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 $\frac{A}{a}\sqrt{\frac{2}{g}}[\sqrt{H_1}-\sqrt{H_2}]$

Let,
$V$ = Rate of decrease of water level in vessel
$v$ = velocity of water coming out of vessel

Applying equation of continuity,
$AV=av$
Here,
$V=-\frac{dh}{dt}$
and $v=\sqrt{2gh}$

So, we get,
$A(-\frac{dh}{dt})=a\sqrt{2gh}$
$\Rightarrow t=\frac{-A}{a\sqrt{2g}}$${\int }_{H_1}^{H_2}\frac{1}{\sqrt{h}}dh$
$t=\frac{A}{a}\sqrt{\frac{2}{g}}[\sqrt{H_1}-\sqrt{H_2}]$