$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{4}$ below the surface of the Earth,what is the new frequency of oscillation? ($R =$ radius of the Earth)

What should be the angular speed of the Earth so that a body at $60^{\circ}$ latitude becomes weightless? (Radius of Earth $R = 6400 \, km$)

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is $0.5\, m$. If the mean density of the moon is two-thirds that of the earth and the radius is one-quarter that of the earth,what is the maximum vertical distance through which he can jump on the moon and the ratio of the duration of the jump on the moon to that on the earth?

$A$ mine is located at depth $\frac{R}{3}$ below the earth's surface. The acceleration due to gravity at that depth in the mine is ($R = \text{radius of earth}$,$g = \text{acceleration due to gravity at surface}$).

The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B$. $A$ man jumps to a height of $2 \,m$ on the surface of $A$. What is the height of jump by the same person on the planet $B$?