If a disc of moment of inertia I and radius r is reshaped into a ring of radius nr, keeping its mass same, its moment of inertia becomes

Options

A. `n^(2)I`
B. `2n^(2)I`
C. `(n^(2))/(2)I`
D. I

Answer

To solve the problem, we need to find the moment of inertia of a ring formed by reshaping a disc while keeping its mass constant. Let's break this down step by step.

### Step-by-Step Solution:

1. **Identify the Moment of Inertia of the Disc**:
The moment of inertia \( I \) of a disc about its center is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass of the disc and \( r \) is its radius.

2. **Mass of the Disc**:
From the formula, we can express the mass \( m \) in terms of the moment of inertia \( I \) and radius \( r \):
\[ m = \frac{2I}{r^2} \]

3. **Reshaping the Disc into a Ring**:
The disc is reshaped into a ring of radius \( nr \). The mass remains the same, which is \( m \).

4. **Moment of Inertia of the Ring**:
The moment of inertia \( I' \) of a ring about its center is given by:
\[ I' = m R^2 \]
where \( R \) is the radius of the ring. In this case, \( R = nr \).

5. **Substituting the Mass**:
Substitute the mass \( m \) from step 2 into the moment of inertia formula for the ring:
\[ I' = \left(\frac{2I}{r^2}\right) (nr)^2 \]

6. **Simplifying the Expression**:
Now simplify the expression:
\[ I' = \frac{2I}{r^2} \cdot n^2 r^2 \]
The \( r^2 \) terms cancel out:
\[ I' = 2I n^2 \]

7. **Final Result**:
Thus, the moment of inertia of the ring is:
\[ I' = 2n^2 I \]

### Conclusion:
The moment of inertia of the ring formed by reshaping the disc is \( 2n^2 I \).

---