Question 32 of 223

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

A$\dfrac{L}{2}$B$4L$C$3L$D$2L$

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Answer: (B) $4L$

Resultant force, ${F}_{net}=\sqrt{2}F+F'$, where
$F=\dfrac{G{m}^{2}}{{a}^{2}}$ and $F'=\dfrac{G{m}^{2}}{{\left(\sqrt{2}a\right)}^{2}}$
Therefore, ${F}_{net}=\sqrt{2}\dfrac{G{m}^{2}}{{a}^{2}}+\dfrac{G{m}^{2}}{2{a}^{2}}$
Comparing with the expression given in the question, we get
$\left(\dfrac{2\sqrt{2}+1}{32}\right)\dfrac{G{m}^{2}}{{L}^{2}}=\dfrac{G{m}^{2}}{{a}^{2}}\left(\dfrac{2\sqrt{2}+1}{2}\right)$
$\Rightarrow a=4L$