Question

Two small drops of mercury, each of radius $R$, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

• A. $1:2^{1/3}$
• B. $2^{1/3}:1$
• C. $2:1$
• D. $1:2$

Answer 1

Answer: (B)

Let $r$ be radius of common drop
$\begin{array}{l}\frac{4}{3}\pi r^3=2\times \frac{4}{3}\pi R^3\\ r=2^{1/3}R\end{array}$
surface tension of 2 drops $2\times 4\pi R^{2}T$
surface tension of common drop $4\pi r^{2}T$
ratio $\frac{2\times 4\pi R^{2}T}{4\pi r^{2}T}=\frac{2R^2}{2^{2/3}{R}^2}=\frac{2^{1/3}}{1}$