The work done in blowing a soap bubble from a radius $10 cm$ to $20 cm$ if the surface tension of soap solution is $25 x {10}^{-3}N/m$. Then compare it with a liquid drop for same radii.

For Liquid Drop: In case of liquid drop there is only one free surface, so extension in area will be half that of the soap bubble

$W_1=T\Delta A=T8\pi \left(R^2-r^2\right)=25\times {10}^{-3}\times 8\pi \left(0.2^2-0.1^2\right)=6\mathrm{\pi }\times {10}^{-3}\mathrm{J}$

$W_2=\frac{W_1}{2}=3\mathrm{\pi } \mathrm{X} {10}^{-3} \mathrm{J}$