In the figure shown an ideal liquid is flowing through the tube which is of uniform area of cross-section. The liquid has velocities vA and vB, and pressures PA and PB at points A and B respectively. Then

Question

(A) vB > vA (B) vB = vA (C) PB < PA (D) PB = PA

Tardigrade Admin · Accepted Answer

For a streamline flow of an ideal liquid, vA = vB. The pressure at B = pressure at A + pressure due to column of liquid of height AB

Q. In the figure shown an ideal liquid is flowing through the tube which is of uniform area of cross-section. The liquid has velocities $v_{A}$ and $v_{B}$ , and pressures $P_{A}$ and $P_{B}$ at points $A$ and $B$ respectively. Then

$v_{B} > v_{A}$

$v_{B} = v_{A}$

$P_{B} < P_{A}$

$P_{B} = P_{A}$

For a streamline flow of an ideal liquid, $v_{A} = v_{B}$ . The pressure at $B =$ pressure at $A +$ pressure due to column of liquid of height $A B$

Q. In the figure shown an ideal liquid is flowing through the tube which is of uniform area of cross-section. The liquid has velocities vA​ and vB​, and pressures PA​ and PB​ at points A and B respectively. Then

vB​>vA​

vB​=vA​

PB​<PA​

PB​=PA​