Question

A siphon tube as shown in figure is discharging a liquid of density $900{\text{kgm}}^{-3}$. Atmospheric pressure is $1.01\times {10}^5{\text{Nm}}^{-2}$.

• A. The speed of liquid through siphon is $10{\text{ms}}^{-1}$
• B. Pressure at point C is $3.2\times {10}^4{\text{Nm}}^{-2}$
• C. Pressure at point B is $4.25\times {10}^4{\text{Nm}}^{-2}$
• D. Pressure at point C is $6.5\times {10}^4{\text{Nm}}^{-2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The speed of liquid through siphon is $10{\text{ms}}^{-1}$
C Pressure at point B is $4.25\times {10}^4{\text{Nm}}^{-2}$
D Pressure at point C is $6.5\times {10}^4{\text{Nm}}^{-2}$

$v$ = speed of liquid through siphon = $\sqrt{2gh}$
$\Rightarrow v=\sqrt{2\times 10\times (4+1)}=10{\text{ms}}^{-1}$
Apply Bernoulli's theorem at $A$ and $B$ , we have ,
$P+0+0=P_B+\frac{1}{2}\rho v^2+\rho g\left(1.5\right)$
$\Rightarrow P_B=4.25\times {10}^4{\text{Nm}}^{-2}$
Apply Bernoulli's theorem, at $B$ & $C$
$P_C=4.25\times {10}^4+900\times 10\times 2.5=6.5\times {10}^4{\text{Nm}}^{-2}$