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

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

The surface energy is directly dependent on the total surface area of the drop.

Initial surface area = $2\times 4\pi R^2=8\pi R^2$

When they merge together to form a single drop, volume of water contained would remain the same.

Hence, $2\times 1.33\pi R^3=1.33\pi r^3$

Therefore, the new radius, r =R $2^{1/3}$

And the new surface area = $4\pi r^2$ = $4\pi 2^{2/3}{R}^2$

Therefore, the ratios of the surface energies = $2^{1/3}:1$