Consider a process in which a soap bubble is blown from 2 cm radius to 5 cm radius. If the surface tension of saop bubble is `0.06 N m^(-1)`, then what is the work done in this process?

Options

A. 3.1 mJ
B. 1.25 mJ
C. 2.51 mJ
D. 4.55 mJ

Answer

A

Explanation

Here, `S= 0.06 "Nm"^(-1), r_(1) = 2 cm = 2 xx 10^(-2) m, r^2 = 5cm = 5 xx 10^(-2)m`
Since bubble has two surfaces, initial surface area of the bubble
`= 2 xx 4pir_(1)^(2) = 2 xx 4pi xx (2 xx 10^(-2))^(2) `
`= 32pi xx 10^(-4) m^2`
Final surface area of the bubble
`= 2 xx 4pir_2^2 = 2 xx 4pi (5 xx 10^(-2))^(2)`
` = 200 pi xx10^(-4)m^2`
Increase is surface area
`= 200 pi xx10^(-4)`
`=168pi xx 10^(-4)m^2`
`:.` Work done = surface tension `xx` increase is surface area
`= 0.06 xx 168pi xx 10^(-4)`
` = 3.12 xx 10^(-3)J = 3.12 mJ`