Question

What will be the change in surface energy when a drop of liquid (S = 0.08N/m) is divided into 10 equal droplets? Radius of initial single drop (R)= 5cm.

(a) 0.1259 J

(b) 0.1141 J

(c) 0.2356 J

(d) 0.1765 J

The question was posed to me in unit test.

My question comes from Fluids Mechanical Properties in portion Mechanical Properties of Fluids of Physics – Class 11

Answer 1

Right option is (b) 0.1141 J

Explanation: Volume of drop = 4/3πR^2.                                                                                                                             Let the radius of droplets be ‘r’. Volume of big drop will be equal to volume of 10 droplets.

10*4/3πr^3  =  4/3πR^3 ⇒ r = 10^1/3 *R.

Surface area of big drop = 4πR^2  & surface area of 10 droplets = 10 * 4πr^2 = 40πR^2(10^2/3).

Surface energy of big drop = 4πR^2 *S = 4π(0.0025)*0.08 = 0.0025

Surface energy of droplets = 40πR^2(10^2/3) *S = 40π(0.0025)(10^2/3)*0.08 = 0.1166.

Change in surface energy = 0.1166 – 0.0025 = 0.1141 J.