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QJEEM14Ph0827: JEE Main 2014 Physics - Class 11 JEEM14Ph0827 Solution: There is a circular tube in a vertical plane. Two liquids which …

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1{d_1} and d2{d_2} are filled in the tube. Each liquid subtends 90∘{90^ \circ } angle at center. Radius joining their interface makes an angle α\alpha with vertical. Radio d1d2{{{d_1}} \over {{d_2}}} is : JEE Main 2014 (Offline) Physics - Properties of Matter Question 239 English

Multiple Choice Options

  • A 1+sin⁡ α1−sin⁡ α{{1 + \sin \,\alpha } \over {1 - \sin \,\alpha }}
  • B 1+cos⁡ α1−cos⁡ α{{1 + \cos \,\alpha } \over {1 - \cos \,\alpha }}
  • C 1+tan⁡ α1−tan⁡ α{{1 + \tan \,\alpha } \over {1 - \tan \,\alpha }}
  • D 1+sin⁡ α1−cos⁡ α{{1 + \sin \,\alpha } \over {1 - \cos \,\alpha }}

Step-by-Step Verified Solution

Pressure at interface A must be same from both the sides to be in equilibrium.
JEE Main 2014 (Offline) Physics - Properties of Matter Question 239 English Explanation
∴\therefore (Rcos⁡α+Rsin⁡α)d2g=(Rcos⁡α−Rsin⁡α)d1g\left( {R\cos \alpha + R\sin \alpha } \right){d_2}g = \left( {R\cos \alpha - R\sin \alpha } \right){d_1}g
⇒d1d2=cos⁡α+sin⁡αcos⁡α−sin⁡α=1+tan⁡α1−tan⁡α\Rightarrow {{{d_1}} \over {{d_2}}} = {{\cos \alpha + \sin \alpha } \over {\cos \alpha - \sin \alpha }} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}

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