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)
A
d=R(nn−1)
B
d=R(n−1n)
C
d=R(n−12n)
D
d=R2(n−1n)
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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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