A wire 1m long has a resistance of `1 Omega`. If it is uniformly stretched, so that its length increases by 25% then its resistance will increase by

Options

A. 0.25
B. 0.5
C. 0.5625
D. 0.7733

Explanation

To solve the problem, we need to determine the new resistance of a wire after it has been stretched, and how much the resistance has increased. Let's break it down step by step.

### Step-by-step Solution:

1. **Initial Parameters**:
- Length of the wire, \( L = 1 \, \text{m} \)
- Initial resistance, \( R = 1 \, \Omega \)

2. **Calculate the New Length**:
- The wire is stretched uniformly, increasing its length by 25%.
- New length, \( L' = L + 0.25L = 1 + 0.25 \times 1 = 1.25 \, \text{m} \)

3. **Volume Conservation**:
- The volume of the wire remains constant during stretching.
- Initial volume, \( V = A \times L \)
- Final volume, \( V' = A' \times L' \)
- Since \( V = V' \), we have:
\[ A \times L = A' \times L' \]
- Substituting the known values:
\[ A \times 1 = A' \times 1.25 \]
- Rearranging gives:
\[ A' = \frac{A}{1.25} \]

4. **Resistance Formula**:
- The resistance of a wire is given by:
\[ R = \frac{\rho L}{A} \]
- For the new resistance \( R' \):
\[ R' = \frac{\rho L'}{A'} \]

5. **Substituting New Values**:
- Substitute \( L' = 1.25 \, \text{m} \) and \( A' = \frac{A}{1.25} \):
\[ R' = \frac{\rho \times 1.25}{\frac{A}{1.25}} = \frac{1.25^2 \rho}{A} \]
- Since \( \frac{\rho}{A} = R \) (initial resistance):
\[ R' = 1.25^2 \times R \]
- Given \( R = 1 \, \Omega \):
\[ R' = 1.25^2 \times 1 = 1.5625 \, \Omega \]

6. **Calculate the Increase in Resistance**:
- The increase in resistance, \( \Delta R \):
\[ \Delta R = R' - R = 1.5625 - 1 = 0.5625 \, \Omega \]

### Final Answer:
The resistance will increase by \( 0.5625 \, \Omega \).

---