The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ $A$ simple pendulum is taken to a planet of mass and radius,$4$ times and $2$ times,respectively,than the Earth. The time period of the pendulum remains the same on Earth and the planet.
Reason $(R) :$ The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements,choose the correct answer from the options given below $:$

$A$ point mass of $10 \,kg$ is placed at the centre of the Earth. The weight of the point mass is

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 $\sqrt{3} = 1.732$]

The ratio of the height $h$ above the surface of the Earth to the depth $d$ below the surface of the Earth,for which the gravitational accelerations are the same (assuming small heights),is:

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$,the radius of the earth,is