The M.I. of a thin ring of mass M and radius R about an axis through the diameter in its plane will be:

Options

A. `(1)/(2)MR^(2)`
B. `(3)/(2)MR^(2)`
C. `MR^(2)`
D. `2MR^(2)`

Answer

To find the moment of inertia (M.I.) of a thin ring of mass \( M \) and radius \( R \) about an axis through the diameter in its plane, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry of the Ring**:
- A thin ring is a circular object with mass \( M \) uniformly distributed along its circumference. The radius of the ring is \( R \).

2. **Identify the Axis of Rotation**:
- We need to calculate the moment of inertia about an axis that passes through the diameter of the ring and lies in its plane.

3. **Use the Perpendicular Axis Theorem**:
- The perpendicular axis theorem states that for a planar object, the moment of inertia about an axis perpendicular to the plane (let’s call it the z-axis) is equal to the sum of the moments of inertia about two perpendicular axes (x and y) lying in the plane of the object. Mathematically, this can be expressed as:
\[ I_z = I_x + I_y \]
- Here, \( I_z \) is the moment of inertia about the z-axis, and \( I_x \) and \( I_y \) are the moments of inertia about the x and y axes, respectively.

4. **Calculate the Moment of Inertia about the z-axis**:
- For a thin ring, the moment of inertia about an axis perpendicular to the plane of the ring (the z-axis) and passing through its center is given by:
\[ I_z = M R^2 \]

5. **Symmetry of the Ring**:
- Due to the symmetry of the ring, the moments of inertia about the x-axis and y-axis are equal:
\[ I_x = I_y \]
- Let’s denote \( I_x = I_y = I_d \) (the moment of inertia about the diameter).

6. **Substituting into the Perpendicular Axis Theorem**:
- Now substituting into the perpendicular axis theorem:
\[ I_z = I_d + I_d = 2I_d \]
- Therefore, we can express \( I_d \) as:
\[ I_d = \frac{I_z}{2} \]

7. **Final Calculation**:
- Substitute \( I_z = M R^2 \) into the equation:
\[ I_d = \frac{M R^2}{2} \]

8. **Conclusion**:
- The moment of inertia of the thin ring about an axis through the diameter in its plane is:
\[ I_d = \frac{M R^2}{2} \]

### Final Answer:
The moment of inertia of a thin ring of mass \( M \) and radius \( R \) about an axis through the diameter in its plane is \( \frac{M R^2}{2} \).

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