A plane is in level flight at constant speed and each of its two wings has an area of $$40$$ m$$^2$$. If the speed of the air is $$180$$ km h$$^{-1}$$ over the lower wing surface and $$252$$ km h$$^{-1}$$ over the upper wing surface, the mass of the plane is ________kg. (Take air density to be $$1$$ kg m$$^{-3}$$ and $$g = 10$$ m s$$^{-2}$$)

We apply Bernoulli’s principle for incompressible, non‐viscous flow in horizontal streamlines: $$p + \frac{1}{2}\rho v^2 = \text{constant} \quad-(1)$$

From $$(1)$$ for upper and lower surfaces, we have:
$$p_{\text{lower}} + \frac{1}{2}\rho v_{\text{lower}}^2 = p_{\text{upper}} + \frac{1}{2}\rho v_{\text{upper}}^2$$

Rearranging gives the pressure difference (lift per unit area):
$$p_{\text{lower}} - p_{\text{upper}} = \frac{1}{2}\,\rho\,(v_{\text{upper}}^2 - v_{\text{lower}}^2)\quad-(2)$$

Convert speeds from km h$$^{-1}$$ to m s$$^{-1}$$:
$$180\;\text{km h}^{-1} = 180 \times \frac{5}{18} = 50\;\text{m s}^{-1}\quad-(3)$$
$$252\;\text{km h}^{-1} = 252 \times \frac{5}{18} = 70\;\text{m s}^{-1}\quad-(4)$$

Substitute $$\rho = 1\;\text{kg m}^{-3}$$, $$v_{\text{upper}}=70\;\text{m s}^{-1}$$, $$v_{\text{lower}}=50\;\text{m s}^{-1}$$ into $$(2)$$:
$$\Delta p = \frac{1}{2} \times 1 \times (70^2 - 50^2) = \frac{1}{2}\,(4900 - 2500) = 1200\;\text{Pa}\quad-(5)$$

Each wing has area $$40\;\text{m}^2$$, so total lifting area is
$$A_{\text{total}} = 2 \times 40 = 80\;\text{m}^2\quad-(6)$$

Lift generated is pressure difference times area:
$$L = \Delta p \times A_{\text{total}} = 1200 \times 80 = 96000\;\text{N}\quad-(7)$$

In level flight at constant speed, lift equals weight: $$L = mg\quad-(8)$$
Using $$g = 10\;\text{m s}^{-2}$$:
$$m = \frac{L}{g} = \frac{96000}{10} = 9600\;\text{kg}\quad-(9)$$

Final Answer: The mass of the plane is $$9600\;\text{kg}$$.