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

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

Time period of simple pendulum on earth's surface is '$$\mathrm{T}$$'. Its time period becomes '$$\mathrm{xT}$$' when taken to a height $$\mathrm{R}$$ (equal to earth's radius) above the earth's surface. Then the value of '$$x$$' will be

The height at which the weight of the body becomes $$\left(\frac{1}{9}\right)^{\text {th }}$$ its weight on the surface of earth is $$(\mathrm{R}=$$ radius of earth)