A single large liquid drop of radius 7 mm is sprayed into N identical smaller droplets. The work done in this process is 2.772 10⁻⁴ J . If the surface tension of the liquid is 0.05 N m ⁻¹ , then the value of N is [Given = 22 7 ]

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 = N ( 4 3 r^3 ) r = R N^ 1/3 The initial surface area is A_i = 4 R^2 . The final total surface area is A_f = N(4 r^2) = N 4 ( R N^ 1/3 )^2 = 4 R^2 N^ 1/3 . The work done is equal to the change in surface energy: W = T A = T(A_f - A_i) = 4 R^2 T (N^ 1/3 - 1) Substituting the given values: W = 4 22 7 (7 10⁻³)^2 0.05 (N^ 1/3 - 1) 2.772 10⁻⁴ = 4 22 7 49 10⁻⁶ 0.05 (N^ 1/3 - 1) 2.772 10⁻⁴ = 88 7 10⁻⁶ 0.0