The depth 'd' at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth)

Assuming that the earth is revolving around the sun in circular orbit of radius R , the angular momentum is directly proportional to $\mathrm{R}^{\mathrm{n}}$. The value of ' $n$ ' is

The height ' h ' from the surface of the earth at which the value of ' $g$ ' will be reduced by $64 \%$ than the value at surface of the earth is ( $\mathrm{R}=$ radius of the earth)

A body starts from rest from a distance $\mathrm{R}_0$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ( $R=$ radius of earth, $M=$ mass of earth)