A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be

Gain in surface energy, $\Delta U=T\Delta A$  

from volume centenary, $\frac{4}{3}\pi R^3=64*\frac{4}{3}\pi r^3$  

$\Rightarrow r=\frac{R}{4}$                                            

Initial surface area, Ai = 4pR²

final surface area, $A_f=64*4\pi r^2$  

$\therefore \Delta U=12\pi R^2.T=0.075*12*3.14*10^{-4}=2.8*10^{-4}J$