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An aircraft of mass 4 xx 10^(5)kg with t...

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)`)

A

0.04

B

0.08

C

0.17

D

0.34

Text Solution

Verified by Experts

The correct Answer is:
C
The weight of the aircraft is balance by the upward force due to the pressure difference.
i.e., `DeltaPxx A=mg`
`DeltaP=(mg)/(A)=((4xx10^(5)kg)(10 m s^(-2)))/(500 m^(2))=(4)/(5)xx10^(4)N m^(-2)`
`8xx10^(3) N m^(-2)`
Let `v_(1) , v_(2)` are the speed of air on the lower and upper surface of the wings of aircraft and `P_(1), P_(2)` are the pressure there.
using Bernoulli's theorem, we get
`P_(1)+(1)/(2) rhov_(1)^(2)=P_(2)rhov_(2)^(2)`
`P_(1)-P_(2)=(1)/(2)rho(v_(2)^(2)-v_(1)^(2))`
`DeltaO=(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, `v_(av)=(v_(1)+v_(2))/(2)=720 Km h^(-1)`
`=720xx(5)/(18)m s^(-1)=200 m s^(-1)`
`therefore (v_(2)-v_(1))/(v_(av))=(DeltaP)/(rhov_("av")^(2))=((4)/(5)xx10^(4))/(1.2xx(200)^(2))`
`=(4xx10^(4))/(5xx1.2xx4xx10^(4))=0.17`
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