Ratio of Time Taken to Empty Water Tanks with Pipe Extensions

For Tank 1: Let $A$ be the cross-sectional area of the tank and $a$ be the cross-sectional area of the hole ($a \ll A$). Let $y$ denote the height of the water level in the tank at time $t$, measured from the bottom of the tank.

Applying Bernoulli's principle between the top surface of the water in Tank 1 and the exit hole: $P_0 + \rho g y + \frac{1}{2}\rho \left(-\frac{dy}{dt}\right)^2 = P_0 + 0 + \frac{1}{2}\rho v_1^2$

Since $a \ll A$, the speed of the water surface is negligible compared to $v_1$, giving Torricelli's law: $v_1 = \sqrt{2gy}$

Using the continuity equation: $-A \frac{dy}{dt} = a \sqrt{2gy}$

Separating variables and integrating from $y = h$ to $y = 0$: $\int_{0}^{h} \frac{dy}{\sqrt{y}} = \frac{a}{A}\sqrt{2g} \int_{0}^{t_1} dt$

$2\sqrt{h} = \frac{a}{A}\sqrt{2g} \, t_1 \implies t_1 = \frac{A}{a}\sqrt{\frac{2h}{g}}$

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For Tank 2: Tank 2 is placed at the top of a building of height $H$. A pipe of cross-sectional area $a$ is connected to the bottom of Tank 2 and extends down to the ground.

Applying Bernoulli's principle between the top surface of the water in Tank 2 (at height $y + H$ relative to the ground) and the exit at the ground level (at height $0$): $P_0 + \rho g (y + H) = P_0 + \frac{1}{2}\rho v_2^2$

$v_2 = \sqrt{2g(y + H)}$

Using the continuity equation for Tank 2: $-A \frac{dy}{dt} = a \sqrt{2g(y + H)}$

Separating variables and integrating from $y = h$ to $y = 0$: $\int_{0}^{h} \frac{dy}{\sqrt{y + H}} = \frac{a}{A}\sqrt{2g} \int_{0}^{t_2} dt$

$\left[ 2\sqrt{y + H} \right]_0^h = \frac{a}{A}\sqrt{2g} \, t_2$

$2\left(\sqrt{h + H} - \sqrt{H}\right) = \frac{a}{A}\sqrt{2g} \, t_2 \implies t_2 = \frac{A}{a}\sqrt{\frac{2}{g}}\left(\sqrt{h + H} - \sqrt{H}\right)$

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Ratio $t_1 / t_2$: $\frac{t_1}{t_2} = \frac{\sqrt{h}}{\sqrt{h + H} - \sqrt{H}}$

Given $H = \frac{16}{9}h$: $\sqrt{H} = \sqrt{\frac{16}{9}h} = \frac{4}{3}\sqrt{h}$

$\sqrt{h + H} = \sqrt{h + \frac{16}{9}h} = \sqrt{\frac{25}{9}h} = \frac{5}{3}\sqrt{h}$

Substituting these values into the ratio: $\frac{t_1}{t_2} = \frac{\sqrt{h}}{\frac{5}{3}\sqrt{h} - \frac{4}{3}\sqrt{h}} = \frac{\sqrt{h}}{\frac{1}{3}\sqrt{h}} = 3$