The radius of gyration of a solid sphere of radius R about a certain axis is also equal to R. If r is the distance between the axis and the centre of the sphere, then r is equal to

Options

A. R
B. 0.5R
C. `sqrt(0.6)R`
D. `sqrt(0.3)R`

Answer

To solve the problem, we need to find the distance \( r \) between the axis and the center of the solid sphere, given that the radius of gyration \( k \) about that axis is equal to the radius \( R \) of the sphere.

### Step-by-Step Solution:

1. **Understanding the Radius of Gyration**:
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 of the object.

2. **Moment of Inertia of the Sphere**:
The moment of inertia \( I \) of a solid sphere about its own center is given by:
\[ I_{C} = \frac{2}{5} M R^2 \]

3. **Using the Parallel Axis Theorem**:
When the axis of rotation is not through the center of mass, we can use the parallel axis theorem:
\[ I = I_{C} + M r^2 \]
where \( r \) is the distance from the center of the sphere to the new axis.

4. **Substituting Values**:
Substituting the moment of inertia of the sphere into the equation:
\[ I = \frac{2}{5} M R^2 + M r^2 \]

5. **Calculating the Radius of Gyration**:
Since \( k = R \), we can set up the equation:
\[ R = \sqrt{\frac{I}{M}} = \sqrt{\frac{\frac{2}{5} M R^2 + M r^2}{M}} \]
Simplifying this gives:
\[ R = \sqrt{\frac{2}{5} R^2 + r^2} \]

6. **Squaring Both Sides**:
Squaring both sides to eliminate the square root:
\[ R^2 = \frac{2}{5} R^2 + r^2 \]

7. **Rearranging the Equation**:
Rearranging the equation to isolate \( r^2 \):
\[ r^2 = R^2 - \frac{2}{5} R^2 \]
\[ r^2 = \left(1 - \frac{2}{5}\right) R^2 \]
\[ r^2 = \frac{3}{5} R^2 \]

8. **Finding \( r \)**:
Taking the square root of both sides gives:
\[ r = \sqrt{\frac{3}{5}} R \]

### Final Answer:
Thus, the distance \( r \) between the axis and the center of the sphere is:
\[ r = \sqrt{0.6} R \]