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In the figure, an ideal flows through th...

In the figure, an ideal flows through the tube, which is 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.
(i) `v_A = v_B`
(ii) `v_B gt v_A` ltbr gt (iii) `p_A = p_B`
(iv) `p_B gt p_A`.
.

A

`v_A=v_B`

B

`v_B gt v_A`

C

`P_A=P_B`

D

`P_B gt P_A`

Text Solution

Verified by Experts

The correct Answer is:
A, D
Since, area does not change hence
`v_A = v_B`
Pressure differ `P_B = P_A+rho g (AB)`.
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