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Question: A plane is in level fight at constant speed and each of its two wings has an area of \(234 \mathrm {...

A plane is in level fight at constant speed and each of its two wings has an area of 234 km h1234 \mathrm {~km} \mathrm {~h} ^ { - 1 } respectively, then the mass of the plane is

(Take the density of the air = 1 kg)

A

1550 kg

B

1750 kg

C

3500 kg

D

3200 kg

Answer

3500 kg

Explanation

Solution

Let P1\mathrm { P } _ { 1 } and are the pressure there

According to Bernoulli’s theorem

Here,

v2=270 km h1=270×518=75 ms1\mathrm { v } _ { 2 } = 270 \mathrm {~km} \mathrm {~h} ^ { - 1 } = 270 \times \frac { 5 } { 18 } = 75 \mathrm {~ms} ^ { - 1 }

Area of wings =2×25 m2=50 m2= 2 \times 25 \mathrm {~m} ^ { 2 } = 50 \mathrm {~m} ^ { 2 }

P1P2=12×1(752652)\therefore \mathrm { P } _ { 1 } - \mathrm { P } _ { 2 } = \frac { 1 } { 2 } \times 1 \left( 75 ^ { 2 } - 65 ^ { 2 } \right)

Upward force on the plane

=(P1P2)A= \left( \mathrm { P } _ { 1 } - \mathrm { P } _ { 2 } \right) \mathrm { A }

=12×1×(752652)×50 m= \frac { 1 } { 2 } \times 1 \times \left( 75 ^ { 2 } - 65 ^ { 2 } \right) \times 50 \mathrm {~m}

As the plane is in level flight therefore upward force balances the weight of the plane.

mg=(P1P2)A\therefore \mathrm { mg } = \left( \mathrm { P } _ { 1 } - \mathrm { P } _ { 2 } \right) \mathrm { A }

Mass of the plane,

=12×1×(752652)10×50= \frac { 1 } { 2 } \times \frac { 1 \times \left( 75 ^ { 2 } - 65 ^ { 2 } \right) } { 10 } \times 50

=(75+65)(7565)×502×10= \frac { ( 75 + 65 ) ( 75 - 65 ) \times 50 } { 2 \times 10 }

=3500 kg= 3500 \mathrm {~kg}