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  • Need explanation for: - Properties of Solids and Liquids - JEE Main

thin uniform tube is bent into a circle of
radius r in the vertical plane. Equal
volumes of two immiscible liquids, whose
densities are p_{1} and p_{2}(p_{1}>p_{2} ), fill half the
circle. The angle θ between the radius
vector passing through the common
interface and the vertical is :

  • Option 1)

    \theta = \tan ^{-1}\pi\left ( \frac{P_{1}}{P_{2}} \right )

     

     

     

  • Option 2)

    \theta = \tan ^{-1}\frac{\pi }{2}\left ( \frac{P_{2}}{P_{1}} \right )

  • Option 3)

    \theta =\tan ^{-1}\left [ \frac{\pi }{2}\left ( \frac{P_{1}-P_{2}}{P_{1}+P_{2}} \right ) \right ]

  • Option 4)

    \theta =\tan ^{-1}\frac{\pi }{2}\left [ \left ( \frac{P_{1}+P_{2}}{P_{1}-P_{2}} \right ) \right ]

 

Answers (1)

best_answer

As we learned

Absolute Pressure -

P= P_{0}+\rho\: gh
 

- wherein

P\rightarrow hydrostatics \: Pressure

P_{0}\rightarrow atmospheric \: Pressure

 

 equating presure at point A

p_{1} g R(\cos \theta -\sin \theta ) = p_{2} gR(\sin \theta +\cos \theta )

\frac{p_{1}}{p_{2}}=\frac{\sin \theta +\cos \theta }{\cos \theta -\sin \theta }=\frac{\tan \theta +1}{1-\tan \theta }

p_{1}-p_{1}\tan \theta =p_{2}+p_{2}\tan \theta

\Rightarrow (p_{1}+p_{2})\tan \theta = p_{1}-p_{2}

\Rightarrow \tan \theta =\frac{p_{1}-p_{2}}{p_{1}++p_{2}}\Rightarrow \theta =\tan ^{-1}\left ( \frac{p_{1}-p_{2}}{p_{1}+p_{2}} \right )


Option 1)

\theta = \tan ^{-1}\pi\left ( \frac{P_{1}}{P_{2}} \right )

 

 

 

Option 2)

\theta = \tan ^{-1}\frac{\pi }{2}\left ( \frac{P_{2}}{P_{1}} \right )

Option 3)

\theta =\tan ^{-1}\left [ \frac{\pi }{2}\left ( \frac{P_{1}-P_{2}}{P_{1}+P_{2}} \right ) \right ]

Option 4)

\theta =\tan ^{-1}\frac{\pi }{2}\left [ \left ( \frac{P_{1}+P_{2}}{P_{1}-P_{2}} \right ) \right ]

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