Question

A plane is in level flight at constant speed and each of its two wings has an area of $ 25\, m^{2} $ . If the speed of the air on the upper and lower surfaces of the wing are $ 270\, km \,h^{-1} $ and $ 234 \,km\, h^{-1} $ respectively, then the mass of the plane is (Take the density of the air $ = 1 \,kg\, m^{-3} $ )

• A. $ 1550\, kg $
• B. $ 1750 \,kg $
• C. $ 3500\, kg $
• D. $ 3200\, kg $

Answer 1

Answer: (C)

Let $ v_{1}, v_{2} $ are the speed of air on the lower and upper surface $ S $ of the wings of the plane $ P_{1} $ and $ P_{2} $ are the pressure there.
According to Bernoulli’s theorem
$ P_{1}+\frac{1}{2} \rho v_{1}^{2}=P_{2}+\frac{1}{2} \rho v_{2}^{2} $
$ P_{1}-P_{2}=\frac{1}{2} \rho\left(v_{2}^{2}-v_{1}^{2}\right) $
Here, $ v_{1}=234 \, km\, h^{-1}=234\times\frac{5}{18}m s^{-1}=65\,m\,s^{-1} $
$ v_{2}=270 \, km\, h^{-1}=270\times\frac{5}{18}=75\, m\, s^{-1} $
Area of wings $ =2\times25\, m^{2}=50\,m^{2} $
$ \therefore P_{1}-P_{2}=\frac{1}{2}\times1\left(75^{2}-65^{2}\right) $
Upward force on the plane $ =\left(P_{1}-P_{2}\right)A $
$ =\frac{1}{2}\times1\times\left(75^{2}-65^{2}\right)\times50\,m $
As the plane is in level flight, therefore upward force balances the weight of the plane.
$ \therefore mg=\left(P_{1}-P_{2}\right)A $
Mass of the plane,
$ m=\frac{\left(P_{1}-P_{2}\right)}{g} A =\frac{1}{2}\times\frac{1\times\left(75^{2}-65^{2}\right)}{10}\times50 $
$ =\frac{\left(75+65\right)\left(75-65\right)\times50}{2\times10}=3500\, kg $