First slide

Fluid dynamics

Question

A plane is in level fight at constant speed and each of its two wings has an area of 25 $m^{2}$ . If the speed of the air on the upper and lower surfaces of the wing are 270 kmph and 234 kmph respectively,then mass of the plane is (take the density of air is 1 kg $m^{3}$

Difficult

A

15550 kg
B

1750 kg
C

3500 kg
D

3200 kg

Solution

Let $v_{1}, v_{2} are the speed of air on the lower and upper surface S of the wings of the plane P_{1} and P_{2} are the Pressure there .$
According to Bernoulli's theorem
$P_{1} + \frac{1}{2} {ρv}_{1}^{2} = P_{2} + \frac{1}{2} {ρv}_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ρ ({v_{2}}^{2} - {v_{1}}^{2})$
Here, $v_{1} = 234 km h^{- 1} = 234 \times \frac{5}{18} {ms}^{- 1} = 65 {ms}^{- 1}$
$v_{2} = 270 km h^{- 1} = 270 \times \frac{5}{18} = 75 {ms}^{- 1}$
Area of wings = 2 $\times 25 m^{2} = 50 m^{2}$
$P_{1} - P_{2} = \frac{1}{2} \times 1 (75^{2} - 65^{2})$
Upward force on the plane = ( $P_{1} - P_{2}$ )A
= $\frac{1}{2} \times 1 \times (75^{2} - 65^{2}) \times 50 m$
As the plane is in level flight, therefore upward force balances the weight of the plane
$mg = (P_{1} - P_{2}) A$
Mass of the plane,
m = $\frac{(P_{1} - P_{2})}{g} A$
= $\frac{1}{2} \times \frac{1 \times (75^{2} - 65^{2})}{10} \times 50$
$= \frac{(75 + 65) (75 - 65) \times 50}{2 \times 10} = 3500 kg$

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