A liquid drop of radius R is broken into N identical smaller drops. If the surface tension of the liquid is T and the work done in this process is 12 R^2 T , then the value of N is :

Step-by-step solution

Let the radius of each newly formed small drop be r . Since the total volume remains constant during the process: 4 3 R^3 = N 4 3 r^3 R^3 = N r^3 r = R N^ 1/3 The initial surface area of the single drop is A_i = 4 R^2 . The final total surface area of the N drops is A_f = N 4 r^2 = 4 N ( R^2 N^ 2/3 ) = 4 R^2 N^ 1/3 . The work done is equal to the product of surface tension and the change in surface area: W = T A = T(A_f - A_i) W = T(4 R^2 N^ 1/3 - 4 R^2) = 4 R^2 T (N^ 1/3 - 1) We are given that the work done is 12