If a copper wire is stretched to make its radius decrease by `0.1%`, then the percentage increase in resistance is approximately -

Options

A. `0.1 %`
B. `0.2%`
C. `0.4%`
D. `0.8%`

Answer

To solve the problem of determining the percentage increase in resistance when a copper wire is stretched, causing its radius to decrease by 0.1%, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Relation of Resistance**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. The area \( A \) for a circular wire is:
\[ A = \pi r^2 \]

2. **Volume Conservation**:
When the wire is stretched, its volume remains constant. The volume \( V \) of the wire can be expressed as:
\[ V = A \times L = \pi r^2 L \]
Since the volume is constant, any change in the radius will affect the length.

3. **Change in Radius**:
The radius decreases by 0.1%. If the initial radius is \( r_1 \), the new radius \( r_2 \) can be calculated as:
\[ r_2 = r_1 - 0.001 \times r_1 = 0.9999 r_1 \]

4. **Calculate the New Area**:
The new cross-sectional area \( A_2 \) can be calculated using the new radius:
\[ A_2 = \pi (r_2)^2 = \pi (0.9999 r_1)^2 = \pi (0.9998 r_1^2) \]

5. **Relate Length and Area**:
Since the volume is constant, we can express the new length \( L_2 \) in terms of the new area:
\[ V = A_1 L_1 = A_2 L_2 \]
Thus,
\[ L_2 = \frac{A_1 L_1}{A_2} \]

6. **Substituting Areas**:
Using \( A_1 = \pi r_1^2 \) and \( A_2 = \pi (0.9998 r_1^2) \):
\[ L_2 = \frac{\pi r_1^2 L_1}{\pi (0.9998 r_1^2)} = \frac{L_1}{0.9998} \]

7. **Calculate the New Resistance**:
The new resistance \( R_2 \) can be expressed as:
\[ R_2 = \frac{\rho L_2}{A_2} = \frac{\rho \left(\frac{L_1}{0.9998}\right)}{\pi (0.9998 r_1^2)} = \frac{\rho L_1}{\pi r_1^2} \cdot \frac{1}{0.9998^2} \]
Thus,
\[ R_2 = R_1 \cdot \frac{1}{0.9998^2} \]

8. **Percentage Increase in Resistance**:
The percentage increase in resistance can be calculated as:
\[ \text{Percentage Increase} = \left(\frac{R_2 - R_1}{R_1}\right) \times 100 = \left(\frac{R_1 \cdot \frac{1}{0.9998^2} - R_1}{R_1}\right) \times 100 \]
Simplifying gives:
\[ \text{Percentage Increase} = \left(\frac{1}{0.9998^2} - 1\right) \times 100 \]
Approximating \( 0.9998^2 \approx 0.9996 \):
\[ \text{Percentage Increase} \approx \left(\frac{1}{0.9996} - 1\right) \times 100 \approx 0.4\% \]

### Final Answer:
The percentage increase in resistance is approximately **0.4%**.