Question

If $T$ is the surface tension of the liquid, the change in surface energy by splitting a liquid drop of radius $R$ into $125$ droplets of radius $r$ is

• A. $16\pi R^{2}T$
• B. $32\pi R^{2}T$
• C. $18\pi R^{2}T$
• D. $36\pi R^{2}T$

Answer 1

Answer: (A)

As volume of $125$ droplets = volume of the drop
$\therefore 125\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$
or $125r^3=R^3$
or $r=\frac{R}{5}$
The surface area of the drop $=4\pi R^2$
and the surface area of $125$ droplets
$=125\left(4\pi r^2\right)=125\left(4\pi {\left(\frac{R}{5}\right)}^2\right)=20\pi R^2$
$\therefore$ The increase in surface area
$=20\pi R^2-4\pi R^2=16\pi R^2$
The change in surface energy
= surface tension $\times$ increase in surface area
$=T\times 16\pi R^2$
$=16\pi R^{2}T$