Question

A cylindrical tank of height $H$ is open at the top end and it has a radius $r$. Water is filled in it up to a height of $h$. The time taken to empty the tank through a hole of radius $r$ in its bottom is :

• A. $\sqrt{\frac{2h}{g}}\frac{r^2}{r^{{}^{\prime }2}}$
• B. $\sqrt{\frac{2H}{g}}\frac{r^2}{r^{{}^{\prime }2}}$
• C. $\sqrt{h}H$
• D. None of these

Answer 1

The correct option is A $\sqrt{\frac{2h}{g}}\frac{r^2}{r^{{}^{\prime }2}}$

From principle of continuity, we know that,

$A_1{V}_2=A_2{V}_2$

We have, $A_1=\pi r^2$ and $V_2=\frac{dh}{dt}$

Also, $A_2=\pi r_{\prime }^2$ and $V_2=\sqrt{2gH}$

Since, $A_1$ is decreasing it will have a negative sign.

Therefore, we have,
$-\pi r^2\times \frac{dh}{dt}=\pi r_{\prime }^2\times \sqrt{2gH}$

or, $\frac{dh}{dt}=\frac{-r_{\prime }^2}{r^2}\sqrt{2gH}$

or, $dt=\frac{-r^2}{r_{\prime }^2}\frac{dh}{\sqrt{2gH}}$

Integrating both sides, we get

${\int }_0^{t}dt=\frac{r_{\prime }^2}{r^2}{\int }_H^0\left(\frac{1}{\sqrt{2gH}}\right)dh$

or, $t=\frac{-r^2}{r_{\prime }^2}\times \frac{1}{\sqrt{g}}2-\sqrt{h}$

or, $t=\frac{r^2}{r_{\prime }^2}\sqrt{\frac{2h}{g}}$ is our required answer.