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A soap bubble is blown to a radius of 3 cm. if it is to be further blown to a radius of 4 cm what is the work done? (surface tension of soap solution `=3.06xx10^(-2)Nm^(-1))`

Answer» Initial radius of soap bubble `R_(1)=3cm=3xx10^(-1)m`
Final radius of soap bubble `R_(2)=4cm=4xx10^(-2)m`
Wirk done in blowing soap buble from radius `R_(1)` is `R_(2)` is `impliesW=8pi(R_(2)^(2)-R_(1)^(2))T`
`=8xx(22)/(7)xx3.06xx10^(-2)(16-9)xx10^(-4)`
`=176xx3.06xx10^(-6)J=539.6xx10^(-6)J`


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