The work done in splitting a water drop of radius 'R' into 64 droplets is (T is the surface tension of water)

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 = 64 4 3 r^3 R^3 = 64 r^3 r = R 4 Initial surface area of the drop: A_i = 4 R^2 Final surface area of 64 droplets: A_f = 64 4 r^2 = 64 4 ( R 4 )^2 = 16 R^2 Change in surface area: A = A_f - A_i = 16 R^2 - 4 R^2 = 12 R^2 Work done in splitting the drop: W = T A = 12 R^2 T Answer: 12 R^2 T