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(C) Let the depth of the lake be $h$ and the radius of the bubble at the bottom be $R$. At the bottom, the total pressure is $P_{1} = P_{atm} + \rho g

Vedclass · Accepted Answer

$7H$

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(C) Let the depth of the lake be $h$ and the radius of the bubble at the bottom be $R$. At the bottom, the total pressure is $P_{1} = P_{atm} + \rho gh = \rho gH + \rho gh = \rho g(H + h)$. The volume at the bottom is $V_{1} = \frac{4}{3} \pi R^{3}$. At the surface, the pressure is $P_{2} = P_{atm} = \rho gH$. The radius of the bubble becomes $2R$, so the volume at the surface is $V_{2} = \frac{4}{3} \pi (2R)^{3} = 8 \cdot \frac{4}{3} \pi R^{3} = 8V_{1}$. Assuming the temperature remains constant, we apply Boyle's Law: $P_{1}V_{1} = P_{2}V_{2}$. $\rho g(H + h) V_{1} = \rho gH (8V_{1})$. $H + h = 8H$. $h = 7H$. Thus, the depth of the lake is $7H$.

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 a column of water of height $H$, then the depth of the lake is:

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