Question

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.
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Answer 1

Answer:

Hey mate here is your answer.....

(A)1×103N(B)1.45×103N(C)1.51×103N(D)2.1×104N(A)1×103N(B)1.45×103N(C)1.51×103N(D)2.1×104N

V1=70m/sV1=70m/s

V2=63m/sV2=63m/s

Area of wing A=2.5m2A=2.5m2

Density of air ρ=1.3Kgm−3ρ=1.3Kgm−3

According to Bernoulli's theorem

P1+12P1+12ρV21ρV12=P2+12=P2+12ρV22ρV22

P2−P1=12P2−P1=12ρ(V21−V22)ρ(V12−V22)

Where P1P1 & P2P2 is the pressure on the upper & lower surface of the ring respectively.

Lift on the wing =(P2−P1)A(P2−P1)A

⇒12⇒12ρ(V21−V22)Aρ(V12−V22)A

⇒12⇒12×1.3(702−632)×2.5×1.3(702−632)×2.5

⇒1.51×103N⇒1.51×103N

Therefore lift on the wing of the aeroplane =1.51×103N1.51×103N

Hence (C) is the correct answer.

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