A large water drop of radius R is split into 10^3 identical smaller droplets, and the work required to do this against surface tension is W₁ . If the same original drop were instead split into 10^6 identical droplets, the work required would be W₂ . The ratio W₂ W₁ is equal to

Step-by-step solution

Let the radius of the original drop be R and the radius of a smaller droplet be r . From the conservation of volume, for N droplets: 4 3 R^3 = N ( 4 3 r^3 ) r = R N^ 1/3 The initial surface energy of the drop is U_i = 4 R^2 T . The final surface energy of N droplets is U_f = N(4 r^2 T) = N 4 ( R N^ 1/3 )^2 T = N^ 1/3 (4 R^2 T) . The work done against surface tension is the change in surface energy: W = U_f - U_i = 4 R^2 T (N^ 1/3 - 1) For N₁ = 10^3 , the work done is: W₁ = 4 R^2 T ((10^3)^ 1/3 - 1) = 4 R^2 T (10 -