1. Liquids acquire a free surface when poured in a container. These surfaces possess some additional energy. This phenomenon is known as surface tension.
2. When a drop is split into tiny droplets, the surface area increases. So work has to be done for splitting the drop.
Let R be radius of the drop and r the radius of the droplets:
R = 1 × 10-3 m
surface area of the drop = 4πR2
= 4π × (1 × 10-3)2
= 4π × 10-6 m2
Volume of the drop = Volume of 106 droplets
∴ Surface area of million droplets =106 × 4π 2
= 106 × 4π(1 × 10-5)2
= 4π × 10-4 m2
∴ Increase in surface area = 4π × 10-4 – 4π × 10-6
= 3.96π × 10-4 m2
∴ Energy expended = 3.967π × 10-4 × S
= 3.96π × 10-4 × 72 × 10-3 J
= 8.95 × 10-5 J.