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)`.

Options

A. `sqrt((2((rho_(m)+rho_(a))gh)/(rho_a))`
B. `sqrt((2((rho_(m)-rho_(a))gh)/(rho_a))`
C. `sqrt(((rho_(m)-rho_(a))gh)/(rho_a))`
D. `sqrt(((rho_(m)+rho_(a))gh)/(rho_a))`

Answer

B

Explanation

Bernoulli's equation between `A` and `B` gives
`(p_A)/(p_B) = (p_B)/(rho_a) + (v^2)/(2) implies 2[(p_(A)-p_(B))/(rho_a)]`
Also equating pressure at horizontal level of `E`
`p_(A)+rho_(a)gy+rho_(a)gh =p_(B)+rho_(a)gy'+rho_(a)gy+rho_(m)gh`
`p_(A) + rho_(a)gh = p_(B)+rho_(m)gh`
`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))`.