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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    If force on a rocket having exhaust velocity of 300 m/sec is 210 N, then rate of combustion of the fuel is       [CBSE PMT 1999; MH CET 2003; Pb. PMT 2004]            

    A)                              0.7 kg/s            

    B)                                       1.4 kg/s            

    C)                         0.07 kg/s            

    D)                           10.7 kg/s

    Correct Answer: A

    Solution :

                                \[F=u\ \left( \frac{dm}{dt} \right)\] Þ \[\frac{dm}{dt}=\frac{F}{u}=\frac{210}{300}=0.7\ kg/s\]            


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