Moment of inertia of a uniform-disc of mass m about an axis `x=a` is `mk^(2)`, where k is the radius of gyration. What is its moment of inertia about an axis `x=a+b`:

Options

A. `mk^(2)+m(a+b)^(2)-ma^(2)`
B. `mk^(2)+m((a+b)^(2))/(2)`
C. `mk^(2)+m(b^(2))/(2)`
D. `mk^(2)+mb^(2)`

Explanation

To find the moment of inertia of a uniform disc of mass \( m \) about an axis \( x = a + b \), we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia \( I \) about any axis parallel to an axis through the center of mass is given by:

\[ I = I_{CM} + m d^2 \]

where:
- \( I_{CM} \) is the moment of inertia about the center of mass,
- \( m \) is the mass of the object,
- \( d \) is the distance between the two axes.

### Step-by-Step Solution:

1. **Identify the given moment of inertia**:
The moment of inertia of the disc about the axis \( x = a \) is given as:
\[ I_a = m k^2 \]

2. **Determine the distance \( d \)**:
The distance \( d \) from the axis \( x = a \) to the new axis \( x = a + b \) is:
\[ d = (a + b) - a = b \]

3. **Apply the parallel axis theorem**:
We need to find the moment of inertia about the new axis \( x = a + b \):
\[ I_{a+b} = I_a + m d^2 \]

4. **Substitute the known values**:
Substitute \( I_a = m k^2 \) and \( d = b \):
\[ I_{a+b} = m k^2 + m b^2 \]

5. **Factor out \( m \)**:
We can factor out \( m \) from the equation:
\[ I_{a+b} = m (k^2 + b^2) \]

### Final Answer:
Thus, the moment of inertia of the uniform disc about the axis \( x = a + b \) is:
\[ I_{a+b} = m (k^2 + b^2) \]