Punjab Medical Punjab - MET Solved Paper-2009

  • question_answer
    A particle is fired with a speed of\[2km{{h}^{-1}}\]. The speed with which it will move in intersteller space is\[({{v}_{e}}=8\sqrt{2}km{{h}^{-1}})\]

    A) \[16.5\,km{{h}^{-1}}\]                  

    B) \[11.2\,km{{h}^{-1}}\]

    C) \[10\,km{{h}^{-1}}\]                      

    D) \[8.8\,km{{h}^{-1}}\]

    Correct Answer: A

    Solution :

    Using law of conservation of energy \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m[{{(20)}^{2}}-v_{e}^{2}]\] Here escape velocity\[{{v}_{e}}=8\sqrt{2}km{{h}^{-1}}\] \[\therefore \]  \[{{v}^{2}}={{(20)}^{2}}-{{(8\sqrt{2})}^{2}}\]                      \[=400-128\]                      \[=272\] So,            \[v=16.5km{{h}^{-1}}\]


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