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  • In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the lower and upper surfaces of the wing are v and <img alt="\mathrm{\sqrt{2} v}" src="https://entrancecorner.onc

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the lower and upper surfaces of the wing are v and \mathrm{\sqrt{2} v} respectively. If the density of air is \mathrm{\rho} and the surface area of the wing is A, the dynamic lift on the wing is given by

Option: 1

\mathrm{\frac{1}{\sqrt{2}} \rho v^2 A}


Option: 2

\mathrm{\frac{1}{2} \rho v^2 A}


Option: 3

\mathrm{\sqrt{2} \rho v^2 A}


Option: 4

\mathrm{2 \rho v^2 A}


Answers (1)

best_answer

Let \mathrm{p_1} and \mathrm{p_2} be the air pressures on the upper and lower surfaces of the wing. Then from Bernoulli's theorem we have

                         \mathrm{ p_1+\frac{1}{2} \rho v_1^2=p_2+\frac{1}{2} \rho v_2^2 }

where \mathrm{v_1=\sqrt{2} v} and \mathrm{v_2=v}. Therefore, the pressure difference is

\mathrm{ p_2-p_1=\frac{1}{2} \rho\left(v_1^2-v_2^2\right)=\frac{\rho}{2}\left(2 v^2-v^2\right)=\frac{\rho v^2}{2} }

\mathrm{ \begin{aligned} \therefore \text { Force of dynamic lift }= & \text { pressure difference } \times \\\\ & \text { surface area of wing } \\\\ & =\frac{1}{2} \rho v^2 A, \end{aligned} }

Posted by

Deependra Verma

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