What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth $=R$ )

If $R$ is the radius of the earth and $g$ the acceleration due to gravity on the earth's surface, the mean density of the earth is

The time period of a simple pendulum on a freely moving artificial satellite is

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

A $90 \mathrm{~kg}$ body placed at $2 \mathrm{R}$ distance from surface of earth experiences gravitational pull of : ( $\mathrm{R}=$ Radius of earth, $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )