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A small aeroplane of mass 1040kg is flying horizontally. The total area of the wings in 10m². Air has a velocity 50m / s below the wings. The velocity of the air above the wings such that the weight of the aeroplane could be balanced is ________ (density of air $= 1 .3 kg / m^{3} g = 10 m / s^{2}$ )

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45m / s

35m / s

30m / s

40m / s

answer is A.

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Detailed Solution

$(P_{2} - P_{1}) A = mg$

$\Rightarrow \frac{(V_{2}^{2} - V_{1}^{2})}{2} ρA = mg$

$\Rightarrow \frac{(50^{2} - V_{1}^{2})}{2} 1.3 (10) = (1040) (10)$

$\Rightarrow V_{1} = 30 m / s$

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