Question

Arrangement of pipes carrying fluids of densities ${\rho }_1,{\rho }_2\&{\rho }_3$ is shown in the figure below. If the pipe carrying fluid of density ${\rho }_1$ makes an angle $\theta$ with the horizontal, then what is the value of ${}^{\prime }{h}^{\prime }$ ?

• A. $h=\frac{({\rho }_1+{\rho }_1)}{{\rho }_2}l\sin \theta$
• B. $h=\left(\frac{{\rho }_3}{{\rho }_2}\right)l\sin \theta$
• C. $h=\left(\frac{{\rho }_3-{\rho }_1}{{\rho }_2}\right)l\sin \theta$
• D. $h=\left(\frac{{\rho }_2}{{\rho }_3}\right)\text{tan}\theta$

Answer 1

The correct option is C $h=\left(\frac{{\rho }_3-{\rho }_1}{{\rho }_2}\right)l\sin \theta$

The vertical height of the left end of pipe from point $A$ is
$h_1=l\sin \theta$
$\Rightarrow P_A=P_0-{\rho }_{1}g{h}_1$
$\Rightarrow P_A=P_0-{\rho }_{1}gl\sin \theta ...\left(i\right)$
Similarly, the vertical height of the right end of the pipe from point $B$ is $h$.
Pressure at point $B$ is given by,
$P_B=P_0+{\rho }_{2}gh...\left(ii\right)$

Also, pressure difference between $A$ and $B$ can be given as:
$P_A=P_B-{\rho }_{3}g\left(l\sin \theta \right)...\left(iii\right)$
From Eq$\left(i\right),\left(ii\right),\left(iii\right)$ equating the difference $(P_A-P_B)$:
$P_0-{\rho }_{1}gl\sin \theta -P_0-{\rho }_{2}gh=-{\rho }_{3}gl\sin \theta$
$\Rightarrow ({\rho }_3-{\rho }_1)gl\sin \theta ={\rho }_{2}gh$
$\therefore h=\frac{({\rho }_3-{\rho }_1)l\sin \theta }{{\rho }_2}$