If a wire of length I has a resistance of R, is stretched by `25 %.` The percentage change in its resistance is ?

Options

A. `25 %`
B. `50 %`
C. `45.5 %`
D. `56.25 %`

Answer

To solve the problem of finding the percentage change in resistance when a wire of length \( L \) and resistance \( R \) is stretched by 25%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The initial length of the wire is \( L \).
- The initial resistance of the wire is \( R \).

2. **Determine the New Length**:
- When the wire is stretched by 25%, the new length \( L' \) can be calculated as:
\[ L' = L + 0.25L = 1.25L \]

3. **Volume Conservation**:
- The volume \( V \) of the wire remains constant during stretching. The volume is given by:
\[ V = A \cdot L \]
where \( A \) is the cross-sectional area of the wire.
- After stretching, the new volume can be expressed as:
\[ V = A' \cdot L' \]
where \( A' \) is the new cross-sectional area.

4. **Set the Volumes Equal**:
- Since the volume remains constant:
\[ A \cdot L = A' \cdot L' \]
- Substituting \( L' \):
\[ A \cdot L = A' \cdot (1.25L) \]
- Rearranging gives:
\[ A' = \frac{A}{1.25} = 0.8A \]

5. **Calculate the New Resistance**:
- The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material.
- The new resistance \( R' \) after stretching is:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (1.25L)}{0.8A} \]
- Simplifying this gives:
\[ R' = \frac{1.25}{0.8} \cdot \frac{\rho L}{A} = 1.5625 R \]

6. **Calculate the Percentage Change in Resistance**:
- The percentage change in resistance is given by:
\[ \text{Percentage Change} = \frac{R' - R}{R} \times 100 \]
- Substituting \( R' \):
\[ \text{Percentage Change} = \frac{1.5625R - R}{R} \times 100 = (0.5625) \times 100 = 56.25\% \]

### Final Answer:
The percentage change in resistance when the wire is stretched by 25% is **56.25%**.