An aircraft of mass `4 xx 10^(5)kg` with total wing area `500m^(2)` in level flight at a speed of 720 km h^(-1)`. The density of a at its height is `1.2 kgm^(-3)`. The fractional increases in the speed of the air on the upper surface of its wings relative to the lower surface is (take `g = 10 ms^(-2)`)

Options

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

Answer

C

Explanation

The weight of the aircraft is balanced by the upward force due to the pressure difference. i.e.,
`DeltaP = (mg)/(A) = ((4xx10^(5)kg)(10ms^(-2)))/(500m^(2)) = 4/5 xx10^(4)Nm^(-2)`
`=8xx10^(3)Nm^(-2)`
Let `v_(1),v_(2)` are the speed of air on the lower and upper surface of the wings of the aircraft and `P_(1),P_(2)` are the pressure there.
Using Bernoulli's theorem, we get
`P_(1)+1/2rhov_(1)^(2)=P_(2)+1/2rhov_(2)^(2)`
`P_(1)-P_(2) = 1/2(rhov_(2)^(2)-rhov_(1)^(2))`
`Delta P = (rho)/(2) (v_(2)+v_(1))(v_(2)-v_(1))`
or, `v_(2)-v_(1) = (Delta P)/(rho v_(av))`
Here, `v_(av) = (v_(1)+v_(2))/(2) = 720 kmh^(-1)`
`=720 xx 5/18 ms^(-1) = 200ms^(-1)`
`(v_(2)-v_(1))/(v_(av)) = (Delta P)/(rho v_(av))=(4/5 xx 10^(4))/(1.2xx(200)^(2))`
`=(4xx10^(4))/(5xx1.2xx4xx10^(4))=0.17`.