The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

The weight of a body on the surface of the earth is $63 \ N$. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth? (in $N$)

The value of $g$ on the Earth's surface is $980 \, cm/s^2$. Its value at a height of $64 \, km$ from the Earth's surface is ........ $cm/s^2$ (Radius of the Earth $R = 6400 \, km$).

Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

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 $............ \ km$.

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is $.......... \, km$ : [Radius of earth $R = 6400 \, km$ and $\sqrt{3} = 1.732$]