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Question

A jet airplane travelling at a speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

A
500 km/h
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B
1000 km/h
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C
1500 km/h
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D
2000 km/h
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Solution

The correct option is B 1000 km/h

Let vj vg and v0 be the velocities of jet, ejected gases and observer on the ground respectively.
Let jet be moving towards right (+ve direction.)
Ejected gases will move towards left (-ve direction).
According to the statement
vjv0=500 km/h(i)
As observer is at rest
vgvj=1500 km/h (given)(i)
Adding Eqs. (i) and (ii), we get the speed of combustion products w.r.t., observer on the ground
(vjv0)+(vgvj)=vgv0=500+(1500)
Or vgv0=1000 km/h
-ve sign shows that relative velocity of the ejected gases w.r.t. observer is towards left or in a direction opposite to the motion of the jet plane.

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