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 center. Radius joining their interface makes an angle α with vertical. Ratio (d₁)/(d₂) is: (a) (1 + sin α)/(1 - sin α) (b) (1 + cos α)/(1 - cos α) (c) (1 + tan α)/(1 - tan α) (d) (1 + sin α)/(1 - cos α)

Answer: (c). Same as a quarter-circle each: heavy liquid d₁ rises r(cos α - sin α) above the interface, light liquid d₂ rises r(cos α + sin α). Equal pressure at the interface: d₁ g r(cos α - sin α) = d₂ g r(cos α + sin α) (d₁)/(d₂) = (cos α + sin α)/(cos α - sin α) = (1 + tan α)/(1 - tan α) So the answer is (c).

There is a circular tube in a vertical plane. Two liquids which do… | JEE Main 2014

Q23JEE Main 2014Difficult

Archimedes' Principle & Floatation

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 center. Radius joining their interface makes an angle $\alpha$ with vertical. Ratio $\frac{d_1}{d_2}$ is:

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Answer: (c)

Same as a quarter-circle each: heavy liquid $d_1$ rises $r(\cos\alpha - \sin\alpha)$ above the interface, light liquid $d_2$ rises $r(\cos\alpha + \sin\alpha)$ .
Equal pressure at the interface: $d_1 g r(\cos\alpha - \sin\alpha) = d_2 g r(\cos\alpha + \sin\alpha)$
$\dfrac{d_1}{d_2} = \dfrac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} = \dfrac{1 + \tan\alpha}{1 - \tan\alpha}$
So the answer is (c).

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