The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m^3

Q: Why do we use Torricelli's law here?

Torricelli's law gives the velocity of fluid flowing out of an orifice under the force of gravity, assuming the tank is open to the atmosphere and the fluid is ideal.

Q: Why is the flow rate constant?

The problem states that the water level is maintained. This means $h$ is constant, so the velocity of efflux $v = \sqrt{2gh}$ is constant, making the flow rate constant.

Q: What if the water level was not maintained?

If the water level wasn't maintained, $h$ would decrease over time, causing the velocity $v$ and the flow rate $Q$ to decrease as well.

Q: Why did we convert minutes to seconds?

Standard SI units for velocity are $\text{m/s}$ and for acceleration due to gravity $g$ is $\text{m/s}^2$. To ensure consistency in our calculations, we must convert the flow rate to $\text{m}^3/\text{s}$.

Q: Does the shape of the opening matter?

The shape only matters for calculating the cross-sectional area $A$. Since it's a circular opening, we use $A = \pi r^2$.

Question

The top of a water tank is open to air and its water level is maintained. It is giving out $0.74 	ext{ m}^3$ water per minute through a circular opening of $2 	ext{ cm}$ radius is its wall. The depth of the centre of the opening from the level of water in the tank is close to

Options:

(A) $4.8 	ext{ m}$

(B) $6.0 	ext{ m}$

(C) $2.9 	ext{ m}$

(D) $9.6 	ext{ m}$

Sigma Percentile · Accepted Answer

Imagine you are standing in front of a massive water tank. The top is completely open to the air, and there's a small circular hole in the side wall. Water is gushing out of this hole. But here's the catch: the water level inside the tank isn't dropping. It's being perfectly maintained. This means the pressure pushing the water out remains absolutely constant. Our mission? To find exactly how deep this hole is from the surface of the water.

### Analyzing the Setup

Let's break down the clues given to us. We know the water is flowing out at a rate of $0.74 	ext{ m}^3$ per minute. This is our volume flow rate, often denoted as $Q$. We also know the radius of the circular opening is $r = 2 	ext{ cm}$.

First things first, we need to speak the language of physics, which means converting everything into standard SI units. 
The flow rate $Q$ needs to be in cubic meters per second:
$$Q = \frac{0.74}{60} 	ext{ m}^3/	ext{s}$$

The radius $r$ needs to be in meters:
$$r = 2 	imes 10^{-2} 	ext{ m}$$

### The Master Equation: Continuity

Now, how do we connect the amount of water flowing out to how fast it's moving? Enter the Equation of Continuity. It tells us that the volume flow rate is simply the cross-sectional area of the opening multiplied by the velocity of the fluid.
$$Q = A \cdot v$$

Since the opening is circular, its area is $A = \pi r^2$. Let's substitute our knowns to find the velocity $v$:
$$v = \frac{Q}{\pi r^2}$$
$$v = \frac{0.74 / 60}{\pi (2 	imes 10^{-2})^2}$$
$$v = \frac{0.74}{60 	imes 4\pi 	imes 10^{-4}}$$
$$v = \frac{7400}{240\pi} \approx 9.81 	ext{ m/s}$$

So, the water is shooting out of the hole at approximately $9.81 	ext{ m/s}$. But how does this help us find the depth?

### Torricelli's Law of Efflux

This is where the magic of fluid dynamics comes in. Torricelli's Law states that the velocity of fluid flowing out of an orifice under gravity is exactly the same as the velocity an object would acquire if it fell freely from the surface of the fluid to the orifice.

Mathematically, it is expressed as:
$$v = \sqrt{2gh}$$
where $h$ is the depth of the hole from the free surface.

### Final Calculation

We have two different ways to express the velocity $v$. Let's equate them!
$$\sqrt{2gh} = \frac{7400}{240\pi}$$

To solve for $h$, we square both sides:
$$2gh = \left( \frac{740}{24\pi} ight)^2$$
$$h = \frac{1}{2g} \left( \frac{740}{24\pi} ight)^2$$

Taking the acceleration due to gravity $g \approx 9.8 	ext{ m/s}^2$ (or $10 	ext{ m/s}^2$ for a quick estimate), we get:
$$h \approx \frac{1}{2 	imes 9.8} (9.81)^2 \approx \frac{96.2}{19.6} \approx 4.9 	ext{ m}$$
If we use $g = 10 	ext{ m/s}^2$, we get $h \approx 4.8 	ext{ m}$. Looking at the options, $4.8 	ext{ m}$ is a perfect match!

**Final Answer:** The depth of the centre of the opening is approximately **$4.8 	ext{ m}$**.

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Visualized Solution

$Visualizing the Tank Setup$

$Volume Flow Rate$

$Equation of Continuity$

$Calculating Velocity$

$Torricelli’s Law of Efflux$

$Equating Velocities$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Flow of Fluid

Analyzing the Setup

The Master Equation

Torricelli's Law of Efflux

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Tank Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Volume Flow Rate

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Continuity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torricelli’s Law of Efflux

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Velocities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward