If the value of acceleration due to gravity on the surface of the Earth is $g$,what will be the value of $g$ at a height equal to the radius of the Earth from the surface?

The density of a planet is $3$ times that of Earth and its radius is $2.5$ times that of Earth. If $g_p$ and $g_e$ represent the acceleration due to gravity on the planet and Earth respectively, then the ratio of $g_p$ to $g_e$ is:

The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth,assumed to be a sphere of radius $R$ of uniform density,is as shown in the figure below. The correct figure is:

Which graph correctly represents the variation of acceleration due to gravity $(g)$ with radial distance $(r)$ from the centre of the earth (radius of the earth $= R_e$)?

The depth $d$ at which 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).

At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?