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A wide tank of cross-sectional area $A$ containing a liquid to a height $H$ , has a small orifice at its base of area $a$ $(a < < A)$ . The time during which liquid level falls to a height $h (= \frac{H}{η})$ $(where η > 1)$ :

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$t = (\frac{A}{a}) [\sqrt{\frac{1}{g}} (\sqrt{H} - \sqrt{\frac{H}{η}})]$

$t = (\frac{a}{A}) [\sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{\frac{H}{η}})]$

$t = (\frac{a}{A}) [\sqrt{\frac{1}{g}} (\sqrt{H} - \sqrt{\frac{H}{η}})]$

$t = (\frac{A}{a}) \sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{\frac{H}{η}})$

answer is D.

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Detailed Solution

Let $h$ be the instantaneous height of the liquid in a tank. Then, the instantaneous velocity of flow through the hole is,

$v = \sqrt{2 g h}$

Using equation of continuity,

$A (- \frac{d h}{d t}) = a v = a \sqrt{2 g h}$

or, $- \int_{+ H}^{H / η} \frac{d h}{\sqrt{h}} = \frac{a}{A} \sqrt{2 g} \int_{0}^{t} d t$

$∴$ $t = (\frac{A}{a}) \sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{\frac{H}{η}})$

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