The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth,is (radius of earth is $6400 \ km$) (in $km$)

Assume there are two identical simple pendulum clocks. Clock-$1$ is placed on the Earth and Clock-$2$ is placed on a space station located at a height $h$ above the Earth's surface. Clock-$1$ and Clock-$2$ operate with time periods of $4\,s$ and $6\,s$ respectively. Then the value of $h$ is $....\,km$ (consider the radius of the Earth $R_E = 6400\,km$ and $g$ on Earth $10\,m/s^2$).

$A$ narrow but tall cabin is falling freely near the earth's surface. Inside the cabin,two small stones $A$ and $B$ are released from rest (relative to the cabin). Initially $A$ is much above the centre of mass and $B$ is much below the centre of mass of the cabin. $A$ close observation of the motion of $A$ and $B$ will reveal that

The height $h$ at which the weight of a body will be the same as that at the same depth $h$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):

$A$ spherical planet has a mass $M$ and diameter $D$. $A$ particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity equal to:

The average density of the Earth: