A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is init

Q: Why does the water stop flowing out of the orifice?

As water flows out, the volume of the trapped air at the top increases, causing its pressure to drop below atmospheric pressure. This creates a partial vacuum. The flow stops when the sum of this reduced air pressure and the hydrostatic pressure of the remaining water column exactly equals the external atmospheric pressure.

Q: Why can we assume the temperature of the trapped air is constant?

The problem states that the process occurs slowly and the system is in thermal contact with the surroundings, allowing us to assume an isothermal process where Boyle's Law ($pV = \text{constant}$) applies.

Q: What is the initial pressure of the trapped air?

Initially, the vessel is open to the atmosphere before being sealed. Therefore, the air inside is at atmospheric pressure $p_0 = 1.0 \times 10^5\text{ Pa}$.

Q: How do we find the final height of the air column?

The total height of the vessel is $500\text{ mm}$, and the final water column height is $200\text{ mm}$. Therefore, the final air column height is $500 - 200 = 300\text{ mm}$.

Q: What would happen if the top cap was not airtight?

If the cap was not airtight, air would continuously enter from the top, keeping the air pressure at $p_0$. The water would continue to flow out completely until the vessel was empty.

Question

A cylindrical vessel of height $500	ext{ mm}$ has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it upto height $H$. Now, the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being $200	ext{ mm}$. Find the fall in height (in mm) of water level due to opening of the orifice.

[Take atmospheric pressure = $1.0 	imes 10^5	ext{ N m}^{-2}$, density of water = $1000	ext{ kg m}^{-3}$ and $g = 10	ext{ ms}^{-2}$. Neglect any effect of surface tension.]

Sigma Percentile · Accepted Answer

### The Magic of Trapped Air: A Journey into Hydrostatics and Gas Laws

Imagine holding a straw filled with water, sealing the top with your thumb, and watching in amazement as the water refuses to fall out.

This simple, everyday phenomenon is governed by a beautiful interplay of hydrostatic pressure and gas laws.

In this problem, we explore a highly elegant variation of this classic setup: a cylindrical vessel filled with water, sealed at the top, and allowed to reach a steady state after opening a small orifice at the bottom.

Let's dive deep into the physics of this system and uncover the elegant mathematics that guides it to equilibrium.

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### Analyzing the Setup: The Initial State

We begin with a cylindrical vessel of total height $L = 500	ext{ mm}$.

Initially, the vessel is open at the top and filled with water up to an unknown height $H$.

Since the top is open, the air column above the water—which has a height of $500 - H$—is in direct contact with the atmosphere.

Therefore, the initial pressure of this trapped air is exactly equal to the atmospheric pressure, $p_0 = 1.0 	imes 10^5	ext{ Pa}$.

Now, we seal the top of the vessel completely with a airtight cap.

At this precise moment, we have trapped a specific mass of air at pressure $p_0$ inside a volume of $A(500 - H)$, where $A$ is the cross-sectional area of the cylinder.

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### The Transition: Opening the Orifice

What happens when we open the small orifice at the bottom of the vessel?

Naturally, gravity pulls the water downwards, and some water begins to flow out of the orifice.

As water leaves the vessel, the volume occupied by the trapped air at the top must increase.

Since the vessel is sealed at the top, no new air can enter.

According to Boyle's Law, as the volume of the trapped air increases, its pressure must decrease below the atmospheric pressure $p_0$.

This creates a partial vacuum at the top of the vessel!

As the water level continues to fall, the air pressure at the top keeps dropping.

Eventually, a point is reached where the combined pressure of the trapped air and the remaining water column at the bottom orifice exactly balances the external atmospheric pressure.

At this moment, the net force driving the water out becomes zero, and the flow stops completely, reaching a steady state.

---

### The Master Equation: Hydrostatic Equilibrium

Let's analyze this final steady state.

The water level has stabilized at a height of $h = 200	ext{ mm} = 0.2	ext{ m}$.

This means the height of the trapped air column is now:
$$500	ext{ mm} - 200	ext{ mm} = 300	ext{ mm} = 0.3	ext{ m}$$

Let the final pressure of the trapped air be $p$.

At the bottom orifice, the pressure inside the vessel is the sum of the air pressure $p$ at the top and the hydrostatic pressure exerted by the water column of height $h$:
$$p_{	ext{inside}} = p + ho g h$$

For the water to remain steady and not flow out, this internal pressure must be perfectly balanced by the external atmospheric pressure $p_0$:
$$p + ho g h = p_0$$

This is our first key equation!

