$A$ hole is drilled halfway to the center of the Earth. $A$ body weighs $300 \ N$ on the surface of the Earth. How much will it weigh at the bottom of the hole (in $N$)?

The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight of an object at the equator becomes $3/5$ of its initial value? (Radius of Earth at the equator is $R = 6400 \, km$,take $g = 10 \, m/s^2$)

The acceleration due to gravity near the surface of a planet of radius $R$ and density $d$ is proportional to

If the earth stops rotating,the value of $g$ at the equator will

$A$ spherical planet far out in space has a mass $M_0$ and diameter $D_0$. $A$ particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to