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JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    Diwali rocket is ejecting \[50g\] of \[gases/\sec \] at a velocity of\[400\,\,m/s\]. The accelerating force on the rocket will be

    A) \[22\,\,dyne\]                  

    B) \[20\,\,N\]

    C)  \[20\,\,dyne\]                 

    D)  \[100\,\,N\]

    Correct Answer: B

    Solution :

    The accelerating force on the rocket = upward thrust\[=\frac{\Delta m}{\Delta t}\cdot u\] Given,\[\frac{\Delta m}{\Delta t}=50\times {{10}^{-3}}kg/s,\,\,v=400\,\,m/s\] So, accelerating force\[=50\times {{10}^{-3}}\times 400\]                                          \[=20\,\,N\]


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