A wind with speed 40 m/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: `(rho_("air")=1.2kg//m^(3))`

Options

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

Answer

A

Explanation

By Bernoulli's theorem,
`underset("inside")ubrace(P_(1)+(1)/(2)rhov_(1)^(2))=underset("outside")ubrace(P_(2)+(1)/(2)rhov_(2)^(2))`
Assuming that the roof width is very small, pressure difference,
`P_(1)-P_(2)=(1)/(2)rho(v_(2)^(2)-v_(1)^(2))`
Here, `rho=1.2kgm^(-3),v_(2)=40ms^(-1),v_(1)=0,A=250m^(2)`
`P_(2)-P_(2)=(1)/(2)xx1.2(40^(2)-0^(2))`
`P_(1)-P_(2)=(1)/(2)xx1.2xx1600=960Nm^(-2)`
Force acting on the roof `F=(P_(1)-P_(2))xxA`
`=960xx250`
`=2.4xx10^(5)N` upwards