A fully loaded boeing aircraft has a mass of \(5.4 \times 10^{5} \mathrm{~kg}\). Its total wing area is \(500 \mathrm{~m}^{2}\). It is in level flight with a speed of \(1080 \mathrm{~km} / \mathrm{h}\). If the density of air \(\rho\) is \(1.2 \mathrm{~kg} \mathrm{~m}^{-3}\), the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be (\(g=10 \mathrm{~m} / \mathrm{s}^{2}\))

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Fluid Mechanics

A fully loaded boeing aircraft has a mass of

5.4 \times 1 0^{5} kg

. Its total wing area is

500 m^{2}

. It is in level flight with a speed of

1080 km / h

. If the density of air

ρ

1.2 kg m^{- 3}

, the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be (

g = 10 m / s^{2}

)

Answer: (D)

Level flight needs the pressure difference across the wing to carry the weight,

(P_{low} - P_{up}) A = m g

, so

P_{low} - P_{up} = \frac{5.4 \times 1 0 ^{5} \times 10}{500} = 1.08 \times 1 0^{4}

Pa. By Bernoulli's theorem this equals

\frac{1}{2} ρ (v_{up}^{2} - v_{low}^{2}) = \frac{1}{2} ρ (v_{up} - v_{low}) (v_{up} + v_{low})

. With the flight speed

v = 1080

km/h

= 300

m/s and

v_{up} + v_{low} \approx 2 v = 600

m/s,

1.08 \times 1 0^{4} = \frac{1}{2} (1.2) (v_{up} - v_{low}) (600)

gives

v_{up} - v_{low} = 30

m/s. As a fraction of the flight speed this is

\frac{30}{300} \times 100 = 10%

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option