1. Use Bernoulli’s principle

For airflow over the upper and lower surfaces of the wing at the same height,

$P + \frac{1}{2}\rho v^2 = \text{constant}$

Hence, the pressure difference between lower and upper surfaces is

$P_{\text{lower}} - P_{\text{upper}} = \frac{1}{2}\rho\left(v_{\text{upper}}^2 - v_{\text{lower}}^2\right)$

Given:

$v_{\text{upper}} = 70\ \text{m s}^{-1}, \quad v_{\text{lower}} = 65\ \text{m s}^{-1}, \quad \rho = 1.2\ \text{kg m}^{-3}$

So,

$\Delta P = \frac{1}{2}(1.2)\left(70^2 - 65^2\right)$

2. Calculate the speed-square difference

$70^2 = 4900, \qquad 65^2 = 4225$

$4900 - 4225 = 675$

Thus,

$\Delta P = 0.6 \times 675 = 405\ \text{Pa}$

3. Compute lift force

Lift is pressure difference multiplied by wing area:

$F = \Delta P \cdot A$

Given wing area,

$A = 2\ \text{m}^2$

Therefore,

$F = 405 \times 2 = 810\ \text{N}$

4. Final answer

$\boxed{810\ \text{N}}$

This matches the stored correct answer.