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

  • question_answer
    A 5000 kg rocket is set for vertical firing. The exhaust speed is \[800\,m{{s}^{-1}}\]. To give an initial upward acceleration of \[20\,m{{s}^{-2}}\], the amount of gas ejected per second to supply the needed thrust will be \[(g=10\,m{{s}^{-2}})\]       [CBSE PMT 1998]

    A)                              \[127.5\,kg\,{{s}^{-1}}\]            

    B)                           \[187.5\,kg\,{{s}^{-1}}\]          

    C)                         \[185.5\,kg\,{{s}^{-1}}\]                  

    D)                         \[137.5\,kg\,{{s}^{-1}}\]

    Correct Answer: B

    Solution :

                              \[F=\frac{udm}{dt}=m(g+a)\]              Þ \[\frac{dm}{dt}=\frac{m(g+a)}{u}\] \[=\frac{5000\times (10+20)}{800}\] \[=187.5\ kg/s\]          


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