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What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth = R)

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Rn2

R(n−1)n

Rn(n−1)

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detailed solution

Correct option is C

Value of acceleration due to gravity at depth d,g′=g1−dRgn=g1−dR1−dR=1ndR=1−1n=n−1nd=Rn−1n

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What is the depth at which the value of acceleration due to gravity becomes 1n times the value that at the surface of earth? (radius of earth = R)

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What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth = R)