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

Options

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

Answer

b

Explanation

`g_(d)=g(1+(d)/(R))`
`(1)/(n)g=g(1-(d)/(R))`
`(1)/(n)-1=-(d)/(R)`
`(1)/(n)-1=-(d)/(R)`
`-1(1)/(n)=(d)/(R)`
`((n-1)/(n))R=d`
`therefore" "d=((n-1)/(n))R.`