A drop of water of radius 1mm is split in to 1000 small droplets of equal size. Find the energy spent if surface tension for water is 0.072 N/m
Let N=1000 be the number of small droplets, R=1mm be the radius of the initial drop, r be the radius of each small droplet, V, S be the volume and surface area of the initial drop, V0, S0 be the volume and surface area of each small droplet, sigma = 0.072 N/m be the surface tension for water is 0.072 N/m.
Then V = 4/3 pi R^3 = N V0 = N (4/3) pi r^3
Hence R^3 = N r^3 and so r = R / N^(1/3)
Notice that the free energy of the initial drop is
E = sigma S = sigma * 4 * pi * R^2
while the total free energy of small droplets is
E0 = N sigma S0 = sigma * N * 4 * pi * r^2 = sigma * N * 4 * pi * R^2/N^(2/3) = sigma * N^(1/3) * 4 * pi * R^2