Question

Two mercury drops (each of radius $r$ ) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

• A. $2^{\frac{5}{3}}\pi r^{2}T$
• B. $4{\pi }^{2}T$
• C. $2\pi r^{2}T$
• D. $2^{\frac{8}{3}}\pi r^{2}T$

Answer 1

Answer: (D)

Let $R$ be the radius of the bigger drop.
Clearly, volume of bigger drop $=2\times$ volume of each smaller drop
$\Rightarrow \frac{4}{3}\pi R^3=2\times \frac{4}{3}\pi r^3\Rightarrow R=2^{1/3}r$
Surface energy of bigger drop,
$U=AT[whereA=surfacearea]]\Rightarrow U=4\pi R^{2}T=4\times 2^{2/3}\pi r^{2}T\Rightarrow U=2^{8/3}\pi r^{2}T$