Two non - viscous, incompressible and immiscible liquids of densities (rho) and (1.5 rho) 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 fig. Each liquid occupies one fourth the cirumference of the tube.
Find the angle (theta) that the radius to the interface makes with the verticles in equilibrium position.

Explanation

To solve the problem, we need to find the angle \( \theta \) that the radius to the interface makes with the vertical in equilibrium position for two immiscible liquids of different densities in a circular tube. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have two liquids with densities \( \rho \) and \( 1.5\rho \) occupying one-fourth of the circumference of a circular tube. The tube is vertical, and we need to find the angle \( \theta \) that the radius to the interface makes with the vertical.

### Step 2: Identify the Geometry
Since each liquid occupies one-fourth of the circumference, the angle at the center for each liquid is \( 90^\circ \). The angle \( \theta \) is the angle made by the radius to the interface with the vertical. Therefore, the other angle at the interface will be \( 90^\circ - \theta \).

### Step 3: Apply Pressure Balance
At equilibrium, the pressure at the same horizontal level in both liquids must be equal. We denote the heights of the liquid columns as \( h_1 \) for the liquid with density \( 1.5\rho \) and \( h_2 \) for the liquid with density \( \rho \).

### Step 4: Determine Heights
From the geometry of the setup:
- The height \( h_1 \) (for the denser liquid) can be expressed as:
\[ h_1 = r \cos \theta - r \sin \theta = r(\cos \theta - \sin \theta) \]
- The height \( h_2 \) (for the less dense liquid) can be expressed as:
\[ h_2 = r \sin \theta + r \cos \theta = r(\sin \theta + \cos \theta) \]

### Step 5: Set Up the Pressure Equation
Using the hydrostatic pressure formula, we equate the pressures at the same horizontal level:
\[ 1.5\rho g h_1 = \rho g h_2 \]
Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ 1.5\rho g [r(\cos \theta - \sin \theta)] = \rho g [r(\sin \theta + \cos \theta)] \]
We can cancel \( \rho g r \) from both sides:
\[ 1.5(\cos \theta - \sin \theta) = \sin \theta + \cos \theta \]

### Step 6: Simplify the Equation
Rearranging gives:
\[ 1.5\cos \theta - 1.5\sin \theta = \sin \theta + \cos \theta \]
Combining like terms:
\[ 1.5\cos \theta - \cos \theta = 1.5\sin \theta + \sin \theta \]
This simplifies to:
\[ 0.5\cos \theta = 2.5\sin \theta \]

### Step 7: Solve for \( \tan \theta \)
Dividing both sides by \( \cos \theta \):
\[ 0.5 = 2.5\tan \theta \]
Thus:
\[ \tan \theta = \frac{0.5}{2.5} = \frac{1}{5} \]

### Step 8: Find \( \theta \)
Finally, we find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]

### Final Answer
The angle \( \theta \) that the radius to the interface makes with the vertical in equilibrium position is:
\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]