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Question

A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ρ1 and ρ2 (ρ1>ρ2) fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is:

A
θ=tan1[π2(ρ1ρ2ρ1+ρ2)]
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B
θ=tan1π2(ρ1ρ2ρ1+ρ2)
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C
θ=tan1π(ρ1ρ2)
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D
None of above
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Solution

The correct option is B None of above
Let us find the pressure at the lowest point 1. Since the liquid has density ρ1 and height h1 on the left hand side of point 1, we have
P1=ρ1gh1.............(1)
Since two liquid columns of height h2 and h2 and densities ρ1 and ρ2 are situated above point 1, on the right hand side, we have
P2=ρ1gh2+ρ2gh2..............(2)
Equating P1andP2, we get
ρ1h2+ρ2h2=ρ1h1
Substituting
h2=Rsinθ+Rcosθ, h2=R(1cosθ) and h1=R(1sinθ)
ρ1R(1cosθ)+ρ2R(sinθ+cosθ)=ρ1R(1sinθ)
This gives
tanθ=ρ1ρ2ρ1+ρ2

807762_868149_ans_4866dac222334e289fe439a3364e7d33.png

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