The work done in splitting a drop of water of 1 mm radius into 10 6 droplets is [S.T. of water $\left.72\times {10}^{-3}\mathrm{J}/{\mathrm{m}}^2\right)$

. In splitting, the volume remains unchanged.
$$\begin{gathered} \therefore \frac{4}{3}\pi R^3=\frac{4}{3}\pi r^3\times n \\ or R=\left(n^{1/3}\right)r or r=R/(n{)}^{1/3} \\ work done W= surface tension x change in area \\ =4\pi r^{2}n-4\pi R^2 \\ =4\pi \left[\frac{n\times R^2}{n^{2/3}}-R^2\right]=4\pi R^2\left[n^{1/3}-1\right] \\ \therefore W=4\pi R^2\left[n^{1/3}-1\right]T \\ Hero n={10}^6,R=1\mathrm{mm}={10}^{-3}\mathrm{m} \\ \mathrm{and} \mathrm{T}=72\times {10}^{-3}\mathrm{J}/{\mathrm{m}}^2 \\ \mathrm{Substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get} \mathrm{W}=8\cdot 95\times {10}^{-5} \mathrm{joule} \end{gathered}$$