Neet Solved Paper 2015 Question 18
Question: A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is . 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}) $
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Options:
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
Show Answer
Answer:
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.
