The variation of acceleration due to gravity $$(g)$$ with distance $$(r)$$ from the center of the earth is correctly represented by (Given $$R$$ = radius of earth)

We need to identify the correct graph showing the variation of acceleration due to gravity $$g$$ with distance $$r$$ from the center of the Earth.

Since inside a uniform solid sphere the acceleration due to gravity varies linearly with distance from the center, $$g = \frac{4}{3}\pi G \rho \cdot r$$, so $$g \propto r$$ (linear increase from 0 at the center to maximum at the surface).

At the surface ($$r = R$$), $$g = \frac{GM}{R^2}$$ represents the maximum value of $$g$$.

For $$r > R$$, outside the Earth, gravity follows the inverse square law, $$g = \frac{GM}{r^2}$$, therefore $$g \propto \frac{1}{r^2}$$, indicating a decrease.

Combining these results, the graph increases linearly from $$r = 0$$ to $$r = R$$, reaches a peak at $$r = R$$, and then decreases as $$1/r^2$$ for $$r > R$$, resulting in a sharp peak at the surface.

The correct answer is Option A.