Fluid Mechanics

JEE Main level, single correct, medium

39 of 192

Question 39 of 192

A fully loaded boeing aircraft has a mass of 5.4×105kg5.4\times {10}^{5}kg. Its total wing area is 500m2500{m}^{2}. It is in level flight with a speed of 1080kmh−11080km{h}^{-1}. If the density of air ρ\rho is 1.2kgm–31.2kg{m}^{–3}, the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be
(g=10ms−2g=10m{s}^{-2})

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Answer: (D) 1010

Required lift force, F=P2A−P1A=5.4×105×gF={P}_{2}A-{P}_{1}A=5.4\times {10}^{5}\times g
⇒P2−P1=5.4×106500=5.4×2×103\Rightarrow {P}_{2}-{P}_{1}=\dfrac{5.4\times {10}^{6}}{500}=5.4\times 2\times {10}^{3}
=10.8×103=10.8\times {10}^{3}
Now applying Bernoulli's Theorem we get
P2+0+12ρV22=P1+0+12ρV12{P}_{2}+0+\dfrac{1}{2}\rho {V}_{2}^{2}={P}_{1}+0+\dfrac{1}{2}\rho {V}_{1}^{2}
⇒P2−P1=12ρ(V12−V22)=12ρ(V1−V2)(V1+V2)[Since for small change,V1=V2=V=1080kmh−1=300ms−1]\Rightarrow {P}_{2}-{P}_{1}=\dfrac{1}{2}\rho \left({V}_{1}^{2}-{V}_{2}^{2}\right)=\dfrac{1}{2}\rho \left({V}_{1}-{V}_{2}\right)\left({V}_{1}+{V}_{2}\right) \left[\text{Since for small change},{V}_{1}={V}_{2}=V=1080km{h}^{-1}=300m{s}^{-1}\right]
⇒10.8×103=12×1.2(V1−V2)×2×300\Rightarrow 10.8\times {10}^{3}=\dfrac{1}{2}\times 1.2\left({V}_{1}-{V}_{2}\right)\times 2\times 300
⇒10.8×10=3.6(V1−V2)\Rightarrow 10.8\times 10=3.6\left({V}_{1}-{V}_{2}\right)
⇒V1−V2=30\Rightarrow {V}_{1}-{V}_{2}=30
Therefore, fractional increase will be(V1−V2V)×100=30300×100=10\left(\dfrac{{V}_{1}-{V}_{2}}{V}\right)\times 100=\dfrac{30}{300}\times 100=10%

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