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. `(pi=3.142)`

Explanation

Given : Initial radius of soap bubble `(r_(1))=1 cm`
`=1xx10^(-2)m`
Final radius of soap bubble `(r_(2))`
`=2cm=2xx10^(-2)m`
Surface tension (T)
`=30 ` dyne/cm
`=(30xx10^(-5)N)/(10^(-2)m)=30xx10^(-3)N//m`
We know that,
Work done = T.dA
`=30xx 10^(-3)xx4pi xx [2^(2)-1^(2)]xx10^(-4)xx2`
`=2.262 xx 10^(-4)` joule