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Air of density $1.2 \mathrm{~kg} \mathrm{~m}^{-3}$ is blowing across the horizontal wings of an aeroplane in such a way that its speeds above and below the wings are $150 \mathrm{~ms}^{-1}$ and $100 \mathrm{~ms}^{-1}$, respectively. The pressure difference between the upper and lower sides of the wings, is :
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Verified Answer
The correct answer is:
$7500 \mathrm{Nm}^{-2}$
$7500 \mathrm{Nm}^{-2}$
Pressure difference
$$
\begin{aligned}
& \mathrm{P}_2-\mathrm{P}_1=\frac{1}{2} \rho\left(\mathrm{v}_2^2-\mathrm{v}_1^2\right) \\
& =\frac{1}{2} \times 1.2\left((150)^2-(100)^2\right) \\
& =\frac{1}{2} \times 1.2(22500-10000) \\
& =7500 \mathrm{Nm}^{-2}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{P}_2-\mathrm{P}_1=\frac{1}{2} \rho\left(\mathrm{v}_2^2-\mathrm{v}_1^2\right) \\
& =\frac{1}{2} \times 1.2\left((150)^2-(100)^2\right) \\
& =\frac{1}{2} \times 1.2(22500-10000) \\
& =7500 \mathrm{Nm}^{-2}
\end{aligned}
$$
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