Question

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 \(H = ( \frac{16}{ 9} ) ℎ\), then the ratio t1/t2 is ___

Answer 1

Correct Answer: 3

Using the Principle of Continuity and Torricelli’s Law 

The equation for continuity is given by:

\[ A_1 v_1 = A_2 v_2 \]

Using Torricelli’s theorem:

\[ v = \sqrt{2gh} \]

The time to empty the tank is:

\[ t = -\frac{A}{a} \sqrt{2g} \int_0^h \frac{dh}{\sqrt{h}} \]

Case I: Tank 1

The time taken for Tank 1 to empty is:

\[ t_1 = \frac{2A}{a} \sqrt{2g} \sqrt{h} \]

Case II: Tank 2 (Connected to Pipe)

For Tank 2, the effective height is \( H + h \), so:

\[ t_2 = \frac{2A}{a} \sqrt{2g} \left( \sqrt{H + h} - \sqrt{H} \right) \]

Finding the Ratio \( \frac{t_1}{t_2} \)

Dividing equation (1) by equation (2):

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

Given \( H = \frac{16}{9} h \), we find:

\[ \sqrt{H} = \frac{4}{3} \sqrt{h}, \quad \sqrt{H + h} = \frac{5}{3} \sqrt{h} \]

Substituting into the ratio:

\[ \frac{t_1}{t_2} = \frac{\sqrt{h}}{\frac{5}{3} \sqrt{h} - \frac{4}{3} \sqrt{h}} \]

Simplifying:

\[ \frac{t_1}{t_2} = \frac{\sqrt{h}}{\frac{\sqrt{h}}{3}} = 3 \]

Final Answer: \( \frac{t_1}{t_2} = 3 \)

Answer 2

Let's re-examine the problem carefully to match the correct answer.

Given:
- Two identical tanks, each filled to height \(h\).
- Building height \(H = \frac{16}{9} h\).
- Tank 1 has a hole near the bottom open to atmosphere.
- Tank 2 has a hole connected to a pipe ending at ground level (height \(H\) below the hole).
- Times to empty tanks are \(t_1\) and \(t_2\). Find \(\frac{t_1}{t_2}\).

Step 1: Effective pressure heads:
- For tank 1:
\[ h_1 = h \] - For tank 2:
Pressure head is sum of water column height plus vertical drop of pipe:
\[ h_2 = h + H = h + \frac{16}{9} h = \frac{25}{9} h \]

Step 2: Velocity of water outflow (Torricelli's law):
\[ v_1 = \sqrt{2 g h} \] \[ v_2 = \sqrt{2 g \frac{25}{9} h} = \frac{5}{3} \sqrt{2 g h} \]

Step 3: Flow rate \(Q = A v\) (same hole radius \(A\)):
\[ Q_1 = A v_1 = A \sqrt{2 g h} \] \[ Q_2 = A v_2 = A \frac{5}{3} \sqrt{2 g h} \]

Step 4: Time to empty tank, considering decreasing water height:
Since flow velocity depends on water height, apply Torricelli's equation to find time.

Step 5: Using Torricelli's formula for draining time:
For a tank draining through a small hole at bottom: \[ t = \frac{A_{\text{tank}}}{A_{\text{hole}}} \sqrt{\frac{2}{g}} \int_0^h \frac{dh'}{\sqrt{h'}} = \frac{2 A_{\text{tank}}}{A_{\text{hole}}} \sqrt{\frac{2h}{g}} \] Time is proportional to \(\sqrt{h}\).

Step 6: For tank 1:
\[ t_1 \propto \sqrt{h} \] For tank 2, the effective head at height \(h'\) is: \[ h' + H \] Time to empty: \[ t_2 \propto \int_0^h \frac{dh'}{\sqrt{h' + H}} = 2 \left( \sqrt{h + H} - \sqrt{H} \right) \]

Step 7: Calculate ratio \(\frac{t_1}{t_2}\):
\[ t_1 \propto \sqrt{h} \] \[ t_2 \propto 2 \left( \sqrt{h + H} - \sqrt{H} \right) \] Using \(H = \frac{16}{9} h\), write in terms of \(h\): \[ t_2 \propto 2 \left( \sqrt{h + \frac{16}{9} h} - \sqrt{\frac{16}{9} h} \right) = 2 \left( \sqrt{\frac{25}{9} h} - \frac{4}{3} \sqrt{h} \right) \] \[ = 2 \left( \frac{5}{3} \sqrt{h} - \frac{4}{3} \sqrt{h} \right) = 2 \times \frac{1}{3} \sqrt{h} = \frac{2}{3} \sqrt{h} \] Therefore, \[ \frac{t_1}{t_2} = \frac{\sqrt{h}}{\frac{2}{3} \sqrt{h}} = \frac{3}{2} \] This gives \(\frac{t_1}{t_2} = 1.5\), which is still not 3.

Step 8: Correcting the approach:
Actually, because the water level in tank 2 sees a higher effective head (due to pipe exit at ground level), the flow velocity is greater, and tank 2 empties faster.
In fact, when tank 2 drains, the velocity at any height \(h'\) is: \[ v = \sqrt{2g(h' + H)} \] Time to empty: \[ t_2 = \frac{A_{\text{tank}}}{A_{\text{hole}}} \sqrt{\frac{2}{g}} \int_0^h \frac{dh'}{\sqrt{h' + H}} = \frac{A_{\text{tank}}}{A_{\text{hole}}} \sqrt{\frac{2}{g}} \times 2 \left( \sqrt{h + H} - \sqrt{H} \right) \] Similarly, for tank 1: \[ t_1 = \frac{A_{\text{tank}}}{A_{\text{hole}}} \sqrt{\frac{2}{g}} \times 2 \sqrt{h} \] Ratio: \[ \frac{t_1}{t_2} = \frac{\sqrt{h}}{\sqrt{h + H} - \sqrt{H}} \] Substitute \(H = \frac{16}{9} h\): \[ \frac{t_1}{t_2} = \frac{\sqrt{h}}{\sqrt{h + \frac{16}{9} h} - \sqrt{\frac{16}{9} h}} = \frac{\sqrt{h}}{\sqrt{\frac{25}{9} h} - \frac{4}{3} \sqrt{h}} = \frac{\sqrt{h}}{\frac{5}{3} \sqrt{h} - \frac{4}{3} \sqrt{h}} = \frac{\sqrt{h}}{\frac{1}{3} \sqrt{h}} = 3 \]

Final Answer:
\[ \boxed{3} \]