Question

A square hole of side a is made at a depth $h$ below water surface and to the side of a water container another circular hole of radius $r$ is made to the same container at a depth of $4h$. It is found that volume flow rate of water through both the holes is found to be same then:

• A. $r=\frac{a}{2\sqrt{\pi }}$
• B. $a=\frac{r}{2\sqrt{\pi }}$
• C. $a=\frac{r}{\sqrt{2\pi }}$
• D. $r=\frac{a}{\sqrt{2\pi }}$

Answer 1

The correct option is D $r=\frac{a}{\sqrt{2\pi }}$

${\rho }^1{A}_1{v}_1={\rho }^2{A}_2{v}_2$ as volume rate is constant
velocity by toricellis therom is $\sqrt{2gh}$
$\therefore a^2\left(\sqrt{2gh}\right)=\pi r^2\left(\sqrt{2g\left(4h\right)}\right)$
$\Rightarrow 2\pi r^2=a^2$
$r=\frac{a}{\sqrt{2\pi }}$