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)

Consider a light planet revolving around a massive star in a circular orbit of radius '$$r$$' with time period '$$T$$'. If the gravitational force of attraction between the planet and the star is proportional to $$\mathrm{r}^{\frac{7}{2}}$$, then $$\mathrm{T}^2$$ is proportional to