Two identical cylindrical vessels are kept on the ground and each contains the same liquid of densit

Q: Why do we use $h/2$ for the height in the potential energy formula instead of just $h$?

Because the liquid is an extended continuous body, not a point mass. Gravity acts on every drop of water from the bottom to the top. To find the total potential energy, we must use the height of the object's center of mass. For a uniform cylinder, the center of mass is exactly in the middle, at $h/2$.

Q: The final energy is less than the initial energy. Doesn't this violate the conservation of energy?

Not at all! The *mechanical* energy (potential + kinetic) is not conserved because non-conservative forces are at play. As the fluid flows, internal friction (viscosity) and turbulence dissipate mechanical energy into thermal energy (heat). The *total* energy of the universe is conserved, but the macroscopic potential energy decreases.

Q: Can we use Bernoulli's equation to solve this problem?

No, Bernoulli's equation is strictly for steady, streamline flow where mechanical energy is conserved. The process of two tanks equalizing is a transient, highly dissipative process. We must rely on basic work-energy principles and center of mass calculations instead.

Q: Does the position of the connecting pipe matter?

Yes, the problem specifies the pipe is 'very close to the bottom'. If the pipe were higher up, only the liquid above the pipe would be able to flow and equalize. The liquid below the pipe would remain trapped, changing the final equilibrium state.

Q: What would happen if the two liquids had different densities?

If the liquids were immiscible and had different densities ($d_1$ and $d_2$), they wouldn't just level out to the same height. Instead, they would reach a hydrostatic equilibrium where the pressure at the bottom of both tanks is equal, meaning $d_1 g h_1 = d_2 g h_2$.

Question

Two identical cylindrical vessels are kept on the ground and each contains the same liquid of density $d$. The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x_1$ and in the other, $x_2$. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is

Options:

(A) $gdS(x_2^2 + x_1^2)$

(B) $\frac{3}{4}gdS(x_2 - x_1)^2$

(C) $gdS(x_2 + x_1)^2$

(D) $\frac{1}{4}gdS(x_2 - x_1)^2$

Sigma Percentile · Accepted Answer

### The Physics of Connecting Vessels: Where Does the Energy Go?

Imagine a classic physics thought experiment: you have two identical cylindrical tanks resting on the ground. One tank is filled with a liquid up to a height $x_1$, and the other is filled with the exact same liquid up to a different height $x_2$. Both tanks share the same base area $S$ and the liquid has a density $d$.

Now, you connect them at the very bottom with a thin pipe. Intuition tells us that the liquid will rush from the higher tank to the lower one until the levels equalize. But what happens to the energy of the system during this chaotic rush? Let's break down the mathematics of this transient process.

### The Quest for Equilibrium (Conservation of Volume)

When the valve opens, the liquid flows until hydrostatic equilibrium is reached, meaning the final height $x_f$ is the same in both vessels. Because the pipe's volume is negligible, the total volume of the liquid must remain strictly conserved.

We can set up a simple volume equation. The initial volume is the sum of the volumes in the two separate tanks:

$V_{	ext{initial}} = S x_1 + S x_2$

The final volume is the sum of the volumes at the new equilibrium height:

$V_{	ext{final}} = S x_f + S x_f = 2 S x_f$

Equating the two, we can easily cancel out the common base area $S$:

$S(x_1 + x_2) = 2 S x_f$

$x_f = \frac{x_1 + x_2}{2}$

Unsurprisingly, the final equilibrium height is simply the arithmetic mean of the two initial heights.

### The Hidden Center of Mass (Potential Energy of Fluids)

To find the change in energy, we must calculate the gravitational potential energy of the liquid columns. A common mistake is to just use $mgh$ where $h$ is the top surface. However, for an extended body like a column of liquid, we must assume its entire mass is concentrated at its **center of mass**.

For a uniform cylinder of height $h$, the center of mass is located exactly halfway up, at $\frac{h}{2}$. Therefore, the potential energy $U$ of a single liquid column is:

$U = M g h_{	ext{COM}} = (	ext{Volume} \cdot 	ext{density}) \cdot g \cdot \left(\frac{h}{2}ight)$

$U = (S \cdot h \cdot d) \cdot g \cdot \left(\frac{h}{2}ight) = \frac{1}{2} d S g h^2$

This quadratic dependence on height is the key to unlocking the energy difference.

### The Mathematical Showdown (Calculating the Change)

Let's calculate the total initial potential energy by summing the energies of the two separate columns:

$U_{	ext{initial}} = \frac{1}{2} d S g x_1^2 + \frac{1}{2} d S g x_2^2 = \frac{1}{2} d S g (x_1^2 + x_2^2)$

Similarly, the final potential energy is the sum of the energies of the two columns at the new height $x_f$:

$U_{	ext{final}} = \frac{1}{2} d S g x_f^2 + \frac{1}{2} d S g x_f^2 = d S g x_f^2$

Now, we substitute our expression for $x_f$ into the final energy equation:

$U_{	ext{final}} = d S g \left( \frac{x_1 + x_2}{2} ight)^2 = \frac{d S g}{4} (x_1^2 + x_2^2 + 2x_1 x_2)$

To find the change in energy $\Delta U$, we subtract the initial energy from the final energy. Taking a common denominator of 4, we get:

$\Delta U = \frac{d S g}{4} (x_1^2 + x_2^2 + 2x_1 x_2) - \frac{2 d S g}{4} (x_1^2 + x_2^2)$

$\Delta U = \frac{d S g}{4} [x_1^2 + x_2^2 + 2x_1 x_2 - 2x_1^2 - 2x_2^2]$

$\Delta U = \frac{d S g}{4} (-x_1^2 - x_2^2 + 2x_1 x_2)$

Factoring out a negative sign reveals a beautiful perfect square:

$\Delta U = -\frac{1}{4} d S g (x_1^2 + x_2^2 - 2x_1 x_2)$

$\Delta U = -\frac{1}{4} d S g (x_2 - x_1)^2$

### The Missing Energy Mystery

The negative sign in our final result is profound. It dictates that $U_{	ext{final}}$ is strictly less than $U_{	ext{initial}}$ (unless $x_1 = x_2$, in which case nothing happens). Energy has been lost from the macroscopic system!

Where did this energy go? It wasn't destroyed; it was dissipated. As the liquid rushed through the connecting pipe, it experienced internal friction (viscosity) and likely formed turbulent eddies. The lost potential energy was converted entirely into work done against these viscous forces, ultimately manifesting as a slight increase in the thermal energy (heat) of the liquid.

Select Answer:

Visualized Solution

Visualizing the Initial Setup

Conservation of Volume

Setting up the Volume Equation

Calculating Final Height $x_{f}$

Potential Energy of a Liquid Column

Initial Potential Energy

Final Potential Energy

Substituting $x_{f}$ into Final Energy

Calculating Change in Energy $Δ U$

Final Simplification

The Way Forward

The Sigma Insight: Fluid Pressure and Pascal's Law

The Physics of Connecting Vessels

The Quest for Equilibrium (Conservation of Volume)

The Hidden Center of Mass (Potential Energy of Fluids)

The Mathematical Showdown (Calculating the Change)

The Missing Energy Mystery

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