The depth at which the value of acceleration due to gravity is `1/n` times the value at the surface, is (R=radius of the earth)

Options

A. `d = R ((n)/(n-1))`
B. `d = R ((n-1)/(2n))`
C. `d = R((n-1)/(n))`
D. `d = R^(2) ((n-1)/(n))`

Answer

C

Explanation

Acceleration due to gravity at depth d is given as,
`g' = g(1-d/R) "…."(i)`
Given, `g' = g/n`
Substituting the values of 'g in Eq. (i), we get
`g/n = g(1-d/R)`
`rArr 1/n = 1 - d/R`
`:. d/R = 1 - (1)/(n) = (n-1)/(n)`
`d = R ((n-1)/(n))`