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(C) Let the depth of the lake be $h$. The atmospheric pressure $P_0$ is equivalent to a water column of height $H = 10 \ m$. At the bottom of the lake

Vedclass · Accepted Answer

$9.53$

Vedclass · Answer

(C) Let the depth of the lake be $h$. The atmospheric pressure $P_0$ is equivalent to a water column of height $H = 10 \ m$. At the bottom of the lake, the pressure $P_1 = P_0 + h ho g = (H + h) ho g$. At the surface, the pressure $P_2 = P_0 = H ho g$. Since the temperature is constant, we use Boyle's Law: $P_1 V_1 = P_2 V_2$. Given $V_1 = \frac{4}{3} \pi r^3$ and $V_2 = \frac{4}{3} \pi (\frac{5r}{4})^3$. Substituting the values: $(H + h) ho g 	imes \frac{4}{3} \pi r^3 = H ho g 	imes \frac{4}{3} \pi (\frac{5r}{4})^3$. $(10 + h) = 10 	imes (\frac{5}{4})^3$. $10 + h = 10 	imes \frac{125}{64} = \frac{1250}{64} = 19.53125$. $h = 19.53125 - 10 = 9.53125 \ m \approx 9.53 \ m$.

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