The acceleration due to gravity at the surface of the planet is same as that at the surface of the earth, but the density of planet is thrice that of the earth. If 'R' is the radius of the earth, the radius of the planet will be

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)