If `T` be the period of revolution of a plant revolving around sun in an orbit of mean radius `R`, then identify the incorrect graph.

Options

A.
B.
C.
D. (d) None of these

Explanation

To solve the problem, we need to analyze the relationship between the period of revolution \( T \) of a planet and the mean radius \( R \) of its orbit around the Sun, based on Kepler's Third Law of Planetary Motion.

### Step-by-Step Solution:

1. **Understanding Kepler's Third Law**:
- According to Kepler's Third Law, the square of the period of revolution \( T \) of a planet is directly proportional to the cube of the mean radius \( R \) of its orbit.
- Mathematically, this can be expressed as:
\[ T^2 \propto R^3 \]
- This implies:
\[ T^2 = k R^3 \]
where \( k \) is a constant.

2. **Identifying the Relationship**:
- From the equation \( T^2 = k R^3 \), we can derive that:
\[ T^2 = k R^3 \implies T^2 = k (R^3) \]
- This indicates that if we plot \( T^2 \) against \( R^3 \), we should get a straight line, which confirms that graph A is correct.

3. **Analyzing Graph B**:
- If we take the cube root of both sides from \( T^2 = k R^3 \), we get:
\[ T^{2/3} \propto R \]
- This means that if we plot \( T^{2/3} \) against \( R \), we should also obtain a linear relationship, confirming that graph B is correct.

4. **Analyzing Graph C**:
- Now, if we consider the relationship between \( T^2 \) and \( \frac{1}{R^3} \):
\[ T^2 \propto R^3 \implies T^2 \propto \frac{1}{\frac{1}{R^3}} \]
- This implies that if we plot \( T^2 \) against \( \frac{1}{R^3} \), we should expect a rectangular hyperbola, which is not a linear relationship. Therefore, graph C is incorrect.

5. **Conclusion**:
- Based on the analysis of the relationships derived from Kepler's Third Law, we conclude that graph C is the incorrect graph.

### Final Answer:
The incorrect graph is **Graph C**.