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### Calculating the Final Air Pressure

Let's substitute the given physical values into our equilibrium equation to find the final pressure $p$ of the trapped air.

We are given:
- Atmospheric pressure, $p_0 = 1.0 	imes 10^5	ext{ Pa}$
- Density of water, $ho = 1000	ext{ kg/m}^3$
- Acceleration due to gravity, $g = 10	ext{ m/s}^2$
- Final water height, $h = 0.2	ext{ m}$

First, let's calculate the hydrostatic pressure exerted by the $200	ext{ mm}$ water column:
$$ho g h = 1000	ext{ kg/m}^3 	imes 10	ext{ m/s}^2 	imes 0.2	ext{ m} = 2000	ext{ Pa}$$

Now, we substitute this back into our equilibrium equation:
$$p + 2000	ext{ Pa} = 1.0 	imes 10^5	ext{ Pa}$$

Solving for $p$:
$$p = 100,000	ext{ Pa} - 2000	ext{ Pa} = 98,000	ext{ Pa}$$

The pressure of the trapped air has decreased from $100,000	ext{ Pa}$ to $98,000	ext{ Pa}$ due to the expansion.

---

### Connecting the States: Boyle's Law

Since the temperature of the system is maintained constant, the expansion of the trapped air is an **isothermal process**.

Therefore, we can apply **Boyle's Law** to connect the initial and final states of the trapped air:
$$p_{	ext{initial}} V_{	ext{initial}} = p_{	ext{final}} V_{	ext{final}}$$

Let's write down the volumes in terms of the cross-sectional area $A$ of the cylinder:
- Initial volume: $V_{	ext{initial}} = A 	imes (500 - H)$
- Final volume: $V_{	ext{final}} = A 	imes 300$

Substituting these into Boyle's Law:
$$p_0 	imes A(500 - H) = p 	imes A(300)$$

Since the cross-sectional area $A$ is constant and non-zero, we can elegantly cancel it from both sides:
$$p_0 (500 - H) = p 	imes 300$$

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### Solving for the Initial Height $H$

Now, let's substitute our known pressure values into this simplified equation:
$$10^5 	imes (500 - H) = 98,000 	imes 300$$

Let's divide both sides by $10^5$ to isolate the term $(500 - H)$:
$$500 - H = \frac{98,000 	imes 300}{100,000}$$

Simplifying the fraction:
$$500 - H = 98 	imes 3 = 294	ext{ mm}$$

Now, we solve for the initial water height $H$:
$$H = 500 - 294 = 206	ext{ mm}$$

This means that initially, the water was filled up to a height of **$206	ext{ mm}$**.

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### Concluding the Journey: The Fall in Water Level

The question asks us to find the **fall in height** of the water level due to the opening of the orifice.

The water level was initially at $H = 206	ext{ mm}$ and dropped to a final steady height of $h = 200	ext{ mm}$.

Therefore, the fall in height is:
$$\Delta H = H - h = 206	ext{ mm} - 200	ext{ mm} = 6	ext{ mm}$$

The water level fell by exactly **$6	ext{ mm}$** before reaching equilibrium.

This is a beautifully clean integer answer!

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### The Way Forward: Deepening Your Intuition

What makes physics so thrilling is that every solved problem is a gateway to new questions.

Let's think about a few fascinating variations of this setup:

1. **What if we did not neglect surface tension?**
   At the bottom orifice, a curved meniscus would form. 
   This meniscus would introduce an additional pressure difference of $\frac{2T}{R}$ due to surface tension, where $T$ is the surface tension and $R$ is the radius of the meniscus. 
   This would slightly alter the equilibrium condition!

2. **What if the temperature of the trapped air changed?**
   If the temperature was raised or lowered, we could no longer use Boyle's Law directly. 
   Instead, we would have to apply the general Ideal Gas Law:
   $$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$$

By mastering these fundamental principles of fluid mechanics and thermodynamics, you build a robust intuition that allows you to conquer even the most complex JEE problems with ease!

Enter Numerical Value:

Visualized Solution

Understanding the Initial State

Key Physical Principles

Pressure Balance at the Orifice

Calculating Final Air Pressure $p$

Applying Boyle's Law

Solving for Initial Water Level $H$

Finding the Fall in Water Level

Exploring Further Variations

The Sigma Insight: Fluid Pressure and Pascal's Law

The Magic of Trapped Air

Analyzing the Setup

The Transition

The Master Equation

Calculating the Final Air Pressure

Connecting the States

Solving for the Initial Height $H$

Concluding the Journey

The Way Forward

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