Question

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

• A. $\frac{1}{\sqrt{2}}\rho v^{2}A$
• B. $\frac{1}{2}\rho v^{2}A$
• C. $\sqrt{2}\rho v^{2}A$
• D. $2\rho v^{2}A$

Answer 1

The correct option is B $\frac{1}{2}\rho v^{2}A$

Let $p_1$ and $p_2$ be the air pressures on the upper and lower surfaces of the wing. Then from Bernoulli's theorem equation,
$p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2$
Since we are taking the point just above and below the wing at the same height $(h=\text{constant})$
where $v_1=\sqrt{2}v$ and $v_2=v$.
Also $v_1>v_2\Rightarrow p_2>p_1$
Therefore, the pressure difference is
$\Delta p=p_2-p_1=\frac{1}{2}\rho (v_1^2-v_2^2)$
$\Delta p=\frac{\rho }{2}(2v^2-v^2)=\frac{\rho v^2}{2}$
Force for dynamic lift is given by,
$F=\Delta p\times A$
$\therefore F=\frac{1}{2}\rho v^{2}A$