Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. if the gravitational force of attraction between the planet and the star is proportational to `R^(-5//2)`, then
(a) `T^(2)` is proportional to `R^(2)`
(b) `T^(2)` is proportional to `R^(7//2)`
(c) `T^(2)` is proportional to `R^(3//3)`
(d) `T^(2)` is proportional to `R^(3.75)`.

Explanation

To solve the problem, we need to establish the relationship between the period of revolution \( T \) and the radius \( R \) of the orbit of a light planet revolving around a massive star, given that the gravitational force of attraction is proportional to \( R^{-5/2} \).

### Step-by-Step Solution:

1. **Understanding the Forces**:
The gravitational force \( F_g \) acting on the planet is given by:
\[ F_g \propto R^{-5/2} \]
This means that:
\[ F_g = k \cdot R^{-5/2} \]
where \( k \) is a constant.

2. **Centripetal Force**:
The centripetal force \( F_c \) required to keep the planet in circular motion is given by:
\[ F_c = \frac{mv^2}{R} \]
where \( m \) is the mass of the planet and \( v \) is its orbital velocity.

3. **Equating Forces**:
Since the gravitational force provides the necessary centripetal force, we can set them equal:
\[ k \cdot R^{-5/2} = \frac{mv^2}{R} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ mv^2 = k \cdot R^{-5/2} \cdot R = k \cdot R^{-3/2} \]
Thus, we have:
\[ v^2 \propto R^{-3/2} \]

5. **Finding the Velocity**:
We can express the velocity \( v \) in terms of \( R \):
\[ v \propto R^{-3/4} \]

6. **Relating Period to Velocity**:
The period \( T \) of revolution is related to the velocity and radius by:
\[ T = \frac{2\pi R}{v} \]
Squaring both sides gives:
\[ T^2 \propto \frac{R^2}{v^2} \]

7. **Substituting for Velocity**:
Substitute \( v^2 \) into the equation:
\[ T^2 \propto \frac{R^2}{R^{-3/2}} = R^2 \cdot R^{3/2} = R^{2 + 3/2} = R^{7/2} \]

8. **Conclusion**:
Therefore, we find that:
\[ T^2 \propto R^{7/2} \]

### Final Answer:
The correct option is (b) \( T^2 \) is proportional to \( R^{7/2} \).