Question

A mercury drop of radius $10^{-3}m$ is broken into $125$ equal size droplets. Surface tension of mercury is $0.45N{m}^{-1}$. The gain in surface energy is:

• A. $5\times 10^{-5}J$
• B. $17.5\times 10^{-5}J$
• C. $2.26\times 10^{-5}J$
• D. $28\times 10^{-5}J$

Answer 1

Answer: (C)

Initial surface energy $=0.45\times 4\pi {\left(10^{-3}\right)}^2$
$\frac{4}{3}\pi {\left(10^{-3}\right)}^3=125\times \frac{4\pi }{3}{R}_{\text{new }}^3$
$\therefore 10^{-3}=5R_{\text{new }}$
$\therefore R_{\text{new }}=\frac{10^{-3}}{5}m$
So, final surface energy $=0.45\times 125\times 4\pi {\left(\frac{10^{-3}}{5}\right)}^2$
Increase in energy $=0.45\times 4\pi \times {\left(10^{-3}\right)}^2\left[\frac{125}{25}-1\right]$
$=4\times 0.45\times 4\pi \times 10^{-6}$
$=2.26\times 10^{-5}J$