Physics - Fluid Question with Solution | TestHub

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 $\begin{array}{rl}& =2\times 4\pi r_1^2=2\times 4\pi (0.02{)}^2\\ & =32\pi \times 10^{-4}{\mathrm{m}}^2\end{array}$ Final surface area of the bubble $\begin{array}{rl}& =2\times 4\pi r_2^2=2\times 4\pi (0.05{)}^2\\ & =200\pi \times 10^{-4}{\mathrm{m}}^2\end{array}$ Final surface area of the bubble $\begin{array}{rl}& =2\times 4\pi {\mathrm{r}}_2^2=2\times 4\pi (0.05{)}^2\\ & =200\pi \times 10^{-4}{\mathrm{m}}^2\end{array}$ Increase in surface area $\begin{array}{rl}& =200\pi \times 10^{-4}-32\pi \times 10^{-4}\\ & =168\pi \times 10^{-4}{\mathrm{m}}^2\end{array}$ $\therefore$ Work done $=\mathrm{S}\times$ Increase in surface area $=0.06\times 168\pi \times 10^{-4}=0.003168\mathrm{J}$