Question

If a drop of radius $r$ breaks up into $27$ small drops then how much will be the change in the surface energy. The surface tension of liquid is $T$.

Answer 1

Let $R=$radius of small drops. Then, by equation the volumes we have.
$\frac{4}{3}\pi r^3=27\left(\frac{4}{3}\pi R^3\right)$
or $R=\frac{r}{3}$
$\Delta U=T\left(\Delta A\right)$
$=T(A_f-A_i)$
$=T[27\left(4\pi R^2\right)-4\pi r^2]$
$=4\pi T[27{\left(\frac{r}{2}\right)}^2-r^2]$
$=8\pi r^{2}T$