Question

Spherical liquid drop of radius r is divided into eight droplets of same size in the surface tension of the liquid is t then the work done in the process will be

Answer 1

its surface tension problems

here you also find change in energy

work done = change in energy

Answer 2

Let us say r is the radius of each eight of the bubble. Since volume of water remains the same, we have

$\frac{4}{3}$ π$R^{3}$= 8×$\frac{4}{3}$π$r^{3}$

⇒r=$\frac{R}{2}$

So the work done in this process will be = change in the surface energy i.e.

(8×4π$r^{2}$×T)-(4π$R^{2}$×T)

⇒(8×4π$(\frac{R}{2}) ^{2}$×T)-(4π$R^{2}$×T)

⇒8π$R^{2}$  T-4π$R^{2}$  T=4π$R^{2}$  T