Let $\mathrm{v}_1, \mathrm{v}_2$ be the speeds on the upper and lower of the wing of aeroplane, and $\mathrm{P}_1$ and $\mathrm{P}_2$ are the on upper and lower surfaces of the wing respectively. Then $\mathrm{v}_1=70 \mathrm{~ms}^{-}$ 1; $\mathrm{v}_2=63 \mathrm{~ms}^{-1}, \rho=1.3 \mathrm{~kg} \mathrm{~m}^{-3}$.
From Bernoulli's theorem
$$
\frac{P_1}{\rho}+g h+\frac{1}{2} v_1^2=\frac{P_2}{\rho}+g h+\frac{1}{2} v_2^2
$$
$$
\therefore \quad \frac{P_1}{\rho}-\frac{P_2}{\rho}=\frac{1}{2}\left(v_2{ }^2-v_1{ }^2\right)
$$
or $\rho_1-\rho_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)$
$$
\begin{aligned}
&=\frac{1}{2} \times 1.3\left[(70)^2-(63)^2\right] \\
&=605.15 \mathrm{~Pa}
\end{aligned}
$$
The difference of pressure provides the lift to the aeroplane. So, lift on the aeroplane $=$ pressure difference $\times$ area of wings $=605.15 \times 2.5=1512.875 \mathrm{~N}$ $=1.51 \times 10^3 \mathrm{~N}$.