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Air is streaming past a horizontal aeroplane wing such that its speed is $120\, m/s$ over the upper surface and $90\, m/s$ at the lower surface. If the density of air is $1.3\, kg/m^3$ and the wing is $10\, m$ long and has an average width of $2\, m$ , then the difference of the pressure on the two sides of the wing is ........ $N/m^2$
$40.95$
$409.5$
$4095$
$40950$
Solution
Using Bernoulli's theorem
$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{v}_{1}^{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{v}_{2}^{2}+\rho \mathrm{gh}_{2}$
For the horizontal motion of the plane $\mathrm{h}_{1}=\mathrm{h}_{2}$, then
$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{v}_{1}^{2}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{v}_{2}^{2}$
or $\quad \mathrm{P}_{2}-\mathrm{P}_{1}=\frac{1}{2} \rho\left(\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}\right)$
$=\frac{1}{2} \times 1.3\left[120^{2}-90^{2}\right]$
$=0.65[210 \times 30]=4095 \mathrm{\,N} / \mathrm{m}^{2}$
