A drop of radius 1mm is sprayed into 1000 droplets of same size. If surface tension of the liquid is 40 dyne (cm^-1), the amount of work done in the process is

Given,

$\mathrm{n}=1000, \mathrm{T}=40\mathrm{dyne}/\mathrm{cm}$

$\mathrm{R}=1\mathrm{mm}={10}^{-3}\mathrm{m} \& \mathrm{R}={\mathrm{n}}^{\frac{1}{3}}\mathrm{r}$

we know that,

$\mathrm{Work} \mathrm{done} = \mathrm{difference} \mathrm{in} \mathrm{surface} \mathrm{area} \times \mathrm{surface} \mathrm{tension}$

$\mathrm{W}= 4\mathrm{\pi }({\mathrm{nr}}^2-{\mathrm{R}}^2)\mathrm{T}=4{\mathrm{\pi R}}^2\mathrm{T}({\mathrm{n}}^{\frac{1}{3}}-1)$

$\mathrm{W}=4(3.14){(10-3)}^2\left(40\right)\times {10}^{-3}[{\left(1000\right)}^{\frac{1}{3}}-1]=4.53\times {10}^{-6}\mathrm{J}$

Hence the correct answer is $4.53\times {10}^{-6}\mathrm{J}.$