Q. A sphere liquid drop of radius $R$ is divided into eight equal to droplets. If surface tension is $T$, then the work done in this process will be
A) $2\pi R^{2}T$
B) $3\pi R^{2}T$
C) $4\pi R^{2}T$
D) $2\pi RT^{2}$
Explanation
(c) Volume $= \frac{4}{3}\pi\ R^{3} = 8 \times \frac{4}{3}\pi\ r^{3}$ or $r = R/2$
Work done $= T \times (4\ \pi r^{2} \times 8 - 4\pi R^{2})$
$= T \times 4\pi\left( \frac{R^{2}}{4} \times 8 - R^{2} \right) = 4\pi R^{2}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.