Physics - Fluid Question with Solution | TestHub

Let the depth of the lake be $h$.

Pressure at half the depth of the lake is given by $P_1=P_0+\frac{\rho gh}{2}$.

Pressure at the bottom of the lake is given by $P_2=P_0+\rho gh$.

According to the given condition, $P_1=\frac{2}{3}{P}_2$.

Substituting the expressions for $P_1$ and $P_2$:

$P_0+\frac{\rho gh}{2}=\frac{2}{3}(P_0+\rho gh)$

$P_0-\frac{2}{3}{P}_0=\frac{2}{3}\rho gh-\frac{1}{2}\rho gh$

$\frac{P_0}{3}=\frac{\rho gh}{6}$

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

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