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The excess pressure inside an air bubble of radius `r` just below the surface of water is `P_(1)`. The excess pressure inside a drop of the same radius just outside the surface is `P_(2)`. If `T` is surface tension thenA. `P_(1)=2P_(2)`B. `P_(1)=P_(2)`C. `P_(2)=2P_(1)`D. `P_(2)=0,P_(1)ne0` |
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Answer» Correct Answer - B Excess pressure inside a bubble just below the surface of water `P_(1)=(2T)/(r )` and excess pressure inside a drop `P_(2) = (2T)/(r ) " "therefore P_(1)=P_(2)` |
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