In the figure shown the velocity and pressure of the liquid at the cross section (2) are given by (If $P_{0 }$is the atmospheric pressure)

Velocity of the liquid just after coming out of section $\text{' }3\text{ ' is }{\mathrm{v}}_3=\sqrt{2\mathrm{gh}}$

Using continuity equation of section ‘2’ and vsection ‘3’

$\frac{\mathrm{A}}{2}{\mathrm{v}}_2=\frac{\mathrm{A}}{4}{\mathrm{v}}_3;{\mathrm{v}}_2=\frac{1}{2}{\mathrm{v}}_3$

${\mathrm{v}}_2=\frac{1}{2}\sqrt{2\mathrm{gh}};{\mathrm{v}}_2=\sqrt{\frac{\mathrm{gh}}{2}}$

Now using Bernoulli’s theorem of section 2 and at the opening end of pipe 3

${\mathrm{p}}_0+\frac{1}{2}{\mathrm{\rho v}}_3^2={\mathrm{p}}_2+\frac{1}{2}{\mathrm{\rho v}}_2^2$

${\mathrm{p}}_2={\mathrm{p}}_0+\frac{1}{2}\mathrm{\rho }\left({\mathrm{v}}_3^2-{\mathrm{v}}_2^2\right)$

$\Rightarrow {\mathrm{p}}_0+\frac{1}{2}\mathrm{\rho }\left(2\mathrm{gh}-\frac{\mathrm{gh}}{2}\right)$

${\mathrm{p}}_2={\mathrm{p}}_0+\frac{3\mathrm{\rho gh}}{4}$