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$\rho_{1}=\rho_{1} g h_{1}$
$\rho_{2}=\rho_{1} g h_{2}+\rho_{2} g h_{2}'$
Now $\rho_{1}+\rho_{2} \Rightarrow \rho_{1} g h_{1}=\rho_{1} g h_{2}+\rho_{2} g h_{2}'$
$\Rightarrow \rho_{1} h_{1}=\rho_{1} h_{2}+\rho_{2} g h_{2}'$
Now $h_{1}=R(1-\sin \theta) ; h_{2}=R(1-\sin \theta) ; h_{2}^{\prime}=R \sin \theta +R \cos \theta$
$\Rightarrow \rho_{1} R(1-\sin \theta)=\rho_{1} R(1-\cos \theta)+\rho_{2}(R \sin \theta-R \cos \theta)$
$\Rightarrow \rho_{1} R-\rho_{1} R \sin \theta=\rho_{1} R-\rho_{1} \cos \theta+\rho_{2} R \sin \theta+\rho_{2} R \cos \theta$
$\Rightarrow \left(\rho_{1}+\rho_{2}\right) R \sin \theta-\left(\rho_{1}-\rho_{2}\right) R \cos \theta=0$
$\Rightarrow \left(\rho_{1}+\rho_{2}\right) R \sin \theta=\left(\rho_{1}-\rho_{2}\right) R \cos \theta$
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{\rho_{1}-\rho_{2}}{\rho_{1}+\rho_{2}}$
$\Rightarrow \tan \theta=\frac{\rho_{1}-\rho_{2}}{\rho_{1}+\rho_{2}}$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{\rho_{1}-\rho_{2}}{\rho_{1}+\rho_{2}}\right)$