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 = tan ⁻¹ π ((ρ₁)/(ρ₂)) (b) θ = tan ⁻¹ π/2 ((ρ₁)/(ρ₂)) (c) θ = tan ⁻¹ ((ρ₁ - ρ₂)/(ρ₁ + ρ₂)) (d) θ = tan ⁻¹ π/2 ((ρ₁ + ρ₂)/(ρ₁ - ρ₂))

Answer: (c). θρ₁ (heavy)ρ₂surface 2surface 1interface Each liquid fills a quarter circle ( 90°). With the interface at angle θ from the vertical, the free surfaces are at angles θ - 90° and θ + 90°. Height of heavy liquid ( ρ₁) above the interface: r(cos θ - sin θ); of light liquid ( ρ₂): r(cos θ + sin θ). Equal pressure at the interface: ρ₁ g r(cos θ - sin θ) = ρ₂ g r(cos θ + sin θ) tan θ = (ρ₁ - ρ₂)/(ρ₁ + ρ₂), so the answer is (c).

A thin uniform tube is bent into a circle of radius r in the vertical… | JEE Main 2018 (15 Apr, Shift 1)

Q20JEE Main 2018 · 15 APR · SHIFT 1Difficult

Archimedes' Principle & Floatation

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:

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Answer: (c)

Each liquid fills a quarter circle ( $90^\circ$ ). With the interface at angle $\theta$ from the vertical, the free surfaces are at angles $\theta - 90^\circ$ and $\theta + 90^\circ$ .
Height of heavy liquid ( $\rho_1$ ) above the interface: $r(\cos\theta - \sin\theta)$ ; of light liquid ( $\rho_2$ ): $r(\cos\theta + \sin\theta)$ .
Equal pressure at the interface: $\rho_1 g r(\cos\theta - \sin\theta) = \rho_2 g r(\cos\theta + \sin\theta)$
$\tan\theta = \dfrac{\rho_1 - \rho_2}{\rho_1 + \rho_2}$ , so the answer is (c).

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