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 $90 °$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $\frac{d_{1}}{d_{2}}$ is

Question

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(3)

Pressure at interface A must be same from both the sides to be in equilibrium

$∴$ $(R \cos α + R \sin α) d_{2} g = (R \cos α - R \sin α) d_{1} g$

$\Rightarrow \frac{d_{1}}{d_{2}} = \frac{\cos α + \sin α}{\cos α - \sin α} = \frac{1 + \tan α}{1 - \tan α}$

Solution

Q.
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 $90 °$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $\frac{d_{1}}{d_{2}}$ is

1 $\frac{1 + \sin α}{1 - \sin α}$

2 $\frac{1 + \cos α}{1 - \cos α}$

3 $\frac{1 + \tan α}{1 - \tan α}$

4 $\frac{1 + \sin α}{1 - \cos α}$

Ans.
(3)

Pressure at interface A must be same from both the sides to be in equilibrium

$∴$ $(R \cos α + R \sin α) d_{2} g = (R \cos α - R \sin α) d_{1} g$

$\Rightarrow \frac{d_{1}}{d_{2}} = \frac{\cos α + \sin α}{\cos α - \sin α} = \frac{1 + \tan α}{1 - \tan α}$

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Solution

Q. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle α with vertical. Ratio d1d2 is

1 1+sinα1-sinα

2 1+cosα1-cosα

3 1+tanα1-tanα

4 1+sinα1-cosα

Ans. (3) Pressure at interface A must be same from both the sides to be in equilibrium ∴ (Rcosα+Rsinα)d2g=(Rcosα-Rsinα)d1g ⇒d1d2=cosα+sinαcosα-sinα=1+tanα1-tanα