Answer to Question #11113 in Mechanics | Relativity for SURJEET GULERIA
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.
V = 4/3 pi R^3 = N V0 = N (4/3) pi r^3
R^3 = N r^3
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.
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!