Thank you for reporting, we will resolve it shortly
Q.
If $ T $ is the surface tension of the liquid, the change in surface energy by splitting a liquid drop of radius $ R $ into $ 125 $ droplets of radius $ r $ is
As volume of $125$ droplets = volume of the drop
$\therefore 125\left(\frac{4}{3}\pi r^{3}\right)=\frac{4}{3}\pi R^{3}$
or $125r^{3}=R^{3}$
or $r=\frac{R}{5}$
The surface area of the drop $= 4\pi R^{2}$
and the surface area of $125$ droplets
$=125\left(4\pi r^{2}\right)=125\left(4\pi\left(\frac{R}{5}\right)^{2}\right)=20 \pi R^{2}$
$\therefore $ The increase in surface area
$=20\pi R^{2}-4\pi R^{2}=16\pi R^{2}$
The change in surface energy
= surface tension $\times$ increase in surface area
$=T\times16\pi R^{2}$
$=16\pi R^{2}T$