Explanation

(a) [image:10_image103.emf]
Vertical height of the liquid in portion $AC$
$h_{1} = DO + OE = R\sin{\theta + R\cos\theta = R(\sin{\theta + \cos{\theta)}}}$
Vertical height of the liquid in portion $CP$
$h_{2} = R - R\cos{\theta = R(1 - \cos{\theta)}}$
Vertical height of the liquid in portion $PB$
$h_{3} = R - R\sin{\theta = R(1 - \sin{\theta)}}$
In equilibrium, the pressure due to liquid on the both sides must be equal at the lowest point $P$
$\delta gh_{1} + \rho gh_{2} = \rho gh_{3}$ [As pressure $= h\rho g$]
$\delta gR(\sin{\theta + \cos\theta) + \rho gR}\left( 1 - \cos\theta \right) = \rho gR(1 - \sin\theta)$
$\delta(\sin{\theta + \cos{\theta) + \rho\left( 1 - \cos\theta \right) = \rho(1 - \sin\theta)}}$
$(\rho + \delta)\sin{\theta = (\rho - \delta)\cos\theta}$
$\tan{\theta = \frac{(\rho - \delta)}{(\rho + \delta)} \Rightarrow \theta = \tan^{- 1}\left( \frac{\rho - \delta}{\rho + \delta} \right)}$