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(B) Let the total depth of the lake be $h$. The pressure at the bottom of the lake is given by $P_{bottom} = P_0 + h\rho g$.The pressure at half the d

Vedclass · Accepted Answer

$20$

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(B) Let the total depth of the lake be $h$. The pressure at the bottom of the lake is given by $P_{bottom} = P_0 + hho g$.The pressure at half the depth $(h/2)$ is given by $P_{half} = P_0 + (h/2)ho g$.According to the problem,$P_{half} = \frac{2}{3} P_{bottom}$.Substituting the expressions: $P_0 + \frac{h}{2}ho g = \frac{2}{3}(P_0 + hho g)$.Multiplying by $3$: $3P_0 + \frac{3}{2}hho g = 2P_0 + 2hho g$.Rearranging the terms: $P_0 = 2hho g - 1.5hho g = 0.5hho g$.Thus,$h = \frac{2P_0}{ho g}$.Substituting the values $P_0 = 10^5 \ Pa$,$ho = 10^3 \ kg/m^3$,and $g = 10 \ m/s^2$:$h = \frac{2 	imes 10^5}{10^3 	imes 10} = \frac{2 	imes 10^5}{10^4} = 20 \ m$.

If the pressure at half the depth of a lake is equal to $2/3$ of the pressure at the bottom of the lake,then what is the depth of the lake in $m$? (Assume atmospheric pressure $P_0 = 10^5 \ Pa$,density of water $\rho = 10^3 \ kg/m^3$,and $g = 10 \ m/s^2$)

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