# Answer on Mechanics | Relativity Question for SURJEET GULERIA

Question #11113

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

Expert's answer

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.

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.

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