Question

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

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

Answer 1

The correct option is A $2^{\frac{1}{3}}:1$


$\frac{4}{3}\pi R^3+\frac{4}{3}\pi R^3=\frac{4}{3}\pi R^{\prime 3}$

$R^{\prime }=2^{\frac{1}{3}}R....\left(1\right)$

Initially, total surface area, $A_i=2\times 4\pi R^2$

Final surface area, $A_f=4\pi R^{\prime 2}=4(2{)}^{2/3}\pi R^2$

Let $\sigma$ be the surface tension of mercury.

$\therefore \frac{U_i}{U_f}=\frac{\sigma A_i}{\sigma A_f}=\frac{2\times 4\pi R^2}{4(2{)}^{2/3}\pi R^2}=2^{\frac{1}{3}}$

$\Rightarrow U_i:U_f=2^{\frac{1}{3}}:1$

Hence, option $\left(\text{A}\right)$ is correct.