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 t1 and t2, respectively. If H = ( 16 9 ) ℎ, then the ratio t1/t2 is _____.

 To determine the ratio 

$\frac{t_1}{t_2}$ for the times taken to empty two identical water tanks with different configurations, we analyze the problem using Torricelli's Law and differential equations.

Key Steps:

1. Tank 1 (Simple Hole):

   • The velocity of efflux is $\sqrt{2gy}$, where $y$ is the height of the water.

   • The differential equation for the water level $y$ is:

     $$\frac{dy}{dt}=-\frac{A_{\text{hole}}}{A_{\text{tank}}}\sqrt{2gy}$$

   • Solving this, the time to empty $t_1$ is:

     $$t_1=\frac{A_{\text{tank}}}{A_{\text{hole}}}\sqrt{\frac{2h}{g}}$$

2. Tank 2 (Pipe Connected to Ground):

   • The effective head is $H+y$, leading to a velocity of $\sqrt{2g(H+y)}$.

   • The differential equation for the water level $y$ is:

     $$\frac{dy}{dt}=-\frac{A_{\text{hole}}}{A_{\text{tank}}}\sqrt{2g(H+y)}$$

   • Solving this, the time to empty $t_2$ is:

     $$t_2=\frac{A_{\text{tank}}}{A_{\text{hole}}}\frac{2(\sqrt{H+h}-\sqrt{H})}{\sqrt{2g}}$$

3. Ratio Calculation:

   • Substituting $H=\frac{16}{9}h$:

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

   • The ratio $\frac{t_1}{t_2}$ simplifies to:

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

Final Answer:

The ratio $\frac{t_1}{t_2}$ is {3}.