A thin uniform tube is bent into a circle of radius r in the...

Pressure at interface A must be same from both the sides to be in equilibrium.
$$\begin{aligned} \therefore \quad &(\mathrm{R} \cos \theta+\mathrm{R} \sin \theta) \rho_2 \mathrm{~g} \\ &=(\mathrm{R} \cos \theta-\mathrm{R} \sin \theta) \rho_1 \mathrm{~g} \end{aligned}$$\Rightarrow \quad \frac{\rho_1}{\rho_2}=\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{1+\tan \theta}{1-\tan \theta} \Rightarrow \rho_1-\rho_1 \tan \theta=\rho_2+\rho_2 \tan \theta \Rightarrow \quad\left(\rho_1+\rho_2\right)$\tan \theta=\rho_1-\rho_2 \therefore \quad \theta=$\tan ^{-1}\left(\frac{\rho_1-\rho_2}{\rho_1+\rho_2}\right), soclosest answer is (a)$