Question

A Spherical liquid drop of radius R is divided into 8 equal droplets .If surface tension is T,then work done in the process will be ?

Answer 1

There is a thing called 'surface energy' which is potential energy per unit area of a liquid surface and it is numerically equal to the value of surface tension.
Hence surface energy = T Joules/m2

Initial volume of drop = \[4/3\pi R ^{3}\]
Let radius of each of the eight drops after it has broken be 'r'

Total volume all eight drops combined =\[8 * 4/3 \pi r ^{3}\]

Initial volume = final volume

This gives r=0.5R

Initial Surface area=\[4 \pi R ^{2}\]

Initial potential energy = Surface energy * Surface area\[U _{i}=T* 4 \pi R ^{2}\]

Final surface area=\[8 * 4\pi (0.5R)^{2}\]

\[U _{f}=T*8*4\pi (0.5R)^{2}\]

Work done will be difference between final and initial potential energies and will be positive.