Question

A large tank filled with water to a height \( h \) is to be emptied through a small hole at the bottom. The ratio of the time taken for the level to fall from \( h \) to \( \frac{h}{2} \) and that taken for the level to fall from \( \frac{h}{2} \) to \( 0 \) is:

• A. \( \sqrt{2} - 1 \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \sqrt{2} \)
• D. \( \frac{1}{\sqrt{2} - 1} \)

Hint:

The time taken to empty a tank follows \( t = \int \frac{dh}{\sqrt{2 g h}} \). When computing ratios, simplify square roots carefully.

Answer 1

The Correct Option is A

Step 1: Using Torricelli’s Law
The velocity of efflux for a liquid flowing out of an orifice is given by Torricelli’s theorem: \[ v = \sqrt{2 g h}. \] The time taken to fall from a height \( h_1 \) to \( h_2 \) is given by: \[ t = \int_{h_2}^{h_1} \frac{dh}{\sqrt{2 g h}}. \] Evaluating the integral: \[ t = \frac{2}{\sqrt{2g}} (\sqrt{h_1} - \sqrt{h_2}). \] Step 2: Computing the Ratio
Let \( t_1 \) be the time taken for the level to fall from \( h \) to \( \frac{h}{2} \): \[ t_1 = \frac{2}{\sqrt{2g}} (\sqrt{h} - \sqrt{h/2}). \] \[ t_1 = \frac{2}{\sqrt{2g}} (\sqrt{h} - \frac{\sqrt{h}}{\sqrt{2}}). \] \[ t_1 = \frac{2}{\sqrt{2g}} \sqrt{h} \left(1 - \frac{1}{\sqrt{2}}\right). \] Let \( t_2 \) be the time taken for the level to fall from \( \frac{h}{2} \) to \( 0 \): \[ t_2 = \frac{2}{\sqrt{2g}} (\sqrt{h/2} - 0). \] \[ t_2 = \frac{2}{\sqrt{2g}} \frac{\sqrt{h}}{\sqrt{2}}. \] Taking the ratio: \[ \frac{t_1}{t_2} = \frac{\sqrt{h} \left(1 - \frac{1}{\sqrt{2}}\right)}{\frac{\sqrt{h}}{\sqrt{2}}}. \] \[ = \sqrt{2} - 1. \] Step 3: Conclusion
Thus, the required ratio is: \[ \sqrt{2} - 1. \]