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← JEE Main 2014 · 6 Apr · Shift 1
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JEE Main 2014 · 6 Apr · Shift 1, Physics question 9

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1d_1 and d2d_2 are filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle α\alpha with vertical. Ratio d1d2\dfrac{d_1}{d_2} is :

A circular tube in a vertical plane. Liquid d₁ (dark) fills a 90° arc at the bottom left and liquid d₂ (dotted) the adjacent 90° arc on the right; the radius to their interface, at the bottom just right of the lowest point, makes angle α with the vertical. A dashed diameter marks the outer ends of the two liquids.
Choose one option
Answer and worked solution

Official answer: C

Measure angles from the lowest point, and let RR be the tube radius. The interface is at angle α\alpha (height R(1−cos⁡α)R(1-\cos\alpha)). The two liquids each span 90∘90^\circ, so their free ends are at α−90∘\alpha-90^\circ and α+90∘\alpha+90^\circ, at heights R(1−sin⁡α)R(1-\sin\alpha) for d1d_1 and R(1+sin⁡α)R(1+\sin\alpha) for d2d_2. Both free surfaces are at the same pressure.

Equate the pressures at the interface:
d1gR(cos⁡α−sin⁡α)=d2gR(sin⁡α+cos⁡α)  ⟹  d1d2=cos⁡α+sin⁡αcos⁡α−sin⁡α=1+tan⁡α1−tan⁡α.d_1gR(\cos\alpha-\sin\alpha)=d_2gR(\sin\alpha+\cos\alpha)\implies\frac{d_1}{d_2}=\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}=\frac{1+\tan\alpha}{1-\tan\alpha}.

Topics:Pressure and Pascal's law
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