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

Hint:

The upward lift on the wing can be derived from Bernoulli's equation, and it depends on the difference in the squares of the velocities on the upper and lower surfaces of the wing.

Answer 1

The Correct Option is C

The upward lift on a wing is given by Bernoulli's principle, which relates the difference in the velocities of air over the upper and lower surfaces of the wing to the lift force. 
The pressure difference \( \Delta P \) between the upper and lower surfaces is related to the velocities \( v_1 \) and \( v_2 \) by: \[ \Delta P = \frac{1}{2} \rho (v_1^2 - v_2^2) \] The upward lift \( L \) is the force exerted by this pressure difference on the cross-sectional area of the wing. Therefore, the lift force is: \[ L = \Delta P \cdot A = \frac{1}{2} \rho A (v_1^2 - v_2^2) \] 
Thus, the correct answer is: \[ \text{(3) } \frac{1}{2} \rho A (v_1^2 - v_2^2) \]