Two non-viscous, incompressible and immiscible liquids of densities rho and 1.5 rho are poured into

Q: Why do we balance the pressure at the interface instead of the bottom-most point of the tube?

You can actually balance the pressure at any point in the tube, including the bottom-most point! However, balancing the pressure at the interface is conceptually cleaner because it directly separates the two immiscible liquids, allowing us to write the height of each column relative to this single boundary.

Q: Why does the heavier liquid occupy the bottom-left part of the tube in equilibrium?

Gravity always seeks to minimize the potential energy of the system. Since the liquid of density $1.5\rho$ is heavier, it naturally sinks to the lowest possible region of the tube, pushing the lighter liquid of density $\rho$ upwards on the other side.

Q: How did we get the vertical displacement of the interface as $h = R\beta \cos\theta$?

When the liquid column is displaced along the arc of the tube by a distance $s = R\beta$, the displacement vector is tangent to the circle at the interface. Since the interface is at an angle $\theta$ to the vertical, the tangent line makes an angle $\theta$ with the horizontal. Therefore, the vertical component of this displacement is $s \cos\theta = R\beta \cos\theta$.

Q: Why is the restoring pressure difference $\Delta p = 2.5 \rho g h$ instead of just $0.5 \rho g h$?

This is a common pitfall! The restoring pressure is not just due to the density difference at the interface. When the column shifts, the heavier liquid rises by $h$ on one side, and the lighter liquid falls by $h$ on the other side. This creates a net head difference of $h$ for both liquids, leading to a total restoring pressure of $\rho_1 g h + \rho_2 g h = 1.5\rho g h + \rho g h = 2.5\rho g h$.

Q: Can we use the torque method instead of the pressure method to find the restoring torque?

Absolutely! In fact, integrating the gravitational torque over the angular distribution of both liquids yields the exact same restoring torque. The pressure method is simply an elegant shortcut that leverages the hydrostatic pressure difference along the tube.

Question

Two non-viscous, incompressible and immiscible liquids of densities $ho$ and $1.5ho$ are poured into the two limbs of a circular tube of radius $R$ and small cross-section kept fixed in a vertical plane as shown in figure. Each liquid occupies one-fourth the circumference of the tube. (1991)

(a) Find the angle $	heta$ that the radius to the interface makes with the vertical in equilibrium position.

(b) If the whole liquid column is given a small displacement from its equilibrium position, show that the resulting oscillations are simple harmonic. Find the time period of these oscillations.

Sigma Percentile · Accepted Answer

### The Magic of Fluids in Motion

Imagine a circular glass tube, bent into a perfect ring of radius $R$, resting vertically in space.

Inside this tube, we pour two immiscible, non-viscous liquids of different densities: $ho$ and $1.5ho$.

Each liquid occupies exactly one-fourth of the tube's circumference, meaning that together they fill exactly half of the tube.

This simple setup leads to a fascinating interplay of gravity, hydrostatic pressure, and rotational dynamics.

Let us embark on a journey to find how this system behaves in equilibrium and how it oscillates when disturbed.

### Part (a): Finding the Equilibrium Configuration

When the liquids are poured into the tube, gravity pulls them downward.

The heavier liquid of density $1.5ho$ naturally sinks to the bottom, but because it is connected to the lighter liquid of density $ho$, they reach a delicate balance.

Let the interface between the two liquids make an angle $	heta$ with the bottom vertical.

To find this angle, we can choose the center of the circle $O$ as our origin and write the vertical heights of the liquid columns.

The interface is at an angular position of $270^\circ + 	heta$.

Its vertical height from the center is:

$$y_{	ext{interface}} = -R \cos	heta$$

The heavier liquid ($1.5ho$) extends by $90^\circ$ to the left, ending at $180^\circ + 	heta$.

Its top surface height is:

$$y_{	ext{top1}} = -R \sin	heta$$

The lighter liquid ($ho$) extends by $90^\circ$ to the right, ending at $360^\circ + 	heta$.

