. 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 \[({{p}_{air}}=1.2kg/{{m}^{3}})\]
A) \[4.8\times {{10}^{5}}N\] N, downwards
B) \[4.8\times {{10}^{5}}N\], upwards
C) \[2.4\times {{10}^{5}}N\], upwards
D) \[2.4\times {{10}^{5}}N\], downwards
Correct Answer: C
Solution :
From Bernoulli's theorem \[{{p}_{1}}+\frac{1}{2}\rho v_{1}^{2}={{p}_{2}}+\frac{1}{2}\rho v_{2}^{2}\] 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 \[{{p}_{1}}-{{p}_{2}}=\frac{1}{2}\rho (v_{2}^{2}-v_{1}^{2})\] \[=\frac{1}{2}\times 1.2({{40}^{2}}-0)=960N/{{m}^{2}}\] Force acting on the roof is given by \[F=({{p}_{1}}+{{p}_{2}})A=960\times 250\] \[=24\times {{10}^{6}}N=24\times {{10}^{5}}N\] As the pressure inside the roof is more than outside to it. So the force will act in the upward direction. i.e. \[F=2.4\times {{10}^{5}}N\] upward.
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