A vessel of cross-sectional area A contains a liquid to a height H1. If a hole having cross-sectional area a is made at the bottom of the vessel, then the time taken by the liquid level to decrease from H₁ and H₂ is

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$\frac{A}{a} \sqrt{\frac{g}{2}} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

$\frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

$\frac{a}{A} \sqrt{\frac{g}{2}} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

$\frac{a}{A} \sqrt{\frac{2}{g}} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

answer is B.

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Detailed Solution

Let h be the height of the water level at any instant. Then the rate of decrease of water level is $- \frac{dh}{dt}$ .Therefore $- A \frac{dh}{dt} = av = a \sqrt{2 gh}$
$\begin{array}{l} \Rightarrow - \frac{dh}{dt} = \frac{a}{A} \sqrt{2 gh} \\ \Rightarrow - \int_{H_{1}}^{H_{2}} \frac{dh}{h^{1 / 2}} = (\frac{a}{A} \sqrt{2 g}) \int_{0}^{1} dt \\ \Rightarrow - 2 {[h^{1 / 2}]}_{H_{1}}^{H_{2}} = (\frac{a}{A} \sqrt{2 g}) t \\ \Rightarrow 2 [\sqrt{H_{1}} - \sqrt{H_{2}}] = (\frac{a}{A} \sqrt{2 g}) t \\ \Rightarrow t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_{1}} - \sqrt{H_{2}}] \end{array}$