Question

A cylindrical vessel with a cross sectional area $A$ has a small hole of area $a$ at the bottom. A liquid is filled in the vessel upto a height $H$. Find the time taken to empty the tank.

• A. $t=\frac{a}{A}\sqrt{\frac{2H}{g}}$
• B. $A\sqrt{\frac{2H}{g}}$
• C. $\frac{A}{a}\sqrt{\frac{2H}{g}}$
• D. $\frac{A}{a}\sqrt{\frac{H}{g}}$

Answer 1

The correct option is C $\frac{A}{a}\sqrt{\frac{2H}{g}}$


Let at any time $t$, the height of the liquid be $y$, and in $(t+dt)$ time, the level of the liquid will be $(y-dy)$.

As per equation of continuity,
$A_1{v}_1=A_2{v}_2$
$\Rightarrow a\sqrt{2gy}=A(-\frac{dy}{dt})$
where, $\frac{dy}{dt}$ is the rate of decrease in the level of liquid.
$\underset{y=H}{\overset{y=0}{\int }}\frac{dy}{\sqrt{y}}=-\frac{a}{A}\sqrt{2g}\underset{t=0}{\overset{t=T}{\int }}dt$
$\Rightarrow {\left[2\sqrt{y}\right]}_H^0=-\frac{a}{A}\sqrt{2g}[t{]}_0^T$
$\Rightarrow T=\frac{A}{a}\sqrt{\frac{2H}{g}}$ is the time taken to empty the tank.
Thus, option (c) is the correct answer.