1. Use Bernoulli’s principle to find pressure difference

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

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

So, the pressure difference between lower and upper surfaces is

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

Given:

• $\rho = 1\ \text{kg m}^{-3}$
• $v_{\text{lower}} = 180\ \text{km/h} = 50\ \text{m/s}$
• $v_{\text{upper}} = 252\ \text{km/h} = 70\ \text{m/s}$

Thus,

$\Delta P = \frac{1}{2}(1)\left(70^2 - 50^2\right)$

$\Delta P = \frac{1}{2}(4900 - 2500) = \frac{1}{2}(2400) = 1200\ \text{Pa}$

2. Find the total lift force

Each wing has area $40\ \text{m}^2$, so total wing area is

$A = 2 \times 40 = 80\ \text{m}^2$

Lift force is

$F_{\text{lift}} = \Delta P \cdot A = 1200 \times 80 = 96000\ \text{N}$

3. In level flight, lift equals weight

Since the plane is in level flight at constant speed,

$F_{\text{lift}} = mg$

So,

$m = \frac{96000}{10} = 9600\ \text{kg}$

4. Final answer

$\boxed{9600}$