Question

A vessel filled with water has holes ‘A’ and ‘B’ at depths \[h\] and \[3h\] from the top respectively. Hole 'A' is a square of side \[L\] and 'B' is circle of radius \[r\]. The water flowing out per second from both the holes is the same. Then \[L\] is equal to
$\begin{align}
  & \text{A}\text{. }{{r}^{\dfrac{1}{2}}}{{\left( \pi \right)}^{\dfrac{1}{2}}}{{3}^{\dfrac{1}{2}}} \\
 & \text{B}\text{. }r{{\left( \pi \right)}^{\dfrac{1}{4}}}{{3}^{\dfrac{1}{4}}} \\
 & \text{C}\text{. }r{{\left( \pi \right)}^{\dfrac{1}{2}}}{{3}^{\dfrac{1}{4}}} \\
 & \text{D}\text{. }{{r}^{\dfrac{1}{2}}}{{\left( \pi \right)}^{\dfrac{1}{3}}}{{3}^{\dfrac{1}{2}}} \\
\end{align}$

Answer 1

Hint: We have two holes at different positions on the wall of the cylinder. We will apply Torricelli’s theorem to calculate the velocity of efflux in both cases. With the velocity of efflux, we can calculate the rate of flowing water, and equating the two will give the relation between the side of the square hole and the radius of the circular hole.

Formula used:
The velocity of fluid leaking from the hole in a container,
$v=\sqrt{2gh}$

Complete step by step answer:
Torricelli’s theorem or Torricelli’s rate of efflux states that the speed of efflux, that is, the speed of water coming out of the hole at some height in the wall of a cylinder is the same speed acquired by a freely falling body from the same height.
We are given that a vessel filled with water has holes ‘A’ and ‘B’ at depths \[h\] and \[3h\] from the top respectively.

We have:
Hole A at depth \[h\] and is a square of side \[L\]
So, the area of hole A is; area of a square of side \[L\]
Area hole A: \[{{A}_{A}}={{L}^{2}}\]
Hole B at depth \[3h\] and is a circle of radius \[r\]
So, the area of hole B is: area of the circle of radius \[r\]
Area hole B: \[{{A}_{B}}=\pi {{r}^{2}}\]
Now, by Torricelli’s theorem, we know that the velocity of fluid coming out of a container from a hole having a fluid column of height \[h\] above it is:
$v=\sqrt{2gh}$
Where \[g\] is the gravitational constant.
As hole A is at depth \[h\], the liquid column of height \[h\] is above it.
So, the velocity of fluid coming out of hole A is:
${{v}_{A}}=\sqrt{2gh}$
And hole B is at depth \[3h\], the liquid column of height \[3h\] is above it.
So, velocity of fluid coming out of hole B is:
${{v}_{B}}=\sqrt{6gh}$
Now, the rate of fluid coming out of a hole is equal to the product of area of hole and velocity of fluid coming out of it.
So rate of water flowing out of hole A per second will be: ${{A}_{A}}{{v}_{A}}$
Or, \[{{L}^{2}}\sqrt{2gh}\]
And, rate of water flowing out of hole B per second will be: ${{A}_{B}}{{v}_{B}}$
Or, \[\pi {{r}^{2}}\sqrt{6gh}\]
As according to the question, water flowing out per second from both the holes is same,
So,
 \[\begin{align}
  & {{L}^{2}}\sqrt{2gh}=\pi {{r}^{2}}\sqrt{6gh} \\
 & {{L}^{2}}\sqrt{2gh}=\pi {{r}^{2}}\sqrt{3}\sqrt{2gh} \\
 & {{L}^{2}}=\sqrt{3}\pi {{r}^{2}} \\
 & L={{\left( \sqrt{3}\pi {{r}^{2}} \right)}^{\dfrac{1}{2}}} \\
 & L=r{{\left( \pi \right)}^{\dfrac{1}{2}}}{{3}^{\dfrac{1}{4}}} \\
\end{align}\]
Hence, the correct option is C.

Note: The speed of water coming out of the hole at some height in the wall of a cylinder is the same speed acquired by a freely falling body from the same height. This speed does not depend upon the area of the hole, area of cross-section of the vessel or cylinder, and the density of fluid or water. The speed of efflux depends only on the height where the hole is situated on the wall of the vessel or cylinder.