A drop of liquid of radius R = 10 - 2 m having surface tension S = 0.1 4 π N m - 1 divides itself into K identical drops. In the process the total change in the surface energy Δ U = 10 - 3 J . If K = 10 α then the value of α is

Step-by-step solution

By mass conservation, ρ . 4 3 π R 3 = ρ . K . 4 3 π r 3 ⇒ R = K 1 3 r ∴ Δ U = T Δ A = T K . 4 π r 2 - 4 π R 2 = T K . 4 π R 2 K - 2 3 - 4 π R 2 Δ U = 4 π R 2 T K 1 3 - 1 Putting the value’s ⇒ 10 - 3 = 10 - 1 4 π × 4 π × 10 - 4 K 1 3 - 1 100 = K 1 3 - 1 ⇒ K 1 3 ≅ 100 = 10 2 Given that K = 10 α ⇒ ∴ 10 α 3 = 10 2 ⇒ α 3 = 2 ⇒ α = 6