Question

A square hole of side length $l$ is made at a depth of $h$ and a circular hole of radius $r$ is made at a depth of $4h$ from the surface of water in a water tank kept on a horizontal surface. If $l<<h$, $r<<h$ and the rate of water flow from the holes is the same, then $r$ is equal to -

• A. $\frac{l}{\sqrt{2\pi }}$
• B. $\frac{l}{\sqrt{3\pi }}$
• C. $\frac{l}{3\pi }$
• D. $\frac{l}{2\pi }$

Answer 1

The correct option is A $\frac{l}{\sqrt{2\pi }}$

Given,
Side length of square shaped hole $=l$
Radius of circular hole $=r$
Rate of flow from the holes is same.
$\Rightarrow Q_1=Q_2$
$\Rightarrow A_1{v}_1=A_2{v}_2$
$[A_1$ is area of square shaped hole and $A_2$ is area of circular hole $]$


$\Rightarrow l^2\sqrt{2gh}=\pi r^2\sqrt{2g\times 4h}$
$\Rightarrow r^2=\frac{l^2}{2\pi }$
$\Rightarrow r=\sqrt{\frac{l^2}{2\pi }}=\frac{l}{\sqrt{2\pi }}$