Two planets $A$ and $B$ of radii $R$ and $1.5 R$ have densities $\rho$ and $\rho / 2$ respectively. The ratio of acceleration due to gravity at the surface of $B$ to $A$ is:

At what depth does the acceleration due to gravity become $\frac{g}{4}$? ($R$ = radius of the Earth)

The value of acceleration due to gravity at a depth $d$ from the surface of the Earth and at an altitude $h$ from the surface of the Earth are in the ratio:

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

$A$ simple pendulum is oscillating with frequency $F$ on the surface of the earth. It is taken to a depth $R/3$ below the surface of the earth ($R =$ radius of earth). The frequency of oscillation at depth $R/3$ is