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

Answer

C

Explanation

Let `v_(1),v_(2)` are the speed of air on the lower and upper surface 5 of the wings of the plane `P_(1)` and `P_2` are the pressure there,
According to Bernoulli.s theorem
`P_(1) + 1/2rhov_(1)^(2) = P_(2) + 1/2rhov_(2)^(2)`
`P_(1)-P_(2) = 1/2(rhov_(2)^(2) - v_(1)^(2))`
Here, `v_(1) = 234 km h^(-1) = 234 xx 5/18 ms^(-1) = 65 ms^(-1)`
`v_(2) = 270 km h^(-1) = 270 xx 5/18 = 75 ms^(-1)`
Area of wings `= 2 xx 25 m^(2) = 50 m^(2)`
`therefore P_(1) - P_(2) = 1/2 xx (75^(2) - 65^(2))`
Upward force on the plane = `(P_(1) - P_(2))A`
`=1/2 xx 1 xx (75^(2) - 65^(2)) xx 50 m^(2)`
As the plane is in level flight, therefore upward force balances the weight of the plane.
`therefore mg = (P_(1) - P_(2))A`
Mass of the plane,
`m = (P_(1) - P_(2))g.A = 1/2 xx (1 xx (75^(2) - 65^(2)))/10 xx 50`
` = ((75 + 65) (75-65) xx 50)/(2 xx 10) = 3500 kg`