The M.I. of a ring of mass M and radius R about a tangential axis perpendicular to its plane is :

Options

A. `2MR^(2)`
B. `MR^(2)`
C. `3//2MR^(2)`
D. `3//4MR^(2)`

Answer

To find the moment of inertia (M.I.) of a ring of mass \( M \) and radius \( R \) about a tangential axis perpendicular to its plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Moment of Inertia about the Center:**
The moment of inertia of a ring about an axis perpendicular to its plane and passing through its center is given by the formula:
\[ I_{C} = M R^2 \]
where \( M \) is the mass of the ring and \( R \) is its radius.

2. **Apply the Parallel Axis Theorem:**
To find the moment of inertia about a tangential axis, we can use the Parallel Axis Theorem. This theorem states that if you know the moment of inertia about a parallel axis through the center of mass, you can find the moment of inertia about any parallel axis by adding the product of the mass and the square of the distance between the two axes.

The formula for the Parallel Axis Theorem is:
\[ I_{T} = I_{C} + M d^2 \]
where \( d \) is the distance between the center of mass axis and the new axis (the tangential axis in this case).

3. **Determine the Distance \( d \):**
For a ring, the distance \( d \) from the center of the ring to the tangential axis is equal to the radius \( R \) of the ring. Thus, we have:
\[ d = R \]

4. **Substitute Values into the Parallel Axis Theorem:**
Now we can substitute the values into the equation:
\[ I_{T} = I_{C} + M R^2 \]
Substituting \( I_{C} = M R^2 \):
\[ I_{T} = M R^2 + M R^2 \]
\[ I_{T} = 2 M R^2 \]

5. **Final Result:**
Therefore, the moment of inertia of the ring about the tangential axis perpendicular to its plane is:
\[ I_{T} = 2 M R^2 \]