The resistance of a wire is 'R' ohm. If it is melted and stretched to `n` times its origianl length, its new resistance will be

Options

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

Explanation

To find the new resistance of a wire when it is melted and stretched to `n` times its original length, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the original resistance of the wire be \( R \) ohms.
- Let the original length of the wire be \( L \).
- Let the original cross-sectional area of the wire be \( A \).

2. **Use the Formula for Resistance**:
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material.

3. **Determine the New Length**:
- When the wire is melted and stretched to \( n \) times its original length, the new length \( L' \) becomes:
\[ L' = nL \]

4. **Calculate the New Volume**:
- The volume of the wire remains constant during the melting and stretching process. The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
- After stretching, the new volume \( V' \) can be expressed as:
\[ V' = A' \cdot L' = A' \cdot (nL) \]
- Since the volume remains constant, we have:
\[ A \cdot L = A' \cdot (nL) \]
- From this, we can solve for the new cross-sectional area \( A' \):
\[ A' = \frac{A}{n} \]

5. **Substitute into the Resistance Formula**:
- Now, we can find the new resistance \( R' \) using the new length and new cross-sectional area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (nL)}{\frac{A}{n}} = \frac{\rho n^2 L}{A} \]

6. **Relate the New Resistance to the Original Resistance**:
- We can express the new resistance \( R' \) in terms of the original resistance \( R \):
\[ R' = n^2 \cdot \frac{\rho L}{A} = n^2 R \]

7. **Conclusion**:
- Therefore, the new resistance of the wire after being melted and stretched to \( n \) times its original length is:
\[ R' = n^2 R \]

### Final Answer:
The new resistance will be \( n^2 R \).

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