Two identical cylindrical vessels with their bases at the same level each contain a liquid of densit

Q: Why is the center of mass of the liquid column at $H/2$?

Since the cylindrical vessel is uniform, the mass of the liquid is distributed symmetrically along its height. Therefore, its center of mass lies exactly at the geometric midpoint, which is $H/2$.

Q: Can we solve this by integrating?

Yes! You can integrate $dU = y \cdot dm \cdot g$ from $0$ to $H$, where $dm = \rho A dy$. This yields the same result: $U = \frac{1}{2} \rho A g H^2$.

Q: Why is the work done by gravity positive?

Gravity does positive work because the overall center of mass of the system moves downwards, releasing gravitational potential energy as the levels equalize.

Q: What happens to this lost potential energy?

In a real fluid, this energy is dissipated as heat due to viscous friction as the fluid flows between the vessels.

Q: Does the connecting tube's volume affect the answer?

In this problem, we assume the connecting tube is extremely thin (negligible volume), so its liquid content does not contribute significantly to the potential energy calculations.

Question

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density $\rho$. The height of the liquid in one vessel is $h_1$ and in the other is $h_2$. The area of either base is $A$. What is the work done by gravity in equalising the levels when the two vessels are connected ?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine two identical cylindrical vessels standing side-by-side on a flat table.

Both vessels have the same cross-sectional area $A$ and are filled with an incompressible liquid of density $ho$.

Initially, a closed valve prevents any flow between them, keeping the liquid in the left vessel at a height $h_1$ and the liquid in the right vessel at a height $h_2$.

Because the heights are different, there is a pressure imbalance at the bottom.

Once the valve is opened, gravity will naturally drive the fluid from the higher column to the lower column until the levels equalize at a common height $h$.

### Finding the Equalized Height

Since the liquid is incompressible, the total volume of the liquid is conserved throughout the process.

Let's write down the conservation of volume:

$$V_{	ext{initial}} = V_{	ext{final}}$$

$$A h_1 + A h_2 = 2 A h$$

Dividing both sides by the area $A$, we find the final equalized height $h$ is simply the arithmetic mean of the initial heights:

$$h = \frac{h_1 + h_2}{2}$$

### Gravitational Potential Energy of a Fluid Column

To calculate the work done by gravity, we must find the change in the gravitational potential energy of the system.

For a continuous, uniform fluid column of height $H$, the mass is distributed uniformly.

We can treat the entire mass $m = ho A H$ as if it were concentrated at its center of mass, which lies exactly at its geometric midpoint:

$$y_{	ext{cm}} = \frac{H}{2}$$

Therefore, the gravitational potential energy $U$ of a single column of height $H$ is:

$$U = m g y_{	ext{cm}} = (ho A H) g \left( \frac{H}{2} ight) = \frac{1}{2} ho A g H^2$$

This is a beautiful and powerful result: the potential energy of a uniform fluid column is proportional to the **square of its height**.

### Calculating Initial and Final Potential Energies

Initially, the two columns are separate. The total initial potential energy $U_i$ is the sum of their individual potential energies:

$$U_i = \frac{1}{2} ho A g h_1^2 + \frac{1}{2} ho A g h_2^2 = \frac{1}{2} ho A g (h_1^2 + h_2^2)$$

In the final state, both vessels are filled to the equalized height $h$. The total final potential energy $U_f$ is:

$$U_f = 2 	imes \left( \frac{1}{2} ho A g h^2 ight) = ho A g h^2$$

Now, substituting $h = \frac{h_1 + h_2}{2}$ into this expression:

$$U_f = ho A g \left( \frac{h_1 + h_2}{2} ight)^2 = \frac{1}{4} ho A g (h_1 + h_2)^2$$

### Work Done by Gravity

According to the work-energy theorem, the work done by gravity $W$ is equal to the loss in gravitational potential energy of the system:

$$W = U_i - U_f$$

Let's substitute our expressions for $U_i$ and $U_f$:

$$W = \frac{1}{2} ho A g (h_1^2 + h_2^2) - \frac{1}{4} ho A g (h_1 + h_2)^2$$

To simplify this algebraically, let's factor out $\frac{ho A g}{4}$:

$$W = \frac{ho A g}{4} \left[ 2(h_1^2 + h_2^2) - (h_1 + h_2)^2 ight]$$

Expanding the terms inside the bracket:

$$W = \frac{ho A g}{4} \left[ 2h_1^2 + 2h_2^2 - (h_1^2 + h_2^2 + 2h_1 h_2) ight]$$

$$W = \frac{ho A g}{4} \left[ h_1^2 + h_2^2 - 2h_1 h_2 ight]$$

Recognizing the perfect square identity $h_1^2 + h_2^2 - 2h_1 h_2 = (h_1 - h_2)^2$, we arrive at the final elegant formula:

$$W = \frac{ho A g}{4}(h_1 - h_2)^2$$

This positive work done by gravity represents the energy released as the fluid levels equalize, which is typically dissipated as heat due to viscous friction within the fluid.

Visualized Solution

Visualizing the Initial State

Finding the Equalized Height

Gravitational Potential Energy of a Fluid Column

Total Initial Potential Energy

Total Final Potential Energy

Substituting the Final Height

Work Done by Gravity

Algebraic Simplification

Final Elegant Formula

The Sigma Insight: Fluid Pressure and Pascal's Law

Analyzing the Setup

Finding the Equalized Height

Gravitational Potential Energy of a Fluid Column

Calculating Initial and Final Potential Energies

Work Done by Gravity

Similar Questions

A U-shaped tube contains a liquid of density $ρ$ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column.

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $\frac{d _{1}}{d _{2}}$ is

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $d_{1} / d_{2}$ is

Visualized Solution

Visualizing the Initial State

Finding the Equalized Height

Gravitational Potential Energy of a Fluid Column

Total Initial Potential Energy

Total Final Potential Energy

Substituting the Final Height

Work Done by Gravity

Algebraic Simplification

Final Elegant Formula

The Sigma Insight: Fluid Pressure and Pascal's Law

Analyzing the Setup

Finding the Equalized Height

Gravitational Potential Energy of a Fluid Column

Calculating Initial and Final Potential Energies

Work Done by Gravity

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