{32}\right) \frac{Gm^2}{L^2}$,we equate the two expressions:$\frac{Gm^2}{a^2} \left(\frac{2\sqrt{2} + 1}{2}\right) = \left(\frac{2\sqrt{2} + 1}{32}\right) \frac{Gm^2}{L^2}$.$\frac{1}{2a^2} = \frac{1}{32L^2} \Rightarrow a^2 = 16L^2 \Rightarrow a = 4L$.

Four identical masses of $m$ are kept at the corners of a square. If the gravitational force exerted on one of the masses by the other masses is $\left(\frac{2 \sqrt{2}+1}{32}\right) \frac{G m^2}{L^2}$,then the length of the side of the square is

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