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
(a) `nR` (b) `R/n` (c) `n^(2)R` (d) `R/(n^(2))`

Options

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

Answer

To solve the problem, we need to determine the new resistance of a wire when it is melted and stretched to `n` times its original length. We will use the relationship between resistance, length, and cross-sectional area.

### Step-by-Step Solution:

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

2. **Initial resistance**:
Let the initial length of the wire be \( l_1 \) and the initial cross-sectional area be \( A_1 \). The initial resistance \( R \) can be expressed as:
\[ R = \frac{\rho l_1}{A_1} \]

3. **New length after stretching**:
When the wire is melted and stretched to \( n \) times its original length, the new length \( l_2 \) becomes:
\[ l_2 = n l_1 \]

4. **Volume conservation**:
The volume of the wire remains constant before and after stretching. Therefore, we can write:
\[ A_1 l_1 = A_2 l_2 \]
where \( A_2 \) is the new cross-sectional area after stretching.

5. **Substituting for \( l_2 \)**:
Substitute \( l_2 \) into the volume equation:
\[ A_1 l_1 = A_2 (n l_1) \]
Dividing both sides by \( l_1 \) (assuming \( l_1 \neq 0 \)):
\[ A_1 = n A_2 \]

6. **Finding new cross-sectional area**:
Rearranging gives:
\[ A_2 = \frac{A_1}{n} \]

7. **New resistance**:
Now we can find the new resistance \( R' \) using the new length and new area:
\[ R' = \frac{\rho l_2}{A_2} = \frac{\rho (n l_1)}{(A_1/n)} = \frac{\rho n^2 l_1}{A_1} \]

8. **Relating new resistance to initial resistance**:
We can relate \( R' \) to the initial resistance \( R \):
\[ R' = n^2 \left(\frac{\rho l_1}{A_1}\right) = n^2 R \]

9. **Final answer**:
Therefore, the new resistance after stretching the wire is:
\[ R' = n^2 R \]
The correct option is (c) \( n^2 R \).