A liquid drop having surface energy E is spread into 512 droplets of same size. The final surface energy of the droplets is nE . What is the value of n ?

Step-by-step solution

The surface area of the liquid drop is A = 4 πR 2 Its surface energy is E When the drop splits in 512 droplets, the surface area of each droplet is 4 πr 2 ∴ Total surface area A 2 = 512 × 4 πr 2 The volume of bigger drop is 4 3 πR 3 and volume of small droplets is 512 × 4 3 πr 3 ∴ 4 3 πR 3 = 512 × 4 3 πr 3 ⇒ r = R 8 ∴ A 2 = 512 × 4 πr 2 = 512 × 4 π R 8 2 = 8 A 1 Surface energy E = A . T ( T is surface tension and A is area) ∴ E n E = A 2 ⋅ T A 1 ⋅ T = 8 ⋅ A 1 A 1 = 8 ∴ E n = 8 E