Moment of inertia of a ring of radius R about a diametric axis is `25 "kg m"^(2)`. The `MI` of the ring about a parallel axis at a distance `R` from the centre is

Options

A. `31.25 "kg-m"^(2)`
B. `75 "kg-m"^(2)`
C. `50 "kg-m"^(2)`
D. `62.5 "kg-m"^(2)`

Explanation

To find the moment of inertia of a ring about a parallel axis at a distance \( R \) from the center, we can use the parallel axis theorem. Here's a step-by-step solution:

### Step 1: Understand the Given Information
We know that the moment of inertia of the ring about a diametric axis (which is in the plane of the ring and passes through the center) is given as:
\[ I_{CM} = 25 \, \text{kg m}^2 \]

### Step 2: Recall the Moment of Inertia Formula
For a ring of radius \( R \), the moment of inertia about its center of mass (CM) for a diametric axis is given by:
\[ I_{CM} = \frac{1}{2} m R^2 \]
where \( m \) is the mass of the ring.

### Step 3: Use the Parallel Axis Theorem
The parallel axis theorem states that the moment of inertia about a parallel axis is given by:
\[ I = I_{CM} + m d^2 \]
where \( d \) is the distance between the two axes. In this case, \( d = R \).

### Step 4: Substitute the Values
We need to find \( I \) about the parallel axis at a distance \( R \) from the center:
\[ I = I_{CM} + m R^2 \]
Substituting \( I_{CM} = 25 \, \text{kg m}^2 \) and \( d = R \):
\[ I = 25 + m R^2 \]

### Step 5: Relate \( m R^2 \) to the Given Moment of Inertia
Since we know:
\[ I_{CM} = \frac{1}{2} m R^2 = 25 \]
We can solve for \( m R^2 \):
\[ m R^2 = 2 \times 25 = 50 \]

### Step 6: Calculate the Moment of Inertia
Now, substitute \( m R^2 = 50 \) into the equation for \( I \):
\[ I = 25 + 50 = 75 \, \text{kg m}^2 \]

### Final Answer
Thus, the moment of inertia of the ring about a parallel axis at a distance \( R \) from the center is:
\[ \boxed{75 \, \text{kg m}^2} \]

---