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 $${ \rho }_{ 1 }$$ and $${ \rho }_{ 2 }$$ $$({ \rho }_{ 1 }> { \rho }_{ 2 })$$ fill half the circle. The angle $$\theta$$ between the radius vector passing through the common interface and the vertical is:

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

Solution

## The correct option is B None of aboveLet us find the pressure at the lowest point 1. Since the liquid has density $$\rho_1$$ and height $$h_1$$ on the left hand side of point 1, we have$$P_1 = \rho_1 g h_1$$.............(1)Since two liquid columns of height $$h_2 \ and \ h'_2$$ and densities $$\rho_1 \ and \ \rho_2$$ are situated above point 1, on the right hand side, we have$$P_2 = \rho_1 g h_2 + \rho_2 g h'_2$$..............(2)Equating $$P_1 and P_2$$, we get$$\rho_1 h_2 + \rho_2 h'_2 = \rho_1 h_1$$Substituting$$h'_2 = R sin\theta +R cos \theta, \ h_2 = R(1 - cos \theta) \ and \ h_1 = R (1 - sin \theta)$$$$\rho_1 R (1 - cos \theta) + \rho_2 R (sin \theta + cos\theta) = \rho_1R (1 - sin \theta)$$ This gives$$tan \theta = \dfrac{\rho_1 - \rho_2}{\rho_1 + \rho_2}$$Physics

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