Its top surface height is:

$$y_{	ext{top2}} = R \sin	heta$$

Now, we calculate the vertical height of each liquid column:

$$h_1 = y_{	ext{top1}} - y_{	ext{interface}} = R(\cos	heta - \sin	heta)$$

$$h_2 = y_{	ext{top2}} - y_{	ext{interface}} = R(\cos	heta + \sin	heta)$$

At equilibrium, the hydrostatic pressure exerted by both columns at the interface must balance perfectly:

$$1.5ho g h_1 = ho g h_2$$

Substituting the expressions for $h_1$ and $h_2$:

$$1.5ho g R(\cos	heta - \sin	heta) = ho g R(\cos	heta + \sin	heta)$$

Canceling the common terms $ho, g, R$:

$$1.5(\cos	heta - \sin	heta) = \cos	heta + \sin	heta$$

$$1.5\cos	heta - 1.5\sin	heta = \cos	heta + \sin	heta$$

$$0.5\cos	heta = 2.5\sin	heta \implies 	an	heta = \frac{0.5}{2.5} = \frac{1}{5}$$

Thus, the equilibrium angle is:

$$	heta = 	an^{-1}\left(\frac{1}{5}ight)$$

### Part (b): The Restoring Force and Simple Harmonic Motion

Now, let us disturb this peaceful equilibrium.

Suppose we displace the entire liquid column by a small angle $\beta$ in the counter-clockwise direction.

This displacement shifts the interface along the arc of the tube by a distance $s = R\beta$.

Because the interface is at an angle $	heta$ to the vertical, the vertical component of this displacement is:

$$h = R\beta \cos	heta$$

This vertical shift creates a restoring pressure difference at the interface.

On one side, the heavier liquid has risen, and on the other side, the lighter liquid has fallen.

The net restoring pressure difference $\Delta p$ is:

$$\Delta p = 1.5ho g h + ho g h = 2.5ho g h = 2.5ho g R\beta \cos	heta$$

This pressure difference acts on the cross-sectional area $A$ of the tube, creating a restoring force along the tube:

$$F = -\Delta p A = -2.5ho g A R\beta \cos	heta$$

The restoring torque $	au$ about the center of the tube is:

$$	au = F R = -2.5ho g A R^2 \cos	heta \beta$$

Notice how the restoring torque is directly proportional to the angular displacement $\beta$.

This is the classic signature of simple harmonic motion!

### Calculating the Moment of Inertia

To find the angular acceleration, we must determine the total moment of inertia $I$ of the liquid column.

Since each liquid occupies exactly one-fourth of the circumference, their masses are:

$$m_1 = 1.5ho A \left(\frac{2\pi R}{4}ight) = 0.75\piho A R$$

$$m_2 = ho A \left(\frac{2\pi R}{4}ight) = 0.5\piho A R$$

The total mass of the liquid column is:

$$M = m_1 + m_2 = 1.25\piho A R$$

Since all parts of the liquid are at a distance $R$ from the center, the moment of inertia is simply:

$$I = M R^2 = 1.25\piho A R^3$$

### The Final Symphony: Finding the Time Period

Using the rotational form of Newton's second law, $I\alpha = 	au$:

$$1.25\piho A R^3 \alpha = -2.5ho g A R^2 \cos	heta \beta$$

Solving for the angular acceleration $\alpha$:

$$\alpha = -\left(\frac{2g\cos	heta}{\pi R}ight)\beta$$

Comparing this with the standard SHM equation $\alpha = -\omega^2 \beta$, we find the angular frequency squared:

$$\omega^2 = \frac{2g\cos	heta}{\pi R}$$

Since $	an	heta = 1/5$, we can find $\cos	heta$ using the right-triangle relationship:

$$\cos	heta = \frac{5}{\sqrt{26}} \approx 0.98$$

Substituting $g = 9.8	ext{ m/s}^2$ and $\cos	heta \approx 0.98$:

$$\omega^2 = \frac{2 	imes 9.8 	imes 0.98}{3.1416 R} \approx \frac{6.11}{R}$$

Finally, the time period of oscillation $T$ is:

$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{6.11}}$$

This elegant result shows that the liquid column oscillates with a time period proportional to the square root of the tube's radius, a beautiful testament to the harmony of fluid mechanics and wave motion!

Visualized Solution

Visualizing the Equilibrium Setup

Expressing the Heights of the Liquid Columns

Applying the Pressure Balance Condition

Solving for the Equilibrium Angle $θ$

Introducing a Small Angular Displacement $β$

Calculating the Restoring Pressure and Force

Finding the Restoring Torque

Determining the Moment of Inertia of the Liquid Column

Setting up the Equation of Motion

Calculating the Time Period of Oscillation

The Sigma Insight: Simple Harmonic Motion (SHM)

The Magic of Fluids in Motion

Part (a)

Part (b)

Calculating the Moment of Inertia

The Final Symphony

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