A wind with speed `40m//s` blows parallel to the roof of a house. The area of the roof is `250 m^(2)`. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be `(P_(air) = 1.2 kg//m^(3))`

Options

A. `4.8xx10^(5)` N, downwards
B. `4.8xx10^(5)` N, upwards
C. `2.4 xx 10^(5)` N, upwards
D. `2.4 xx 10^(5)` N, downwards

Answer

C

Explanation

From Bernolli's theorem
`p_(1)+1/2rhov_(1)^(2)=p_(2)+1/2rhov_(2)^(2)`
Where, `p_(p) p_(2)` are pressure inside and outside the roof and `v_(b) v_(2)` are velocities of wind inside and outside the roof. Neglect the width of the roof. Pressure difference is:
`p_(1)-p_(2)=1/2rho(v_(2)^(2)-v_(1)^(2)) = 1/2 x 12(40^(2)-0)=960 Nm^(-2)`
Force acting on the roof is given be
`F=(p_(1)-p_(2))A= 960 xx 250`
`=24 xx 10^(4)N=2.4 x 10^(5)`N