The resistance of a wire is r ohm. If it is melted and stretched to n times its original length , its new resistance will be

Options

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

Answer

To solve the problem of finding the new resistance of a wire that has been melted and stretched to n times its original length, we can follow these steps:

### Step-by-Step Solution:

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

2. **Volume Conservation:**
When the wire is melted and stretched, its volume remains constant. The volume \( V \) of the wire can be expressed as:
\[ V = A \times L \]
After stretching, the new length becomes \( nL \) (where \( n \) is the factor by which the length is increased). Let the new cross-sectional area be \( A' \). The volume after stretching is:
\[ V' = A' \times (nL) \]
Since the volume remains constant, we have:
\[ A \times L = A' \times (nL) \]

3. **Relate the Areas:**
From the volume conservation equation, we can simplify:
\[ A = A' \times n \]
Therefore, the new cross-sectional area \( A' \) can be expressed as:
\[ A' = \frac{A}{n} \]

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

5. **Relate New Resistance to Original Resistance:**
We know that the original resistance \( R \) is:
\[ R = \frac{\rho L}{A} \]
Therefore, we can express \( R' \) in terms of \( R \):
\[ R' = n^2 \cdot R \]

### Final Answer:
The new resistance of the wire after being melted and stretched to n times its original length is:
\[ R' = n^2 R \]