Question 19 of 192

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

A$\theta ={\tan }^{-1}⁡\dfrac{\pi }{2}\left(\dfrac{{\rho }_{1+}{\rho }_{2}}{{\rho }_{1}-{\rho }_{2}}\right)$B$\theta ={\tan }^{-1}⁡\left[\left(\dfrac{{\rho }_{1-}{\rho }_{2}}{{\rho }_{1}+{\rho }_{2}}\right)\right]$C$\theta ={\tan }^{-1}⁡\dfrac{\pi }{2}\left(\dfrac{{\rho }_{2}}{{\rho }_{1}}\right)$D$\theta ={\tan }^{-1}⁡\pi \left(\dfrac{{\rho }_{1}}{{\rho }_{2}}\right)$

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Answer: (B) $\theta ={\tan }^{-1}⁡\left[\left(\dfrac{{\rho }_{1-}{\rho }_{2}}{{\rho }_{1}+{\rho }_{2}}\right)\right]$

Pressure at the point $A$ is the same on both sides of the tube,
${\rho }_{1}gr(\cos \theta -\sin \theta )={\rho }_{2}gr(\sin \theta +\cos \theta )$
$\Rightarrow \dfrac{{\rho }_{1}}{{\rho }_{2}}=\dfrac{\sin \theta +\cos \theta }{\cos \theta -\sin \theta }=\dfrac{\tan \theta +1}{1-\tan \theta }$
$\Rightarrow {\rho }_{1}-{\rho }_{1}\tan \theta ={\rho }_{2}+{\rho }_{2}\tan \theta$
$\Rightarrow \left({\rho }_{1}+{\rho }_{2}\right)\tan \theta ={\rho }_{1}-{\rho }_{2}$

$\theta ={\tan }^{-1}⁡\left(\dfrac{{\rho }_{1}-{\rho }_{2}}{{\rho }_{1}+{\rho }_{2}}\right)$