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(D) The gravitational force exerted by particle $B$ on particle $A$ is $F = \frac{G m^2}{L^2}$ directed towards $B$.Similarly,the gravitational force

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$\sqrt{3} \frac{G m}{L^2}$

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(D) The gravitational force exerted by particle $B$ on particle $A$ is $F = \frac{G m^2}{L^2}$ directed towards $B$.Similarly,the gravitational force exerted by particle $C$ on particle $A$ is $F = \frac{G m^2}{L^2}$ directed towards $C$.The angle between these two forces is $60^{\circ}$ because the triangle is equilateral.The resultant force $F_{	ext{net}}$ on particle $A$ is given by the vector sum:$F_{	ext{net}} = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^{\circ}} = \sqrt{2 F^2 + 2 F^2 (0.5)} = \sqrt{3 F^2} = \sqrt{3} F$Substituting $F = \frac{G m^2}{L^2}$,we get $F_{	ext{net}} = \sqrt{3} \frac{G m^2}{L^2}$.According to Newton's second law,the acceleration $a$ of particle $A$ is $a = \frac{F_{	ext{net}}}{m}$.$a = \frac{\sqrt{3} G m^2}{L^2 m} = \sqrt{3} \frac{G m}{L^2}$.

Three particles $A, B$ and $C$ each of mass $m$ are lying at the corners of an equilateral triangle of side $L$. If the particle $A$ is released keeping the particles $B$ and $C$ fixed,the magnitude of instantaneous acceleration of $A$ is ........

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