CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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 $$(\rho_{air}=1.2 \ kg/m^3)$$


A
2.4×105N, upward
loader
B
2.4×105N, downward
loader
C
4.8×105N, downward
loader
D
4.8×105N, upward
loader

Solution

The correct option is D $$2.4\times 10^5N$$, upward
Applying Bernoulli's theorem to two points just inside the house and outside the house,

$$P + \dfrac{1}{2}\rho v^2 = constant$$

Inside the pressure is $$P_{atm}$$

The pressure outside is $$P$$

v is zero inside.

$$ \therefore P - P_{atm} = \frac{1}{2}\rho v^2= \frac{1}{2} \times 1.2  \times 40^2=960$$

Force is : $$F = (P - P_{atm}) \times Area= 960 \times 250 = 2.4 \times 10^5\:N$$ in upward direction as pressure goes from higher to lower pressure area.

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image