Three particles each having a mass of 100 gm are placed on the vertices of an equilateral triangle of side 20 cm. The work done in increasing the side of the triangle to 40 cm is  $\left[G=6.67 \times 10^{-11} \frac{N-m^2}{kg^2}\right]$

5.0 × 10^-12 J

W = U_final − U_initial

$=3\left[-\left\{\frac{\mathrm{G}(\mathrm{m})(\mathrm{m})}{\mathrm{r}_{\mathrm{f}}}\right\}-\left\{\frac{\mathrm{G}(\mathrm{m})(\mathrm{m})}{\mathrm{r}_{\mathrm{i}}}\right\}\right]$

Putting $\mathrm{r}_{\mathrm{f}}=0.4 \mathrm{~m}, \quad \mathrm{r}_{\mathrm{i}}$ = 0.2 m and

G = $6.67 \times 10^{-11} \frac{\mathrm{N}-\mathrm{m}^2}{\mathrm{~kg}^2}$ and m = 0.1 kg

We get W = 5.0 × 10^-12 J