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 find the new moment of inertia (M.I.) of a ring about a tangential axis that lies in the plane of the ring, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The moment of inertia of a ring about a tangential axis normal to its surface is given as \( I \).
- We need to find the moment of inertia about a tangential axis that lies in the plane of the ring.

2. **Use the Parallel Axis Theorem:**
- The Parallel Axis Theorem states that:
\[ I_t = I_{cm} + m h^2 \]
- Here, \( I_t \) is the moment of inertia about the tangential axis, \( I_{cm} \) is the moment of inertia about the center of mass, \( m \) is the mass of the ring, and \( h \) is the perpendicular distance between the two axes.

3. **Calculate \( I_{cm} \):**
- For a ring, the moment of inertia about its center of mass (which is along the axis through the center and perpendicular to the plane) is:
\[ I_{cm} = m r^2 \]
- From the problem, we know that the moment of inertia about the tangential axis normal to its surface is \( I \). Therefore, we can express this as:
\[ I = I_{cm} + m r^2 \]
- Rearranging gives:
\[ I = m r^2 + m r^2 = 2m r^2 \]
- Thus, we can express \( m r^2 \) as:
\[ m r^2 = \frac{I}{2} \quad \text{(Equation 1)} \]

4. **Find the Moment of Inertia About the Diameter:**
- Using the Perpendicular Axis Theorem, we know:
\[ I_z = I_x + I_y \]
- For a ring, the moment of inertia about the diameter (which is in the plane of the ring) is:
\[ I_d = \frac{1}{2} I_z \]
- Given that \( I_z = m r^2 \), we find:
\[ I_d = \frac{1}{2} m r^2 = \frac{1}{2} \cdot \frac{I}{2} = \frac{I}{4} \]

5. **Apply the Parallel Axis Theorem Again:**
- Now, we apply the Parallel Axis Theorem to find the moment of inertia about the tangential axis in the plane of the ring:
\[ I_t' = I_d + m h^2 \]
- Here, \( h = r \) (the radius of the ring), thus:
\[ I_t' = I_d + m r^2 = \frac{I}{4} + m r^2 \]
- Substituting \( m r^2 \) from Equation 1:
\[ I_t' = \frac{I}{4} + \frac{I}{2} \]
- Combine the terms:
\[ I_t' = \frac{I}{4} + \frac{2I}{4} = \frac{3I}{4} \]

6. **Conclusion:**
- The new moment of inertia about the tangential axis in the plane of the ring is:
\[ I_t' = \frac{3I}{4} \]

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