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Answer to Question #22934 in Mechanics | Relativity for Anitha

Question #22934
What amount of energy is released if two mercury droplets of radii 0.1 & 0.2cm collapse into one single drop. The surface tension of mercury is 500x10^-3 Nm^-1?
Expert's answer
Energy of surface tension equal:
E=sigma*S
sigma - surface tension
S - area
Amount of energy
dE = (E1+E2) - E
E1 - energy of first drop
E2 - energy of second drop
E - energy of drop after collapse
E1 = sigma*4*pi*R1^2
R1 - radius of first drop
E2 = sigma*4*pi*R2^2
R1 - radius of second drop
Mass is uniform:
M = M1+ M2
4/3*pi*R^3*density = 4/3*pi*R1^3*density + 4/3*pi*R2^3*density
R^3 = R1^3 + R2^3
E = sigma*4*pi*(R1^3 + R2^3)^(2/3)
dE = (E1+E2) - E = sigma*4*pi*(R1^2+R2^2-(R1^3 + R2^3)^(2/3)) =
= 500*10^-3Hm^-1*4*Pi*(0.1^2+0.2^2-(0.1^3 + 0.2^3)^(2/3))*10^-4m^2 =& 4.23*10^-6 J
Answer: dE = 4.23*10^-6 J

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