Q. A certain number of spherical drops of liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then :
A) Energy = $4VT\left(\frac{1}{r} - \frac{1}{R}\right)$ is released.
B) Energy = $3VT\left(\frac{1}{r} + \frac{1}{R}\right)$ is absorbed
C) Energy = $3VT\left(\frac{1}{r} - \frac{1}{R}\right)$ is released
D) Energy is neither released nor absorbed
Explanation
[Solution] With n drops, $n\frac{4}{3}\pi r^3 = V$. Surface area falls from $\frac{3V}{r}$ to $\frac{3V}{R}$, so energy released = $3VT\left(\frac{1}{r} - \frac{1}{R}\right)$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.