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 times taken for the level of water to fall from \(h\) to \(\frac{h}{2}\) and from \(\frac{h}{2}\) to zero is

• A. \( \sqrt2 \)
• B. \( \dfrac1{\sqrt2} \)
• C. \( \sqrt2-1 \)
• D. \( \dfrac1{\sqrt2-1} \)

Hint:

In Torricelli's theorem based problems: \[ v=\sqrt{2gh} \] and emptying time depends on square root of liquid height.

Answer 1

The Correct Option is C

Step 1: Use the draining tank idea.
Water leaves through a hole at the bottom, and the speed of outflow depends on the height of water above the hole (Torricelli's law). Working through the time taken, the time for the level to drop from one height to another comes out as \[ t \propto \sqrt{h_1} - \sqrt{h_2} \]

Step 2: Time for the first part.
The level falls from $h$ to $\tfrac{h}{2}$. \[ t_1 \propto \sqrt{h} - \sqrt{\tfrac{h}{2}} = \sqrt{h}\left(1 - \frac{1}{\sqrt2}\right) \]
Step 3: Time for the second part.
The level falls from $\tfrac{h}{2}$ to $0$. \[ t_2 \propto \sqrt{\tfrac{h}{2}} - 0 = \frac{\sqrt{h}}{\sqrt2} \]
Step 4: Set up the ratio.
\[ \frac{t_1}{t_2} = \frac{\sqrt{h}\left(1 - \frac{1}{\sqrt2}\right)}{\frac{\sqrt{h}}{\sqrt2}} \] The $\sqrt{h}$ cancels.
Step 5: Simplify the fraction.
Multiply top and bottom by $\sqrt2$. \[ = \sqrt2\left(1 - \frac{1}{\sqrt2}\right) = \sqrt2 - 1 \]
Step 6: State the result.
\[ \frac{t_1}{t_2} = \sqrt2 - 1 \] \[ \boxed{\sqrt2 - 1} \]