Question

If S is the surface tension of a liquid, the energy needed to break a liquid drop of radius R into 64 drops is

• A. $6\pi R^{2}S$
• B. $4\pi R^{2}S$
• C. $12\pi R^{2}S$
• D. $8\pi R^{2}S$

Answer 1

Answer: (C)

Let $r$ be the radius of each small drop. Then
Volume of 64 small drops = Volume of big drop
or $64\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$ or $r=\frac{R}{4}\dots (i)$
Surface area of big drop $=4\pi R^2$
Surface area of 64 small drops $=64\times 4\pi r^2$
$\therefore$ Increase in surface area $=64\times 4\pi r^2-4\pi R^2=4\pi \left[64r^2-R^2\right]$
$=4\pi \left[4R^2-R^2\right]=12\pi R^2$ (Using (i))
Energy needed surface tension $\times$ increase in surface area
$=S\times 12\pi R^2=12\pi R^{2}S$