If a disc of mass m and radius r is reshaped into a ring a radius 2r,the mass remaining the same, the radius of gyration about centroidal axis perpendicular to plane goes up by a factor of `sqrt(x)` . Find the value of x.

Answer

To solve the problem, we need to find the value of \( x \) when a disc of mass \( m \) and radius \( r \) is reshaped into a ring of radius \( 2r \), and the radius of gyration about the centroidal axis perpendicular to the plane increases by a factor of \( \sqrt{x} \).

### Step-by-Step Solution:

1. **Moment of Inertia of the Disc**:
The moment of inertia \( I \) of a solid disc about an axis perpendicular to its plane through its center is given by the formula:
\[ I_{\text{disc}} = \frac{1}{2} m r^2 \]

2. **Moment of Inertia of the Ring**:
The moment of inertia \( I \) of a ring about an axis perpendicular to its plane through its center is given by:
\[ I_{\text{ring}} = m (2r)^2 = 4 m r^2 \]

3. **Radius of Gyration**:
The radius of gyration \( k \) is related to the moment of inertia by the formula:
\[ I = m k^2 \]
Therefore, we can express the radius of gyration for the disc and the ring as follows:
- For the disc:
\[ k_{\text{disc}} = \sqrt{\frac{I_{\text{disc}}}{m}} = \sqrt{\frac{\frac{1}{2} m r^2}{m}} = \sqrt{\frac{1}{2}} r = \frac{r}{\sqrt{2}} \]
- For the ring:
\[ k_{\text{ring}} = \sqrt{\frac{I_{\text{ring}}}{m}} = \sqrt{\frac{4 m r^2}{m}} = \sqrt{4} \cdot r = 2r \]

4. **Finding the Factor of Increase**:
We need to find the factor by which the radius of gyration increases when going from the disc to the ring:
\[ \text{Factor} = \frac{k_{\text{ring}}}{k_{\text{disc}}} = \frac{2r}{\frac{r}{\sqrt{2}}} = 2r \cdot \frac{\sqrt{2}}{r} = 2\sqrt{2} \]

5. **Relating to \( \sqrt{x} \)**:
According to the problem, this factor of increase is equal to \( \sqrt{x} \):
\[ \sqrt{x} = 2\sqrt{2} \]

6. **Squaring Both Sides**:
To find \( x \), we square both sides:
\[ x = (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]

### Final Answer:
The value of \( x \) is \( 8 \).