The M.I. of a ring about a tangential, axis normal to its surface is I. If the axis is now tangential along the plane of ring, its new M.I. will be

Options

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

Answer

To solve the problem, we need to find the moment of inertia (M.I.) of a ring about a tangential axis that lies in the plane of the ring, given that the moment of inertia about a tangential axis normal to its surface is \( I \).

### Step-by-step Solution:

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

2. **Using the Parallel Axis Theorem:**
- The moment of inertia about a tangential axis (normal to the surface) is given as \( I \). By the Parallel Axis Theorem, we can express this as:
\[ I = I_{\text{cm}} + m d^2 \]
- Where \( d \) is the distance from the center of mass to the new axis. For a tangential axis, \( d = r \):
\[ I = m r^2 + m r^2 = 2 m r^2 \]

3. **Finding the Moment of Inertia in the Plane of the Ring:**
- Now, we need to find the moment of inertia about a tangential axis in the plane of the ring. The moment of inertia about this axis can also be found using the Parallel Axis Theorem:
\[ I_{\text{plane}} = I_{\text{cm}} + m d^2 \]
- For this case, the moment of inertia about the center of mass in the plane is:
\[ I_{\text{cm}} = \frac{1}{2} m r^2 \]
- The distance \( d \) from the center of mass to the tangential axis in the plane is still \( r \):
\[ I_{\text{plane}} = \frac{1}{2} m r^2 + m r^2 \]
- Simplifying this gives:
\[ I_{\text{plane}} = \frac{1}{2} m r^2 + 1 m r^2 = \frac{3}{2} m r^2 \]

4. **Relating to the Given Moment of Inertia \( I \):**
- From the earlier step, we know that \( I = 2 m r^2 \). Thus, we can express \( m r^2 \) in terms of \( I \):
\[ m r^2 = \frac{I}{2} \]
- Substituting this into the equation for \( I_{\text{plane}} \):
\[ I_{\text{plane}} = \frac{3}{2} \left(\frac{I}{2}\right) = \frac{3I}{4} \]

5. **Final Result:**
- Therefore, the moment of inertia of the ring about the tangential axis in the plane of the ring is:
\[ I_{\text{plane}} = \frac{3I}{4} \]

### Conclusion:
The new moment of inertia of the ring about the tangential axis along the plane of the ring is \( \frac{3I}{4} \).