The height at which the weight of a body becomes $\left(\frac{1}{9}\right)^{\text{th}}$ of its weight on the surface of the Earth is $(R = \text{radius of Earth})$: (in $R$)

The diameters of two planets are in the ratio $4 : 1$ and their mean densities in the ratio $1 : 2$. The acceleration due to gravity on the planets will be in the ratio:

$A$ body thrown on Earth reaches a height of $90\,m$. If the same body is thrown on a planet with $\frac{1}{10}$ of the mass and $\frac{1}{3}$ of the radius of Earth,it will reach a height of ....... $m$.

$Assertion$ : In a free fall,the weight of a body becomes effectively zero.
$Reason$ : The acceleration due to gravity acting on a body in free fall is zero.

At a certain depth $d$ below the surface of the earth,the value of acceleration due to gravity becomes four times its value at a height $3R$ above the earth's surface. Where $R$ is the radius of the earth (Take $R = 6400 \ km$). The depth $d$ is equal to $............ \ km$.

Earth is assumed to be a sphere of radius $R$. If $g_{\phi}$ is the value of effective acceleration due to gravity at latitude $30^{\circ}$ and $g$ is the value at the equator,then the value of $|g - g_{\phi}|$ is ($\omega$ is the angular velocity of rotation of the Earth,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$).