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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 vj−v0=500 km/h……(i) As observer is at rest vg−vj=−1500 km/h (given)……(i) ∴ Adding Eqs. (i) and (ii), we get the speed of combustion products w.r.t., observer on the ground (vj−v0)+(vg−vj)=vg−v0=500+(−1500) Or vg−v0=−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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