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Question

(C) The time $t$ taken for the water level to fall from an initial height $H_1$ to a final height $H_2$ in a tank with cross-sectional area $A$ and or

Vedclass · Accepted Answer

$\sqrt{2} - 1$

Vedclass · Answer

(C) The time $t$ taken for the water level to fall from an initial height $H_1$ to a final height $H_2$ in a tank with cross-sectional area $A$ and orifice area $A_0$ is given by $t = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{H_1} - \sqrt{H_2})$.For the first interval,the level falls from $h$ to $\frac{h}{2}$. Thus,$t_1 = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{h} - \sqrt{\frac{h}{2}}) = \frac{A}{A_0} \sqrt{\frac{2h}{g}} (1 - \frac{1}{\sqrt{2}})$.For the second interval,the level falls from $\frac{h}{2}$ to $0$. Thus,$t_2 = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{\frac{h}{2}} - 0) = \frac{A}{A_0} \sqrt{\frac{2h}{g}} (\frac{1}{\sqrt{2}})$.The ratio is $\frac{t_1}{t_2} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1$.

$A$ 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 of water to fall from $h$ to $\frac{h}{2}$ and from $\frac{h}{2}$ to zero is:

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