A drop of liquid of diametre 2.8mm breaks up inti 125 identical drops The change in energy is nearly

let the radius of each droplet be r m

Volume of bigger drop = total volume of 125 droplets

(4/3)* pi * (1.4*10^-3)³ = 125 * (4/3) * pi * r³

=> r = (1.4/5)* 10^-3 = 0.28*10^-3

change in surface area = ds = 4*pi* [125* (0.28*10^-3) ² - (1.4*10^-3)²] m

                                       = 0.0984 m

if surface tension of the liquid ne T Nm-¹ then work done W = Tds = 0.098T J.