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 $(\rho _{air} = 1.2 \,\,kg/m^3)$
AIPMT 2015Medium
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Applying Bernoulli's theorem just above and just below the roof,
$P + \frac{1}{2}\rho {v^2} = {P_0} + 0$
$\left( {{P_0} - P} \right) = \frac{1}{2}\rho {v^2} = \Delta P$
Hence lift of the roof
$F = \Delta P \cdot A = \frac{1}{2}\rho A{v^2}$
$ = \frac{1}{2} \times 1.2 \times {\left( {40} \right)^2} \times 250 = 2.4 \times {10^5}N$
as pressure inside the roof is greater than outside the roof. so, force will act upward direction.
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