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A circular tube lies in a vertical plane. Two immiscible liquids of densities d1 and d2 fill the tube, each occupying a quarter of the circle (subtending 90 deg at the centre). The radius drawn to their common interface makes an angle alpha with the vertical. Find the ratio d1/d2.
- (1 + tan(alpha))/(1 - tan(alpha))
- (1 + sin(alpha))/(1 - cos(alpha))
- (1 + sin(alpha))/(1 - sin(alpha))
- (1 + cos(alpha))/(1 - cos(alpha))
Correct answer: (1 + tan(alpha))/(1 - tan(alpha))
Solution
Equating hydrostatic pressures of the two quarter columns at the bottom junction leads to d1*(cos(alpha) - sin(alpha))... carrying out the standard derivation gives d1/d2 = (1 + tan(alpha))/(1 - tan(alpha)). This is the classic JEE result.
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