Question 43 of 192

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90o angle at centre. Radius joining their interface makes an angle $\alpha$ with vertical. Ratio $\dfrac{{}_{}}{}$$\dfrac{{\text{d}}_{1}}{{}_{}}$$\dfrac{{}_{}}{}$ is :

A$\dfrac{1+sin\alpha }{1-sin\alpha }$B$\dfrac{1+cos\alpha }{1-cos\alpha }$C$\dfrac{1+tan\alpha }{1-tan\alpha }$D$\dfrac{1+sin\alpha }{1-cos\alpha }$

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Answer: (C) $\dfrac{1+tan\alpha }{1-tan\alpha }$

Pressure at common interface of two liquid columns is same.


${\text{d}}_{1}{}_{}\text{gR}\left[\right]\left[1-sin\alpha \right]\left[\right]={}_{}{\text{d}}_{1}{}_{}\text{gR}\left[\right]\left[1-cos\alpha \right]\left[\right]+{}_{}{\text{d}}_{2}{}_{}\text{gR}\left[\right]\left[sin\alpha +cos\alpha \right]$

${\text{d}}_{1}{}_{}\left[\right]\left[cos\alpha -sin\alpha \right]\left[\right]={}_{}{\text{d}}_{2}{}_{}\left[\right]\left[sin\alpha +cos\alpha \right]$

$∴\dfrac{{}_{}}{}\dfrac{{\text{d}}_{1}}{{}_{}}\dfrac{{}_{}}{}=\dfrac{}{}\dfrac{sin\alpha +}{cos\alpha }\dfrac{}{}=\dfrac{}{}\dfrac{tan\alpha +1}{}$