The work done in splitting a water drop of radius R in to 64 droplets is ( T = surface tension of water)

Step-by-step solution

Work done in splitting a water drop equals the increase in surface energy due to the change in surface area. Given a large drop of radius R splits into 64 smaller droplets, and volume is conserved: V_ large = 64 V_ small . Since volume scales as R^3 , we have R^3 = 64 r^3 , so r = R/4 . Initial surface area is A_i = 4 R^2 . Final surface area is A_f = 64 4 r^2 = 256 r^2 . Substituting r = R/4 gives A_f = 256 (R^2/16) = 16 R^2 . Increase in surface area is A = A_f - A_i = 16 R^2 - 4 R^2 = 12 R^2 . Work done is W = T