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)

3500 kg

Let $v_1$, $v_2$ are the speed of air on the lower and upper surfaces of the wings of the plane $P_1$ and $P_2$ are the pressure there.

According to Bernoulli's theorem

$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$

$P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)$

Here, $v_1$ = 234 km h${}^{-1}$ = 234 x 58 m s${}^{-1}$ =65 m s${}^{-1}$

$v_2$ = 270 km h$-1$ = 270 x $\frac{5}{18}$ = 75 m s${}^{-1}$

Area of wings = 2 x 25 m${}^2$ = 50 m${}^2$

$\therefore P_1-P_2=\frac{1}{2}\times 1({75}^2-{65}^2)$ Upward force on the plane = $(P_1-P_2)A$

$=\frac{1}{2}\times 1\times ({75}^2-{65}^2)\times 50m$

As the plane is in level flight, therefore upward force balances the weight of the plane.

$\therefore mg=(P_1-P_2)A$

Mass of the plane,

$\therefore m=\frac{(P_1-P_2)}{g}\times =\frac{1}{2}\times \frac{1\times ({75}^2-{65}^2)}{10}\times 50$

$=\frac{(75+65)(75-65)\times 50}{2\times 10}=3500kg$