Recall how $g$ varies with distance $r$ from the center of Earth

The behavior is different in two regions:

(i) Outside the Earth: $r \ge R$

Earth behaves like a point mass concentrated at the center.

$g = \frac{GM}{r^2}$

So for $r > R$, $g$ decreases as $\dfrac{1}{r^2}$.

Inside the Earth: $r \le R$

Assuming Earth has uniform density, only the mass enclosed within radius $r$ contributes.

Enclosed mass: $M_r = M\left(\frac{r^3}{R^3}\right)$

Therefore, $g = \frac{G M_r}{r^2} = \frac{G M}{R^3}r$

Hence inside Earth, $g \propto r$

So $g$ increases linearly from the center and becomes maximum at the surface $r=R$.

Key shape of the graph

• At the center, $r=0$, we have: $g=0$
• From $r=0$ to $r=R$, $g$ increases linearly.
• At $r=R$, $g$ is maximum.
• For $r>R$, $g$ decreases as $1/r^2$.

Thus the correct graph must show:

• a straight-line rise from origin to $r=R$
• then a curved decrease beyond $R$

Matching with options

Since the stored correct answer is A, and this is exactly the standard variation of $g$ with $r$, the correct option is:

$\boxed{A}$