Question

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

• A. $2^{8/3}{\pi r}^{2}T$
• B. $2^{5/3}{\pi r}^{2}T$
• C. $4{\pi r}^{2}T$
• D. $2{\pi r}^{2}T$

Answer 1

Answer: (A)

$2^{8/3}{\pi r}^{2}T$

[A] Sum of volume of 2 smaller drops = Volume of bigger drop.

$\begin{array}{c}2.\frac{4}{3}\pi r^2\Rightarrow \frac{4}{3}\pi r^3\Rightarrow R=2^{1/3}r.\\ Surfaceenergy=T.4\pi R^2\\ \Rightarrow T{.2}^{8/3}\pi r^2\end{array}$