Question

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 ${\rho }_2({\rho }_1>{\rho }_2)$, fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is :

• A. $\theta ={\tan }^{-1}\pi \left(\frac{{\rho }_1}{{\rho }_2}\right)$
• B. $\theta ={\tan }^{-1}\frac{\pi }{2}\left(\frac{{\rho }_1}{{\rho }_2}\right)$
• C. $\theta ={\tan }^{-1}\left[\frac{\pi }{2}\left(\frac{{\rho }_1-{\rho }_2}{{\rho }_1+{\rho }_2}\right)\right]$
• D. None of the above

Answer 1

Answer: (D)

We have${\rho }_1={\rho }_{1}g{h}_1$
${\rho }_2={\rho }_{1}g{h}_2+{\rho }_{2}g{h}_2^{\prime }$
Now ${\rho }_1+{\rho }_2\Rightarrow {\rho }_{1}g{h}_1={\rho }_{1}g{h}_2+{\rho }_{2}g{h}_2^{\prime }$
$\Rightarrow {\rho }_1{h}_1={\rho }_1{h}_2+{\rho }_{2}g{h}_2^{\prime }$
Now $h_1=R(1-\sin \theta );h_2=R(1-\sin \theta );h_2^{\prime }=R\sin \theta +R\cos \theta$
$\Rightarrow {\rho }_{1}R(1-\sin \theta )={\rho }_{1}R(1-\cos \theta )+{\rho }_2(R\sin \theta -R\cos \theta )$
$\Rightarrow {\rho }_{1}R-{\rho }_{1}R\sin \theta ={\rho }_{1}R-{\rho }_1\cos \theta +{\rho }_{2}R\sin \theta +{\rho }_{2}R\cos \theta$
$\Rightarrow \left({\rho }_1+{\rho }_2\right)R\sin \theta -\left({\rho }_1-{\rho }_2\right)R\cos \theta =0$
$\Rightarrow \left({\rho }_1+{\rho }_2\right)R\sin \theta =\left({\rho }_1-{\rho }_2\right)R\cos \theta$
$\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{{\rho }_1-{\rho }_2}{{\rho }_1+{\rho }_2}$
$\Rightarrow \tan \theta =\frac{{\rho }_1-{\rho }_2}{{\rho }_1+{\rho }_2}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{{\rho }_1-{\rho }_2}{{\rho }_1+{\rho }_2}\right)$