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JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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 ........ \(Nm^{-2}\)
- A \(60\)
- B \(180\)
- C \(7500\)
- D \(12500\)
Answer & Solution
Correct Answer
(C) \(7500\)
Step-by-step Solution
Detailed explanation
Pressure difference \({P_2} - {P_1} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)\) \( = \frac{1}{2} \times 1.2\left( {{{\left( {150} \right)}^2} - {{\left( {100} \right)}^2}} \right)\) \( = \frac{1}{2} \times 1.2\left( {22500 - 10000} \right)\) \( = 7500\,N{m^{ - 2}}\)
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