Question

As shown in figure a liquid of density $\rho$ is standing in a sealed container to a height $h$. The container contains compressed air at gauge pressure of $p$. The horizontal outlet pipe at E has a cross sectional area half of that at C. After the outlet is open, choose the correct option(s)

• A. The velocity of liquid at C will be ${\left(\frac{\left(}{p+\rho gh}\right]}^{\frac{1}{2}}$
• B. The velocity of liquid at C will be ${\left(\frac{2\left(p+\rho gh\right)}{\rho }\right]}^{\frac{1}{2}}$
• C. $h_1=3\frac{V_C^2}{2g},where{V}_C$ is the velocity at point C
• D. $p_C>p_E$, where $p_{C}and{p}_E$ are pressures at point C and E

Answer 1

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

The correct options are
A The velocity of liquid at C will be ${\left(\frac{\left(}{p+\rho gh}\right]}^{\frac{1}{2}}$
C $h_1=3\frac{V_C^2}{2g},where{V}_C$ is the velocity at point C
D $p_C>p_E$, where $p_{C}and{p}_E$ are pressures at point C and E

$p_A=p+p_o+\rho gh.................\left(i\right)$
Apply Bernoulli's theorem at A, C and E
$(p+p_o)+\rho gh=p_C+\rho \frac{V_C^2}{2}=p_o+\rho \frac{V_E^2}{2}...................\left(ii\right)p_C=p_o+\rho g{h}_1..................\left(iii\right)A_C{V}_C=A_E{V}_E{A}_C{V}_C=\left(\frac{A_C}{2}\right)V_E......................\left(iv\right)Fromeq.\left(i\right),(p_o+\rho g{h}_1)+\rho \frac{V_C^2}{2}=p_o+\rho \frac{{\left(2V_C\right)}^2}{2}\rho g{h}_1=2\rho V_C^2-\frac{\rho }{2}{V}_C^2=\frac{3}{2}\rho V_C^{2}or{h}_1=\frac{3V_C^2}{2g}Fromeq.\left(ii\right),(p+p_o)+\rho gh=(p_o+\rho g{h}_1)+\rho \frac{V_C^2}{2}p+\rho gh=\rho g\left(\frac{3V_C^2}{2g}\right)+\rho \frac{V_C^2}{2}{V}_C^2=\frac{(p+\rho gh)}{2\rho }$