The pressure at half the depth of a lake is e | vedclass

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Question

(A) The pressure at a depth '$y$' in a liquid is given by $P = P_0 + \rho gy$.At half the depth, $y = h/2$, so the pressure is $P_{h/2} = P_0 + \rho g

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$\frac{2P_0}{\rho g}$

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(A) The pressure at a depth '$y$' in a liquid is given by $P = P_0 + \rho gy$.At half the depth, $y = h/2$, so the pressure is $P_{h/2} = P_0 + \rho g(h/2) = P_0 + \frac{\rho gh}{2}$.At the bottom of the lake, $y = h$, so the pressure is $P_h = P_0 + \rho gh$.According to the problem, $P_{h/2} = \frac{2}{3} P_h$.Substituting the expressions: $P_0 + \frac{\rho gh}{2} = \frac{2}{3} (P_0 + \rho gh)$.Multiply both sides by $6$ to clear the fractions: $6P_0 + 3\rho gh = 4P_0 + 4\rho gh$.Rearranging the terms: $6P_0 - 4P_0 = 4\rho gh - 3\rho gh$.$2P_0 = \rho gh$.Therefore, the depth $h = \frac{2P_0}{\rho g}$.

The pressure at half the depth of a lake is equal to two-third pressure at the bottom of the lake. So the depth '$h$' of the lake is ($\rho$ = density of water in the lake, $g$ = acceleration due to gravity, $P_0$ = atmospheric pressure).

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