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A Pitot tube is shown in figure. Wind bl...

A Pitot tube is shown in figure. Wind blows in the direction shown. Air at inlet `A` is brought to rest, whereas its speed just outside of opening `B` is unchanged. The `U` tube contains mercury of density `rho_(m)`. Find the speed of wind respect to Pitot tube. Neglect the height difference between `A` and `B` and take the density of air as `rho_(a)`.

Text Solution

Verified by Experts

The correct Answer is:
`v = sqrt((2(rho_(m) - rho_(a))gh)/(rho_(a)))`
Bernoulli's equation between `A` and `B` gives
`(P_(A))/(rho_(a)) = (P_(B))/(rho_(a)) + (v^(2))/(2) rArr v^(2) = 2[(p_(A) - p_(B))/(rho_(a))]`
Also equating pressures at horizontal level of `E`
`P_(A) + rho_(a)gy + rho_(a)gh = p_(B) + rho_(a)gy' + rho_(a)gy + rho_(m)gh`.
`rArr p_(A) + rho_(a)gh = p_(B) + rho_(m)gh [:' y' = 0]`
`p_(A) - p_(B) = (rho_(m) - rho_(a))gh`.
`v^(2) = (2(rho_(m) - rho_(a))gh)/(rho_(a))`
`v = sqrt((2(rho_(m) - rho_(a))gh)/(rho_(a)))`
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