A liquid drop of radius 1 cm and surface tension 0.07 N m ⁻¹ is sprayed into N identical smaller droplets. If the total work done in this process is 7.92 10⁻⁴ J , the value of N is (Take = 22 7 )

Step-by-step solution

Let the radius of the initial large drop be R and the radius of each of the N smaller droplets 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 increase 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 = T A = T 4 R^2 (N^ 1/3 - 1) Substitute the given values ( R = 1 cm = 10⁻² m , T = 0.07 N m ⁻¹ , W = 7.9