Ncert Exercises & Solutions

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^-1respectively. 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.

Given,

Flow speed on the upper surface of the wing, v₁ = 70 m/s

Flow speed on the lower surface of the wing, v₂ = 63 m/s

Area of the wing, A = 2.5 m²

The density of air, ρ = 1.3 kg m^–3

According to Bernoulli’s theorem,

$P_{1} + \frac{1}{2} {ρV}_{1}^{2} = P_{2} + \frac{1}{2} {pV}_{2}^{2}$
$P_{2} - P_{1} = \frac{1}{2} ρ (V_{1}^{2} - V_{2}^{2})$

Where,

P₁ = Pressure on the upper surface of the wing

P₂ = Pressure on the lower surface of the wing

Lift on the wing $= (P_{2} - P_{1}) A$

$= \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) A = \frac{1}{2} 1.3 ({(70)}^{2} - {(63)}^{2}) \times 2.5 = 1512.87 = 1.51 \times 10^{3} N$

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