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: \[ \sqrt{2} - 1. \]

Answer 2

To solve this problem, we must understand the principles of fluid dynamics, specifically Torricelli's Law, which states the speed \( v \) of efflux of a fluid under gravity through a hole is \( v = \sqrt{2gh} \). Using this, we can determine the time taken for water to fall from one height to another.
First, the time \( t \) taken to fall from height \( h_1 \) to \( h_2 \) in a tank can be represented by the formula:
\[ t = \frac{A}{a\sqrt{2g}} \left(\sqrt{h_1} - \sqrt{h_2}\right) \]
where \( A \) is the cross-sectional area of the tank, \( a \) is the area of the hole, and \( g \) is the acceleration due to gravity.
Applying the given sections:

• The time \( t_1 \) to fall from \( h \) to \( \frac{h}{2} \):
  \[ t_1 = \frac{A}{a\sqrt{2g}} \left(\sqrt{h} - \sqrt{\frac{h}{2}}\right) = \frac{A}{a\sqrt{2g}} \left(\sqrt{h} - \frac{\sqrt{2}}{2}\sqrt{h}\right) = \frac{A}{a\sqrt{2g}} \left(\sqrt{h}(1 - \frac{\sqrt{2}}{2})\right) \]
• The time \( t_2 \) to fall from \( \frac{h}{2} \) to \( 0 \):
  \[ t_2 = \frac{A}{a\sqrt{2g}} \left(\sqrt{\frac{h}{2}} - 0\right) = \frac{A}{a\sqrt{2g}} \cdot \frac{\sqrt{2}}{2}\sqrt{h} \]

Now, the ratio of \( t_1 \) to \( t_2 \) is:
\[ \frac{t_1}{t_2} = \frac{\left(\sqrt{h}\left(1 - \frac{\sqrt{2}}{2}\right)\right)}{\frac{\sqrt{2}}{2}\sqrt{h}} = \frac{2(1 - \frac{\sqrt{2}}{2})}{\sqrt{2}} \]
Simplifying the expression:
\[ \frac{t_1}{t_2} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1 \]
Thus, the ratio of the time taken is \(\sqrt{2} - 1\).