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The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane $arefv_1$ and $v_2$, respectively. If $A$ is the cross-sectional area of the wing and $\rho$ is the density of air, then the upward lift is
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The correct answer is:
$\frac{1}{2} \rho A\left(v_1^2-v_2^2\right)$
Due to the specific shape of wings, when the aeroplane runs, air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called 'dynamic lift' acts on the plate.
$\therefore$ Upward lift $=$ pressure difference $\times$ area of the wings
$[\because F=p \times A]$
From Bernoulli's equation,
$\begin{aligned}
p_1+\frac{1}{2} \rho v_1{ }^2 & =p_2+\frac{1}{2} \rho v_2{ }^2 \\
\because \quad p_2-p_1 & =\frac{1}{2} \rho\left(v_1{ }^2-v_2{ }^2\right) \\
\text { Hence, upward lift } & =\frac{1}{2} \rho A\left(v_1^2-v_2^2\right)
\end{aligned}$
$\therefore$ Upward lift $=$ pressure difference $\times$ area of the wings
$[\because F=p \times A]$
From Bernoulli's equation,
$\begin{aligned}
p_1+\frac{1}{2} \rho v_1{ }^2 & =p_2+\frac{1}{2} \rho v_2{ }^2 \\
\because \quad p_2-p_1 & =\frac{1}{2} \rho\left(v_1{ }^2-v_2{ }^2\right) \\
\text { Hence, upward lift } & =\frac{1}{2} \rho A\left(v_1^2-v_2^2\right)
\end{aligned}$
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