A wire of resistance `R` is elongated `n-fold` to make a new uniform wire. The resistance of new wire.

Options

A. n R
B. `n^2 R`
C. 2 nR
D. `2n^2 R`

Explanation

To solve the problem of finding the resistance of a new wire that has been elongated `n-fold`, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the original wire have a resistance \( R \), length \( L \), and cross-sectional area \( A \).
- The resistance of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material.

2. **Determine the New Length and Area**:
- When the wire is elongated `n-fold`, the new length \( L' \) becomes:
\[ L' = nL \]
- Since the volume of the wire remains constant during elongation, we have:
\[ \text{Volume} = \text{Cross-sectional Area} \times \text{Length} \]
\[ A \cdot L = A' \cdot L' \]
- Substituting \( L' \):
\[ A \cdot L = A' \cdot (nL) \]
- Rearranging gives:
\[ A' = \frac{A}{n} \]

3. **Calculate the New Resistance**:
- The resistance of the new wire \( R' \) can be calculated using the new length \( L' \) and new area \( A' \):
\[ R' = \frac{\rho L'}{A'} \]
- Substituting the values of \( L' \) and \( A' \):
\[ R' = \frac{\rho (nL)}{\frac{A}{n}} = \frac{\rho n^2 L}{A} \]

4. **Relate New Resistance to Original Resistance**:
- From the original resistance formula \( R = \frac{\rho L}{A} \), we can express \( \rho L/A \) as \( R \):
\[ R' = n^2 \cdot R \]

5. **Final Result**:
- Therefore, the resistance of the new wire after being elongated `n-fold` is:
\[ R' = n^2 R \]

### Conclusion:
The resistance of the new wire is \( n^2 R \).

---