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 use the Parallel Axis Theorem. Here’s a step-by-step solution:

### Step 1: Understand the Geometry
We have a ring with mass \( M \) and radius \( R \). The axis we are interested in is tangential to the ring and perpendicular to its plane.

### Step 2: Identify the Moment of Inertia about the Center
The moment of inertia of a ring about an axis that passes through its center and is perpendicular to its plane is given by:
\[ I_{\text{cm}} = M R^2 \]

### Step 3: Apply the Parallel Axis Theorem
The Parallel Axis Theorem states that if you want to find the moment of inertia about an axis that is parallel to an axis through the center of mass, you can use the formula:
\[ I = I_{\text{cm}} + M d^2 \]
where \( d \) is the distance between the two axes.

### Step 4: Determine the Distance \( d \)
In this case, the distance \( d \) between the center of the ring and the tangential axis is equal to the radius \( R \) of the ring:
\[ d = R \]

### Step 5: Substitute into the Formula
Now substituting the values into the Parallel Axis Theorem:
\[ I = I_{\text{cm}} + M d^2 = M R^2 + M R^2 \]
\[ I = M R^2 + M R^2 = 2 M R^2 \]

### Conclusion
Thus, the moment of inertia of the ring about the tangential axis perpendicular to its plane is:
\[ \boxed{2 M R^2} \]