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(D) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_{atm} + \rho gh$,where $P_{atm} = \rho g(10)$.Thus,$P_1 = \rh

Vedclass · Accepted Answer

$9.5$

Vedclass · Answer

(D) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_{atm} + \rho gh$,where $P_{atm} = \rho g(10)$.Thus,$P_1 = \rho g(10 + h)$.The pressure at the surface is $P_2 = P_{atm} = \rho g(10)$.Using Boyle's Law,$P_1 V_1 = P_2 V_2$,assuming temperature is constant.$\rho g(10 + h) \cdot \frac{4}{3} \pi r^3 = \rho g(10) \cdot \frac{4}{3} \pi \left( \frac{5r}{4} \right)^3$.$(10 + h) = 10 \cdot \frac{125}{64}$.$10 + h = \frac{1250}{64} = 19.53$.$h = 19.53 - 10 = 9.53 \ m$.Rounding to the nearest value,the depth is $9.5 \ m$.

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