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When a large bubble rises from the bottom of a lake to the surface. Its radius doubles. If atmospheric pressure is equal to that of column of water height H , then the depth of lake is

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Detailed Solution

$P_{1} V_{1} = P_{2} V_{2} \Rightarrow (P_{0} + hρg) \times \frac{4}{3} {πr}^{3} = P_{0} \frac{4}{3} π {(2 r)}^{3}$

Where, h=depth of lake

$(H + h) = (H) (8)$

$\Rightarrow h = 7 H$

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