Question

In the figure shown, the velocity and pressure of the liquid at the cross-section $\left(2\right)$ are given by


$(P_0$ is the atmospheric pressure$)$
$(\rho$ is the density of the liquid$)$

• A. $\sqrt{2hg},P_0+\frac{\rho hg}{2}$
• B. $\sqrt{hg},P_0+\frac{\rho hg}{2}$
• C. $\sqrt{\frac{hg}{2}},P_0+\frac{3\rho hg}{4}$
• D. $\frac{\sqrt{hg}}{2},P_0+\frac{3\rho hg}{4}$

Answer 1

The correct option is C $\sqrt{\frac{hg}{2}},P_0+\frac{3\rho hg}{4}$

Velocity of the liquid just after coming out of section $\left(3\right)$ is

$v_3=\sqrt{2gh}...........\left(1\right)$

Using equation of continuity in section $\left(2\right)$ and section $\left(3\right)$,

$\frac{A}{2}{v}_2=\frac{A}{4}{v}_3$

$\Rightarrow v_2=\frac{1}{2}{v}_3$

$\Rightarrow v_2=\sqrt{\frac{hg}{2}}$

Considering horizontal pipe, using Bernoulli's theorem at section $\left(2\right)$ and at the opening end of pipe $\left(3\right)$,

$P_0+\frac{1}{2}\rho v_3^2=P_2+\frac{1}{2}\rho v_2^2$

$\Rightarrow P_2=P_0+\frac{1}{2}\rho (2gh-\frac{gh}{2})$

$\therefore P_2=P_0+\frac{3\rho gh}{4}$

Hence, option $\left(\text{C}\right)$ is correct.