A spherical liquid drop of radius R is broken into 512 identical small droplets. If the work done against surface tension in this process is W and the initial surface energy of the drop is E , then the ratio W E is ________

Step-by-step solution

Let the radius of each small droplet be r . By conservation of volume: 4 3 R^3 = 512 4 3 r^3 R^3 = 512 r^3 R = 8r r = R 8 The initial surface energy of the large drop is: E = 4 R^2 T The final total surface energy of the 512 small droplets is: E_f = 512 4 r^2 T E_f = 512 4 ( R 8 )^2 T E_f = 512 64 4 R^2 T = 8 4 R^2 T = 8E The work done against surface tension is the change in surface energy: W = E_f - E = 8E - E = 7E Thus, the ratio W E = 7 . Answer: 7