Answer to Question #11113 in Mechanics | Relativity for SURJEET GULERIA

Question

A drop of water of radius 1mm is split in to 1000 small droplets of equal size. Find the energy spent if surface tension for water is 0.072 N/m

Answer 1

Let N=1000 be the number of small droplets,
R=1mm be the radius of the initial drop,
r be the radius of each small droplet,
V, S be the volume and surface area of the initial drop,
V0, S0 be the volume and surface area of each small droplet,
sigma = 0.072 N/m be the surface tension for water is 0.072 N/m.


Then
V = 4/3 pi R^3 = N V0 = N (4/3) pi r^3

Hence
R^3 = N r^3
and so
r = R / N^(1/3)


Notice that the free energy of the initial drop is

E = sigma S = sigma * 4 * pi * R^2

while the total free energy of small droplets is

E0 = N sigma S0
= sigma * N * 4 * pi * r^2
= sigma * N * 4 * pi * R^2/N^(2/3)
= sigma * N^(1/3) * 4 * pi * R^2


Hence the energy spent is equal to

E0-E = sigma * N^(1/3) * 4 * pi * R^2 - sigma * 4 * pi * R^2
= sigma * 4 * pi * R^2 ( N^(1/3) - 1 )
= 0.072 * 4 * 3.14159 * 0.001^2 * (10-1)
= 8.143 * 10^(-6) J.