A wire is stretched to n times its length. Then the resistance now will be increase by

Options

A. n times
B. 1/n times
C. `n^(2)` times
D. `1// n^(2)` times

Explanation

To solve the problem of how the resistance of a wire changes when it is stretched to n times its original length, we can follow these steps:

### Step-by-Step Solution

1. **Define Initial Parameters:**
- Let the initial length of the wire be \( L \).
- Let the initial cross-sectional area of the wire be \( A \).
- Let the resistivity of the material be \( \rho \).

2. **Calculate Initial Resistance:**
- The resistance \( R \) of the wire can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]

3. **Determine New Length and Area After Stretching:**
- When the wire is stretched to \( n \) times its original length, the new length \( L' \) becomes:
\[ L' = nL \]
- Since the volume of the wire remains constant during stretching, we can express the volume before and after stretching:
\[ \text{Initial Volume} = L \cdot A \]
\[ \text{New Volume} = L' \cdot A' = nL \cdot A' \]
- Setting these equal gives:
\[ L \cdot A = nL \cdot A' \]
- Canceling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ A = nA' \]
- Rearranging gives:
\[ A' = \frac{A}{n} \]

4. **Calculate New Resistance:**
- The new resistance \( R' \) can be calculated using the new length and new cross-sectional area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (nL)}{(A/n)} = \frac{\rho nL \cdot n}{A} = \frac{n^2 \rho L}{A} \]
- Thus, we can express \( R' \) in terms of the original resistance \( R \):
\[ R' = n^2 R \]

5. **Determine the Increase in Resistance:**
- The increase in resistance can be calculated as:
\[ \text{Increase} = R' - R = n^2 R - R = (n^2 - 1) R \]

### Conclusion
The resistance of the wire increases by a factor of \( n^2 \), meaning the new resistance \( R' \) is \( n^2 \) times the original resistance \( R \).