JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Question Bank Acceleration Due to Gravity

  • question_answer
    The depth d at which the value of acceleration due to gravity becomes \[\frac{1}{n}\] times the value at the surface, is [R = radius of the earth]      [MP PMT 1999; Kerala PMT 2005]

    A)             \[\frac{R}{n}\]           

    B)             \[R\,\left( \frac{n-1}{n} \right)\]

    C)             \[\frac{R}{{{n}^{2}}}\]     

    D)             \[R\,\left( \frac{n}{n+1} \right)\]

    Correct Answer: B

    Solution :

                    \[{g}'=g\left( 1-\frac{d}{R} \right)\,\Rightarrow \,\frac{g}{n}=g\left( 1-\frac{d}{R} \right)\]\[\Rightarrow \,\,d=\left( \frac{n-1}{n} \right)\ R\]


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