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(A) Let the four particles be at the corners of a square inscribed in a circle of radius $R$. For any one particle,the gravitational forces exerted by

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$\sqrt{\frac{G}{2}(1+2 \sqrt{2})}$

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(A) Let the four particles be at the corners of a square inscribed in a circle of radius $R$. For any one particle,the gravitational forces exerted by the other three particles are:$1$. Force $F_1$ due to the particle diametrically opposite: $F_1 = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2}$.$2$. Forces $F_2$ and $F_3$ due to the two adjacent particles: $F_2 = F_3 = \frac{G m^2}{(\sqrt{2}R)^2} = \frac{G m^2}{2R^2}$.The net gravitational force $F_{	ext{net}}$ acting towards the center of the circle is the sum of the components of these forces along the radius:$F_{	ext{net}} = F_1 + F_2 \cos 45^{\circ} + F_3 \cos 45^{\circ}$$F_{	ext{net}} = \frac{G m^2}{4R^2} + \frac{G m^2}{2R^2} \left(\frac{1}{\sqrt{2}}ight) + \frac{G m^2}{2R^2} \left(\frac{1}{\sqrt{2}}ight)$$F_{	ext{net}} = \frac{G m^2}{R^2} \left(\frac{1}{4} + \frac{1}{\sqrt{2}}ight) = \frac{G m^2}{4R^2} (1 + 2\sqrt{2})$.This net force provides the necessary centripetal force for circular motion: $F_{	ext{net}} = \frac{m v^2}{R}$.Equating the two: $\frac{m v^2}{R} = \frac{G m^2}{4R^2} (1 + 2\sqrt{2})$.$v^2 = \frac{G m}{4R} (1 + 2\sqrt{2})$.Given $m = 1 \, kg$ and $R = 1 \, m$,we get $v^2 = \frac{G}{4} (1 + 2\sqrt{2})$.$v = \sqrt{\frac{G}{4} (1 + 2\sqrt{2})} = \frac{\sqrt{G(1+2\sqrt{2})}}{2}$.Comparing with the options,this is equivalent to $\sqrt{\frac{G}{4}(1+2\sqrt{2})}$,which matches option $A$.

$\text{LIST-I}$	$\text{LIST-II}$
$A$. Gravitational constant	$I$. $[LT^{-2}]$
$B$. Gravitational potential energy	$II$. $[L^2 T^{-2}]$
$C$. Gravitational potential	$III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity	$IV$. $[M^{-1} L^3 T^{-2}]$

Four identical particles of equal masses $m = 1 \, kg$ are made to move along the circumference of a circle of radius $R = 1 \, m$ under the action of their own mutual gravitational attraction. The speed of each particle will be

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