At what height over the earth's pole,the free fall acceleration decreases by one percent? (Assume the radius of earth to be $6400 \, km$)

Acceleration due to gravity at a height $h$ is equal to that at a depth $d$ below the surface of the earth,if

The depth $d$ below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is ($R$ = radius of the earth).

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$,the radius of the earth,is

When one moves from a point $16 \text{ km}$ below the earth's surface to a point $16 \text{ km}$ above the earth's surface,the change in $g$ is approximately $\alpha \%$. The value of $\alpha$ is . . . . . . . (Take radius of the earth $R = 6400 \text{ km}$.)

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