A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is 10⁻⁶ J. The value of is _______. (Take surface tension of liquid = 0.08 N/m)

Step-by-step solution

Let the radius of the initial drop be R and the radius of each small droplet be r . Given diameter D = 2 mm, so R = 1 mm = 10⁻³ m. By conservation of volume: 4 3 R^3 = 512 4 3 r^3 R^3 = 512 r^3 R = 8r r = R 8 Initial surface area, A_i = 4 R^2 Final surface area, A_f = 512 4 r^2 = 512 4 ( R 8 )^2 = 32 R^2 Change in surface area, A = A_f - A_i = 32 R^2 - 4 R^2 = 28 R^2 Change in surface energy, U = T A U = 0.08 28 R^2 Substituting R = 10⁻³ m and 22 7 : U = 0.08 28 22 7 (10⁻³)^2 U = 0.08 4 22 10⁻⁶ U = 7.04 10⁻⁶ J Comp