Four bodies of each mass ‘M’ are kept at the vertices of a square of side ‘x’ m. The resultant force on any mass is

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$F = \frac{G M^{2}}{2 x^{2}} (2 \sqrt{2} + 1)$

$F = \frac{G M^{2}}{x^{2}} (2 \sqrt{2} + 1)$

$F = \frac{G M^{2}}{2 x^{2}} (2 \sqrt{2} - 1)$

$F = \frac{G M^{2}}{x^{2}} (2 \sqrt{2} - 1)$

answer is A.

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Detailed Solution

The force on ‘C’ due to mass at A, B and D are $F_{3}, F_{1}$ and $F_{2}$ respectively
$F_{1} = F_{2} = \frac{G M^{2}}{x^{2}}; F^{1} = \sqrt{F_{1}^{2} + F_{2}^{2}}$
$F_{3} = \frac{G M^{2}}{2 x^{2}} F^{1} = \sqrt{2} F_{1}$ $∴ θ = 90^{0}$
$∴$ Net force on ‘M’ at ‘c’ is

$\begin{array}{l} F = F_{3} + F^{1} \\ F = F_{3} + \sqrt{2} F_{1} \\ F = \frac{G M^{2}}{2 x^{2}} + \frac{\sqrt{2} G M^{2}}{x^{2}} \\ F = \frac{G M^{2}}{2 x^{2}} (\sqrt{2} + \frac{1}{2}) \\ F = \frac{G M^{2}}{2 x^{2}} (2 \sqrt{2} + 1) \end{array}$