RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    The gravitational constant is G and radius of earth is\[{{R}_{e}},\]The relation between average density \[\rho \] of earth to gravitational acceleration will be

    A)  \[\rho =\left[ \frac{g}{G} \right]/\left[ \frac{4\pi }{3}R_{e}^{2} \right]\]

    B)  \[\rho =\left[ \frac{g}{G} \right]/\left[ \frac{4\pi }{3}R_{e}^{3} \right]\]

    C)  \[\rho =\left[ \frac{g}{G} \right]/\left[ \frac{4\pi }{3}R_{e}^{3} \right]\]

    D)  \[\rho =\left[ \frac{g}{G} \right]/\left[ \frac{4\pi }{3}R_{e}^{{}} \right]\]

    Correct Answer: D

    Solution :

     We know that \[g=\frac{GM}{R_{e}^{2}}\] \[\Rightarrow \] \[GM=gR_{e}^{2}\] \[\Rightarrow \] \[G\rho :\frac{4}{3}\pi R_{e}^{3}=gR_{e}^{2}\] \[\Rightarrow \] \[\rho =\frac{\left[ \frac{g}{G} \right]}{\frac{4}{3}\pi {{R}_{e}}}\]


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