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Q 10.27 A plane is in level flight at constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass.

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Speed of the wind above the upper wing surface is v1 = 234 km h-1

\\v_{1}=234\times \frac{1000}{3600}\\v_{1}=65ms^{-1}

Speed of the wind over the lower wing is v2 = 180 km h-1

\\v_{2}=180\times \frac{1000}{3600}\\ v_{2}=50ms^{-1}

Let the pressure over the upper and lower wing be P1 and P2

Let the plane be flying at a height of h

The density of air is \rho =1\ kg\ m^{-3}

Applying Bernoulli's Principle at two points over the upper and lower wing we get

\\P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gh=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gh\\ P_{2}-P_{1}=\frac{1}{2}\rho (v_{1}^{2}-v_{2}^{2})\\ \Delta P=\frac{1}{2}(65^{2}-50^{2})\\ \Delta P=862.5\ Pa

Area of each wing is a = 25 m2

The net upward force on the plane is F

\\F=2a\Delta P\\ F=2\times 25\times 862.5\\ F=43125\ N

This upward force is equal to the weight m of the plane.

\\mg=F\\ m=\frac{F}{g}\\ m=\frac{43125}{9.8}\\ m=4400.51\ kg

The mass of the plane is about 4400 kg

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Sayak

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