Question

A spherical liquid drop of radius R is divided into two equal droplets of radius r
(1)Find increase in surface area
(2)If T is surface tension, find increase in surface energy

Answer 1

Volume of the big drop = 4/3 π R³
Volume of two small droplets:  2 × 4/3 π r³
So 2 r³ = R³    => r = R/∛2

Surface area of large drop: 4 πR² 
Surface area of two droplets:  2*4 π r² = 8 π R²/ ∛4 = 4 ∛2 π R²
Increase in Surface area = ΔS = (∛2 - 1) 4 πR²

Increase in Surface Energy :  T * ΔS = 4 π (∛2 - 1) R² T

This energy comes either from the environment or from the internal thermal energy.