Moment of inertia of a uniform circular disc about a diameter is `I`. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be.

Options

A. `5I`
B. `3I`
C. `6l`
D. `4I`

Answer

To find the moment of inertia of a uniform circular disc about an axis perpendicular to its plane and passing through a point on its rim, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Information**:
- Moment of inertia of the disc about a diameter: \( I \).
- We need to find the moment of inertia about an axis perpendicular to the plane of the disc and passing through a point on its rim.

2. **Use the Moment of Inertia Formula for a Disc**:
- The moment of inertia of a uniform circular disc about its diameter is given by:
\[ I = \frac{1}{4} M R^2 \]
- From this, we can express \( M R^2 \) in terms of \( I \):
\[ M R^2 = 4I \]

3. **Calculate Moment of Inertia About the Center**:
- The moment of inertia of the disc about an axis perpendicular to its plane and passing through the center (center of mass) is given by:
\[ I_{\text{cm}} = \frac{1}{2} M R^2 \]
- Substituting \( M R^2 = 4I \) into this equation:
\[ I_{\text{cm}} = \frac{1}{2} (4I) = 2I \]

4. **Apply the Parallel Axis Theorem**:
- The parallel axis theorem states:
\[ I' = I_{\text{cm}} + M d^2 \]
- Here, \( d \) is the distance from the center of mass to the new axis. For the disc, this distance is equal to the radius \( R \):
\[ I' = I_{\text{cm}} + M R^2 \]
- Substitute \( I_{\text{cm}} = 2I \) and \( M R^2 = 4I \):
\[ I' = 2I + 4I = 6I \]

5. **Conclusion**:
- The moment of inertia of the disc about the axis perpendicular to its plane and passing through a point on its rim is:
\[ I' = 6I \]

### Final Answer:
The moment of inertia about the axis perpendicular to the plane and passing through a point on its rim is \( 6I \).

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