Question

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 a radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water the quantities of water flowing out per second from the holes are the same then the value of $R$ is:

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

Answer 1

Answer: (A)

$\frac{L}{\sqrt{2\pi }}$

Speed of water leaking out at a depth y will be:
$v_1$ = $\sqrt{2gy}$
Similarly, for the hole at a depth 4y speed of water will be:
$v_2$ = $\sqrt{2g\times 4y}$
Now that the quantities of water flowing out per second from the holes are the same.
This implies: $A_1{v}_1=A_2{v}_2$
(Where, $A_1$ and $A_2$ are the area of the holes at height y and 4y respectively)
Hence,
$L^2\sqrt{2gy}=\pi R^2\sqrt{2g\times 4y}$
$L^2=2\pi R^2$
$R=\frac{L}{\sqrt{2\pi }}$