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 circular hole of 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 two holes are the same. Then value of $R$ is:
A. $\dfrac{L}{{\sqrt {2\pi } }}$
B. $2\pi L$
C. $L\sqrt {\dfrac{2}{\pi }}$
D. $\dfrac{L}{{2\pi }}$

Answer 1

Use the expression for principle of continuity for the liquid. This formula gives the relation between the areas of the ends of the pipe and velocities of the liquid flowing from both the ends of the pipe. Determine the areas and velocities for both the holes in the open tank and derive the expression for radius of the second hole.

Formulae used:
The expression for principle of continuity is
${A_1}{v_1} = {A_2}{v_2}$ …… (1)
Here, ${A_1}$ and ${A_2}$ are the areas of the two ends of pipe and ${v_1}$ and ${v_2}$ are the velocities of the liquid from these two ends.

Complete step by step answer:
We have given that the shape of the first hole is square with side $L$.
Hence, the area ${A_1}$ of this square hole is
${A_1} = {L^2}$
We have given that the shape of the first hole is circular with radius $R$.
Hence, the area ${A_2}$ of this square hole is
${A_2} = \pi {R^2}$
The minimum speed ${v_1}$ of the water coming out of the square shaped hole is
${v_1} = \sqrt {2gy}$

The minimum speed ${v_2}$ of the water coming out of the circular hole is
${v_2} = \sqrt {2g\left( {4y} \right)}$
$\Rightarrow {v_2} = \sqrt {8gy}$
Substitute ${L^2}$ for ${A_1}$, $\pi {R^2}$ for ${A_2}$, $\sqrt {2gy}$ for ${v_1}$ and $\sqrt {8gy}$ for ${v_2}$ in equation (1).
${L^2}\sqrt {2gy} = \pi {R^2}\sqrt {8gy}$
$\Rightarrow {R^2} = \dfrac{{{L^2}}}{\pi }\dfrac{{\sqrt {2gy} }}{{\sqrt {8gy} }}$
$\therefore R = \dfrac{L}{{\sqrt {2\pi } }}$
Therefore, the value of the radius of the second hole is $\dfrac{L}{{\sqrt {2\pi } }}$.

Hence, the correct option is A.

Note: The students may think how we have derived the formula for velocity of the water flowing out of the square shaped and circular hole. We have used the expression for Bernoulli’s equation or law of conservation of energy for the water and then determine the value of velocity of the liquid flowing out of the holes.