A large liquid drop is broken into N identical smaller droplets. If the total surface energy of the system increases by 300 % during this process, the value of N is:

Step-by-step solution

Let the radius of the large drop be R and the radius of each small droplet be r . By conservation of volume: 4 3 R^3 = N ( 4 3 r^3 ) R = N^ 1/3 r The initial surface energy is U_i = 4 R^2 T . The final total surface energy is: U_f = N(4 r^2 T) = N (4 ( R N^ 1/3 )^2 T ) = N^ 1/3 (4 R^2 T) = N^ 1/3 U_i Given that the surface energy increases by 300 % , the final surface energy is: U_f = U_i + 300 100 U_i = 4 U_i Equating the two expressions for U_f : N^ 1/3 U_i = 4 U_i N^ 1/3 = 4 Cubing both sides, we obtain: N = 64