A liquid drop having surface energy E is sprayed into 512 droplets of the same size. Then the final surface energy is

Step-by-step solution

Let the radius of the initial drop be R and the surface tension of the liquid be T . The initial surface energy is given by E = 4 R^2 T . Let the radius of each of the 512 smaller droplets be r . Since the total volume remains constant during the process: 4 3 R^3 = 512 4 3 r^3 R^3 = 512 r^3 R = 8r r = R 8 The final surface energy of the 512 droplets is: E' = 512 4 r^2 T Substituting r = R 8 : E' = 512 4 ( R 8 )^2 T E' = 512 4 R^2 T 64 E' = 8 (4 R^2 T) E' = 8E Answer: 8E