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Q.

In a horizontal tube with area of cross-section A1and A2 as shown in fig (5) , the liquid is flowing  with velocities v1 and v2 respectively. The difference in the levels of the liquid in the two vertical tubes is h. Then

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a

v2−v1=(2gh)

b

v22−v12=2gh

c

the volume of the liquid flowing through the tube in unit time is A1 v1

d

the energy per unit mass of the liquid is the same in both sections of the tube

answer is B.

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Detailed Solution

P1ρ+V122=P2ρ+V222 ∴ P1−P2=ρgh=ρ2v22−v12 or v22−v12=2gh
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