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

The acceleration due to gravity at a depth (d) below the surface of the earth is
`g'=g(1-(d)/(R ))` but `g'=(g)/(n)` (given)
`therefore (g)/(n)=g(1-(d)/(R ))`
`therefore (1)/(n)=(1-(d)/(R ))`
`therefore (d)/(R )=1-(1)/(n)=(n-1)/(n)`
`therefore d = R((n-1)/(n))`