A) 0.7 kg/s
B) 1.4 kg/s
C) 0.07 kg/s
D) 10.7 kg/s
Correct Answer: A
Solution :
Thrust force on the rocket |
\[{{F}_{t}}={{v}_{r}}\,\left( -\frac{dm}{dt} \right)\] (upwards) |
Rate of combustion of fuel |
\[-\frac{dm}{dt}=\frac{{{F}_{t}}}{{{v}_{r}}}\] |
Given, \[{{F}_{t}}=210\,N,\,\,{{v}_{r}}=300\,m/s\] |
\[\therefore \] \[-\frac{dm}{dt}=\frac{210}{300}=0.7\,kg/s\] |
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