1. Given setup

• Two identical tanks, each filled with water up to height $h$.
• Each has a small hole near the bottom.
• Tank 1 discharges directly to air.
• Tank 2 has a pipe attached to the hole, of same radius as the hole, ending at ground level.
• Building height is $H$ and given: $H=\frac{16}{9}h$

We need the ratio $\dfrac{t_1}{t_2}$.

2. Time to empty a tank through a small orifice

Let the cross-sectional area of each tank be $A$ and the area of the hole/pipe be $a$.

If the water level above the hole at any time is $y$, then: $A\,\frac{dy}{dt}=-a v$ where $v$ is the speed of efflux.

So, for a given $v(y)$, $dt=-\frac{A}{a}\frac{dy}{v(y)}$ Hence, $t=\frac{A}{a}\int_0^h \frac{dy}{v(y)}$

Thus we only need the effective efflux speed in the two cases.

3. Tank 1: direct discharge

For tank 1, water comes out directly to atmosphere. By Torricelli’s law, $v_1=\sqrt{2gy}$

Therefore, $t_1=\frac{A}{a}\int_0^h \frac{dy}{\sqrt{2gy}}$

Compute the integral: $t_1=\frac{A}{a\sqrt{2g}}\int_0^h y^{-1/2}dy$ $t_1=\frac{A}{a\sqrt{2g}}\left[2\sqrt{y}\right]_0^h$ $t_1=\frac{2A\sqrt{h}}{a\sqrt{2g}}$

4. Tank 2: pipe attached, outlet at ground

Let at some instant the water surface be at height $y$ above the hole inside the tank.

Since the pipe outlet is at ground level, which is $H$ below the hole, apply Bernoulli between the free surface and the pipe outlet:

• Pressure at free surface = atmospheric
• Pressure at outlet = atmospheric
• Height drop from free surface to outlet = $y+H$

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

Hence, $t_2=\frac{A}{a}\int_0^h \frac{dy}{\sqrt{2g(y+H)}}$

Compute: $t_2=\frac{A}{a\sqrt{2g}}\int_0^h \frac{dy}{\sqrt{y+H}}$ $t_2=\frac{A}{a\sqrt{2g}}\left[2\sqrt{y+H}\right]_0^h$ $t_2=\frac{2A}{a\sqrt{2g}}\left(\sqrt{H+h}-\sqrt{H}\right)$

5. Take the ratio

Now, $\frac{t_1}{t_2}=\frac{\sqrt{h}}{\sqrt{H+h}-\sqrt{H}}$

Rationalize: $\frac{t_1}{t_2}=\frac{\sqrt{h}(\sqrt{H+h}+\sqrt{H})}{(H+h)-H}$ $\frac{t_1}{t_2}=\frac{\sqrt{H+h}+\sqrt{H}}{\sqrt{h}}$

Now use $H=\frac{16}{9}h$

Then $H+h=\frac{16}{9}h+h=\frac{25}{9}h$

So, $\sqrt{H}=\frac{4}{3}\sqrt{h}, \qquad \sqrt{H+h}=\frac{5}{3}\sqrt{h}$

Therefore, $\frac{t_1}{t_2}=\frac{\frac{5}{3}\sqrt{h}+\frac{4}{3}\sqrt{h}}{\sqrt{h}}=\frac{9}{3}=3$

6. Final answer

$\boxed{\frac{t_1}{t_2}=3}$