Question

The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are $v_1$ and $v_2,$ respectively. If $A$ is the cross-sectional area of the wing and $\rho$ is the density of air, then the upward lift is

• A. $\frac{1}{2}\rho A\left(v_1-v_2\right)$
• B. $\frac{1}{2}\rho A\left(v_1+v_2\right)$
• C. $\frac{1}{2}\rho A\left(v_1^2-v_2^2\right)$
• D. $\frac{1}{2}\rho A\left(v_1^2+v_2^2\right)$

Answer 1

Answer: (C)

Due to the specific shape of wings when the aeroplane runs, air passes at higher speed over it is compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called 'dynamic lift' acts on the plate.
$\therefore$ Upward lift $=$ pressure difference $\times$ area of wing
$[\because F=\rho \times A]$
From Bernoulli's theorem,
$p_1+\frac{1}{2}\rho V_1^2=p_2+\frac{1}{2}\rho V_2^2$
Pressure difference is
$\because p_2-p_1=\frac{1}{2}\rho \left(V_1^2-V_2^2\right)$
upward lift $=\left(p_2-p_1\right)\times A$
$=\frac{1}{2}\rho \left(V_1^2-V_2^2\right)A$