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

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the given mass density function and determine how it affects the gravitational acceleration of a test particle as a function of the distance \( r \) from the center of the spherical body.

### Step-by-Step Solution:

1. **Understanding the Mass Density**:
The mass density of the spherical body is given by:
\[ \rho(r) = \frac{k}{r} \quad \text{for } r \leq R \]
and
\[ \rho(r) = 0 \quad \text{for } r > R \]
Here, \( k \) is a constant, and \( R \) is the radius of the sphere.

2. **Finding the Mass of a Shell**:
To find the mass of a thin spherical shell of thickness \( dr \) at a distance \( r \) from the center, we use the formula for the volume of the shell:
\[ dV = 4\pi r^2 dr \]
The mass of this shell is:
\[ dm = \rho(r) \cdot dV = \frac{k}{r} \cdot 4\pi r^2 dr = 4\pi k r \, dr \]

3. **Calculating the Total Mass Inside Radius \( r \)**:
To find the total mass \( M(r) \) inside a radius \( r \), we integrate the mass of the shell from \( 0 \) to \( r \):
\[ M(r) = \int_0^r 4\pi k r' \, dr' = 4\pi k \int_0^r r' \, dr' = 4\pi k \left[\frac{(r')^2}{2}\right]_0^r = 4\pi k \frac{r^2}{2} = 2\pi k r^2 \]

4. **Finding the Gravitational Field \( g(r) \)**:
The gravitational field \( g(r) \) at a distance \( r \) from the center is given by:
\[ g(r) = \frac{G M(r)}{r^2} \]
Substituting \( M(r) \):
\[ g(r) = \frac{G (2\pi k r^2)}{r^2} = 2\pi G k \]
This shows that \( g(r) \) is constant for \( r \leq R \).

5. **Behavior for \( r > R \)**:
For \( r > R \), the mass inside the sphere is constant (equal to \( M(R) \)). Thus, the gravitational field becomes:
\[ g(r) = \frac{G M(R)}{r^2} \]
Since \( M(R) = 2\pi k R^2 \), we have:
\[ g(r) = \frac{G (2\pi k R^2)}{r^2} \]
This indicates that \( g(r) \) decreases with \( r^2 \) for \( r > R \).

6. **Conclusion**:
- For \( r \leq R \), \( g(r) \) is constant.
- For \( r > R \), \( g(r) \) decreases as \( \frac{1}{r^2} \).

### Correct Graph:
The graph of the gravitational acceleration \( g(r) \) as a function of \( r \) will be constant for \( r \leq R \) and will decrease for \( r > R \). Thus, the correct option is the one that shows a constant value for \( g \) up to \( R \) and a decreasing curve after \( R \).