Two particles each of mass m are placed at po | vedclass

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(C) Let the distance $RS = x$. From the figure,$RS = l$. The distance from $P$ to $S$ is $PS = \sqrt{PR^2 + RS^2} = \sqrt{(l/2)^2 + l^2} = \sqrt{l^2/4

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$\frac{16 G m^2}{5 \sqrt{5} l^2}$

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(C) Let the distance $RS = x$. From the figure,$RS = l$. The distance from $P$ to $S$ is $PS = \sqrt{PR^2 + RS^2} = \sqrt{(l/2)^2 + l^2} = \sqrt{l^2/4 + l^2} = \sqrt{5l^2/4} = \frac{\sqrt{5}}{2} l$.The gravitational force exerted by particle at $P$ on particle at $S$ is $F_1 = \frac{G m^2}{PS^2} = \frac{G m^2}{5l^2/4} = \frac{4 G m^2}{5 l^2}$.Similarly,the gravitational force exerted by particle at $Q$ on particle at $S$ is $F_2 = \frac{G m^2}{QS^2} = \frac{4 G m^2}{5 l^2}$.Let $	heta$ be the angle between $RS$ and $PS$. Then $\cos 	heta = \frac{RS}{PS} = \frac{l}{(\sqrt{5}/2) l} = \frac{2}{\sqrt{5}}$.The horizontal components of the forces $F_1$ and $F_2$ cancel each other out. The resultant force is the sum of the vertical components:$F = F_1 \cos 	heta + F_2 \cos 	heta = 2 F_1 \cos 	heta$.Substituting the values:$F = 2 	imes \left( \frac{4 G m^2}{5 l^2} ight) 	imes \left( \frac{2}{\sqrt{5}} ight) = \frac{16 G m^2}{5 \sqrt{5} l^2}$.

Two particles each of mass $m$ are placed at points $P$ and $Q$ as shown in the figure. $R$ is the mid-point of $PQ = l$. The gravitational force on the third particle of mass $m$ placed at point $S$ on the perpendicular bisector of $PQ$ is

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