A circular tube of uniform cross sectional is filled with two liquids of density $p_1 and p_2$ such that half of each liquid occupies a quarter of volume of the tube. If the line joining the free surfaces of the liquid makes an angle  $\theta$ with horizontal, find the value of ‘ $\theta$’.

$$\begin{gathered} {\mathrm{h}}_1=n(1-\cos \theta )R\cos \theta P_1=e_{1}g{h}_1\to \left(1\right) \\ \text{ Since two liquid columns of height }{h}_1\text{ and }{h}_2\text{ and density }{e}_1\text{ and }{e}_2\text{ are } \\ \text{ situated point }1\text{, on the left hand side, we have } \\ \begin{array}{l}{\mathrm{P}}_2={\mathrm{e}}_1{\mathrm{gh}}_2+{\mathrm{e}}_2{\mathrm{gh}}_2^1\to \left(2\right)\text{ From (1) and (2) }\\ {\mathrm{e}}_1{\mathrm{h}}_2+{\mathrm{e}}_2{\mathrm{h}}_2^1=e_1{\mathrm{h}}_1{\mathrm{h}}_2^1={\mathrm{R}}^1\sin \theta +R\cos \theta ,h_2=R(1-\cos \theta ) \\ {e}_1\mathrm{R}(1-\cos \theta )\text{ in equation (3) we get } \\ \begin{array}{l}{e}_1\mathrm{R}(1-\cos \theta )+e_{2}R(\sin \theta +\cos \theta )=e_{1}R(1-\cos \theta )\\ \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }=\frac{e_1}{e_2}\tan \theta =\frac{e_1-e_2}{e_1+e_2};\theta ={\tan }^{-1}\left(\frac{e_1-e_2}{e_1+e_2}\right)\end{array}\end{array} \end{gathered}$$