Q. What change in surface energy will be noticed when a drop of radius $R$ splits up into 1000 droplets of radius$r$, surface tension$T$?
A) $4\ \pi\ R^{2}T$
B) $7\ \pi\ R^{2}T$
C) $16\ \pi\ R^{2}T$
D) $36\ \pi\ R^{2}T$
Explanation
(d) Increase in surface energy = surface tension $\times$ increase in surface area
$= S\left( 1000 \times 4\pi r^{2} - 4\pi R^{2} \right)$
$\left( 100 \times \frac{4}{3}\pi r^{3} = \frac{4}{3}R^{3}\ or\ r = R/10 \right)$
$= S \times 4\pi\left( 1000 \times \frac{R^{2}}{100} - R^{2} \right) = 36\ \pi\ R^{2}S$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.