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JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2024 · 29 January · Morning Shift · Physics · Mechanical Properties of Fluids

In a test experiment on a model aeroplane in wind tunnel, the flow...

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})

Correct answer

0

Step-by-step solution

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}ρ{\left({v}_{1}\right)}^{2} + ρg{h}_{1} = {P}_{2} + \dfrac{1}{2}ρ{\left({v}_{2}\right)}^{2} + ρg{h}_{2}

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

⇒P2−P1=12ρ(v1)2−12ρ(v2)2⇒ {P}_{2} - {P}_{1} = \dfrac{1}{2}ρ{\left({v}_{1}\right)}^{2} - \dfrac{1}{2}ρ{\left({v}_{2}\right)}^{2}

Therefore, upthrust acting due to pressure difference:

⇒FA=12ρ(v12−v22)⇒F=12×1.2×(702−652)×2⇒ \dfrac{F}{A} = \dfrac{1}{2}ρ\left({v}_{1}^{2} - {v}_{2}^{2}\right) ⇒ F = \dfrac{1}{2} \times 1 . 2 \times \left({70}^{2} - {65}^{2}\right) \times 2

=810N= 810N

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