Eight mercury drops, each of radius r , coalesce to form a bigger drop. The surface energy released in this process is _______. ( S is the surface tension of mercury).

Step-by-step solution

Let the radius of the bigger drop be R . Since the total volume remains conserved during the process: 8 4 3 r^3 = 4 3 R^3 R^3 = 8r^3 R = 2r Initial surface energy of the 8 drops is: U_i = 8 (4 r^2 S) = 32 r^2 S Final surface energy of the bigger drop is: U_f = 4 R^2 S = 4 (2r)^2 S = 16 r^2 S The surface energy released in the process is: U = U_i - U_f = 32 r^2 S - 16 r^2 S = 16 r^2 S Answer: 16 r^2 S