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Q.
A mercury drop of radius $10^{-3} m$ is broken into $125$ equal size droplets. Surface tension of mercury is $0.45\, Nm ^{-1}$. The gain in surface energy is:
JEE MainJEE Main 2023Mechanical Properties of Fluids
Solution:
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}=5 R_{\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 $=0.45 \times 4 \pi \times\left(10^{-3}\right)^2\left[\frac{125}{25}-1\right]$
$=4 \times 0.45 \times 4 \pi \times 10^{-6}$
$=2.26 \times 10^{-5} J$