Animated Solution for Physics - Properties of Solids and Liquids: 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 $\frac{d_1}{d_2}$ is

Hydrostatics in a Circular Tube

Imagine you are standing at the lowest point of this circular tube, looking up at the two liquid columns. The beauty of hydrostatics is that it doesn't care about the curved shape of the container; it only cares about the vertical heights of the liquid columns above you.

Let's break down this elegant problem by tracing the journey of each liquid and applying the fundamental principle of equal pressure.

Visualizing the Geometry

We have a circular tube containing two immiscible liquids of densities $d_1$ and $d_2$. Each liquid subtends exactly $90^{\circ }$ at the center of the tube. The interface between them is shifted by an angle $\alpha$ from the lowest vertical point.

Because the interface is shifted, Liquid 1 (with density $d_1$) crosses the lowest point of the tube, occupying a portion of the right side before extending up the left side. Liquid 2 (with density $d_2$) sits entirely on the right side, resting above the interface.

For the fluid to be in static equilibrium, the pressure exerted by the liquid column on the left side of the lowest point (let's call it point $A$) must perfectly balance the pressure exerted by the liquid column on the right side.

To find these pressures, we need to determine the vertical heights of the liquid columns relative to point $A$.

Let's analyze the left side of point $A$. Liquid 1 extends from the bottom up to its free surface. Since the interface is at an angle $\alpha$ from the bottom, and Liquid 1 subtends $90^{\circ }$, its top surface is at an angle of $180^{\circ }+\alpha$. Using basic trigonometry, the vertical height of this left column from point $A$ is $R(1-\sin \alpha )$. Therefore, the pressure from the left is:

Now, let's focus on the right side. This side contains a small portion of Liquid 1 (from point $A$ up to the interface) and the entire $90^{\circ }$ column of Liquid 2. The interface is at a height of $R(1-\cos \alpha )$ from the bottom. The top of Liquid 2 reaches a height of $R\sin \alpha$ above the center, making its total column height $R(\sin \alpha +\cos \alpha )$. Adding these contributions, we get:

Equating the pressures from both sides, we notice that the atmospheric pressure $P_{atm}$ cancels out immediately.

We can also cancel out the acceleration due to gravity $g$ and the radius $R$ from all terms. Now, let's group the terms with density $d_1$ on one side:

Notice how the $1$s cancel out beautifully on the left side, leaving us with:

Finally, we rearrange the terms to find the ratio $\frac{d_1}{d_2}$:

To match the format of the given options, we divide both the numerator and the denominator by $\cos \alpha$:

And there we have it! A perfect geometric harmony translated into a clean algebraic ratio. Always remember, in hydrostatics, vertical height is the ultimate kingmaker.