Question

A big water drop is formed by the combination of ${}^{\prime }{n}^{\prime }$ small water drops of equal radii. The ratio of the surface energy of ${}^{\prime }{n}^{\prime }$ drops to the surface energy of big drop is

• A. $n^2:1$
• B. $n:1$
• C. $\sqrt{n}:1$
• D. $\sqrt[3]{3n}:1$

Answer 1

Answer: (D)

$\sqrt[3]{3n}:1$

By mass conservation new radius R is $R=n^{\frac{1}{3}}r$

now ratio of energy is $\frac{nT\times 4\pi r^2}{T\times 4\pi R^2}=n^{1/3}$