A plane is in level flight at constant speed and each of its two wings has an area of 20 m². If the speed of the air is 198 km/h over the lower wing and 270 km/h over the upper wing. Determine the plane's mass? (Take air density = 1 kg/m³)

Concept:

Using Bernoulli's Equation:

\(P + \frac{1}{2}ρ v^2 + ρ gh = C\)

Also, ΔP = F /A, F = mg

Where, A is the total area of the two wings of the plane,
Calculation:

Given: A = 40m² ,since the plane is two wings, v = 198 km/h = 55m/s, v = 270 km/h = 75m/s, ρ = 1kg/m³

Now since the plane is in a horizontal plane therefore there is no change in h i.e. Δh = 0

We can apply Bernoulli's theorem between any two points of the flow region as the flow is irrotational.

Applying Bernoulli's Equation between the upper and lower sides of the plane

\(P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2\)

\(P_1 -P_2 = \frac{1}{2} × 1 × [ 75^2- 55^2]\)

\( Δ P = 1300 \ Pascal\)

Now ΔP × A = F = mg

⇒ 1300 × (2 × 20) = mg 

Note, that we have to take the total area ie the area of the two wings.

⇒ mg = 52000

⇒  m = 52000/9.8 = 5306.1 kg

Hence the correct option is m = 5306.1 kg

Additional InformationBernoulli's equation in terms of head

\(\frac{P}{\rho g} + \frac{v^2}{2g} + z = constant\)