Question

An aircraft of mass 4 $\times$ 10${}^5$ kg with total wing area 500 m${}^2$ in level flies at a speed of 720 km h${}^{-1}$. The density of air at its height is 1.2 kg m${}^{-3}$. The fractional increase in the speed of the air on the upper surface of its wings relative to the lower surface is: (Take g = 10 m s${}^{-2}$)

• A. 0.04
• B. 0.08
• C. 0.17
• D. 0.34

Answer 1

Answer: (C)

0.17

Given, $mass=4\times {10}^{5}kg,Area=500m^2,speed=720km/h,height=1.2kg/m^3,g=10m/s^2$

So,

The weight of the aircraft is balanced by the upward force due to the pressure difference $i;e\Delta p=\frac{mg}{A}=\frac{4\times {10}^{5}kg\times {10}^m/s^2}{500m^2}=\frac{4}{5}\times {10}^{4}N/m^2$

Let $v_1,v_2$ be the speed of air on the lower and upper of the aircraft $p_1,p_2$ are the pressure,

By the Bernoullis theorem,

$p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2$

$p_1-p_2=\frac{\rho }{2}(v_2^2-v_1^2)$

$\Rightarrow \Delta p=\frac{\rho }{2}(v_2+v_1)(v_2-v_1)=\rho v_{av}(v_2-v_1)$

$\Rightarrow v_2-v_1=\frac{\Delta p}{\rho v_{av}}$

here : $v_{av}=\frac{v_1+v_2}{2}=720km/h=200m/s$

$\Rightarrow \frac{v_2-v_1}{v_{av}}=\frac{\Delta p}{\rho v_{av}}=\frac{\frac{4}{5}\times {10}^4}{1.2\times {200}^2}=0.17$