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 xx A= mg`
`DeltaP = (mg)/A= ((4xx10^(5)kg)(10ms^(-2)))/(500m^2) =4/5xx10^(4)Nm^(-2)`
`= 8 xx10^3 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 pressures there.
Using Bernoulli.s theorem, we get
`P_1+1/2rhov_1^2=P_2+1/2rhov_2^2`
`P_1-P_2=1/2rho(v_2^2-v_1^2)`
`DeltaP=rho/2(v_2+v_1)(v_2-v_1)`
`=rhov_(av)(v_2-v_1)`
or `v_2-v_1=(DeltaP)/(rhov_(av))`
Here, `a_(av)=(v_1+v_2)/(2)= 720 km h^(-1)`
`=720 xx5/18 ms^(-1)= 200 ms^(-1)`
`:. (v_2-v_1)/(v_(av))=(DeltaP)/(rhov_(av)^2)=(4/5xx10^4)/(1.2 xx(200)^2)`
`= (4xx10^4)/(5xx1.254 xx10^4)=0.17`