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(C) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_0 + h\rho g$,where $P_0$ is the atmospheric pressure and $\rh

Vedclass · Accepted Answer

$7H$

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(C) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_0 + hho g$,where $P_0$ is the atmospheric pressure and $ho$ is the density of water.Given that atmospheric pressure $P_0 = Hho g$.So,$P_1 = Hho g + hho g = (H + h)ho g$.The volume of the bubble at the bottom is $V_1 = \frac{4}{3}\pi r^3$.At the surface,the pressure is $P_2 = P_0 = Hho g$ and the radius becomes $2r$,so the volume is $V_2 = \frac{4}{3}\pi (2r)^3 = 8 	imes \frac{4}{3}\pi r^3$.Assuming the temperature remains constant,we use Boyle's Law: $P_1V_1 = P_2V_2$.$(H + h)ho g 	imes \frac{4}{3}\pi r^3 = Hho g 	imes 8 	imes \frac{4}{3}\pi r^3$.$(H + h) = 8H$.$h = 7H$.

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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