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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{\sigma} is the surface tension of soap solution, the energy spent in doing so is

Option: 1

\mathrm{3 \pi \sigma r^2}


Option: 2

\mathrm{6 \pi \sigma r^2}


Option: 3

\mathrm{12 \pi \sigma r^2}


Option: 4

\mathrm{24 \pi \sigma r^2}


Answers (1)

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Surface area of bubble of radius \mathrm{r=4 \pi r^2}. Surface area of bubble of radius \mathrm{2 r=4 \pi(2 r)^2=16 \pi r^2}. Therefore, increase in surface area \mathrm{=16 \pi r^2-4 \pi r^2 =12 \pi r^2.} Since a bubble has two surfaces, the total increase in surface area \mathrm{=24 \pi r^2.}

\mathrm{ \therefore \quad \text { Energy spent }=\text { work done }=24 \pi \sigma r^2 }

Posted by

Pankaj Sanodiya

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