In this figure,an ideal liquid flows through | vedclass

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(D) According to the equation of continuity,$A v = 	ext{constant}$. Since the cross-sectional area $A$ is uniform,the velocity of the liquid remains

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$P_B  >  P_A$

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(D) According to the equation of continuity,$A v = 	ext{constant}$. Since the cross-sectional area $A$ is uniform,the velocity of the liquid remains the same at all points,so $v_A = v_B$.Applying Bernoulli's theorem between points $A$ and $B$:$P_A + \frac{1}{2} ho v_A^2 + ho g h_A = P_B + \frac{1}{2} ho v_B^2 + ho g h_B$Since $v_A = v_B$,the kinetic energy terms cancel out:$P_A + ho g h_A = P_B + ho g h_B$$P_B - P_A = ho g (h_A - h_B)$Since point $B$ is at a lower level than point $A$ $(h_A  >  h_B)$,the term $(h_A - h_B)$ is positive.Therefore,$P_B  >  P_A$.

In this figure,an ideal liquid flows through a tube of uniform cross-section. The liquid has velocities $v_A$ and $v_B$,and pressures $P_A$ and $P_B$ at points $A$ and $B$ respectively.

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