A large spherical liquid drop of radius 5 cm is broken into 1000 identical smaller droplets. If the surface tension of the liquid is 0.04 N m ⁻¹ , the work done against surface tension in this process is:

Step-by-step solution

Let the radius of the large drop be R and the radius of each smaller droplet be r . By conservation of volume: 4 3 R^3 = 1000 4 3 r^3 R^3 = 1000 r^3 r = R 10 Initial surface area of the single drop (which has only one free surface): A_i = 4 R^2 Final total surface area of the 1000 droplets: A_f = 1000 4 r^2 = 1000 4 ( R 10 )^2 = 1000 4 R^2 100 = 40 R^2 Change in surface area: A = A_f - A_i = 40 R^2 - 4 R^2 = 36 R^2 Work done is the change in surface energy: W = T A = T 36 R^2 Substitute R = 5 cm = 5 10⁻² m and T =