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.

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