In the figure, an ideal liquid flows through the tube, which is of uniform cross - section. The liquid has velocities ${\mathrm{\upsilon }}_{\mathrm{A}} \mathrm{and} {\mathrm{\upsilon }}_{\mathrm{B}}$, and pressures P_A and P_B at points A and b respectively

Av = constant according to the continuity equation. Since the cross-sectional area is constant, we have $v_A=v_B.$

And according to Bernoulli's theorem, point B is at the lower level.

${\mathrm{P}}_{\mathrm{B}}+\frac{1}{2}{\mathrm{\rho v}}^2={\mathrm{P}}_{\mathrm{A}}+\mathrm{h\rho g}+\frac{1}{2}{\mathrm{\rho v}}^2$

$\mathrm{Hence} {\mathrm{P}}_{\mathrm{B}}>{\mathrm{P}}_{\mathrm{A}}$.