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Question
Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2 cm. The surface tension of soap solution is 30 dyne/cm. (Π = 3.142).
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Solution
Whenever a liquid film is expanded, the work done gets stored as energy in the film. Hence the increase in energy of the film is equal to the work done.
Work done = (increase in the surface ares of bubble) x Surface Tension
Given : r1 = 1 cm, r2 = 2cm, T = 30 dynes/cm
To find : Work (W)
Formula : W = TΔA
Calculation : `A_1 = 4pi r_1^2 = 4pi xx 1^2 = 4pi cm^2`
`A_2 = 4pi r_2^2 = 4pi xx 2^2 = 16 pi cm^2`
`DeltaA = A_2 - A_1 = (16pi - 4pi) = 12 pi cm^2`
Since soap bubble has two surfaces
From formula,
W = 2T x ΔA
= 2 x 30 x 12π
= 2 x 30 x 12 x 3.14
W = `2.26 xx 10^3` erg
`therefore` The work done in increaseing the radius of the soap bubble is 2.26 x 103 erg
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