VMMC VMMC Medical Solved Paper-2014

  • question_answer
    Assuming, the earth to be sphere of uniform density, what is the value of acceleration due to gravity at a point 100 km below the earth surface? Given \[R=6380\times {{10}^{3}}m\])

    A) \[3.10\,m/{{s}^{2}}\]                     

    B) \[5.06\,m/{{s}^{2}}\]

    C) \[7.64\,m/{{s}^{2}}\]                     

    D) \[9.66\,m/{{s}^{2}}\]

    Correct Answer: D

    Solution :

    Here, Depth \[=100\,km=100\times {{10}^{3}}m\] Radius of the earth \[R=6380\times {{10}^{3}}m\] Acceleration due to gravity below the earth's surface is given by \[g'=g\left( 1-\frac{d}{R} \right)\] \[=9.8\left[ 1-\frac{100\times {{10}^{3}}}{6380\times {{10}^{3}}} \right]\] \[=9.8\left[ 1-\frac{1}{63.8} \right]\] \[=9.8\times \frac{62.8}{63.8}\] \[=9.66\,\text{m/}{{\text{s}}^{2}}\]


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