A thin uniform tube is bent into a circle of radius r in the virtical plane. Equal volumes of two immiscible liquids, whose densities are ρ 1 and ρ 2 ( ρ 1 ρ 2 ) fill half the circle. The an

Question

A thin uniform tube is bent into a circle of radius r in the virtical plane. Equal volumes of two immiscible liquids, whose densities are ρ 1 ​ and ρ 2 ​ ( ρ 1 ​ > ρ 2 ​ ) fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is

TestHub · Accepted Answer

Pressure at interface A must be same from both the sides to be in equilibrium.
$∴ (R cos θ + R sin θ) ρ_{2} g = (R cos θ - R sin θ) ρ_{1} g$ $\Rightarrow \frac{ρ _{1}}{ρ _{2}} = \frac{c o s θ + s i n θ}{c o s θ - s i n θ} = \frac{1 + t a n θ}{1 - t a n θ}$ $\Rightarrow ρ_{1} - ρ_{1} tan θ = ρ_{2} + ρ_{2} tan θ$ $\Rightarrow (ρ_{1} + ρ_{2}) tan θ = ρ_{1} - ρ_{2}$ $∴ θ = tan^{- 1} (\frac{ρ _{1} - ρ _{2}}{ρ _{1} + ρ _{2}})$ , soclosest answer is (a)

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