A drop of mercury of radius 2 mm is split into 8 identical droplets. Determine the increase in surface energy. The surface tension of mercury is 0.465 J/m ^2 .

Step-by-step solution

Let the radius of the large drop be R = 2 mm = 2 10⁻³ m and the radius of each small droplet be r . From the conservation of volume: 4 3 R^3 = 8 4 3 r^3 R^3 = 8r^3 r = R 2 Initial surface area of the drop is A₁ = 4 R^2 . Final surface area of the 8 droplets is A₂ = 8 4 r^2 = 32 ( R 2 )^2 = 8 R^2 . Increase in surface area is A = A₂ - A₁ = 8 R^2 - 4 R^2 = 4 R^2 . Increase in surface energy is given by U = T A , where T is the surface tension. Substituting the given values: U = 0.465 4 (2 10⁻³)^2 U = 0.465 16 10⁻⁶ U