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Four particles $A, B, C$ and $D$ each of mass $m$ are kept at the corners of a square of side $L$. Now the particle $D$ is taken to infinity by an external agent keeping the other particles fixed at their respective positions. The work done by the gravitational force acting on the particle $D$ during its movement is ..........
$2 \frac{G m^2}{L}$
$-2 \frac{G m^2}{L}$
$\frac{G m^2}{L}\left(\frac{2 \sqrt{2}+1}{\sqrt{2}}\right)$
$-\frac{G m^2}{L}\left(\frac{2 \sqrt{2}+1}{\sqrt{2}}\right)$
Solution
(d)
Work done by the gravitational force acting on the particle $D$ during its movement
$=-\Delta U$
$=-\left(U_{\text {final }}-U_{\text {initial }}\right)$
$=U_{\text {initial }}-U_{\text {final }}$
Now, when the particle is at infinity, $U=0$
$\Rightarrow U_{\text {final }}=0$
$\Rightarrow \text { Work done }=U_{\text {initial }}$
$U_{\text {initial }}=-\frac{G m^2}{L}-\frac{G m^2}{L}-\frac{G m^2}{\sqrt{2} L}$
$=-\frac{G m^2}{L}\left(2+\frac{1}{\sqrt{2}}\right)$
$=-\frac{G m^2}{L}\left(\frac{2 \sqrt{2}+1}{\sqrt{2}}\right)$
