Question

A rocket of mass 1000 kg is to be projected vertically upward the gaseous are existed vertically downwards with velocity so metre per second with respect to the rocket what is the minimum rate of burning fuel so as to just leave the rocket of words against the habitation attraction

Answer 1

Q. A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity 100 ms−1ms−1 with respect to the rocket. What is the minimum rate of burning of fuel, so as to just lift the rocket upwards against the gravitational attraction? (Take g = 10 ms−2)ms−2)



 



A 

50 kgs−1kgs−1

correct answerB 

100 kgs−1kgs−1

C 

200 kgs−1kgs−1

D 

400 kgs−1kgs−1

Solution:
The velocity of exhaust gases with respect to rocket 
             =100ms−1             =100ms−1 
The minimum force on the rocket to lift it 
    Fmin=mg=l000×10=10000N    Fmin=mg=l000×10=10000N
Hence, minimum rate of burning of fuel is given by 
      dmdt=fminv=10000100      dmdt=fminv=10000100 
      100kgs−1