The radius of gyration of a body about an axis at a distance of `4cm` from its centre of mass is `5cm`. The radius of gyration about a parallel axis through centre of mass is

Options

A. `2cm`
B. `5 cm`
C. `4 cm`
D. `3 cm`

Explanation

To find the radius of gyration about a parallel axis through the center of mass, we can use the parallel axis theorem. The steps are as follows:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The radius of gyration (K) about an axis at a distance (D) of 4 cm from the center of mass is given as 5 cm.
- We need to find the radius of gyration (K2) about a parallel axis through the center of mass.

2. **Apply the Parallel Axis Theorem**:
The parallel axis theorem states:
\[ I = I_{cm} + M \cdot D^2 \]
where:
- \(I\) is the moment of inertia about the new axis,
- \(I_{cm}\) is the moment of inertia about the center of mass,
- \(M\) is the mass of the body,
- \(D\) is the distance between the two axes.

3. **Relate Moment of Inertia to Radius of Gyration**:
The moment of inertia can also be expressed in terms of the radius of gyration:
\[ I = M \cdot K^2 \]
For the axis at a distance of 4 cm from the center of mass:
\[ I = M \cdot K^2 = M \cdot (5 \, \text{cm})^2 = M \cdot 25 \, \text{cm}^2 \]

4. **Substitute into the Parallel Axis Theorem**:
Substitute \(I\) into the parallel axis theorem:
\[ M \cdot 25 = I_{cm} + M \cdot (4 \, \text{cm})^2 \]
\[ M \cdot 25 = I_{cm} + M \cdot 16 \]

5. **Solve for \(I_{cm}\)**:
Rearranging gives:
\[ I_{cm} = M \cdot 25 - M \cdot 16 = M \cdot (25 - 16) = M \cdot 9 \]

6. **Express \(I_{cm}\) in terms of \(K2\)**:
Since \(I_{cm} = M \cdot K2^2\), we have:
\[ M \cdot K2^2 = M \cdot 9 \]
Dividing both sides by \(M\) (assuming \(M \neq 0\)):
\[ K2^2 = 9 \]

7. **Calculate \(K2\)**:
Taking the square root gives:
\[ K2 = \sqrt{9} = 3 \, \text{cm} \]

### Final Answer:
The radius of gyration about a parallel axis through the center of mass is **3 cm**.