Question

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:

• A. $\frac{A}{a}\sqrt{\frac{2}{g}}[{\sqrt{H}}_1-{\sqrt{H}}_2]$
• B. $\sqrt{2gh}$
• C. $\sqrt{2gh(H_1-H_2)}$
• D. $\frac{A}{a}\sqrt{\frac{g}{2}}[{\sqrt{H}}_1-{\sqrt{H}}_2]$

Answer 1

The correct option is A $\frac{A}{a}\sqrt{\frac{2}{g}}[{\sqrt{H}}_1-{\sqrt{H}}_2]$

At height $h$ the rate of discharge of liquid through hole at bottom is $\sqrt{2gh}\times a$ (by Bernoulli's principle).

Let in time $dt$ level of water in tank decreases by $dh$.

Equating volumes, $Adh=dt\sqrt{2gh}\times a$

$\Rightarrow dt=\frac{A}{a\sqrt{2g}}\frac{dh}{\sqrt{h}}$

Integrating above we get,

${\int }_0^{t}dt=\frac{A}{a\sqrt{2g}}{\int }_{H_2}^{H_1}\frac{dh}{\sqrt{h}}$

$\Rightarrow t=\frac{A}{a\sqrt{2g}}\frac{h^{1/2}}{1/2}|{}_{H_2}^{H_1}=\frac{A}{a}\sqrt{2/g}[\sqrt{H_1}-\sqrt{H_2}]$