JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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
- A \(\theta = {\tan ^{ - 1}}\left[ {\frac{\pi }{2}\left( {\frac{{{\rho _1} - {\rho _2}}}{{{\rho _1} + {\rho _2}}}} \right)} \right]\)
- B \(\theta = {\tan ^{ - 1}}\frac{\pi }{2}\left( {\frac{{{\rho _1} + {\rho _2}}}{{{\rho _1} - {\rho _2}}}} \right)\)
- C \(\theta = {\tan ^{ - 1}}\pi \left( {\frac{{{\rho _1}}}{{{\rho _2}}}} \right)\)
- D \(\theta = {\tan ^{ - 1}}\frac{\pi }{2}\left( {\frac{{{\rho _2}}}{{{\rho _1}}}} \right)\)
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. \(\therefore \left( {R\cos \theta + R\sin \theta } \right){\rho _2}g\) \( = \left( {R\cos \theta - R\sin \theta } \right){\rho _1}g\)…
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