Assertion : If we draw a circle around the centre of mass of a rigid body, then moment of inertia about all parallel axes passing through any point on this circle has a constant value.
Reason : Dimensions of radius gyration are `[M^(0)L T^(0)]`

Options

A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion
B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion
C. If Assertion is true but Reason is fasle
D. If Assertion is false but Reason is true

Explanation

To solve the given question, we need to analyze both the assertion and the reason provided.

### Step 1: Understanding the Assertion
The assertion states that if we draw a circle around the center of mass of a rigid body, then the moment of inertia about all parallel axes passing through any point on this circle has a constant value.

1. **Identify the Center of Mass**: The center of mass (COM) is a specific point in the body where the mass can be considered to be concentrated for the purpose of analyzing translational motion.
2. **Draw a Circle**: When a circle is drawn around the COM, any point on this circle can be considered as a new axis of rotation.
3. **Moment of Inertia Calculation**: According to the parallel axis theorem, the moment of inertia \( I \) about any axis parallel to an axis through the COM is given by:
\[ I = I_{COM} + Md^2 \]
where \( I_{COM} \) is the moment of inertia about the COM, \( M \) is the mass of the body, and \( d \) is the distance from the COM to the new axis.

### Step 2: Analyzing the Circle
Since all points on the circle are equidistant from the center of mass (the radius \( R \) of the circle), the distance \( d \) from the COM to any point on the circle is constant and equal to \( R \).

4. **Moment of Inertia for Points on the Circle**: Therefore, for any point \( A \) on the circle, the moment of inertia can be expressed as:
\[ I_A = I_{COM} + M R^2 \]
Similarly, for any other point \( B \) on the circle, we have:
\[ I_B = I_{COM} + M R^2 \]
Since \( R \) is constant for all points on the circle, \( I_A \) and \( I_B \) will have the same value. Thus, the assertion is true.

### Step 3: Understanding the Reason
The reason states that the dimensions of the radius of gyration are \( [M^0 L^0 T^0] \).

5. **Analyzing the Dimensions**: The radius of gyration \( k \) is defined as:
\[ k = \sqrt{\frac{I}{M}} \]
where \( I \) is the moment of inertia and \( M \) is the mass. The dimensions of \( I \) are \( [M L^2] \) and the dimensions of \( M \) are \( [M] \). Therefore, the dimensions of \( k \) are:
\[ [k] = \sqrt{\frac{[M L^2]}{[M]}} = \sqrt{[L^2]} = [L] \]
Hence, the dimensions of the radius of gyration are \( [M^0 L^1 T^0] \), which means the reason provided is incorrect.

### Conclusion
- The assertion is true.
- The reason is false because the dimensions of the radius of gyration are \( [M^0 L^1 T^0] \), not \( [M^0 L^0 T^0] \).

### Final Answer
Both the assertion and reason are true, but the reason does not correctly explain the assertion. Therefore, the correct answer is that the assertion is true, and the reason is false.