A mercury drop of radius 10 ^ – 3 m is broken into 125 equal size...

As the volume of the mercury will remain same, therefore

$\dfrac{4}{3}\pi{R}^{3} = 125\left(\dfrac{4}{3}\pi{r}^{3}\right) ⇒ R = 5r$

Now, increase in surface energy

$∆ U = 125 \times \left(S \times 4\pi{r}^{2}\right) - \left(S \times 4\pi{R}^{2}\right)$

$= 0 . 45 \times 4\pi \times {\left({10}^{- 3}\right)}^{2}\left[\dfrac{125}{25} - 1\right]$

$= 4 \times 0 . 45 \times 4\pi \times {10}^{- 6}$

$= 2 . 26 \times {10}^{- 5}J$