The variation of acceleration due to gravity $g$ with distance $x$ from the centre of the Earth is best represented by ($R \rightarrow$ Radius of the Earth):

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (both $x$ and $h$ being much smaller than the radius of the earth)

At what height from the ground will the value of $g$ be the same as that in a $10 \, km$ deep mine below the surface of the Earth?

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is [$R =$ radius of the earth].

What effect occurs on the frequency of a pendulum if it is taken from the earth's surface to deep into a mine?

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R = \text{radius of the earth})$