Q. Two mercury drop (each of radius $r)$ merge to form a 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) $2^{8/3}\ \pi r^{2}T$
Explanation
(d) Let R be the radius of the bigger drop, then
$Volume\ of\ bigger\ drop = 2 \times volume\ of\ small\ drop$
$\frac{4}{3}\pi R^{3} = 2 \times \frac{4}{3}\pi r^{3}$
$R = 2^{\frac{1}{3}}r$
Surface energy of bigger drop,
$E = 4\pi R^{2}T$
$= 4 \times 2^{\frac{2}{3}}\pi r^{2}T$
$= 2^{2}\pi r^{2}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.