A vessel of area of cross-section A has liquid to a height $H$ . There is a hole at the bottom of vessel having area of cross-section a. The time taken to decrease the level from ${H_1}$ to ${H_2}$ will be
Diffcult
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(a) At height $h$ the rate of discharge of liquid through hole at bottom is $\sqrt{2 g h} \times a$ (by Bernoulli's principle).
Let in time $d t$ level of water in tank decreases by $d h$
Equating volumes, $A d h=d t \sqrt{2 g h} \times a$
$\Rightarrow d t=\frac{A}{a \sqrt{2 g}} \frac{d h}{\sqrt{h}}$
Integrating above we get,
$\int_{0}^{t} d t=\frac{A}{a \sqrt{2 g}} \int_{H_{2}}^{H_{1}} \frac{d h}{\sqrt{h}}$
$\Rightarrow t=\left.\frac{A}{a \sqrt{2 g}} \frac{h^{1 / 2}}{1 / 2}\right|_{H_{2}} ^{H_{1}}=\frac{A}{a} \sqrt{2 / g}[\sqrt{H_{1}}-\sqrt{H_{2}}]$
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