The variation of acceleration due to gravity $g$ with distance $x$ from the centre of the Earth is best represented by ($R \rightarrow$ Radius of the Earth):

$A$ body weighs $45 \ N$ on the surface of the earth. The gravitational force on a body due to the earth at a height equal to half the radius of the earth will be: (in $N$)

At what altitude will the acceleration due to gravity be $25\%$ of that at the earth's surface (given radius of earth is $R$)?

$A$ simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4 \ m$,then the time period of small oscillations will be . . . . . . $s$. [Take $g = \pi^2 \ ms^{-2}$]

If the earth were to cease rotating about its own axis,the increase in the value of $g$ in the $C.G.S.$ system at a place of latitude of $45^{\circ}$ will be ........ $cm/sec^{2}$.

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