Question

A tank full of water has a small hole at the bottom. If one-fourth of the tank is emptied in $t_1$ seconds and the remaining three-fourths of the tank is emptied in $t_2$ seconds, then the ratio $t_1/t_2$ is

• A. $\sqrt{3}$
• B. $\sqrt{2}$
• C. $\dfrac{1}{\sqrt{2}}$
• D. $\dfrac{2}{\sqrt{3}}-1$

Hint:

For tanks with small orifices, time of emptying is proportional to the difference of square roots of water heights, not the heights themselves.

Answer 1

The Correct Option is D

Step 1: Let the initial height of water in the tank be $H$. According to Torricelli’s law, the rate of fall of water level is: \[ \frac{dh}{dt} \propto -\sqrt{h} \] Hence, time taken for the water level to fall from height $h_1$ to $h_2$ is: \[ t \propto \left(\sqrt{h_1}-\sqrt{h_2}\right) \]
Step 2: One-fourth of the tank emptied means the height falls from $H$ to $\dfrac{3H}{4}$. So, \[ t_1 \propto \sqrt{H}-\sqrt{\frac{3H}{4}} = \sqrt{H}\left(1-\frac{\sqrt{3}}{2}\right) \]
Step 3: The remaining three-fourths empties when the height falls from $\dfrac{3H}{4}$ to $0$. So, \[ t_2 \propto \sqrt{\frac{3H}{4}}=\sqrt{H}\,\frac{\sqrt{3}}{2} \]
Step 4: Find the ratio: \[ \frac{t_1}{t_2} = \frac{1-\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = \frac{2-\sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3}}-1 \]