The diameters of two planets are in the ratio $4 : 1$ and their mean densities in the ratio $1 : 2$. The acceleration due to gravity on the planets will be in the ratio:

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 $,R$)

$A$ simple pendulum performing small oscillations at a height $R$ above the Earth's surface has a time period of $T_1 = 4 \ s$. What would be its time period $T_2$ if it is brought to a point at a height $2R$ from the Earth's surface? Choose the correct relation ($R =$ radius of Earth).

If the acceleration due to gravity experienced by a point mass at a height $h$ above the surface of the Earth is the same as the acceleration due to gravity at a depth $d = \alpha h$ $(h \ll R_{e})$ from the Earth's surface,then the value of $\alpha$ will be: (use $R_{e} = 6400 \ km$)

The depth at which acceleration due to gravity becomes $\frac{g}{2n}$ is ($R=$ radius of earth,$g=$ acceleration due to gravity on earth's surface,$n$ is an integer).

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