The M.I. of a ring and disc of same mass and radius about their geometric axes will be:

Options

A. more for the disc
B. more for the ring
C. eual in both cases
D. none of these.

Answer

To solve the problem of comparing the moment of inertia (M.I.) of a ring and a disc of the same mass and radius about their geometric axes, we can follow these steps:

### Step 1: Understand the Shapes
- A ring is a hollow circular object with mass distributed along its circumference.
- A disc is a solid circular object with mass distributed throughout its area.

### Step 2: Identify the Moment of Inertia Formulas
- The moment of inertia \( I \) of a ring about its geometric axis (which is perpendicular to the plane of the ring and passes through its center) is given by:
\[ I_{\text{ring}} = m R^2 \]
where \( m \) is the mass and \( R \) is the radius of the ring.

- The moment of inertia \( I \) of a disc about its geometric axis (which is also perpendicular to the plane of the disc and passes through its center) is given by:
\[ I_{\text{disc}} = \frac{1}{2} m R^2 \]

### Step 3: Substitute the Values
- Since both the ring and the disc have the same mass \( m \) and radius \( R \), we can directly compare their moments of inertia:
- For the ring:
\[ I_{\text{ring}} = m R^2 \]
- For the disc:
\[ I_{\text{disc}} = \frac{1}{2} m R^2 \]

### Step 4: Compare the Moments of Inertia
- Now, we can compare the two expressions:
\[ I_{\text{ring}} = m R^2 \quad \text{and} \quad I_{\text{disc}} = \frac{1}{2} m R^2 \]
- Clearly, \( I_{\text{ring}} > I_{\text{disc}} \).

### Conclusion
- The moment of inertia of the ring is greater than that of the disc. Therefore, the answer is:
\[ I_{\text{ring}} > I_{\text{disc}} \]