The wire is stretched to increase the length by 1%. Find the percentage change in the resistance.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem of finding the percentage change in resistance when a wire is stretched to increase its length by 1%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Resistance**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance,
- \( \rho \) = resistivity of the material (constant),
- \( L \) = length of the wire,
- \( A \) = cross-sectional area of the wire.

2. **Consider the Change in Length**:
When the wire is stretched, its length increases. Given that the length increases by 1%, we can express this as:
\[ \Delta L = 0.01L \]
where \( L \) is the original length.

3. **Volume Conservation**:
The volume \( V \) of the wire remains constant during stretching. The volume is given by:
\[ V = A \cdot L \]
After stretching, the new length \( L' \) becomes:
\[ L' = L + \Delta L = L + 0.01L = 1.01L \]
Let the new cross-sectional area be \( A' \). Since volume is constant:
\[ A \cdot L = A' \cdot L' \]
Substituting for \( L' \):
\[ A \cdot L = A' \cdot (1.01L) \]
Simplifying gives:
\[ A' = \frac{A}{1.01} \]

4. **Calculate the New Resistance**:
The new resistance \( R' \) can be expressed using the new length and area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (1.01L)}{\frac{A}{1.01}} = \frac{\rho (1.01^2)L}{A} = R \cdot 1.01^2 \]

5. **Percentage Change in Resistance**:
To find the percentage change in resistance, we can use the formula:
\[ \text{Percentage Change} = \frac{R' - R}{R} \times 100 \]
Substituting \( R' \):
\[ \text{Percentage Change} = \frac{R \cdot 1.01^2 - R}{R} \times 100 = (1.01^2 - 1) \times 100 \]
Calculating \( 1.01^2 \):
\[ 1.01^2 = 1.0201 \]
Thus:
\[ \text{Percentage Change} = (1.0201 - 1) \times 100 = 0.0201 \times 100 = 2.01\% \]

6. **Conclusion**:
The percentage change in resistance when the wire is stretched to increase its length by 1% is approximately **2.01%**.