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Question: Air of density \[1.3kg{m^{ - 3}}\]blows horizontally with a speed of\(108km{h^{ - 1}}\). A house has...

Air of density 1.3kgm−31.3kg{m^{ - 3}}blows horizontally with a speed of108kmh−1108km{h^{ - 1}}. A house has a plain roof of area40m240{m^2}. The magnitude of aerodynamic lift on the roof is.

  1. 2.34×104Dynes2.34 \times {10^4}Dynes
    2)0.234×104N0.234 \times {10^4}N
  2. 2.34×104N2.34 \times {10^4}N
  3. 23.4×104N23.4 \times {10^4}N
Explanation

Solution

Hint:-
The reason behind the lifting of the roof is Bernoulli’s principle. According to the principle, the pressure on one side of the surface is equal to the other side of the surface. Here, at the top of the roof there is low pressure as the wind is blowing while below the roof there is high pressure as there is no wind flowing inside the house. So, the high pressure in the inside and low pressure on the outside of the roof causes the roof to lift up.

Complete step-by-step solution
p1+ρgh+12ρ1v12=p2+ρgh+12ρ2v22{p_1} + \rho gh + \dfrac{1}{2}{\rho _1}v_1^2 = {p_2} + \rho gh + \dfrac{1}{2}{\rho _2}v_2^2;
Cancel out the common variables, put v1=v{v_1} = vandv2=0{v_2} = 0;
p2=p1+12ρ1v12{p_2} = {p_1} + \dfrac{1}{2}{\rho _1}v_1^2;
p2−p1=12ρ1v12{p_2} - {p_1} = \dfrac{1}{2}{\rho _1}v_1^2;
Here: Δp=p2−p1\Delta p = {p_2} - {p_1};
Δp=12ρ1v12\Delta p = \dfrac{1}{2}{\rho _1}v_1^2;
Putv1{v_1}= v in the above equation:
Δp=12ρv2\Delta p = \dfrac{1}{2}\rho {v^2};
p2−p1=12ρv2{p_2} - {p_1} = \dfrac{1}{2}\rho {v^2};
Apply the relation between force and pressure:
(P1−P2)=FA({P_1} - {P_2}) = \dfrac{F}{A};
F=(P1−P2)×AF = ({P_1} - {P_2}) \times A;
Put the value p2−p1=12ρv2{p_2} - {p_1} = \dfrac{1}{2}\rho {v^2}in the above equation:
F=12ρv2×AF = \dfrac{1}{2}\rho {v^2} \times A;
Here v=108kmh−1v = 108km{h^{ - 1}}is equal tov=108×100060×60=30m/sv = \dfrac{{108 \times 1000}}{{60 \times 60}} = 30m/s;
F=12×1.3×900×40F = \dfrac{1}{2} \times 1.3 \times 900 \times 40;
Do the necessary calculation:
F=468002F = \dfrac{{46800}}{2};
The lift force is given as:
F=23,400NF = 23,400N;
F=2.34×104NF = 2.34 \times {10^4}N;

Final Answer: Option”3” is correct. The magnitude of aerodynamic lift on the roof is2.34×104N2.34 \times {10^4}N.

Note:- Here we have to find the difference in pressure by applying Bernoulli’s theorem and then apply the formulaF=(P1−P2)×AF = ({P_1} - {P_2}) \times A. After applying Bernoulli’s equation the common terms ρgh\rho ghwill cancel out and the final velocity v2{v_2}will be zero. In the end we will get the force which is also known as lift force and is equal to the magnitude of aerodynamic lift on the roof.