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

Answer

c

Explanation

From bernoulli's theorem
`p_(1) + (1)/(2)rhov_(1)^(2) = p_(2) + (1)/(2)rho v_(2)^(2)`
where, `p_(1), p_(2)` are pressure inside and outside the roof and `v_(1), 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)/(2)rho(v_(2)^(2)-v_(1)^(2)) = (1)/(2) xx 1.2(40^(2)-0)`
`= 960 N//m^(2)`
Pressure acting on the roof is given by
`F = (p_(1) - p_(2))A = 960 xx 250`
`= 24 xx 10^(4)N = 2.4 xx 10^(5)N`
As the pressure inside the roof is more than outside to it. So the force will act in the upward direction,
i.e., `F = 2.4 xx 10^(5)N` -upwards.