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Physics Question on mechanical properties of fluid

Air of density 1.2 kg m−31.2\, 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−1150\, ms^{-1} and 100 ms−1100\, ms^{-1}, respectively. The pressure difference between the upper and lower sides of the wings, is :

A

60 Nm−260\, Nm^{-2}

B

180 Nm−2180\, Nm^{-2}

C

7500 Nm−27500\, Nm^{-2}

D

12500 Nm−212500\, Nm^{-2}

Answer

7500 Nm−27500\, Nm^{-2}

Explanation

Solution

Pressure difference
P2−P+1=12ρ(v22−v12)P_{2}-P+_{1}=\frac{1}{2}\rho\left(v^{2}_{2}-v^{2}_{1}\right)
=12×1.2((150)2−(100)2)=\frac{1}{2}\times1.2\left(\left(150\right)^{2}-\left(100\right)^{2}\right)
=12×1.2(22500−10000)=\frac{1}{2}\times1.2\left(22500-10000\right)
=7500 Nm−2=7500\,Nm^{-2}