A) 50 kg/s
B) 100 kg/s
C) 200 kg/s
D) 400 kg/s
Correct Answer: B
Solution :
Given: Rockets mass\[m=100\text{ }kg\] The velocity of exhaust gases with respect to rocket\[=10\text{ }m/s\] \[g=10\text{ }m/{{s}^{2}}\] The minimum force on the rocket to lift it \[{{F}_{\min }}=mg=1000\times 10=10000\text{ }N\] Hence, minimum rate of burning of fuel is given by \[\frac{dm}{dt}=\frac{{{f}_{\min }}}{\upsilon }=\frac{10000}{100}\] \[=100kg/s\]You need to login to perform this action.
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