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Question

A pitot's tube is attached at one of the wing's of an aeroplane in order to determine its velocity with respect to air. If the difference of two liquid levels in manometer is 10 cm and density of liquid is 0.8gm/cm3, then the velocity of plane with respect to air will be (given density of air = 1.293×103gm/cm3).

A
34.82cm/s.
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B
3.48cm/s.
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C
348.2cm/s.
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D
3482cm/s.
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Solution

The correct option is D 3482cm/s.
Velocity of liquid/gas as measured from pitot's tube is given by
v=2hρlgρ, where
h= difference in the liquid levels in manometer attached to pitot's tube
ρl= density of liquid is pitot's tube
ρ= density of moving fluid (liquid/gas )
v=2×10×0.8×9801.293×103=3482cm/s

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