Question

Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?

• A. $2^{1/2}:1$
• B. $1:1$
• C. $2^{2/3}:1$
• D. None of these

Answer 1

The correct option is C $2^{2/3}:1$

let two drops of radii " r " combines together to form a drop of radii " R ".

sum of the volumes of two smaller drops of raddi "r" would be equal to the volume of bigger drop of radii "R" .

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

so 2$r^3$ = $R^3$

hence R = $2^{\frac{2}{3}}$r

now surface energy = surface tension (surface area) =$S.4\pi r^2$ where S is the surface tension of the drop and r is the radius of the drop.

so we can say surface energy is proportional to $r^2$.

hence the ratio of surface energy of bigger drop to that of smaller drop = $\frac{R^2}{r^2}$ = $\frac{2^{\frac{2}{3}}}{1}$

hence option (C) is correct.