JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer           The value of acceleration due to gravity at Earth's surface is \[9.8m{{s}^{-2}}\]. The altitude above its surface at which the acceleration due to gravity decreases to \[4.9\,m{{s}^{-2}}\], is close to :
    (Radius of earth \[=6.4\times {{10}^{6}}m\])
    [JEE Main 10-4-2019 Morning]

    A) \[1.6\times {{10}^{6}}m\]   

    B) \[6.4\times {{10}^{6}}m\]

    C) \[9.0\times {{10}^{6}}m\]               

    D) \[2.6\times {{10}^{6}}m\]

    Correct Answer: D

    Solution :

    \[\frac{GM}{{{(R+h)}^{2}}}=\frac{GM}{2{{R}^{2}}}\]                         \[R+h=\sqrt{2}R\]                         \[h=(\sqrt{2}-1)R\]                         \[\simeq 2.6\times {{10}^{6}}m\]


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