Question

A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy given surface tension of mercury O.465J/m^2....

Answer 1

The increase in surface energy is 23.376

Explanation:

Radius "r" v= 2 mm = 2 x 10^-3 m

4/3 π (R)^3 = 8 x 4/3 πr^3

(R)^3 = 8 r^3

(R)^3 / 8 = r^3

( R /2 )^3 = r^3

r = 1 mm = 1 x 10^-3 m

A1 = 4 πR^2

A1 = 4 π (2 x 10^-3)^2

A1 = 4π x 4 x 10^-6 = 16 π x 10^-6 m

A2 = 8 x 4 πr^2

A2 = 32 π (10^-3)^2 = 32 π x 10^-6

ΔA = A2 - A1 = [ 32π - 16π ] x 10^-6

ΔA = 16π x 10^-6

TΔA = 23.376  

Thus the increase in surface energy is 23.376

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