Fluid Mechanics

JEE Main level, numerical answer, medium

54 of 192

Question 54 of 192

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70ms−170m{s}^{-1} and 65ms−165m{s}^{-1} respectively. If the wing area is 2m22{m}^{2}, the lift of the wing is _______NN. (Given density of air =1.2kgm−3=1.2kg{m}^{-3})

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Answer: 810

Applying Bernoulli's principle on upper & lower surfaces, we get
P1+12ρ(v1)2+ρgh1=P2+12ρ(v2)2+ρgh2{P}_{1}+\dfrac{1}{2}\rho {\left({v}_{1}\right)}^{2}+\rho g{h}_{1}={P}_{2}+\dfrac{1}{2}\rho {\left({v}_{2}\right)}^{2}+\rho g{h}_{2}
Since, height difference between upper and lower surface is negligible.
⇒P2−P1=12ρ(v1)2−12ρ(v2)2\Rightarrow {P}_{2}-{P}_{1}=\dfrac{1}{2}\rho {\left({v}_{1}\right)}^{2}-\dfrac{1}{2}\rho {\left({v}_{2}\right)}^{2}
Therefore, upthrust acting due to pressure difference:
⇒FA=12ρ(v12−v22)⇒F=12×1.2×(702−652)×2\Rightarrow \dfrac{F}{A}=\dfrac{1}{2}\rho \left({v}_{1}^{2}-{v}_{2}^{2}\right) \Rightarrow F=\dfrac{1}{2}\times 1.2\times \left({70}^{2}-{65}^{2}\right)\times 2
=810N=810N

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