Two large, identical water tanks, 1 and 2, kept on the top of a building of height $$H$$, are filled with water up to height $$h$$ in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2, and the pipe ends at the ground level. When the water flows the tanks 1 and 2 through the holes, the times taken to empty the tanks are $$t_1$$ and $$t_2$$, respectively. If $$H = \left(\frac{16}{9}\right)h$$, then the ratio $$t_1/t_2$$ is ______.

Let the cross-sectional area of each tank be $$A$$ (large) and the cross-sectional area of the small hole (and of the pipe) be $$a$$.

At any instant let $$y$$ be the height of water column above the hole inside the tank (so $$y$$ decreases from $$h$$ to $$0$$ as the tank empties).

Case 1: Tank 1 (hole open to atmosphere at height $$H$$ above ground)
The exit point of water is the hole itself. The instantaneous speed of efflux is given by Torricelli’s theorem: $$v_1 = \sqrt{2 g y}$$

Volume flow rate: $$a v_1$$. Fall of water level in the tank: $$A \, dy = -\,a v_1\, dt$$.

Hence $$A \frac{dy}{dt} = -\,a \sqrt{2 g y}$$ $$\Rightarrow dt = -\frac{A}{a\sqrt{2 g}} \frac{dy}{\sqrt{y}}$$

Total time $$t_1$$: integrate $$y$$ from $$h$$ to $$0$$, changing sign to keep time positive. $$t_1 = \frac{A}{a\sqrt{2 g}}\int_{0}^{h} \frac{dy}{\sqrt{y}} = \frac{A}{a\sqrt{2 g}}\left[\,2\sqrt{y}\,\right]_{0}^{h} = \frac{2A}{a\sqrt{2 g}}\sqrt{h}$$

Case 2: Tank 2 (hole connected to a vertical pipe ending at ground level)
The exit point is now at ground (height $$0$$). The free surface of water is at height $$H+y$$ above ground. Applying Bernoulli between the free surface and the ground exit (both open to atmosphere): $$v_2 = \sqrt{2 g\,(H+y)}$$

Again, $$A \, dy = -\,a v_2\, dt$$ gives $$dt = -\frac{A}{a\sqrt{2 g}} \frac{dy}{\sqrt{H+y}}$$

Integrate from $$y=h$$ to $$0$$: $$t_2 = \frac{A}{a\sqrt{2 g}}\int_{0}^{h} \frac{dy}{\sqrt{H+y}}$$

Put $$s = H+y \; \Rightarrow \; ds = dy,$$ and when $$y=0,\,s=H$$; when $$y=h,\,s=H+h$$. $$t_2 = \frac{A}{a\sqrt{2 g}}\int_{H}^{H+h} \frac{ds}{\sqrt{s}} = \frac{A}{a\sqrt{2 g}}\left[\,2\sqrt{s}\,\right]_{H}^{H+h} = \frac{2A}{a\sqrt{2 g}}\left(\sqrt{H+h}-\sqrt{H}\right)$$

Ratio of times
$$\frac{t_1}{t_2} = \frac{\sqrt{h}}{\sqrt{H+h}-\sqrt{H}}$$

Given $$H = \frac{16}{9}\,h$$.

Compute the square roots:
$$\sqrt{H} = \sqrt{\tfrac{16}{9}h} = \tfrac{4}{3}\sqrt{h},$$
$$\sqrt{H+h} = \sqrt{\left(\tfrac{16}{9}+1\right)h} = \sqrt{\tfrac{25}{9}h} = \tfrac{5}{3}\sqrt{h}.$$

Denominator:
$$\sqrt{H+h}-\sqrt{H} = \left(\tfrac{5}{3}-\tfrac{4}{3}\right)\sqrt{h} = \tfrac{1}{3}\sqrt{h}.$$

Therefore $$\frac{t_1}{t_2} = \frac{\sqrt{h}}{\tfrac{1}{3}\sqrt{h}} = 3.$$

Hence the required ratio is 3.

Final Answer: 3