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

At depth `g =g(1-(h)/(R )) or (1-(d)/(R ))`
`(g)/(g)=(1)/(n)=(1-(d)/(R ))=(1)/(n)=(1-(d)/(R )) rarr d=R(n-1)/(n)`