# 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. - Physics

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 1

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:

P_1 + 1/2rhoV_1^2 = P_2 + 1/2 rhoV_2^2

P_2 - P_1 = 1/2 rho(V_1^2 - V_2^2)

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 = (P_2 - P_1)A

=1/2 rho(V_1^2 - V_2^2)A

= 1/2 1.3((70)^2-(63)^2)xx 2.5

= 1512.87

= 1.51 × 103 N

Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.

#### Solution 2

Let v1, v2 be the speeds on the upper and lower surfaces of the wing of the aeroplane, and P1 and P2 be the pressure on upper and lower surfaces of the wing respectively.

Then v_1 = 70 ms^(-1), v_2 = 63 ms^(-1); rho = 1.3 kg m^(-3)

From Berniulli's theorem

P_1/rho + gh + 1/2v_1^2 = P_2/rho + gh + 1/2 v_2^2

:.P_1/rho - P_2/rho = 1/2 (v_2^2 -v_1^2)

or P_1 -P_2 = 1/2 rho(v_2^2 - v_1^2) = 1/2 xx 1.3[(70)^2 - (63)^2] Pa = 605.15 Pa

This difference of pressure provides the lift to the aeroplane. So lift on the aeroplane = pressiure difference x area of wing

= 605.15 x 2.5 N = 1512.875 N

= 1.15 x 103 N

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 10 Mechanical Properties of Fluids
Q 14 | Page 269