Manipal Medical Manipal Medical Solved Paper-2002

  • question_answer
    Assuming 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)  \[3.10\text{ }m/s\]          

    B)  \[5.06\text{ }m/{{s}^{2}}\]

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

    D)  \[9.66\text{ }m/{{s}^{2}}\]

    Correct Answer: D

    Solution :

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


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