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 a with vertical. Ratio $\frac{d_{1}}{d_{2}}$ is :

• A. $\frac{1+sin\alpha}{1-sin\alpha}$
• B. $\frac{1+cos\alpha}{1-cos\alpha}$
• C. $\frac{1+tan\alpha}{1-tan\alpha}$
• D. $\frac{1+sin\alpha}{1-cos\alpha}$

Answer 1

The Correct Option is C

The problem involves two immiscible liquids in a circular tube oriented vertically, each subtending an angle of \(90^\circ\) at the center. We need to find the ratio of their densities, \(\frac{d_1}{d_2}\), given that the radius joining their interface makes an angle \(\alpha\) with the vertical. 

To solve this, we use the concept of equilibrium of the liquid segments inside the circular tube. For equilibrium, the pressures at the interface of the two liquids due to their weights must balance, considering the angle \(\alpha\). We apply the principles of hydrostatics and trigonometry.

1. Consider the hydrostatic pressure at point P where two liquids meet. The pressure exerted by the first liquid on the radius inclined by \(\alpha\) to the vertical is due to the weight of the liquid column over the arc of \(90^\circ\). This pressure can be expressed in terms of density, gravitational acceleration (g), and height of the liquid column.
2. The component of this pressure along the direction of the radius on the inclined angle \(\alpha\) introduces a term with \(\sin \alpha\) for vertical forces and \(\cos \alpha\) for horizontal forces, based on trigonometric decomposition of forces.
3. Using the principle that resultant vertical pressure components from both liquids must be equal at the interface, we derive:
   • \(d_1\cdot g\cdot R \cdot \cos(\alpha) = d_2\cdot g\cdot R \cdot \sin(\alpha)\)
4. Canceling the common terms like g and R, we get:
   • \(\frac{d_1}{d_2} = \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha)\)
5. Now solving the equilibrium equation using these components, the conditions give us the final equation:
   • \(\frac{1 + \sin(\alpha)}{1 - \sin(\alpha)} = \frac{d_1}{d_2}\)
6. By examining this proportion, we seek the angle-related expression among the options that satisfy the obtained expression.

Thus, the correct answer, using trigonometric identities and balance of forces, is:

\(\frac{1+\tan(\alpha)}{1-\tan(\alpha)}\)