The moment of inertia of a solid disc made of thin metal of radius R and mass M about one of its diameters is given by `(MR^(2))/(4)` . What will be the moment of inertia about this axis if the disc is folded in half about this diameter .

Options

A. `(MR^(2))/(8)`
B. `(MR^(2))/(2)`
C. `(MR^(2))/(4)`
D. `MR^(2)`

Explanation

To solve the problem of finding the moment of inertia of a solid disc folded in half about one of its diameters, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Moment of Inertia**:
The moment of inertia (I) of a solid disc about one of its diameters is given by the formula:
\[ I = \frac{1}{4} MR^2 \]
where \( M \) is the mass of the disc and \( R \) is the radius.

2. **Concept of Folding the Disc**:
When the disc is folded in half about its diameter, we are effectively bringing the two halves of the disc together. This means that the mass distribution changes, but the total mass remains the same.

3. **Moment of Inertia and Mass Distribution**:
The moment of inertia depends on how the mass is distributed relative to the axis of rotation. In this case, even though we are folding the disc, we are not changing the axis of rotation, and we are not removing any mass; we are simply overlapping half of the mass onto the other half.

4. **Conclusion on Moment of Inertia After Folding**:
Since the mass distribution relative to the axis of rotation remains the same, the moment of inertia will not change. Therefore, the moment of inertia of the folded disc about the same diameter is still:
\[ I = \frac{1}{4} MR^2 \]

5. **Final Answer**:
Thus, the moment of inertia of the disc after it is folded in half about its diameter remains:
\[ I = \frac{1}{4} MR^2 \]