Question

A square side 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 h'. It is found that volume flow rate of water through both the holes is founded to be same then :

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

Answer 1

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

Ar the same depth 'h' water flows same rate from are direction

ar pressure is same at that depth.

ar flow rate $(=\frac{dv}{dt})$ is equal, the holes must have equal area

$\Rightarrow \pi r^2=a^2\Rightarrow r=\frac{a}{\sqrt{\pi }}$ (Read)