At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $$\mathrm{3 R}$$ above earth surface. Where $$\mathrm{R}$$ is Radius of earth (Take $$\mathrm{R}=6400 \mathrm{~km}$$ ). The depth $$\mathrm{d}$$ is equal to

If the gravitational field in the space is given as $$\left(-\frac{K}{r^{2}}\right)$$. Taking the reference point to be at $$\mathrm{r}=2 \mathrm{~cm}$$ with gravitational potential $$\mathrm{V}=10 \mathrm{~J} / \mathrm{kg}$$. Find the gravitational potential at $$\mathrm{r}=3 \mathrm{~cm}$$ in SI unit (Given, that $$\mathrm{K}=6 \mathrm{~Jcm} / \mathrm{kg}$$)