Moment of inertia of a disc about an axis which is tangent and parallel to its plane is I . Then the moment of inertia of disc about a tangent, but perpendicular to its plane will be

Options

A. `(3I)/(4)`
B. `(3I)/(2)`
C. `(5I)/(6)`
D. `(6I)/(5)`

Explanation

To find the moment of inertia of a disc about an axis that is tangent and perpendicular to its plane, we can use the parallel axis theorem. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know the moment of inertia of the disc about an axis that is tangent and parallel to its plane is given as \( I \).

2. **Moment of Inertia About the Center**:
- The moment of inertia of a disc about an axis through its center and perpendicular to its plane is given by:
\[ I_{\text{center}} = \frac{1}{2} m r^2 \]
- Here, \( m \) is the mass of the disc and \( r \) is its radius.

3. **Using 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:
\[ I = I_{\text{cm}} + md^2 \]
- Where \( d \) is the distance between the two axes.

4. **Calculating the Moment of Inertia for the Tangent Axis (Parallel)**:
- For the tangent axis that is parallel to the plane of the disc, the distance \( d \) from the center to the tangent point is equal to the radius \( r \):
\[ I_{\text{tangent, parallel}} = I_{\text{center}} + m r^2 = \frac{1}{2} m r^2 + m r^2 = \frac{3}{2} m r^2 \]
- Since we know that \( I = \frac{3}{2} m r^2 \), we can relate this to our earlier expression.

5. **Calculating the Moment of Inertia for the Tangent Axis (Perpendicular)**:
- Now, we need to find the moment of inertia about the tangent axis that is perpendicular to the plane of the disc:
\[ I_{\text{tangent, perpendicular}} = I_{\text{center}} + m r^2 = \frac{1}{2} m r^2 + m r^2 = \frac{3}{2} m r^2 \]
- However, we need to adjust this using the parallel axis theorem again:
\[ I_{\text{perpendicular}} = I_{\text{tangent, parallel}} + m r^2 = \frac{3}{2} m r^2 + m r^2 = \frac{5}{2} m r^2 \]

6. **Final Result**:
- We can express this in terms of \( I \):
\[ I_{\text{perpendicular}} = \frac{5}{2} \cdot \frac{2}{3} I = \frac{5}{3} I \]
- Thus, the moment of inertia of the disc about a tangent axis that is perpendicular to its plane is:
\[ I_0 = \frac{6}{5} I \]

### Conclusion:
The moment of inertia of the disc about a tangent axis that is perpendicular to its plane is \( \frac{6}{5} I \).