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