A large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top…
A large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4 y$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from the two holes are the same. Then value of $R$ is
$\frac{L}{\sqrt{2 \pi}}$
$2 \pi L$
$L \sqrt{\frac{2}{\pi}}$
$\frac{L}{2 \pi}$
Solution
By the principle of continuity
$A_1 v_1=A_2 v_2$
According to question, $A_1=L^2$
$\begin{aligned}
& v_1=\sqrt{2 g y} \\
& \text { and } \\
& A_2=\pi R^2 \\
& v_2=\sqrt{2 g 4 y} \\
& \therefore \quad L^2 \sqrt{2 g y}=\pi R^2 \sqrt{2 g 4 y} \\
&
\end{aligned}$
or $L^2=2 \pi R^2$
or $R=\frac{L}{\sqrt{2 \pi}}$