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 :

Question

$28\times10^{-5}~\mathrm{J}$

$17.5\times10^{-5}~\mathrm{J}$

$5\times10^{-5}~\mathrm{J}$

$2.26\times10^{-5}~\mathrm{J}$

Solution

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