Question

A tank is filled with a liquid upto a height \( H \). A small hole is made at the bottom of this tank. Consider \( t_1 \) be the time taken to empty the first half of the tank and \( t_2 \) be the time taken to empty the rest half of the tank. Then, determine the ratio \( \frac{t_1}{t_2} \).

• A. 1.33
• B. 1.5
• C. 2
• D. 0.414

Hint:

The time required to empty a tank through a hole at the bottom depends on the square root of the height of the liquid.

Answer 1

The Correct Option is D

The rate of flow of liquid out of the hole is proportional to the square root of the height of liquid, as given by Torricelli's law.
Therefore, the time required to empty the liquid is inversely proportional to the square root of the height.
Let the total time to empty the tank be \( t \).
The time to empty the first half of the tank is given by: \[ t_1 \propto \sqrt{\frac{H}{2}} = \frac{\sqrt{H}}{\sqrt{2}} \] The time to empty the second half of the tank is: \[ t_2 \propto \sqrt{H} \] The ratio of \( t_1 \) to \( t_2 \) is: \[ \frac{t_1}{t_2} = \frac{\frac{\sqrt{H}}{\sqrt{2}}}{\sqrt{H}} = \frac{1}{\sqrt{2}} \approx 0.414 \]
Thus, the correct answer is (d).