Air of density 1.2 kg 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 ⁻¹ and 100 ms ⁻¹, respectively. The pressure difference between the upper and lower sides of the wings, is: (a) 60 Nm ⁻² (b) 180 Nm ⁻² (c) 7500 Nm ⁻² (d) 12500 Nm ⁻²

Answer: (c). Bernoulli: Δ P = 1/2ρ(v₁² - v₂²) Δ P = 1/2(1.2)(150² - 100²) = 0.6 × 12500 = 7500 N m⁻² So the answer is (c) 7500 N m ⁻².

Air of density 1.2 kg m ⁻³ is blowing across the horizontal wings of… | JEE Main 2013 (22 Apr, Shift 1)

Q10JEE Main 2013 · 22 APR · SHIFT 1Medium

Applications of Bernoulli's Theorem

Air of density 1.2 kg 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 $^{-1}$ and 100 ms $^{-1}$ , respectively. The pressure difference between the upper and lower sides of the wings, is:

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Answer: (c)

Bernoulli: $\Delta P = \frac{1}{2}\rho(v_1^2 - v_2^2)$
$\Delta P = \frac{1}{2}(1.2)(150^2 - 100^2) = 0.6 \times 12500 = 7500\text{ N m}^{-2}$
So the answer is (c) 7500 N m $^{-2}$ .

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