Mechanical Properties of Fluids NCERT Solutions Class 11 Physics - Solved Exercise Question 10.14

Question 10.14:

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^–1and 63 m s^–1 respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m^–3.

Solution:

Speed of wind on the upper surface of the wing, V₁ = 70 m/s
Speed of wind on the lower surface of the wing, V₂ = 63 m/s
Area of the wing, A = 2.5 m²
Density of air, ρ = 1.3 kg m^–3
According to Bernoulli’s theorem, we have the relation:
P₁ + (1/2)ρV₁² = P₂ + (1/2)ρV₂²
P₂ - P₁ = (1/2) ρ (V₁² - V₂²)
Where,
P₁ = Pressure on the upper surface of the wing
P₂ = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing = (P₂ - P₁) A
= (1/2) ρ (V₁² - V₂²) A
= (1/2) X 1.3 X [ 70² - 63² ] X 2.5
= 1512.87 N  =  1.51 X 10³ N
Therefore, the lift on the wing of the aeroplane is 1.51 × 10³ N.