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Question

A manometer connected to a closed tap reads 3.5×105 N/m2. When the valve is opened, the reading of the manometer falls to 3.0×105 N/m2, then the velocity of flow of water is:
(consider the water pipe to be horizontal)

A
100 m/s
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B
10 m/s
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C
1 m/s
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D
1010 m/s
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Solution

The correct option is B 10 m/s
Applying Bernoulli's theorem per unit mass of liquid,
P1ρ+v212=P2ρ+v222
Since the pipe carrying the water is horizontal, the elevation is same everywhere.
v1=0 as initially, tap was closed
& v2=v when tap opens and water starts to flow.
As the liquid starts flowing, its kinetic energy increases and pressure will decrease.
12v2=P1P2ρ
12v2=(3.5×105)(3×105)103
v2=2×0.5×105103
v=100=10 m/s

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