Q. Two mercury drops (each of radius $r$ ) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is
NTA AbhyasNTA Abhyas 2022
Solution:
Let $R$ be the radius of the bigger drop.
Clearly, volume of bigger drop $= 2 \times $ volume of each smaller drop
$\Rightarrow \frac{4}{3}\pi R^{3}=2\times \frac{4}{3}\pi r^{3}\Rightarrow R=2^{1 / 3}r$
Surface energy of bigger drop,
$U=AT\left[\right.whereA=surfacearea\left]\right.\left]\right.\Rightarrow U=4\pi R^{2}T=4\times 2^{2 / 3}\pi r^{2}T\Rightarrow U=2^{8 / 3}\pi r^{2}T$
Clearly, volume of bigger drop $= 2 \times $ volume of each smaller drop
$\Rightarrow \frac{4}{3}\pi R^{3}=2\times \frac{4}{3}\pi r^{3}\Rightarrow R=2^{1 / 3}r$
Surface energy of bigger drop,
$U=AT\left[\right.whereA=surfacearea\left]\right.\left]\right.\Rightarrow U=4\pi R^{2}T=4\times 2^{2 / 3}\pi r^{2}T\Rightarrow U=2^{8 / 3}\pi r^{2}T$