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

Answer 1

Answer: (D)

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

Answer 2

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