Question

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 ${90}^o$ angle at centre. Radius joining their interface makes an angle $\alpha$ with vertical . The ratio $d_1/d_2$ is

• A. $\frac{1+\tan \alpha }{1-\tan \alpha }$
• B. $\frac{1+\sin \alpha }{1-\cos \alpha }$
• C. $\frac{1+\sin \alpha }{1-\sin \alpha }$
• D. $\frac{1+\cos \alpha }{1-\cos \alpha }$

Answer 1

The correct option is A $\frac{1+\tan \alpha }{1-\tan \alpha }$

Answer is A
Let the point be O of tube touching the ground.
Pressure due to liquid of density, $d_1$, on the left side of O = Pressure due to liquid of density, $d_1$ + Pressure due to liquid of density, $d_2$, on the right side of O. Let $P_o$ be the pressure due to air above liquid, then
$P_o+d_{1}g(R-R\cos \alpha )+d_{2}gR\sin (\frac{\pi }{2}-\alpha )+d_{2}gR\sin \alpha =P_o+d_{1}g(R-R\cos (\frac{\pi }{2}-\alpha )$
Therefore,
$d_1(\cos \alpha -\sin \alpha )=d_2(\cos \alpha +\sin \alpha )$
$\frac{d_1}{d_2}=\frac{1+\tan \alpha }{1-\tan \alpha }$ .