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JEE Main & Advanced AIEEE Solved Paper-2009

  • question_answer
    The height at which the acceleration due to gravity becomes\[\frac{g}{9}\](where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is:     AIEEE  Solved  Paper-2009

    A) 2R                                          

    B) \[\frac{R}{\sqrt{2}}\]                     

    C)        R/2                        

    D)        \[\sqrt{2}R\]

    Correct Answer: A

    Solution :

    \[\frac{GM}{9{{R}^{2}}}=\frac{GM}{{{(R+h)}^{2}}}\] \[\Rightarrow \]\[3R=R+h\] \[\Rightarrow \]\[h=2R\]


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