The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is $..........km$ : [Radius of earth $R =6400 \,km$ and $\sqrt{3}=1.732]$

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400\, km$. At the poles $g = 10\,m{s^{ - 2}})$

Taking density of Earth constant.Which graph correctly presents the variation of acceleration due to gravity with the distance from the centre of the earth (radius of the earth $= R_E$ )?

The acceleration due to gravity at pole and equator can be related as

Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $d=\frac{R}{2}$ from the surface of earth, if its werght on the surface of earth is $200\,N$, will be $...........\,N$ ( $Given R =$ Radrus of earth)

Assuming the earth to be a sphere of uniform density, the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to