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In a test experiment on a model aeroplane in a wind tunnel,the flow speeds on the upper and lower surfaces of the wing are $70ms^{-1}$ respectively.What is the lift on the wing if its area is $2.5m^2$?Take the density of air to be $1.3Kgm^{-3}$

$\begin{array}{1 1}(A)\;1\times 10^3N\\(B)\;1.45\times 10^3N\\(C)\;1.51\times 10^3N\\(D)\;2.1\times 10^4N\end{array} $

1 Answer

Area of wing $A=2.5m^2$
Density of air $\rho=1.3Kgm^{-3}$
According to Bernoulli's theorem
$P_1+\large\frac{1}{2}$$\rho V_1^2$$=P_2+\large\frac{1}{2}$$\rho V_2^2$
$P_2-P_1=\large\frac{1}{2}$$\rho (V_1^2-V_2^2)$
Where $P_1$ & $P_2$ is the pressure on the upper & lower surface of the ring respectively.
Lift on the wing =$(P_2-P_1)A$
$\Rightarrow \large\frac{1}{2}$$\rho (V_1^2-V_2^2)A$
$\Rightarrow \large\frac{1}{2}$$\times 1.3 (70^2-63^2)\times 2.5$
$\Rightarrow 1.51\times 10^3N$
Therefore lift on the wing of the aeroplane =$1.51\times 10^3N$
Hence (C) is the correct answer.
answered May 9, 2014 by sreemathi.v

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