The M.I of a uniform thin circular ring about an axis through its centre and normal to the plane of ring is I. Its M.I about a tangent in its plane is equal to

Options

A. `1.5I`
B. `2.5I`
C. `I//2`
D. `2I`

Answer

To find the moment of inertia (M.I) of a uniform thin circular ring about a tangent in its plane, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The moment of inertia of a uniform thin circular ring about an axis through its center and normal to the plane of the ring is given as \( I \).

2. **Identify the Axes**:
- We need to find the moment of inertia about a tangent in the plane of the ring. Let's denote the moment of inertia about the center as \( I_z = I \) and the moment of inertia about the tangent as \( I_t \).

3. **Use the Perpendicular Axis Theorem**:
- For a planar object, the perpendicular axis theorem states that the moment of inertia about an axis perpendicular to the plane (here, the center axis) is the sum of the moments of inertia about two perpendicular axes in the plane.
- Therefore, we can express this as:
\[ I_z = I_x + I_y \]
- Since the ring is symmetric, \( I_x = I_y \). Let \( I_x = I_y = I_p \). Thus:
\[ I = I_p + I_p = 2I_p \implies I_p = \frac{I}{2} \]

4. **Apply the Parallel Axis Theorem**:
- The parallel axis theorem states that:
\[ I_t = I_p + m d^2 \]
- Here, \( I_p \) is the moment of inertia about the axis through the center, \( m \) is the mass of the ring, and \( d \) is the distance from the center to the tangent. For a ring, this distance is equal to the radius \( r \).
- Therefore, we have:
\[ I_t = \frac{I}{2} + m r^2 \]

5. **Substitute \( m r^2 \)**:
- The moment of inertia about the center for the ring is given as \( I = m r^2 \). Thus, we can replace \( m r^2 \) in the equation:
\[ I_t = \frac{I}{2} + I \]
- Simplifying this gives:
\[ I_t = \frac{I}{2} + \frac{2I}{2} = \frac{3I}{2} \]

6. **Final Result**:
- The moment of inertia of the ring about a tangent in its plane is:
\[ I_t = \frac{3I}{2} \]