JEE Main 2024PhysicsMechanical Properties of FluidsMediumNumerical

JEE Main 2024 — Mechanical Properties of Fluids Question with Solution

JEE Main 2024 (29 Jan Shift 1)

Question

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 m s-1 and 65 m s-1 respectively. If the wing area is 2 m2, the lift of the wing is _______N. (Given density of air =1.2 kg m-3)

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Show full solutionCorrect answer: 810
Correct answer
810

Step-by-step explanation

Applying Bernoulli's principle on upper & lower surfaces, we get

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2

Since, height difference between upper and lower surface is negligible.

⇒P2-P1=12ρv12-12ρv22

Therefore, upthrust acting due to pressure difference:

⇒FA=12ρv12-v22⇒F=12×1.2×702-652×2

=810 N

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About this question

This is a previous-year question from JEE Main 2024, covering the Mechanical Properties of Fluids chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.