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)

$2^{1/3}:1$

Answer 2

As $R = n^{1/3}r$ $= 2^{1/3}r$ ⇒ $R^{2} = 2^{2/3}r^{2}$⇒ $\frac{r^{2}}{R^{2}} = 2^{- 2/3}$

$\frac{\text{Initial surface energy}}{\text{Final surface energy}}\mathbf{=}\frac{\mathbf{2(4\pi}\mathbf{r}^{\mathbf{2}}\mathbf{T)}}{\mathbf{(4\pi}\mathbf{R}^{\mathbf{2}}\mathbf{T)}} = 2\left( \frac{r^{2}}{R^{2}} \right) = 2 \times 2^{- 2/3}$

= 2^1/3