A plane is in level flight at constant speed and each of the two wings has an area of `25m^(2)`. If the speed of the air on the upper and lower surfaces of the wing are `270kmh^(-1)` and `234 kmh^(-1)` respectively, then the mass of the plane is (take the density of the air = `1kg m^(-3)`)

Options

A. `1550kg`
B. `1750kg`
C. `3500kg`
D. `3200kg`

Answer

C

Explanation

Let `v_(1), v_(2)` are the speed of air on the lower and upper surfaces `S` of the wings of the plane `P_(1)` and `P_(2)` are the pressure there.
According to Bernoulli's theorem
`P_(1) +1/2 rho v_(1)^(2) = P_(2+1/2rhov_(2)^(2)`
`P_(1)-P_(2) = 1/2 rho (v_(2)^(2)-v_(1)^(2))`
Here, `v_(1) = 234 kmh^(-1) = 234 xx 5/18 ms^(-1)= 65ms^(-1)`
`v_(2) = 270 kmh^(-1) = 270 xx 5/18 = 75 ms^(-1)`
Area of wings =`2 xx 25m^(2) = 50m^(2)`
`:. P_(1)-P_(2)= 1/2 xx 1 (75^(2)-65^(2))`
upward force on the plane = `(P_(1)-P_(2))A`
`=1/2 xx 1 xx (75^(2)-65^(2)) xx 50m`
As the plane is in level flight, therefore upward force balances the weight of the plane.
`:. mg = (P_(1)-P_(2))A`
Mass of the plane
`m = ((P_(1)-P_(2))/(g)A`
`=1/2xx(1xx(75^(2)-65^(2)))/(10)xx50`
`=((75+65)(75-65)xx50)/(2xx10) = 3500kg`.