A single liquid drop of radius R and surface tension T is sprayed into n identical smaller droplets. If the mechanical work done in this process is exactly three times the initial surface energy of the original drop, the value of n is :

Step-by-step solution

The initial surface energy of the single drop is given by: U_ i = 4 R^2 T The mechanical work done to spray the drop is given as 3 times the initial surface energy: W = 3U_ i = 12 R^2 T The final surface energy of the n droplets is: U_ f = U_ i + W = 4 R^2 T + 12 R^2 T = 16 R^2 T Let r be the radius of each smaller droplet. By conservation of volume: 4 3 R^3 = n 4 3 r^3 r = R n^ 1/3 The final surface energy can also be written as: U_ f = n(4 r^2 T) = n 4 ( R n^ 1/3 )^2 T = 4 R^2 T n^ 1/3 Equating the two expression