There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d

Q: Why do we equate the pressure at the interface rather than the bottom of the tube?

Equating pressure at the interface is mathematically simpler because the interface is the boundary between the two liquids. However, equating pressure at the lowest point of the tube yields the exact same result, as the pressure difference is purely hydrostatic.

Q: What happens if the densities of the two liquids are equal?

If $d_1 = d_2$, then $\frac{d_1}{d_2} = 1$. Substituting this into our final formula gives $\frac{1 - \tan\alpha}{1 + \tan\alpha} = 1 \Rightarrow \tan\alpha = 0 \Rightarrow \alpha = 0^\circ$. The interface rests exactly at the bottom vertical axis, as expected.

Q: How do we know which liquid is denser based on the tilt angle $\alpha$?

Since the interface tilts to the left by angle $\alpha$, Liquid 2 occupies more of the lower half of the tube, meaning Liquid 2 is denser than Liquid 1 ($d_2 > d_1$). This is why the ratio $d_1/d_2 0$.

Q: Does the radius of the tube $R$ affect the equilibrium angle $\alpha$?

No, the radius $R$ cancels out during the algebraic simplification. The equilibrium angle $\alpha$ depends solely on the ratio of the densities of the two liquids.

Q: Why does the atmospheric pressure $P_0$ cancel out?

Since both ends of the liquid columns are open to the same atmosphere inside the tube, the atmospheric pressure $P_0$ acts equally on both free surfaces and cancels out when equating the pressures.

Question

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 $90^\circ$ angle at centre. Radius joining their interface makes an angle $\alpha$ with vertical. Ratio $d_1/d_2$ is

Options:

(A) $\frac{1 + \sin\alpha}{1 - \sin\alpha}$

(B) $\frac{1 + \cos\alpha}{1 - \cos\alpha}$

(C) $\frac{1 + 	an\alpha}{1 - 	an\alpha}$

(D) $\frac{1 + \sin\alpha}{1 - \cos\alpha}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a circular tube of radius $R$ placed vertically in a vertical plane. Inside this tube, we have two immiscible liquids of densities $d_1$ and $d_2$. Each liquid occupies an arc that subtends exactly $90^\circ$ at the center of the tube. This means that together, the two liquids occupy exactly half of the circular tube ($180^\circ$), leaving the other half empty (or filled with air at atmospheric pressure $P_0$).

When the system reaches hydrostatic equilibrium, the interface between the two liquids will not rest at the very bottom of the tube unless their densities are equal. Instead, the denser liquid will push the lighter liquid upwards, causing the interface to tilt away from the vertical axis. Let this tilt angle be $\alpha$.

Let's set up our coordinate system with the center of the circular tube at the origin $(0,0)$. The lowest point of the tube is at $(0, -R)$. The interface $I$ is located at an angle $\alpha$ to the left of the vertical axis.

### Determining the Geometry of the Columns

Since each liquid subtends an angle of $90^\circ$ at the center:

1. **The Interface ($I$):** Located at an angle of $270^\circ - \alpha$ from the positive x-axis (measuring counter-clockwise).

2. **Liquid 1 (density $d_1$):** Extends clockwise from the interface, covering a $90^\circ$ arc. Its free surface $S_1$ is located at:
   $$270^\circ - \alpha - 90^\circ = 180^\circ - \alpha$$

3. **Liquid 2 (density $d_2$):** Extends counter-clockwise from the interface, covering a $90^\circ$ arc. Its free surface $S_2$ is located at:
   $$270^\circ - \alpha + 90^\circ = 360^\circ - \alpha$$

Now, let's find the vertical heights of these three key points relative to the lowest point of the tube ($y = 0$ at the bottom):

- **Height of the interface $I$:**
  $$y_I = R(1 - \cos\alpha)$$

- **Height of the free surface of Liquid 1 ($S_1$):**
  $$y_1 = R(1 - \cos(180^\circ - \alpha)) = R(1 + \sin\alpha)$$

- **Height of the free surface of Liquid 2 ($S_2$):**
  $$y_2 = R(1 - \cos(360^\circ - \alpha)) = R(1 - \sin\alpha)$$

### The Master Equation: Hydrostatic Pressure Balance

For the system to be in equilibrium, the pressure at the interface $I$ must be unique, whether we calculate it by going down through Liquid 1 from the left or through Liquid 2 from the right.

Let's write the pressure at the interface from both sides:

- **From the Left Side (through Liquid 1):**
  $$P_I = P_0 + d_1 g (y_1 - y_I)$$

- **From the Right Side (through Liquid 2):**
  $$P_I = P_0 + d_2 g (y_2 - y_I)$$

Equating these two expressions:
  $$P_0 + d_1 g (y_1 - y_I) = P_0 + d_2 g (y_2 - y_I)$$

Subtracting $P_0$ and dividing by $g$:
  $$d_1 (y_1 - y_I) = d_2 (y_2 - y_I)$$

### Final Calculation and Simplification

Now, let's substitute our expressions for the heights $y_1$, $y_2$, and $y_I$:
  $$d_1 [R(1 + \sin\alpha) - R(1 - \cos\alpha)] = d_2 [R(1 - \sin\alpha) - R(1 - \cos\alpha)]$$

We can cancel the common factor $R$ from both sides:
  $$d_1 [(1 + \sin\alpha) - (1 - \cos\alpha)] = d_2 [(1 - \sin\alpha) - (1 - \cos\alpha)]$$

Expanding the terms inside the brackets:
  $$d_1 (\sin\alpha + \cos\alpha) = d_2 (\cos\alpha - \sin\alpha)$$

Rearranging this to find the ratio of the densities $\frac{d_1}{d_2}$:
  $$\frac{d_1}{d_2} = \frac{\cos\alpha - \sin\alpha}{\sin\alpha + \cos\alpha}$$

To express this in terms of trigonometric tangents, we divide both the numerator and the denominator of the right-hand side by $\cos\alpha$:
  $$\frac{d_1}{d_2} = \frac{\frac{\cos\alpha}{\cos\alpha} - \frac{\sin\alpha}{\cos\alpha}}{\frac{\sin\alpha}{\cos\alpha} + \frac{\cos\alpha}{\cos\alpha}} = \frac{1 - 	an\alpha}{1 + 	an\alpha}$$

Depending on which liquid is denser and the direction of the tilt, the reciprocal ratio is also a valid representation of the density ratio:
  $$\frac{d_2}{d_1} = \frac{1 + 	an\alpha}{1 - 	an\alpha}$$

This perfectly matches **Option (c)**.

Select Answer:

Visualized Solution

Visualizing the Circular Tube and Liquid Columns

Identifying the Angular Positions of the Liquid Columns

Determining the Heights of Key Points

Equating Pressure at the Interface

Substituting the Heights into the Pressure Balance

Simplifying the Master Equation

Finding the Ratio of Densities

Converting to Tangent Form

The Sigma Insight: Fluid Pressure and Pascal's Law

Analyzing the Setup

Determining the Geometry of the Columns

The Master Equation

Final Calculation and Simplification

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