Question

As shown in the figure a liquid of density $\rho$ is kept in a sealed container. The height of liquid column is $h$. The container contains compressed air at a gauge pressure of $p$. The horizontal pipe has a outlet with cross-sectional area $A/2$ at $E$, where $A$ is the cross-sectional area of the horizontal pipe. Choose the correct option(s).

• A. The velocity of liquid at $C$ will be ${\left[\frac{(p+\rho gh)}{2\rho }\right]}^{1/2}$
• B. The velocity of liquid at $C$ will be ${\left[\frac{2(p+\rho gh)}{\rho }\right]}^{1/2}$
• C. The discharge rate is given by $\frac{A}{2\rho }(p+\rho gh{)}^{1/2}$
• D. The discharge rate is given by $\frac{A}{\sqrt{2\rho }}(p+\rho gh{)}^{1/2}$

Answer 1

Answer: (A), (D)

The discharge rate is given by $\frac{A}{\sqrt{2\rho }}(p+\rho gh{)}^{1/2}$

Pressure at point $A$
$p_A=p+p_0+\rho gh...\left(i\right)$
Applying Bernoulli's theorem at $A$ and $C$ and $E$,
$p+\rho gh+\frac{1}{2}\rho v^2=\text{constant}$
$(p+p_0)+\rho gh=p_C+\rho \frac{v_C^2}{2}=p_0+\rho \frac{v_E^2}{2}...\left(ii\right)$
Pressure at $C$ can be given as, taking $h_1$ as height of liquid above $C$
$p_c=p_0+\rho g{h}_1...\left(iii\right)$
From continuity equation
$A_C{v}_C=A_E{v}_E$
$A_C{v}_C=\left(\frac{A_c}{2}\right)v_E...\left(iv\right)$
Applying Bernoulli's equation at $A$ and $E$,
$(p+p_0)+\rho gh=p_0+\rho \frac{v_E^2}{2}$
$p+\rho gh=\rho \frac{v_E^2}{2}$
$\therefore v_E={\left[\frac{2(p+\rho gh)}{\rho }\right]}^{\frac{1}{2}}$
$\therefore v_C=\frac{v_E}{2}={\left[\frac{(p+\rho gh)}{2\rho }\right]}^{\frac{1}{2}}$

Discharge rate will be,
$Q=A_C{v}_C=A{\left(\frac{(p+\rho gh)}{2\rho }\right)}^{1/2}$
$\therefore Q=\frac{A}{\sqrt{2\rho }}{\left(p+\rho gh\right)}^{1/2}$
Hence correct options are $\left(a\right),\left(d\right)$