A number of identical small liquid drops coalesce to form a single large drop. If the total surface energy of the small drops is 6 times the surface energy of the large drop, the number of small drops is _________

Step-by-step solution

Let there be N small drops, each of radius r , coalescing to form a large drop of radius R . By conservation of volume: N 4 3 r^3 = 4 3 R^3 R = N^ 1/3 r The total surface energy of the N small drops is: E₁ = N 4 r^2 T The surface energy of the large drop is: E₂ = 4 R^2 T = 4 (N^ 1/3 r)^2 T = N^ 2/3 4 r^2 T Given that the ratio of the surface energies is 6 : E₁ E₂ = N N^ 2/3 = N^ 1/3 = 6 Cubing both sides: N = 6^3 = 216 Answer: 216