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 from the tanks 1 and 2 through the holes, the times taken to empty the tanks are $t_{1}$ and $t_{2},$ respectively. If $H = (\frac{16}{9}) h,$ then the ratio $t_{1}/t_{2}$ is _____.

Step 1: Identify the given parameters.

The building has height $H$, each tank is filled with water to height $h$, and the relation between $H$ and $h$ is given.

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

Step 2: Determine the velocity of efflux for both tanks.

Let $y$ be the instantaneous height of water in the tank. By Torricelli's Law, the efflux velocities $v_1$ and $v_2$ are determined by the total head of water above the exit point.

$$v_1 = \sqrt{2gy}$$ $$v_2 = \sqrt{2g(H+y)}$$

Step 3: Formulate the differential equation for the emptying time.

By the equation of continuity, the rate of volume flow out of the hole equals the rate of decrease of water volume in the tank.

$$a v = -A \frac{dy}{dt}$$ $$dt = -\frac{A}{a} \frac{dy}{v}$$

Step 4: Calculate the time $t_1$ to empty Tank 1.

Integrate the differential equation from $y = h$ to $y = 0$ using the velocity $v_1$.

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

Step 5: Calculate the time $t_2$ to empty Tank 2.

Integrate the differential equation from $y = h$ to $y = 0$ using the velocity $v_2$.

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

Step 6: Express the ratio of the times.

Divide $t_1$ by $t_2$ to eliminate the common constants.

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

Step 7: Substitute the value of $H$ to find the ratio.

Substitute $H = \frac{16}{9}h$ into the ratio expression.

$$\sqrt{H} = \frac{4}{3}\sqrt{h}$$ $$\sqrt{H+h} = \sqrt{\frac{16}{9}h + h} = \frac{5}{3}\sqrt{h}$$ $$\frac{t_1}{t_2} = \frac{\sqrt{h}}{\frac{5}{3}\sqrt{h} - \frac{4}{3}\sqrt{h}}$$ $$\frac{t_1}{t_2} = 3$$

Answer: 3