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Punjab Medical Punjab - MET Solved Paper-2004

  • question_answer
    If a force on a rocket moving with a velocity of \[300\,\,m/s\] is\[210\,\,N\]. Then the rate of combustion of the fuel, will be:

    A) \[\text{1}\text{.7}\,\,\text{kg/sec}\]                     

    B) \[\text{0}\text{.7}\,\,\text{kg/sec}\]

    C) \[\text{2}\text{.7}\,\,\text{kg/sec}\]                     

    D) \[\text{1}\text{.5}\,\,\text{kg/sec}\]

    Correct Answer: B

    Solution :

    Here: Velocity of rocket\[u=300m/s\] Force     \[F=210\,\,N\] The force,\[F=u\left( \frac{dm}{dt} \right)\] or            \[\frac{dm}{dt}=\frac{F}{u}=\frac{210}{300}=0.7\,\,kg/\sec \]


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