The force on a rocket moving with a veloctiy 300 m/s is 210N. The rate of consumption of fuel of rocket is

Options

A. `0.07 kg s^(-1)`
B. `1.4kg s^(-1)`
C. `0.7kg s^(-1)`
D. `10.7kg s^(-1)`

Answer

C

Explanation

Force `=(d)/(dt)` (momentum)
`(d)/(dt)("mv")=v((dm)/(dt)) Rightarrow 210=300((dm)/(dt))`
rate of combustion, `(dm)/(dt)=(210)/(300)=0.7 kg s^(-1)`