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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 $\mathrm{T}$ is the surface tension of soap solution, the energy spent in the blowing
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Verified Answer
The correct answer is:
$24 \pi \operatorname{Tr}^{2}$
Initially area of soap bubble, $A_{1}=4 \pi r^{2}$ Under isothermal condition radius becomes $2 r$.
$\therefore$ Area $A_{2}=4 \pi(2 r)^{2}=16 \pi r^{2}$
Increase in surface area
$$
\Delta A=2\left(A_{2}-A_{1}\right)=2\left(16 \pi r^{2}-4 \pi r^{2}\right)=24 \pi r^{2}
$$
Energy spent,
$$
W=T \times \Delta A=T \cdot 24 \pi r^{2}=24 \pi T r^{2} \mathrm{~J}
$$
$\therefore$ Area $A_{2}=4 \pi(2 r)^{2}=16 \pi r^{2}$
Increase in surface area
$$
\Delta A=2\left(A_{2}-A_{1}\right)=2\left(16 \pi r^{2}-4 \pi r^{2}\right)=24 \pi r^{2}
$$
Energy spent,
$$
W=T \times \Delta A=T \cdot 24 \pi r^{2}=24 \pi T r^{2} \mathrm{~J}
$$
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