Consider a water tank as shown in the figure. It's cross-sectional area is 0.4 m^2. The tank has an

Q: Why do we assume the velocity at the top is zero?

Because the cross-sectional area of the tank ($0.4 \text{ m}^2$) is 4000 times larger than the hole ($1 \text{ cm}^2$). By the equation of continuity ($A_1 v_1 = A_2 v_2$), the top surface moves down so slowly that its kinetic energy is practically zero.

Q: Why does atmospheric pressure cancel out?

Both the top surface of the tank and the water exiting the hole are exposed to the atmosphere. Since $P_0$ is added to both sides of Bernoulli's equation, it cancels out.

Q: How do we calculate the pressure exerted by the load?

Pressure is force divided by area. The force is the weight of the load ($mg$), and the area is the cross-sectional area of the tank ($A$). So, the extra pressure is $mg/A$.

Q: What is the reference level for potential energy?

We chose the bottom of the tank (where the hole is) as our reference level ($h=0$). This makes the potential energy at the hole zero, and the potential energy at the top surface $\rho g h$.

Q: Why don't we use Torricelli's law directly?

Torricelli's law ($v = \sqrt{2gh}$) assumes the only pressure driving the fluid is from its own weight. Here, the $24 \text{ kg}$ load adds extra pressure, so we must use the full Bernoulli's equation to account for it.

Question

Consider a water tank as shown in the figure. It's cross-sectional area is $0.4 	ext{ m}^2$. The tank has an opening $B$ near the bottom whose cross-section area is $1 	ext{ cm}^2$. A load of $24 	ext{ kg}$ is applied on the water at the top when the height of the water level is $40 	ext{ cm}$ above the bottom, the velocity of water coming out the opening $B$ is $v 	ext{ ms}^{-1}$. The value of $v$, to the nearest integer, is ............... .
(Take value of $g$ to be $10 	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

### The Setup: More Than Just Gravity

Imagine a large water tank, filled to a height of $40 	ext{ cm}$. At the bottom, there is a tiny opening where water is rushing out. If this were a simple tank, the water would flow out purely due to the pressure created by its own weight.

But there is a twist! A heavy $24 	ext{ kg}$ load is placed right on top of the water surface. This load acts like a piston, squeezing the water and forcing it out of the bottom hole even faster.

Our goal is to find the exact velocity of the water as it exits the tank. To do this, we need a powerful tool that connects pressure, velocity, and height in a flowing fluid.

### The Master Equation: Bernoulli's Principle

When dealing with fluid flow, **Bernoulli's Principle** is our best friend. It states that for an incompressible, non-viscous fluid, the total energy per unit volume remains constant along a streamline.

We can write this mathematically by comparing two points. Let's choose Point $A$ at the top surface of the water, and Point $B$ just outside the exit hole.

$$ P_A + \frac{1}{2}ho v_A^2 + ho g h_A = P_B + \frac{1}{2}ho v_B^2 + ho g h_B $$

This equation balances the pressure energy, kinetic energy, and potential energy at both points. Now, we need to carefully evaluate each term.

### Evaluating the Physical Conditions

Let's look at Point $A$. The pressure here is not just the atmospheric pressure ($P_0$). The $24 	ext{ kg}$ load is spread over an area of $0.4 	ext{ m}^2$, creating an additional pressure.

$$ P_A = P_0 + \frac{mg}{A} $$

Because the tank's cross-sectional area is massive compared to the tiny $1 	ext{ cm}^2$ hole, the water level drops incredibly slowly. For all practical purposes, the velocity at the top surface is zero ($v_A \approx 0$).

Now, let's look at Point $B$. The water is exiting into the open air, so the pressure is simply atmospheric ($P_B = P_0$). If we take the bottom of the tank as our reference level, the height at $B$ is zero ($h_B = 0$), and the height at $A$ is $0.4 	ext{ m}$ ($h_A = 0.4 	ext{ m}$).

### Crunching the Numbers

Let's substitute all these conditions into Bernoulli's equation. Notice how the atmospheric pressure $P_0$ appears on both sides and beautifully cancels out!

$$ \left(P_0 + \frac{24 	imes 10}{0.4}ight) + 0 + (1000 	imes 10 	imes 0.4) = P_0 + \frac{1}{2}(1000) v_B^2 + 0 $$

Now, we perform the arithmetic. The extra pressure from the load is $\frac{240}{0.4} = 600 	ext{ Pa}$. The potential energy term gives us $4000$.

$$ 600 + 4000 = 500 v_B^2 $$

$$ 4600 = 500 v_B^2 $$

Dividing both sides by $500$, we isolate the velocity squared.

$$ v_B^2 = \frac{4600}{500} = 9.2 $$

Taking the square root, we find the efflux velocity.

$$ v_B = \sqrt{9.2} \approx 3.03 	ext{ m/s} $$

The question asks for the nearest integer, which gives us our final answer: **$3 	ext{ m/s}$**.

### The Physical Insight

What if the $24 	ext{ kg}$ load wasn't there? We would simply use Torricelli's Law, $v = \sqrt{2gh}$.

$$ v = \sqrt{2 	imes 10 	imes 0.4} = \sqrt{8} \approx 2.82 	ext{ m/s} $$

By comparing $2.82 	ext{ m/s}$ to our result of $3.03 	ext{ m/s}$, we can clearly see the effect of the load. The extra pressure acts exactly like an additional column of water, increasing the pressure head and driving the fluid out with greater kinetic energy. Always look beyond the numbers to understand the physical story!

Enter Numerical Value:

Visualized Solution

$Analyzing the Setup$

$Bernoulli’s Principle$

$Evaluating the Terms$

$Substituting the Values$

$Simplifying the Equation$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Flow of Fluid

The Setup

The Master Equation

Evaluating the Physical Conditions

Crunching the Numbers

The Physical Insight

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bernoulli’s Principle

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

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

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

.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