Fluid Mechanics

JEE Main level, single correct, medium

22 of 192

Question 22 of 192

Air of density 1.2 kg m−31.2 \mathrm{~kg} \mathrm{~m}^{-3} is blowing across the horizontal wings of an aeroplane in such a way that its speeds above and below the wings are 150 ms−1150 \mathrm{~ms}^{-1} and 100 ms−1100 \mathrm{~ms}^{-1}, respectively. The pressure difference between the upper and lower sides of the wings, is :

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Answer: (C) 7500Nm−27500 \mathrm{Nm}^{-2}

Pressure difference
P2−P1=12ρ(v22−v12)=12×1.2((150)2−(100)2)=12×1.2(22500−10000)=7500Nm−2\begin{aligned} & \mathrm{P}_2-\mathrm{P}_1=\frac{1}{2} \rho\left(\mathrm{v}_2^2-\mathrm{v}_1^2\right) \\ & =\frac{1}{2} \times 1.2\left((150)^2-(100)^2\right) \\ & =\frac{1}{2} \times 1.2(22500-10000) \\ & =7500 \mathrm{Nm}^{-2} \end{aligned}

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