Question

In a test experiment on a model aeroplane in a wind tunnel, the flow speed on the upper and lower surfaces of the wings are $v_1$ and $v_2$ respectively. If $\rho$ is the density of air, then the upward lift is ( Given: A = cross-sectional area of the wing)

• A. $\frac{1}{2}\rho A(v_1^2-v_2^2)$
• B. $\frac{1}{2}\rho A(v_2^2-v_1^2)$
• C. $A{v}_1-A{v}_2$
• D. $v_1-v_2$

Answer 1

The correct option is A $\frac{1}{2}\rho A(v_1^2-v_2^2)$

Applying Bernoulli's principle, we have
$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2$
$\Rightarrow P_2-P_1=\frac{1}{2}\rho (v_1^2-v_2^2)$
The difference in pressure provides the lift to the aeroplane.

So lift on the aeroplane $=$ pressure difference $\times$ area of wings
$=\frac{1}{2}\rho (v_1^2-v_2^2)\times A=\frac{1}{2}\rho A(v_1^2-v_2^2)$