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)

$2^{1/3}:1$

Let r be radius of common drop

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

$r=2^{1/3}R$

surface tension of 2 drops $2\times 4\pi R^{2}T$

surface tension of common drop $4\pi r^{2}T$

ratio $\frac{2\times 4\pi R^{2}T}{4\pi r^{2}T}=\frac{2R^2}{2^{2/3}{R}^2}=\frac{2^{1/3}}{1}$