Question

Pipes are arranged as shown in the figure below. ${\rho }_1,{\rho }_2$ and ${\rho }_3$ are the densities of the fluids in these pipes where ${\rho }_3=2{\rho }_1$ and ${\rho }_1={\rho }_2$. Find the angle $\theta$ shown in the figure below.

• A. $\sin \theta =\frac{2h}{l}$
• B. $\sin \theta =\frac{h}{l}$
• C. $\sin \theta =\frac{h}{2l}$
• D. $\sin \theta =1$

Answer 1

The correct option is B $\sin \theta =\frac{h}{l}$

For pressure difference, only vertical height matters.

Pressure at point $A$
$P_A=P_0-{\rho }_{1}g\left(l\sin \theta \right)$ .....(i) [from left]
[$l\sin \theta$ is vertical distance of $A$ from the base]
Similarly, Pressure at point $B$
$P_B=P_0+{\rho }_{2}gh$ ...(ii) (from right)

Also $P_A=P_B-{\rho }_{3}gl\sin \theta$ ...(iii)

From eqn (i) (ii) and (iii) :
$P_0-{\rho }_{1}gl\sin \theta =(P_0+{\rho }_{2}gh)-{\rho }_{3}gl\sin \theta$
$\Rightarrow ({\rho }_3-{\rho }_1)l\sin \theta ={\rho }_{2}h$
[$\because {\rho }_3=2{\rho }_1$ and ${\rho }_1={\rho }_2$]
$\Rightarrow (2{\rho }_1-{\rho }_1)l\sin \theta ={\rho }_{1}h$
i.e $\sin \theta =\frac{h}{l}$