If the radius of cross-section of the conductor is increased by `0.1 %` keeping volume constant, then percentage change in the resistance of the conductor is

Options

A. `- 0.2 %`
B. ` - 0.1 %`
C. `-0.3 %`
D. `-0.4%`

Answer

To solve the problem, we need to find the percentage change in the resistance of a conductor when the radius of its cross-section is increased by 0.1%, while keeping the volume constant.

### Step-by-Step Solution:

1. **Understand the relationship between resistance, length, and area**:
The resistance \( R \) of a conductor is given by the formula:
\[ R = \frac{L}{A} \]
where \( L \) is the length of the conductor and \( A \) is the cross-sectional area.

2. **Express the cross-sectional area in terms of the radius**:
For a cylindrical conductor, the cross-sectional area \( A \) can be expressed as:
\[ A = \pi r^2 \]
where \( r \) is the radius of the conductor.

3. **Calculate the new radius after the increase**:
If the radius is increased by 0.1%, the new radius \( r_2 \) can be calculated as:
\[ r_2 = r + 0.001r = 1.001r \]

4. **Calculate the new cross-sectional area**:
The new cross-sectional area \( A_2 \) will be:
\[ A_2 = \pi r_2^2 = \pi (1.001r)^2 = \pi (1.002001) r^2 \]

5. **Set up the volume condition**:
Since the volume \( V \) of the conductor remains constant, we have:
\[ V_1 = A_1 L_1 = A_2 L_2 \]
Thus,
\[ A_1 L_1 = A_2 L_2 \]
Rearranging gives:
\[ L_2 = \frac{A_1}{A_2} L_1 \]

6. **Substitute the areas**:
Substituting \( A_1 = \pi r^2 \) and \( A_2 = \pi (1.002001) r^2 \):
\[ L_2 = \frac{\pi r^2}{\pi (1.002001) r^2} L_1 = \frac{L_1}{1.002001} \]

7. **Calculate the new resistance**:
The new resistance \( R_2 \) can be expressed as:
\[ R_2 = \frac{L_2}{A_2} = \frac{\frac{L_1}{1.002001}}{\pi (1.002001) r^2} = \frac{L_1}{\pi r^2} \cdot \frac{1}{1.002001^2} = R_1 \cdot \frac{1}{1.002001^2} \]

8. **Find the ratio of the new resistance to the old resistance**:
Thus, we have:
\[ \frac{R_2}{R_1} = \frac{1}{1.002001^2} \]

9. **Calculate the percentage change in resistance**:
The percentage change in resistance is given by:
\[ \text{Percentage Change} = \left( \frac{R_2 - R_1}{R_1} \right) \times 100 = \left( \frac{\frac{R_1}{1.002001^2} - R_1}{R_1} \right) \times 100 \]
Simplifying gives:
\[ = \left( \frac{1 - \frac{1}{1.002001^2}}{1} \right) \times 100 \]
Calculating \( 1.002001^2 \approx 1.004004 \):
\[ = \left( 1 - \frac{1}{1.004004} \right) \times 100 \approx (1 - 0.996) \times 100 \approx 0.4\% \]
Since the resistance decreases, we have:
\[ \text{Percentage Change} \approx -0.4\% \]

### Final Answer:
The percentage change in the resistance of the conductor is approximately **-0.4%**.