RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • 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 m/s                             

    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 earth \[R=6380\,\times {{10}^{3}}\,m\]. Acceleration due to gravity below the earths surface 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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