Which graph correctly presents the variation of acceleration due to gravity with the distance from the centre of the earth (radius of the earth $= R_E$)?

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

If the radius of the Earth were to shrink by half,its mass remaining the same,what would be the weight of an object on its surface?

$A$ planet has the same density as that of the Earth,and the universal gravitational constant $G$ is twice that of the Earth. The ratio of the acceleration due to gravity on the planet to that on the Earth is:

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

$A$ body weighs $49 \, N$ on a spring balance at the North Pole. What will be its weight recorded on the same weighing machine if it is shifted to the equator (in $, N$)? (Use $g = \frac{GM}{R^2} = 9.8 \, m/s^2$ and radius of the Earth,$R = 6400 \, km$.)