The mass density of a spherical body is given by $$\rho(r) = \frac{k}{r}$$ for $$r \le R$$ and $$\rho(r) = 0$$ for $$r > R$$, where $$r$$ is the distance from the center. The correct graph that describes qualitatively the acceleration, $$a$$ of a test particle as a function of $$r$$ is:

We are given a spherical body with mass density $$\rho(r) = \frac{k}{r}$$ for $$r \le R$$ and $$\rho(r) = 0$$ for $$r > R$$. We need to find the acceleration $$a$$ of a test particle as a function of $$r$$.

Step 1: Find the mass enclosed within radius $$r$$ (for $$r \le R$$).

Using the mass element in a spherical shell of thickness $$dr$$:

$$dM = \rho(r) \cdot 4\pi r^2 \, dr = \frac{k}{r} \cdot 4\pi r^2 \, dr = 4\pi k \, r \, dr$$

Integrating from 0 to $$r$$:

$$M(r) = \int_0^r 4\pi k \, r' \, dr' = 4\pi k \cdot \frac{r^2}{2} = 2\pi k r^2$$

Step 2: Find the gravitational acceleration for $$r \le R$$.

By Newton's law of gravitation (using the shell theorem, only mass enclosed within radius $$r$$ contributes):

$$a(r) = \frac{G \cdot M(r)}{r^2} = \frac{G \cdot 2\pi k r^2}{r^2} = 2\pi G k$$

This is a constant, independent of $$r$$. So for $$r \le R$$, the acceleration is constant.

Step 3: Find the gravitational acceleration for $$r > R$$.

The total mass of the sphere is $$M(R) = 2\pi k R^2$$.

For $$r > R$$, the entire sphere acts as a point mass at the center:

$$a(r) = \frac{G \cdot M(R)}{r^2} = \frac{2\pi G k R^2}{r^2}$$

This decreases as $$\frac{1}{r^2}$$.

Step 4: Describe the graph.

The acceleration $$a$$ is constant for $$r \le R$$, and then drops as $$\frac{1}{r^2}$$ for $$r > R$$.

The correct answer is Option C.