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A uniform long tube is bent into a circle of radius $\mathrm{R}$ and it lies in vertical plane. Two liquids of same volume but densities $\rho$ and $\delta$ fill half the tube. The angle $\theta$ is
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(\frac{\rho-\delta}{\rho+\delta}\right)$
$$
\begin{aligned}
& \text { Hints: } \delta g R(\cos \theta+\sin \theta)=\rho g R(\cos \theta-\sin \theta) \\
& \delta \cos \theta+\delta \sin \theta=\rho \cos \theta-\rho \sin \theta \\
& \sin \theta(\delta+\rho)=\cos \theta(\rho-\delta) \\
& \tan \theta=\frac{\rho-\delta}{\rho+\delta}
\end{aligned}
$$
\begin{aligned}
& \text { Hints: } \delta g R(\cos \theta+\sin \theta)=\rho g R(\cos \theta-\sin \theta) \\
& \delta \cos \theta+\delta \sin \theta=\rho \cos \theta-\rho \sin \theta \\
& \sin \theta(\delta+\rho)=\cos \theta(\rho-\delta) \\
& \tan \theta=\frac{\rho-\delta}{\rho+\delta}
\end{aligned}
$$
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