A fully loaded boeing aircraft has a mass of 5.4×10⁵ kg. Its total wing area is 500 m². It is in level flight with a speed of 1080 km/h. If the density of air ρ is 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/s²)

Solution

Physics sir · 4 steps

Lift equals weight, so the pressure difference is mg over wing area. Bernoulli does the rest.

1. Lift equals weight. In level flight the upward lift balances the weight mg = 5.4 × 10⁵ × 10 = 5.4 × 10⁶ N. Dividing by the wing area gives the pressure difference: Δ P = (5.4 × 10⁶)/500 = 1.08 × 10⁴ Pa.
2. Apply Bernoulli. Faster air on the upper surface means lower pressure there. With upper speed v₂ and lower speed v₁, write v₂² - v₁² = (v₂ - v₁)(v₂ + v₁). Both speeds are close to the flight speed v, so v₂ + v₁ ≈ 2v and v₂ - v₁ = Δ v.

   Δ P = 1/2ρ(v₂² - v₁²) ≈ ρ v Δ v ⇒ (Δ v)/v = (Δ P)/(ρ v²)

3. Put in numbers. v = 1080 km/h = 1080 × 5/18 = 300 m s⁻¹. So (Δ v)/v = (1.08 × 10⁴)/(1.2 × 300²) = 10800/108000 = 0.1.
4. So. The air on the upper surface is faster by a fraction of 0.1, that is 10%, which is option (d).

The idea to keep. For lift problems, take Δ P = mg/A, then use Δ P ≈ ρ v Δ v by treating v₂ + v₁ ≈ 2v. Always convert km/h to m/s before you substitute.

Answer from the NTA final answer key. Solution by PrepWiser.