JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2018
- Hard
- MCQ
A thin uniform tube is bent into a circle of radius \(r\) in the virtical plane. Equal volumes of two immiscible liquids, whose densities are \({\rho _1}\) and \({\rho _2}\left( {{\rho _1} > {\rho _2}} \right)\) fill half the circle. The angle \(\theta\) between the radius vector passing through the common interface and the vertical is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\theta = {\tan ^{ - 1}}\left[ {\frac{\pi }{2}\left( {\frac{{{\rho _1} - {\rho _2}}}{{{\rho _1} + {\rho _2}}}} \right)} \right]\)
Step-by-step Solution
Detailed explanation
Pressure at interface \(A\) must be same from both the sides to be in equilibrium.
\(\therefore \left( {R\cos \theta + R\sin \theta } \right){\rho _2}g\)
\( = \left( {R\cos \theta - R\sin \theta } \right){\rho _1}g\)
\( \Rightarrow \frac{{{\rho _1}}}{{{\rho _2}}} = \frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }} = \frac{{1 + \tan \theta }}{{1 - \tan \theta }}\)
\( \Rightarrow {\rho _1} - {\rho _1}\,\tan \,\theta = {\rho _2} + {\rho _2}\,\tan \,\theta \)
\( \Rightarrow \left( {{\rho _1} + {\rho _2}} \right)\tan \theta = {\rho _1} - {\rho _2}\)
\(\therefore \theta = {\tan ^{ - 1}}\left( {\frac{{{\rho _1} - {\rho _2}}}{{{\rho _1} + {\rho _2}}}} \right),so\,closest\,answer\,is\left( a \right)\)
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