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${}^{\circ }$ angle at centre. Radius joining their interface makes an angle $\alpha$ with vertical. Ratio $\frac{d_1}{d_2}$ is:

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

Answer 1

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

$\begin{array}{c}\text{As we know pressure on line AB}⟹P=P^0+\rho gh\\ as\angle XOD={90}^{\circ }\&\angle DOE=\alpha \end{array}$

$\therefore \angle XOC\text{also}\alpha$ $\begin{array}{c}\text{So by}\text{trignometry}{y}_1=R\sin \alpha \\ ⟹\cos \alpha =\frac{y_2}{R}\end{array}$

$\begin{array}{c}P=P_0+d_{2}gR\sin \alpha +d_{2}gR\cos \alpha \\ P=P_0+d_{2}gR\{\sin \alpha +\cos \alpha \}-\left(i\right)\end{array}$

$\begin{array}{c}{y}_3=R\cos \alpha -R\sin \alpha \\ P=P^0+d_{1}g\{R\cos \alpha -R\sin \alpha \}\\ P=P_0+d_{1}gR\{\cos \alpha -\sin \alpha \}\end{array}$

$\text{From eq (1) \& (2)}$

$P_0+d_{2}gR\{\sin \alpha +\cos \alpha \}=P_0+d_{1}gR\{\cos \alpha -\sin \alpha \}$

$\therefore \frac{d_1}{d_2}=\frac{\sin \alpha +\cos \alpha }{\cos \alpha -\sin \alpha }$

$\frac{d_1}{d_2}=\frac{1+\tan \alpha }{1-\tan \alpha }$