The mass of a circular ring is `M` and its radius is `R`. Its moment of inertia about an axis in the plane of ring at perpendicular distance `R//2` from centre of ring is

Options

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

Explanation

To find the moment of inertia of a circular ring about an axis in its plane at a perpendicular distance of \( \frac{R}{2} \) from the center, we can use the parallel axis theorem. Here’s the step-by-step solution:

### Step 1: Understand the Moment of Inertia of a Ring
The moment of inertia \( I \) of a circular ring about an axis passing through its center and perpendicular to its plane is given by:
\[ I_{\text{cm}} = MR^2 \]
where \( M \) is the mass of the ring and \( R \) is its radius.

### Step 2: Use the Parallel Axis Theorem
The parallel axis theorem states that if you know the moment of inertia about an 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:
\[ I = I_{\text{cm}} + Md^2 \]
where \( d \) is the distance between the two axes.

### Step 3: Identify the Values
In this case:
- \( I_{\text{cm}} = MR^2 \) (moment of inertia about the center)
- \( d = \frac{R}{2} \) (the distance from the center to the new axis)

### Step 4: Calculate \( Md^2 \)
Now, we need to calculate \( Md^2 \):
\[ Md^2 = M\left(\frac{R}{2}\right)^2 = M\frac{R^2}{4} \]

### Step 5: Substitute into the Parallel Axis Theorem
Now substitute \( I_{\text{cm}} \) and \( Md^2 \) into the parallel axis theorem:
\[ I = MR^2 + M\frac{R^2}{4} \]

### Step 6: Simplify the Expression
Combine the terms:
\[ I = MR^2 + \frac{MR^2}{4} = MR^2\left(1 + \frac{1}{4}\right) = MR^2\left(\frac{4}{4} + \frac{1}{4}\right) = MR^2\left(\frac{5}{4}\right) \]

### Final Answer
Thus, the moment of inertia of the circular ring about the specified axis is:
\[ I = \frac{5}{4} MR^2 \]

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