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

The escape velocity from earth surface is $11 \mathrm{~km} / \mathrm{s}$. The escape velocity from a planet having twice the radius and same mean density as earth is

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

The height ' $h$ ' above the earth's surface at which the value of acceleration due to gravity $(\mathrm{g})$ becomes $\left(\frac{\mathrm{g}}{3}\right)$ is ( $\mathrm{R}=$ radius of the earth)