In S.I. system, the total energy of the free surface of a liquid drop is 2$\mathrm{\pi }$ times the surface tension of the liquid the diameter of the drop is __________

The total energy in free space E=SA, where A is the drop's surface area and S is the liquid's surface tension. 

$A=4{\mathrm{\pi R}}^2; E=2\mathrm{\pi S}$

$\mathrm{we} \mathrm{get} 2\mathrm{\pi S}=\left(4{\mathrm{\pi R}}^2\right)\mathrm{S}$

$\mathrm{R}=\frac{1}{\sqrt{2}}$

$\mathrm{Diameter} \mathrm{D}=2\mathrm{RD}=2\times \frac{1}{\sqrt{2}}$

$\mathrm{D}=\sqrt{2}\mathrm{m}$

Hence the correct answer is $\sqrt{2}m.$