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Air of density \[1.3kg{m^{ - 3}}\]blows horizontally with a speed of$108km{h^{ - 1}}$. A house has a plain roof of area$40{m^2}$. The magnitude of aerodynamic lift on the roof is.
1) $2.34 \times {10^4}Dynes$
2)$0.234 \times {10^4}N$
3) $2.34 \times {10^4}N$
4) $23.4 \times {10^4}N$

Last updated date: 13th Jun 2024
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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
${p_1} + \rho gh + \dfrac{1}{2}{\rho _1}v_1^2 = {p_2} + \rho gh + \dfrac{1}{2}{\rho _2}v_2^2$;
Cancel out the common variables, put ${v_1} = v$and${v_2} = 0$;
${p_2} = {p_1} + \dfrac{1}{2}{\rho _1}v_1^2$;
\[{p_2} - {p_1} = \dfrac{1}{2}{\rho _1}v_1^2\];
Here: $\Delta p = {p_2} - {p_1}$;
$\Delta p = \dfrac{1}{2}{\rho _1}v_1^2$;
Put${v_1}$= v in the above equation:
$\Delta p = \dfrac{1}{2}\rho {v^2}$;
${p_2} - {p_1} = \dfrac{1}{2}\rho {v^2}$;
Apply the relation between force and pressure:
$({P_1} - {P_2}) = \dfrac{F}{A}$;
$F = ({P_1} - {P_2}) \times A$;
Put the value ${p_2} - {p_1} = \dfrac{1}{2}\rho {v^2}$in the above equation:
$F = \dfrac{1}{2}\rho {v^2} \times A$;
Here $v = 108km{h^{ - 1}}$is equal to$v = \dfrac{{108 \times 1000}}{{60 \times 60}} = 30m/s$;
\[F = \dfrac{1}{2} \times 1.3 \times 900 \times 40\];
Do the necessary calculation:
\[F = \dfrac{{46800}}{2}\];
The lift force is given as:
\[F = 23,400N\];
\[F = 2.34 \times {10^4}N\];

Final Answer: Option”3” is correct. The magnitude of aerodynamic lift on the roof is\[2.34 \times {10^4}N\].

Note:- Here we have to find the difference in pressure by applying Bernoulli’s theorem and then apply the formula$F = ({P_1} - {P_2}) \times A$. After applying Bernoulli’s equation the common terms $\rho gh$will cancel out and the final velocity ${v_2}$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.