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Question: If the force on a rocket, moving with a velocity of \(300ms^{- 1}\)is 210, then the rate of combusti...

If the force on a rocket, moving with a velocity of 300ms−1300ms^{- 1}is 210, then the rate of combustion of the fuel is :

A

0.07kgs−10.07kgs^{- 1}

B

1.4kgs−11.4kgs^{- 1}

C

0.7kgs−10.7kgs^{- 1}

D

10.7kgs−110.7kgs^{- 1}

Answer

0.7kgs−10.7kgs^{- 1}

Explanation

Solution

Force =ddt= \frac{d}{dt} (momentum)

=ddt(mv)=v(dmdt)⇒210=300(dmdt)= \frac{d}{dt}(mv) = v\left( \frac{dm}{dt} \right) \Rightarrow 210 = 300\left( \frac{dm}{dt} \right)

Rate of combustion, , dmdt=210300=0.7kgs−1\frac{dm}{dt} = \frac{210}{300} = 0.7kgs^{- 1}