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(C) The time $t$ taken for the water level to drop from height $H_1$ to $H_2$ is given by the formula: $t = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{H_

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$\sqrt{2} - 1$

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(C) The time $t$ taken for the water level to drop from height $H_1$ to $H_2$ is given by the formula: $t = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{H_1} - \sqrt{H_2})$,where $A$ is the cross-sectional area of the container and $A_0$ is the area of the hole.For the first interval (from $h$ to $\frac{h}{2}$): $t_1 = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{h} - \sqrt{\frac{h}{2}})$.For the second interval (from $\frac{h}{2}$ to $0$): $t_2 = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{\frac{h}{2}} - \sqrt{0})$.Taking the ratio: $\frac{t_1}{t_2} = \frac{\sqrt{h} - \sqrt{\frac{h}{2}}}{\sqrt{\frac{h}{2}}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1$.

$A$ container is filled with water up to a height $h$. $A$ small hole is made at the bottom. What is the ratio of the time taken for the water level to fall from $h$ to $\frac{h}{2}$ to the time taken for it to fall from $\frac{h}{2}$ to $0$?

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