Question

If $\sigma$ is surface tension, the work done in breaking a big drop of radius, $R$, into $n$ drops of equal radius is:

• A. $R{n}^{2/3}\sigma$
• B. $(n^{2/3}-1)\sigma R^2$
• C. $4\pi R^2(n^{1/3}-1)\sigma$
• D. $\pi R^2(n^{1/3}-1)\sigma$

Answer 1

Answer: (C)

$4\pi R^2(n^{1/3}-1)\sigma$

Volume will be conserved
$\frac{4}{3}\pi R^3=n\frac{4}{3}\pi r^3$
$r=n^{-1/3}R$
workdone = final energy - initial energy
= $[n4\pi n^{-2/3}{R}^2-4\pi R^2]\sigma$
= $4\pi R^2(n^{1/3}-1)\sigma$