A mercury drop of radius $10^{-3}~\mathrm{m}$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45~\mathrm{Nm}^{-1}$. The gain in surface energy is :
Solution
Initial surface energy $=0.45 \times 4 \pi\left(10^{-3}\right)^2$
<br/><br/>$$
\begin{aligned}
& \frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text {new }}^3 \\\\
\therefore & 10^{-3}=5 R_{\text {new }} \\\\
\therefore & R_{\text {new }}=\frac{10^{-3}}{5} \mathrm{~m}
\end{aligned}
$$
<br/><br/>So, final surface energy $=0.45 \times 125 \times 4 \pi\left(\frac{10^{-3}}{5}\right)^2$
<br/><br/> Increase in energy $=0.45 \times 4 \pi \times\left(10^{-3}\right)^2\left[\frac{125}{25}-1\right]$
<br/><br/>$$
\begin{aligned}
& =4 \times 0.45 \times 4 \pi \times 10^{-6} \\\\
& =2.26 \times 10^{-5} \mathrm{~J}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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