A single liquid drop of radius 2 cm is broken into N identical smaller droplets. If the surface tension of the liquid is 0.05 N m ⁻¹ and the total work done in this process is 5.6 10⁻⁴ J , the value of N is:

Step-by-step solution

Let the radius of the initial large drop be R and the radius of each smaller droplet be r . By conservation of volume: 4 3 R^3 = N ( 4 3 r^3 ) r = R N^ 1/3 The initial surface area of the drop is A_i = 4 R^2 . The final total surface area of the N droplets is A_f = N(4 r^2) = N 4 ( R N^ 1/3 )^2 = 4 R^2 N^ 1/3 . The change in surface area is: A = A_f - A_i = 4 R^2 (N^ 1/3 - 1) The work done is equal to the increase in surface energy: W = S A = S 4 R^2 (N^ 1/3 - 1) Substitute the given values ( R = 2 cm = 0.02 m , S