Calculate the amount of work done in increasing the diameter of a soap bubble from 2cm to 5cm (Given, surface tension of the soap solution `=3 xx 10^(-2) Nm^(-1))`

Explanation

Given `R_(1)=2 xx 10^(-2)m, R_(2)=5 xx 10^(-2)m`
`T=3 xx 10^(-2)Nm^(-1)`
Increase in S.A. `=4piR_(2)^(2)-4piR_(1)^(2)=4 xx 3.142 (10^(-2))^(2) (25-4)=2.639 xx 10^(-2) m^(2)`
Work done =(2T) (increase in SA) `=2 xx 3 xx 10^(-2) (2.639 xx 10^(-2))=1.5834 xx 10^(-3)J`