A condutor of length l is stretched such that its length increases by 10 % calculate the percentage change in the resistance of the conductor.

Answer

To solve the problem of finding the percentage change in the resistance of a conductor when its length is increased by 10%, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the initial length and the change in length**:
- Let the initial length of the conductor be \( L \).
- The length increases by 10%, so the new length \( L' \) can be calculated as:
\[ L' = L + 0.10L = 1.10L = \frac{11}{10}L \]

2. **Understand the relationship between volume and area**:
- The volume of the conductor remains constant when it is stretched. The volume \( V \) can be expressed as:
\[ V = A \times L \]
- After stretching, the new volume is:
\[ V' = A' \times L' \]
- Since the volume remains constant, we have:
\[ A \times L = A' \times L' \]

3. **Substituting the new length into the volume equation**:
- Substitute \( L' = \frac{11}{10}L \) into the volume equation:
\[ A \times L = A' \times \left(\frac{11}{10}L\right) \]
- Cancelling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ A = A' \times \frac{11}{10} \]
- Rearranging gives:
\[ A' = A \times \frac{10}{11} \]

4. **Calculate the initial and new resistance**:
- The resistance \( R \) of the conductor is given by:
\[ R = \frac{\rho L}{A} \]
- The new resistance \( R' \) after stretching is:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{11}{10}L\right)}{A \times \frac{10}{11}} = \frac{\rho \cdot \frac{11}{10}L \cdot 11}{10A} = \frac{121 \rho L}{100 A} \]
- Thus, we can express \( R' \) in terms of the original resistance \( R \):
\[ R' = \frac{121}{100} R = 1.21 R \]

5. **Calculate the percentage change in resistance**:
- The percentage change in resistance is calculated as:
\[ \text{Percentage Change} = \frac{R' - R}{R} \times 100 \]
- Substituting the values:
\[ \text{Percentage Change} = \frac{1.21R - R}{R} \times 100 = \frac{0.21R}{R} \times 100 = 21\% \]

### Final Result:
The percentage change in the resistance of the conductor is **21%**.