Question

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

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

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, $v_1=234km{h}^{-1}=234\times \frac{5}{18}m{s}^{-1}=65m{s}^{-1}$
$v_2=270km{h}^{-1}=270\times \frac{5}{18}=75m{s}^{-1}$
Area of wings $=2\times 25m^2=50m^2$
$\therefore P_1-P_2=\frac{1}{2}\times 1\left(75^2-65^2\right)$
Upward force on the plane $=\left(P_1-P_2\right)A$
$=\frac{1}{2}\times 1\times \left(75^2-65^2\right)\times 50m$
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,
$m=\frac{\left(P_1-P_2\right)}{g}A=\frac{1}{2}\times \frac{1\times \left(75^2-65^2\right)}{10}\times 50$
$=\frac{\left(75+65\right)\left(75-65\right)\times 50}{2\times 10}=3500kg$