Question

Two mercury drops of radius r merge to form bigger drop. Surface energy of bigger drop is (T is surface tension of Hg)

Answer 1

We know that,

Sum of volumes of 2 smaller drops = volume of the bigger drop

$2\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

$R=2^{\frac{1}{3}}r$

Now,

Surface energy

$=T4\pi R^2$

$=T\pi 2^{\frac{8}{3}}{r}^2$

Hence, the surface energy is $T\pi 2^{\frac{8}{3}}{r}^2$