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)=1.2 kg//m^(3))`

Options

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

Answer

A

Explanation

By Bernoulli.s theorem, `ubrace(P_(1) + (1)/(2)rho v_(1)^(2))_("inside")= ubrace(P_(2) = (1)/(2) rho v_(2)^(2))_("outside")`
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 = 1.2 kg m^(-3), v_(2) = 40 ms^(-1), v_(1) =0, A = 250 m^(2)` Force acting on the roof `F= (P_(1) - P_(2)) xx A = 960 xx 250`
`=2.4 xx 10^(5)N` upwards