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A soap bubble of radius $r$ is blown up to form a bubble of radius $2 r$ under isothermal conditions. If $T$ is the surface tension of soap solution, the energy spent in the blowing
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Verified Answer
The correct answer is:
$6 \pi T r^2$
Initially area of soap bubble
$A_1=4 \pi r^2$
Under isothermal condition radius becomes $2 r$,
Then, $\quad$ area $A_2=4 \pi(2 r)^2$
$\begin{aligned}
& =4 \pi \cdot 4 r^2 \\
& =16 \pi r^2
\end{aligned}$
Increase in surface area
$\begin{aligned}
\Delta A & =2\left(A_2-A_1\right) \\
& =2\left(16 \pi r^2-4 \pi r^2\right)=24 \pi r^2
\end{aligned}$
Energy spent
$\begin{aligned}
W & =T \times \Delta A \\
& =T \cdot 24 \pi r^2 \\
or \quad W=24 \pi \operatorname{Tr}^2 \mathrm{~J}
\end{aligned}$
$A_1=4 \pi r^2$
Under isothermal condition radius becomes $2 r$,
Then, $\quad$ area $A_2=4 \pi(2 r)^2$
$\begin{aligned}
& =4 \pi \cdot 4 r^2 \\
& =16 \pi r^2
\end{aligned}$
Increase in surface area
$\begin{aligned}
\Delta A & =2\left(A_2-A_1\right) \\
& =2\left(16 \pi r^2-4 \pi r^2\right)=24 \pi r^2
\end{aligned}$
Energy spent
$\begin{aligned}
W & =T \times \Delta A \\
& =T \cdot 24 \pi r^2 \\
or \quad W=24 \pi \operatorname{Tr}^2 \mathrm{~J}
\end{aligned}$
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