JEE Main · 2011 · 10 Apr 2011 · Offline · Fluid Mechanics

A soap bubble has two surfaces, an inner one and an outer one, so the area created against surface tension is twice the area of a single sphere. The work done is the surface tension times the increase in total area, $W=T\Delta A=T\times 2\times 4\pi \left(r_2^2-r_1^2\right)$. With $T=0.03\mathrm{N}{\mathrm{m}}^{-1}$, $r_1=0.03\mathrm{m}$ and $r_2=0.05\mathrm{m}$, the change in squared radii is $0.05^2-0.03^2=0.0016{\mathrm{m}}^2$, so $W=0.03\times 8\pi \times 0.0016=3.84\times 10^{-4}\pi \mathrm{J}$, which is about $0.4\pi \mathrm{mJ}$.