The M.I. of a uniform disc about a diameter is I. Its M.I. about an axis perpendiçular to its plane and passing through a point on its rim is

Options

A. 51
B. 61
C. I
D. 4I

Explanation

To find the moment of inertia (M.I.) of a uniform 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**:
- The moment of inertia of the uniform disc about a diameter is given as \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass of the disc and \( r \) is its radius.

2. **Use the Parallel Axis Theorem**:
- The Parallel Axis Theorem states that if you know the moment of inertia about an axis through the center of mass, you can find the moment of inertia about any parallel axis by using the formula:
\[ I' = I + m d^2 \]
- Here, \( I' \) is the moment of inertia about the new axis, \( I \) is the moment of inertia about the center, \( m \) is the mass, and \( d \) is the distance between the two axes.

3. **Determine the Distance \( d \)**:
- For our case, the distance \( d \) is equal to the radius \( r \) of the disc, since the new axis is at the rim of the disc.

4. **Substitute Values**:
- Substitute \( I = \frac{1}{2} m r^2 \) and \( d = r \) into the Parallel Axis Theorem:
\[ I' = \frac{1}{2} m r^2 + m r^2 \]

5. **Combine Terms**:
- Combine the terms:
\[ I' = \frac{1}{2} m r^2 + 1 m r^2 = \frac{1}{2} m r^2 + \frac{2}{2} m r^2 = \frac{3}{2} m r^2 \]

6. **Express in Terms of \( I \)**:
- Since we know that \( I = \frac{1}{2} m r^2 \), we can express \( m r^2 \) in terms of \( I \):
\[ m r^2 = 2I \]
- Substitute this back into the equation for \( I' \):
\[ I' = \frac{3}{2} (2I) = 3I \]

7. **Final Result**:
- Thus, the moment of inertia of the uniform disc about an axis perpendicular to its plane and passing through a point on its rim is:
\[ I' = 3I \]