Question

Two non-viscous, incompressible and immiscible liquids of densities $p$ and $1.5$ $p$ are poured into the two limbs of a circular tube of radius $R$ and small cross-section kept fixed in a vertical plane as shown in Figure. Each liquid occupies one-fourth the circumference of the tube.

Answer 1

Since each liquid occupies on fourth the circumference of the tube,

$\angle AOC={90}^0$

$\angle BOC={90}^0$

The pressure $P_1$ at $D$ due to liquid on the left limb is $P_1=(R-R_1\sin \theta )\times 1.5pg$

The pressure $P_2$ at $D$ due to liquid on the right limb is $P_2=(R-R\cos \theta )1.5pg+(R\sin \theta +R\cos \theta )pg$

At equilibrium, $P_1=P_2$ thus, we have

$(1-\sin \theta )1.5=(1-\cos \theta )1.5+\sin \theta +\cos \theta$

Solving the equation, we get $2.5\sin \theta =0.5\cos \theta$

$\tan \theta =\frac{0.5}{2.5}=0.2$

$\theta ={11.3}^0$.