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(C) Let the depth of the lake be $h$. The atmospheric pressure $P_0$ is equivalent to $10 \ m$ of water column,so $P_0 = 10 \rho g$.At the bottom of t

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$9.53$

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(C) Let the depth of the lake be $h$. The atmospheric pressure $P_0$ is equivalent to $10 \ m$ of water column,so $P_0 = 10 \rho g$.At the bottom of the lake,the pressure $P_1$ is the sum of atmospheric pressure and the pressure due to the water column of depth $h$:$P_1 = P_0 + h \rho g = 10 \rho g + h \rho g = \rho g(10 + h)$.The volume of the bubble at the bottom is $V_1 = \frac{4}{3} \pi r^3$.At the surface,the pressure $P_2$ is equal to the atmospheric pressure:$P_2 = P_0 = 10 \rho g$.The volume of the bubble at the surface is $V_2 = \frac{4}{3} \pi (\frac{5r}{4})^3$.Since the temperature is constant,we apply Boyle's Law $(P_1 V_1 = P_2 V_2)$:$\rho g(10 + h) \cdot \frac{4}{3} \pi r^3 = 10 \rho g \cdot \frac{4}{3} \pi (\frac{5r}{4})^3$.$(10 + h) = 10 \cdot \frac{125}{64}$.$10 + h = \frac{1250}{64} = 19.53125$.$h = 19.53125 - 10 = 9.53125 \ m$.Thus,the depth of the lake is approximately $9.53 \ m$.

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