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(C) Let the radius of the bubble at the bottom be $r_1$ and at the surface be $r_2$. Given $r_2 = 2r_1$.Since the volume of a spherical bubble $V = \f

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$7H$

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(C) Let the radius of the bubble at the bottom be $r_1$ and at the surface be $r_2$. Given $r_2 = 2r_1$.Since the volume of a spherical bubble $V = \frac{4}{3}\pi r^3$,we have $V \propto r^3$.Thus,$\frac{V_2}{V_1} = \left(\frac{r_2}{r_1}\right)^3 = (2)^3 = 8$,which implies $V_2 = 8V_1$.According to Boyle's law,for a constant temperature,$P_1V_1 = P_2V_2$,where $P_1$ is the pressure at the bottom and $P_2$ is the atmospheric pressure at the surface.Given $P_2 = P_H$ (pressure due to water column of height $H$),so $P_2 = \rho g H$.The pressure at depth $d$ is $P_1 = P_2 + \rho g d = \rho g H + \rho g d$.Substituting these into Boyle's law: $(\rho g H + \rho g d) V_1 = (\rho g H)(8V_1)$.Dividing both sides by $\rho g V_1$,we get $H + d = 8H$.Therefore,$d = 7H$.

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

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