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:

\( \frac{t_1}{t_2} = 3 \)