CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    A body is projected with a velocity of\[2\times 11.2\] \[km{{s}^{-1}}\]from the surface of earth. The velocity of the body when it escapes the gravitational pull of earth is

    A)  \[\sqrt{3}\times 11.2km{{s}^{-1}}\]

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

    C)         \[\sqrt{2}\times 11.2\,km{{s}^{-1}}\]

    D)         \[6.5\times 11.2\,km{{s}^{-1}}\]

    E)  \[2\times 11.2\,km{{s}^{-1}}\]

    Correct Answer: B

    Solution :

    \[KE=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{(11.2)}^{2}}\] \[=\frac{1}{2}m{{(2\times 11.2)}^{2}}-\frac{1}{2}m{{(11.2)}^{2}}\] \[\frac{1}{2}m{{v}^{2}}=3\times \frac{1}{2}m\times {{(11.2)}^{2}}\] \[v=\sqrt{3}\times 11.2\]            


You need to login to perform this action.
You will be redirected in 3 sec spinner