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In this figure, an ideal liquid flows through the tube, which is of uniform cross-section. The liquid has velocities $v_A$ and $V_B$, and pressure $P_A$ and $P_B$ at points $A$ and $B$ respectively
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$v_A=v_B$, $P_B \gt P_A$
As per the continuity equation, $\mathrm{Av}=$ constant. As the crosssectional area remains same we get $\mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{B}}$.
And as point $B$ is at the lower level, by Bernoulli's theorem $P_B+\frac{1}{2} \rho v^2=P_A+h+\frac{1}{2} \rho v^2$ hence $P_B \gt P_A$.
And as point $B$ is at the lower level, by Bernoulli's theorem $P_B+\frac{1}{2} \rho v^2=P_A+h+\frac{1}{2} \rho v^2$ hence $P_B \gt P_A$.
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