The depth ′d′ at which the value of acceleration due to gravity becomes 1n times the value at the earth's surface is (R= radius of earth)

d=R(nn−1)

d=R(n−1n)

d=R(n−12n)

d=R2(n−1n)

A

d=R(n−1n)

B

d=R(nn−1)

C

d=R2(n−1n)

D

d=R(n−12n)

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Solution

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Acceleration due to gravity at a depth d under the earth surface g′=gs(1−dR)

Given : g′=gsn

∴gsn=gs(1−dR)

Or 1n=(1−dR)

⟹d=R(n−1n)

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