The work done in blowing a soap bubble of surface tension 0.06 Nm^...

Here, $S=0.06 \mathrm{Nm}^{-1}$, $\mathrm{r}_1=2 \mathrm{~cm}=0.02 \mathrm{~m}, \mathrm{r}_2=5 \mathrm{~cm}=0.05 \mathrm{~m}$ Since, bubble has two surfaces, initial surface area of the bubble $$\begin{aligned} & =2 \times 4 \pi r_1^2=2 \times 4 \pi(0.02)^2 \\ & =32 \pi \times 10^{-4} \mathrm{~m}^2\end{aligned}$$ Final surface area of the bubble $$\begin{aligned} & =2 \times 4 \pi r_2^2=2 \times 4 \pi(0.05)^2 \\ & =200 \pi \times 10^{-4} \mathrm{~m}^2\end{aligned}$$ Final surface area of the bubble $$\begin{aligned} & =2 \times 4 \pi \mathrm{r}_2^2=2 \times 4 \pi(0.05)^2 \\ & =200 \pi \times 10^{-4} \mathrm{~m}^2\end{aligned}$$ Increase in surface area $$\begin{aligned} & =200 \pi \times 10^{-4}-32 \pi \times 10^{-4} \\ & =168 \pi \times 10^{-4} \mathrm{~m}^2\end{aligned}$$\therefore$Work done$=\mathrm{S} \times$Increase in surface area$=0.06 \times 168 \pi \times 10^{-4}=0.003168 \mathrm{~J}$