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The work done in blowing a soap bubble of surface tension $0.06 \mathrm{Nm}^{-1}$ from 2 cm radius to 5 cm radius is
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Verified Answer
The correct answer is:
$0.003168 J$
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
$=0.06 \times 168 \pi \times 10^{-4}=0.003168 \mathrm{~J}$
$\mathrm{r}_1=2 \mathrm{~cm}=0.02 \mathrm{~m}, \mathrm{r}_2=5 \mathrm{~cm}=0.05 \mathrm{~m}$
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
$=0.06 \times 168 \pi \times 10^{-4}=0.003168 \mathrm{~J}$
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