The moment of inertia of a solid sphere about an axis passing through the centre radius is `2/5MR^(2)` , then its radius of gyration about a parallel axis t a distance `2R` from first axis is

Options

A. `5R`
B. `sqrt(22/5)R`
C. `5/2R`
D. `sqrt(12/5)R`

Explanation

To solve the problem, we will use the parallel axis theorem and the formula for the moment of inertia of a solid sphere.

### Step-by-Step Solution:

1. **Identify the Moment of Inertia about the Center Axis**:
The moment of inertia (I_C) of a solid sphere about an axis passing through its center is given as:
\[ I_C = \frac{2}{5} MR^2 \]

2. **Apply the Parallel Axis Theorem**:
The parallel axis theorem states:
\[ I = I_C + Md^2 \]
where:
- \( I \) is the moment of inertia about the new axis,
- \( I_C \) is the moment of inertia about the center axis,
- \( M \) is the mass of the sphere,
- \( d \) is the distance between the two axes.

In this case, the distance \( d \) is given as \( 2R \).

3. **Substitute the Values into the Equation**:
Substitute \( I_C \) and \( d \) into the parallel axis theorem:
\[ I = \frac{2}{5} MR^2 + M(2R)^2 \]
Simplifying \( (2R)^2 \):
\[ (2R)^2 = 4R^2 \]
Therefore, the equation becomes:
\[ I = \frac{2}{5} MR^2 + M \cdot 4R^2 \]

4. **Combine the Terms**:
Now, combine the terms:
\[ I = \frac{2}{5} MR^2 + 4MR^2 = \frac{2}{5} MR^2 + \frac{20}{5} MR^2 = \frac{22}{5} MR^2 \]

5. **Find the Radius of Gyration (k)**:
The moment of inertia can also be expressed in terms of the radius of gyration \( k \):
\[ I = Mk^2 \]
Setting the two expressions for \( I \) equal gives:
\[ Mk^2 = \frac{22}{5} MR^2 \]
Dividing both sides by \( M \):
\[ k^2 = \frac{22}{5} R^2 \]

6. **Take the Square Root**:
To find \( k \), take the square root of both sides:
\[ k = \sqrt{\frac{22}{5}} R \]

### Final Answer:
The radius of gyration about the parallel axis is:
\[ k = \sqrt{\frac{22}{5}} R \]