The correct variation of gravitational potential `V` with radius `r` measured from the centre of earth of radius `R` is given by

Options

A.
B.
C.
D. None of these

Explanation

To solve the question regarding the variation of gravitational potential \( V \) with radius \( r \) measured from the center of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Gravitational Potential**:
The gravitational potential \( V \) at a distance \( r \) from the center of a solid sphere (like Earth) is given by the formula:
\[ V = -\frac{GM}{r} \]
for \( r \geq R \) (outside the sphere), where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

2. **Potential Inside the Sphere**:
For points inside the sphere (where \( r < R \)), the gravitational potential is constant and given by:
\[ V = -\frac{3GM}{2R} \]
This means that as we move from the center of the Earth to the surface, the potential does not change.

3. **At the Surface of the Earth**:
At the surface of the Earth (where \( r = R \)), we can substitute \( r \) in the formula for outside the sphere:
\[ V = -\frac{GM}{R} \]

4. **Graphical Representation**:
- For \( r < R \): The gravitational potential remains constant at \( -\frac{3GM}{2R} \).
- For \( r = R \): The potential changes to \( -\frac{GM}{R} \).
- For \( r > R \): The potential decreases as \( -\frac{GM}{r} \).

5. **Conclusion**:
The variation of gravitational potential \( V \) with radius \( r \) can be summarized as:
- Constant value of \( V = -\frac{3GM}{2R} \) for \( r < R \).
- \( V \) decreases as \( r \) increases for \( r \geq R \).

### Final Answer:
The correct variation of gravitational potential \( V \) with radius \( r \) measured from the center of the Earth is that it remains constant for \( r < R \) and decreases as \( -\frac{GM}{r} \) for \( r \geq R \). Since all options provided are in the positive axis, the answer is "none of these".