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A soap bubble of radius $r$ is blown up to form a bubble of radius $3 r$ under isothermal conditions. If $\sigma$ is the surface tension of soap solution, then the energy spent in blowing is
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The correct answer is:
$64 \pi \sigma r^{2}$
Surface area of bubble of radius, $r=4 \pi r^{2}$
Surface area of bubble of radius, $3 r=4 \pi(3 r)^{2}=36 \pi r^{2}$
Therefore, increase in surface area
$$
=36 \pi r^{2}-4 \pi r^{2}=32 \pi r^{2}
$$
Since, a bubble has two surfaces, the total increase in surface area $=2 \times 32 \pi r^{2}=64 \pi r^{2}$
$\therefore$ Energy spent $=$ Work done $=$ Surface tension $\times$ Total increase in surface area $=64 \pi \sigma r^{2}$
Surface area of bubble of radius, $3 r=4 \pi(3 r)^{2}=36 \pi r^{2}$
Therefore, increase in surface area
$$
=36 \pi r^{2}-4 \pi r^{2}=32 \pi r^{2}
$$
Since, a bubble has two surfaces, the total increase in surface area $=2 \times 32 \pi r^{2}=64 \pi r^{2}$
$\therefore$ Energy spent $=$ Work done $=$ Surface tension $\times$ Total increase in surface area $=64 \pi \sigma r^{2}$
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