If the radius of the Earth shrinks to $1/n$ of its present value while its mass remains constant,what will be the value of $g'_e$ on its surface?

The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals

Given below are two statements:
Statement-$I$: Acceleration due to gravity is different at different places on the surface of Earth.
Statement-$II$: Acceleration due to gravity increases as we go down below the Earth's surface.
In the light of the above statements,choose the correct answer from the options given below:

The value of $g$ at the surface of the earth is $9.8 \, m/s^2$. Then the value of $g$ at a place $480 \, km$ above the surface of the earth will be nearly .......... $m/s^2$ (radius of the earth is $6400 \, km$).

What is the value of acceleration due to gravity $(g)$ at the centre of the Earth? What will be the variation of $g$ below and above the surface of the Earth?