Answer

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**Hint:**According to Newton, Gravity decreases by the square of the increased distance divided by the original distance from an object's centre of mass.

**Complete step by step solution:**

We know that \[g = \dfrac{{GM}}{{{r^2}}}\] where M is the mass of the earth r is the radius of the earth(or the distance between the centre of the earth and you standing on its surface) G is the gravitational constant.

\[g = \dfrac{{GM}}{{{r^2}}}\]

Now by multiplying and dividing the RHS by volume ‘V’ we get,

\[g = \dfrac{{GM}}{V} \times V \times \dfrac{1}{{{r^2}}}\]

Now we know that \[\dfrac{M}{V}\]is density ‘\[\rho \]’

Therefore,

\[g = \dfrac{{\rho GV}}{{{r^2}}}\]

So now that for a sphere \[V = \dfrac{4}{3}\pi {r^3}\]then we get,

\[\begin{array}{l}

g = \dfrac{4}{3}\pi {r^3} \times \dfrac{{\rho G}}{{{r^2}}}\\

g = \dfrac{{4\pi r\rho G}}{3}

\end{array}\]

At the centre of the earth r=0 then the whole expression becomes 0

Thus g=0 at the centre of the earth

**Therefore, option D is correct.**

**Note:**As you approach the core, the pull from all sides of Earth gets stronger and stronger, once you reach the very centre, the pull is roughly the same form all sides. This would mean that at the centre of the Earth the mass of the earth is equally distributed in all directions so pulling equally in all directions all the gravity cancels for a net zero pull. However, it is not a perfect sphere and has bumps and different masses in different areas. Therefore, you would feel slight gravitational attraction to some areas but it would be very minimal. As the distance from the centre decreases, the acceleration due to gravity also decreases. As the distance from the centre decreases, the acceleration due to gravity also decreases.

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