The angular speed of the Earth,such that an object on the equator may appear weightless,is ($g = 10\,m/s^2$,radius of the Earth $R = 6400\,km$).

The height above the surface of the earth where acceleration due to gravity becomes $\frac{g}{9}$ is ( $R$ is the radius of the earth,$g$ is the acceleration due to gravity at the surface).

$A$ simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface,where $R$ is the radius of the earth. The value of $T_2/T_1$ is

The length of a seconds pendulum at a height $h=2R$ from the earth's surface will be. (Given: $R =$ radius of earth and acceleration due to gravity at the surface of earth $g = \pi^{2} \ m/s^{2}$)

Two planets $A$ and $B$ have the same mass $M$ and radius $R$. The variation of density $\rho$ of the planets with distance $r$ from the center is shown in the following diagrams. The ratio of acceleration due to gravity at the surface of the planets $A$ and $B$ will be:

Select the correct statement out of the following: