Question

A small uniform tube is hent into a circle of radius r whose plane is vertical. Equal volumes of two fluids whose densities are p and $\sigma (\rho >\sigma )$ fill half the circle. Find the angle that the radius passing through the interface makes with the vertical.

Answer 1

$h_{AB}=r-r\cos (90-\theta )=r-r\sin \theta$
$h_{BC}=r-r\cos \theta$
$h_{CD}=r\sin (90-\theta )=r\cos \theta$
$h_{DE}=r\sin \theta$
Writing pressure equation between points A and E we have
$P_A+(r-r\sin \theta )\rho g(r-r\cos \theta )\rho g-\left(r\cos \theta \right)\left(\sigma \right)g-\left(r\sin \theta \right)\sigma g=P_E$
But $P_A=P_E$
Solving this equation we get
$\tan \theta =\left(\frac{\rho -\sigma }{\rho -\sigma }\right)$