Physics | VITEEE 2017 | Discussion Page

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) $2^{5/3}\pi r^{2}T$

B) $4\pi r^{2}T$

C) $2\pi r^{2}T$

D) $2^{8/3}\pi r^{2}T$

Option D

Let R be the radius of the bigger drop, then Volume of bigger drop = 2 x volume of small drop

$\frac{4}{3}\pi R^{3}=2\times\frac{4}{3}\pi r^{3}\Rightarrow R =2^{1/3} r$

Surface energy of bigger drop,

$E=4\pi R^{2}T = 4\times2^{2/3}\pi r^{2}T$ = $2^{8/3}\pi r^{2}T$

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