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AMU Medical AMU Solved Paper-2005

  • question_answer
    A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity 100 m/s 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? \[\left( \text{Take g}=\text{ 1}0\text{ m}/{{\text{s}}^{\text{2}}} \right).\]

    A)  50 kg/s                                

    B)  100 kg/s

    C)   200 kg/s                            

    D)  400 kg/s

    Correct Answer: B

    Solution :

                     The force (F) lifting the rocket is given by                                 \[F={{v}_{r}}\left( \frac{\Delta m}{\Delta t} \right)\]        Where \[{{v}_{r}}\] velocity of the gas relative to rocket, \[\frac{\Delta m}{\Delta t}\] is, is rate of change of mass.                 Given, \[F=Mg=1000\times 10,000\,N\]                 \[{{v}_{r}}=100\,m/s\]                 \[\therefore \]  \[\frac{\Delta m}{\Delta t}=\frac{F}{{{v}_{r}}}\]                                 \[=\frac{10,000}{100}=100\,kg/s\]


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