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

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In Torricelli's theorem based problems: \[ v=\sqrt{2gh} \] and emptying time depends on square root of liquid height.
Updated On: Jun 17, 2026
  • \( \sqrt2 \)
  • \( \dfrac1{\sqrt2} \)
  • \( \sqrt2-1 \)
  • \( \dfrac1{\sqrt2-1} \)
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The Correct Option is C

Solution and Explanation

Concept: Time taken for water level to fall from height \(h_1\) to \(h_2\): \[ t\propto \sqrt{h_1}-\sqrt{h_2} \]

Step 1: Time from \(h\) to \(\frac h2\). \[ t_1\propto \sqrt h-\sqrt{\frac h2} \] \[ =\sqrt h\left(1-\frac1{\sqrt2}\right) \]

Step 2: Time from \(\frac h2\) to \(0\). \[ t_2\propto \sqrt{\frac h2}-0 \] \[ =\frac{\sqrt h}{\sqrt2} \]

Step 3: Take ratio. \[ \frac{t_1}{t_2} = \frac{\sqrt h\left(1-\frac1{\sqrt2}\right)} {\frac{\sqrt h}{\sqrt2}} \] \[ =\sqrt2-1 \] Hence, \[ \boxed{\sqrt2-1} \]
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