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(B) Let the total depth of the lake be $h$ and atmospheric pressure be $P_0$.Pressure at half the depth $(h/2)$ is given by: $P_{half} = P_0 + \frac{h

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Let the total depth of the lake be $h$ and atmospheric pressure be $P_0$.Pressure at half the depth $(h/2)$ is given by: $P_{half} = P_0 + \frac{h}{2} ho g$.Pressure at the bottom of the lake is given by: $P_{bottom} = P_0 + h ho g$.According to the given condition: $P_{half} = \frac{2}{3} P_{bottom}$.Substituting the expressions: $P_0 + \frac{h}{2} ho g = \frac{2}{3} (P_0 + h ho g)$.Multiplying both sides by $3$: $3P_0 + \frac{3h}{2} ho g = 2P_0 + 2h ho g$.Rearranging the terms: $3P_0 - 2P_0 = 2h ho g - 1.5h ho g$.$P_0 = 0.5 h ho g$.$h = \frac{2P_0}{ho g}$.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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