1. Understand the geometry

A circular tube lies in a vertical plane. Two immiscible liquids of densities $d_1$ and $d_2$ each occupy an arc of $90^\circ$.

So there are two interfaces between the liquids, diametrically separated in the circular tube.

Let the radius to one interface make angle $\alpha$ with the vertical. Since each liquid occupies $90^\circ$, the other interface for the same liquid is $90^\circ$ away along the circle.

2. Condition for hydrostatic equilibrium

Take the two interfaces as points $A$ and $B$. Pressure must be continuous at each interface.

If we move from $A$ to $B$ through the liquid of density $d_1$, and also from $A$ to $B$ through the liquid of density $d_2$, the pressure difference between $A$ and $B$ must be the same in both paths.

Thus,

where $\Delta h_1$ and $\Delta h_2$ are the vertical height differences between the two interfaces measured through the respective liquid columns.

Since $g$ cancels,

3. Find vertical height differences

Let the tube radius be $R$.

For a point on the circle whose radius makes angle $\theta$ with vertical, its vertical coordinate is

if angle is measured from upward vertical.

One interface is at angle $\alpha$, so its vertical coordinate is

The other interface is $90^\circ$ away. Depending on orientation, its vertical coordinate becomes

Hence the vertical difference between the two interfaces through one liquid is

For the other liquid, the path is through the other $90^\circ$ arc, whose endpoints are the same interfaces but the effective vertical difference corresponds to the remaining geometry:

A cleaner way is to note the two arcs joining the same interfaces correspond to central angles $90^\circ$ and $270^\circ$, but each liquid occupies $90^\circ$. The correct pair of interface coordinates are actually diametrically opposite ends of each $90^\circ$ liquid segment.

Let us assign the two interfaces at angular positions such that one liquid occupies from $\alpha$ to $\alpha+90^\circ$. Then for this liquid,

The other liquid occupies the opposite $90^\circ$ arc, from $\alpha+180^\circ$ to $\alpha+270^\circ$. So

Using identities,

therefore

This symmetric approach does not distinguish densities, so instead we should use the standard pressure balance between the two interfaces around the full loop.

4. Pressure balance around the loop

Suppose the two interfaces are at vertical heights corresponding to radii making angles $\alpha$ and $90^\circ-\alpha$ in effect. For a circular tube with two liquids each occupying quarter-circle arcs, the equilibrium condition gives

or equivalently,

5. Match with options

Thus,

So the correct option is:

6. Compare with stored answer

Stored correct answer is C, i.e.

which does not match the hydrostatic geometry result.

Hence I disagree with the stored answer.