If `I` moment of inertia of a thin circular plate about an axis passing through tangent of plate in its plane. The moment of inertia of same circular plate about an axis perpendicular to its plane and passing through its centre is

Options

A. `(4I)/5`
B. `(2I)/5`
C. `(4I)/3`
D. `(2I)/3`

Explanation

To solve the problem, we need to find the moment of inertia of a thin circular plate about an axis perpendicular to its plane and passing through its center, given the moment of inertia about a tangent axis in its plane.

### Step-by-Step Solution:

1. **Identify Given Information**:
- Let \( I \) be the moment of inertia of the circular plate about the tangent axis in its plane.
- We need to find the moment of inertia about an axis perpendicular to its plane and passing through its center.

2. **Use the Moment of Inertia for Circular Plates**:
- The moment of inertia of a circular plate about an axis perpendicular to its plane and passing through its center (let's denote it as \( I_{cm} \)) is given by the formula:
\[ I_{cm} = \frac{1}{2} m r^2 \]
where \( m \) is the mass of the plate and \( r \) is its radius.

3. **Apply 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:
\[ I_{tangent} = I_{cm} + m d^2 \]
where \( d \) is the distance between the two axes. In this case, the distance \( d \) is equal to the radius \( r \) of the plate.

4. **Substituting Values**:
- From the previous step, we have:
\[ I_{tangent} = I_{cm} + m r^2 \]
- Substituting \( I_{cm} = \frac{1}{2} m r^2 \):
\[ I_{tangent} = \frac{1}{2} m r^2 + m r^2 \]
- Simplifying this gives:
\[ I_{tangent} = \frac{1}{2} m r^2 + \frac{2}{2} m r^2 = \frac{3}{2} m r^2 \]

5. **Relate \( I \) to \( I_{tangent} \)**:
- We know from the problem that \( I = I_{tangent} \). Thus, we can write:
\[ I = \frac{3}{2} m r^2 \]

6. **Find \( I_{cm} \)**:
- We can rearrange the equation to find \( I_{cm} \):
\[ I_{cm} = \frac{1}{2} I \]
- Since we have \( I = \frac{3}{2} m r^2 \), we can substitute this back to find:
\[ I_{cm} = \frac{1}{2} \left(\frac{3}{2} m r^2\right) = \frac{3}{4} m r^2 \]

7. **Final Result**:
- Thus, the moment of inertia of the circular plate about the axis perpendicular to its plane and passing through its center is:
\[ I_{cm} = \frac{1}{2} m r^2 \]

### Conclusion:
The moment of inertia of the circular plate about the axis perpendicular to its plane and passing through its center is \( \frac{2}{5} I \).