A plane is in level flight at constant speed and each of its two wings has an area of $40\ \text{m}^2$. If the speed of the air is $180\ \text{km h}^{-1}$ over the lower wing surface and $252\ \text{km h}^{-1}$ over the upper wing surface, the mass of the plane is ______ kg. (Take air density to be $1\ \text{kg m}^{-3}$ and $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 9600
$180\ \text{km h}^{-1} = 50\ \text{m s}^{-1}$ and $252\ \text{km h}^{-1} = 70\ \text{m s}^{-1}$.
By Bernoulli's principle, the pressure difference is
$$\Delta P = \frac12\rho(v_2^2 - v_1^2) = \frac12\times1\times(4900 - 2500) = 1200\ \text{Pa}$$
Lift on both wings (total area $80\ \text{m}^2$) balances the weight:
$$mg = 1200\times80 = 96000\ \text{N} \Rightarrow m = 9600\ \text{kg}$$
Solution by Sreeraj P, M.Sc Physics