Equal volumes of two non viscous incompressible and immiscible liquids of densities ρ₁ and ρ₂ and are poured into two limbs of a circular tube of radius R kept vertical such that half of the tube is filled with liquid. The angular position of interface from vertical is

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\large tan^{-1}\left ( \frac {\rho_2} {\rho_1} \right )

\large tan^{-1}\left ( \frac {\rho_1+\rho_2}{\rho_1-\rho_2} \right )

$\large sin^{-1}\left ( \frac {\rho_1-\rho_2}{\rho_1+\rho_2} \right )$

answer is A.

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Detailed Solution

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h₁ = R(1 - sinθ); h₂ = R(1 - cosθ); h₃ = R(1 + sinθ)
$\large \therefore$ Pressure at the interface
$\large \Rightarrow \rho_1g(h_1-h_2)=\rho_2g(h_3-h_2)$
$\large \Rightarrow \rho_1(cos\theta-sin\theta)=\rho_2(sin\theta+cos\theta)$
$\large \Rightarrow \frac {cos\theta+sin\theta}{cos\theta-sin\theta}=\frac {\rho_1}{\rho_2}$
$\large \Rightarrow cot\theta = \frac {\rho_1+\rho_2}{\rho_1-\rho_2}\Rightarrow \theta=tan^{-1}\left ( \frac {\rho_1-\rho_2}{\rho_1+\rho_2} \right )$