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  4. The work done in blowing a soap bubble of surface tension 0.06 N m-1 from 2 cm radius to 5 cm radius is

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Q. The work done in blowing a soap bubble of surface tension $ 0.06 \,N \,m^{-1} $ from $ 2\, cm $ radius to $ 5\, cm $ radius is

Mechanical Properties of Fluids

Solution:

Here, $ S=0.06\,N \,m^{-1}, r_{1}=2\,cm=2\times10^{-2}\,m $ ,
$ r_{2}=5\,cm=5\times10^{-2}\,m $
Since bubble has two surfaces, initial surface area of the bubble
$ =2\times4\pi r_{1}^{2}=2\times4\pi\times\left(2\times10^{-2}\right)^{2} $
$ =32\pi \times10^{-4}\, m^{2} $
Final surface area of the bubble
$ =2 \times4\pi r_{2}^{2}=2\times4\pi\left(5\times10^{-2}\right)^{2} $
$ =200\,\pi \times10^{-4}\, m^{2} $
Increase in surface area $ =200\,\pi \times10^{-4}-32\pi\times10^{-4} $
$ =168\,\pi \times10^{-4}\,m^{2} $
$ \therefore $ Work done = surface tension $ \times $ increase in surface area
$ =0.06\times168\pi\times10^{-4} $
$ =3.12\times10^{-3}\,J=3.12\,mJ $