# 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

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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