Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

Step-by-step solution

Let the radius of each small drop be r and the radius of the big drop be R . Equating the volumes before and after coalescing: 4 3 R^3 = 2 4 3 r^3 R = 2^ 1/3 r The total surface energy of the two small drops before coalescing is: E_i = 2 (4 r^2 T) = 8 r^2 T The total surface energy of the big drop after coalescing is: E_f = 4 R^2 T = 4 (2^ 1/3 r)^2 T = 4 2^ 2/3 r^2 T The ratio of the total surface energies after and before the change is: E_f E_i = 4 2^ 2/3 r^2 T 8 r^2 T = 2^ 2/3 2 = 2^ -1/3 The ratio is 2^ -1/3 : 1