Let $v_1 \text{ and } v_2$ be the speeds on the upper and lower surfaces of the wings of the aeroplane respectively, $P_1 \text{ and } P_2$ be the pressures on the upper and lower surfaces of the wings respectively.Here, $v_{1}=70\,\text{m}\,\text{s}^{-1};$$v_{2}=63\,\text{m}\,\text{s}^{-1};$$\rho=1.3\,\text{kg}\,\text{m}^{-3}$The level of the upper and lower surfaces of the wings from the ground may be taken same.$\therefore h_1 = h_2$area of wing, $A = 2.5\,\text{m}^2$ Thus from Bernoulli’s Theorem,or $P_{2}-P_{1}=\frac{1}{2}\rho(v_{1}^{2}-v_{2}^{2})\ldots(i)$This pressure difference provides the lift to the aeroplane. If F be the lift on the wing, then$=\frac{1}{2}\times 1.3\times(70^{2}-63^{2})\times 2.5$$=\frac{1}{2}\times 1.3\times 931\times 2.5=1512.9\,\text{N}$$=1.5\times 10^{3}\,\text{N}$