The depth $d$ at which the acceleration due to gravity becomes $\frac{g}{n}$ is (where $R$ is the radius of the Earth,$g$ is the acceleration due to gravity at the surface,and $n$ is an integer).

What should be the angular speed with which the Earth must rotate on its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as their present weight?

Suppose that the angular velocity of rotation of the earth is increased. Then,as a consequence:

Which of the following statements are true about acceleration due to gravity?
$(a)$ '$g$' decreases in moving away from the centre if $r > R$
$(b)$ '$g$' decreases in moving towards the centre if $r < R$
$(c)$ '$g$' is zero at the centre of the Earth
$(d)$ '$g$' decreases if the Earth stops rotating on its axis

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

The height at which the weight of a body becomes $\frac{1}{9}$ of its weight on the surface of the earth (radius of the earth is $R$) is: