If the length of a conductor is increased by `100%` keeping its area of cross-section constant, the percentage increase in its resistance is

Options

A. `100%`
B. `50%`
C. `200%`
D. `25%`

Answer

To solve the problem, we need to determine the percentage increase in resistance when the length of a conductor is increased by 100%, while keeping its area of cross-section constant.

### Step-by-Step Solution:

1. **Define Initial Length and Area**:
Let the initial length of the conductor be \( L \) and the area of cross-section be \( A \).

2. **Calculate Final Length**:
Since the length is increased by 100%, the final length \( L' \) can be calculated as:
\[ L' = L + 100\% \text{ of } L = L + L = 2L \]

3. **Area of Cross-Section**:
The area of cross-section remains constant, so:
\[ A' = A \]

4. **Resistance Formula**:
The resistance \( R \) of a conductor is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material.

5. **Calculate Initial Resistance**:
The initial resistance \( R_{\text{initial}} \) is:
\[ R_{\text{initial}} = \frac{\rho L}{A} \]

6. **Calculate Final Resistance**:
The final resistance \( R_{\text{final}} \) with the new length \( L' \) and the same area \( A \) is:
\[ R_{\text{final}} = \frac{\rho L'}{A} = \frac{\rho (2L)}{A} = 2 \cdot \frac{\rho L}{A} = 2 R_{\text{initial}} \]

7. **Determine Change in Resistance**:
The change in resistance \( \Delta R \) is:
\[ \Delta R = R_{\text{final}} - R_{\text{initial}} = 2 R_{\text{initial}} - R_{\text{initial}} = R_{\text{initial}} \]

8. **Calculate Percentage Increase in Resistance**:
The percentage increase in resistance is given by:
\[ \text{Percentage Increase} = \left( \frac{\Delta R}{R_{\text{initial}}} \right) \times 100 = \left( \frac{R_{\text{initial}}}{R_{\text{initial}}} \right) \times 100 = 100\% \]

### Conclusion:
The percentage increase in resistance when the length of the conductor is increased by 100% is **100%**.