JEE MAIN - Physics (2014 (Offline) - No. 20)
There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $${d_1}$$ and $${d_2}$$ are filled in the tube. Each liquid subtends $${90^ \circ }$$ angle at center. Radius joining their interface makes an angle $$\alpha $$ with vertical. Radio $${{{d_1}} \over {{d_2}}}$$ is :
_en_20_1.png)
$${{1 + \sin \,\alpha } \over {1 - \sin \,\alpha }}$$
$${{1 + \cos \,\alpha } \over {1 - \cos \,\alpha }}$$
$${{1 + \tan \,\alpha } \over {1 - \tan \,\alpha }}$$
$${{1 + \sin \,\alpha } \over {1 - \cos \,\alpha }}$$
Explanation
Pressure at interface A must be same from both the sides to be in equilibrium.
_en_20_2.png)
$$\therefore$$ $$\left( {R\cos \alpha + R\sin \alpha } \right){d_2}g = \left( {R\cos \alpha - R\sin \alpha } \right){d_1}g$$
$$ \Rightarrow {{{d_1}} \over {{d_2}}} = {{\cos \alpha + \sin \alpha } \over {\cos \alpha - \sin \alpha }} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}$$
$$\therefore$$ $$\left( {R\cos \alpha + R\sin \alpha } \right){d_2}g = \left( {R\cos \alpha - R\sin \alpha } \right){d_1}g$$
$$ \Rightarrow {{{d_1}} \over {{d_2}}} = {{\cos \alpha + \sin \alpha } \over {\cos \alpha - \sin \alpha }} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}$$
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