A mercury drop of radius I cm is broken into ${10}^6$ droplets of equal size. The work done is ($\mathrm{T} = 35 \times {10}^{-2}$ N/m)

Initial radius of drop,

 R = 1 cm = $1\times {10}^{-2}\mathrm{m}$

Suppose, r is the radius of droplets.

Since, volume remain constant.

   $\frac{4}{3}{\mathrm{\pi R}}^3 = {10}^6\times \frac{4}{3}{\mathrm{\pi r}}^3$

[Since, number of droplets = ${10}^6$]

$\therefore \mathrm{r} = \frac{1\times {10}^{-2}}{{10}^2}={10}^{-4} [\because \mathrm{R} = 1 \times {10}^{-2} \mathrm{m}]$

Increase in surface area,

$∆S = {10}^6\times 4{\mathrm{\pi r}}^2-4{\mathrm{\pi R}}^2$

      = ${10}^6\times 4\times \mathrm{\pi }\times {10}^{-8}-4\mathrm{\pi }\times {10}^{-4}$

       = $4\times 9.9\times \mathrm{\pi }\times {10}^{-3}$

$\therefore \mathrm{Work} \mathrm{done} = \mathrm{Increase} \mathrm{in} \mathrm{surface} \mathrm{area} \times \mathrm{Surface} \mathrm{tension}$

      = $4 \times 9.9 \times \mathrm{\pi }\times {10}^{-3}\times 35\times {10}^{-2}$

       = $4.35\times {10}^{-2} \mathrm{J}$