Three particles each of mass $'m'$ are at the vertices of an equilateral triangle of side length $'l'$. Then the work done is moving one particle to infinity is:

Solution.

Initial gravitation potentialenergy $=U_{i}=\frac{-G_{m m}}{2}-\frac{G_{m m}}{2}-\frac{G_{m m}}{2}=\frac{-3 G_{m m}}{2}$.

$\begin{array}{l} \text { Workdore }= \text { of }-V_{i}=\frac{-3 G m m}{2 L}-\left(-\frac{3 G_{m m}}{L}\right)= \\ =\frac{3 G m m}{2 L}+\frac{6 G_{m m}}{2 L} \\ =\frac{3 a-}{2 L} \\ \text { If } 2 L=\text { Infitity }=l . \end{array}$

If $2 L=$ Infitity $=l$

Then workdone $=\frac{3 \mathrm{~Gm}^{2}}{l}$

$\therefore$ option $A$ is correct