Q. A bigger drop of radius $R$ is converted into $n$ smaller drops of radius $r$, the required energy is
A) $\left( 4\ \pi r^{2}n - 4\ \pi\ R^{2} \right)T$
B) $\left( \frac{4}{3}\ \pi r^{2}n - \frac{4}{3}\ \pi\ R^{3} \right)T$
C) $\left( 4\ \pi R^{2} - 4\ \pi\ r^{2} \right)nT$
D) $\left( n4\ \pi r^{2} - n4\ \pi\ R^{2} \right)T$
Explanation
(a) Work done = surface tension $\times$ increase in surface area
$= T(n\ 4\ \pi r^{2} - 4\ \pi\ R^{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.