The length of a conductor is doubled and its radius is halved, its specific resistance is

Options

A. halved
B. doubled
C. quadrupled
D. unchanged

Explanation

To solve the problem, we need to understand how the specific resistance (also known as resistivity) of a conductor is affected by its dimensions. The specific resistance of a material is a property that depends only on the material itself and not on its shape or size.

### Step-by-Step Solution:

1. **Understand the Concept of Specific Resistance**:
Specific resistance (or resistivity) is denoted by the symbol \( \rho \) and is defined as the resistance of a unit length of a material with a unit cross-sectional area. It is a property of the material and does not change with the dimensions of the conductor.

2. **Identify the Changes in Dimensions**:
- The length of the conductor is doubled: If the original length is \( L \), the new length \( L' = 2L \).
- The radius of the conductor is halved: If the original radius is \( r \), the new radius \( r' = \frac{r}{2} \).

3. **Calculate the Cross-Sectional Area**:
The cross-sectional area \( A \) of a circular conductor is given by the formula:
\[ A = \pi r^2 \]
For the original conductor, the area is:
\[ A = \pi r^2 \]
For the new conductor with radius \( r' \):
\[ A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} = \frac{1}{4} \pi r^2 \]

4. **Understand the Relationship of Resistance**:
The resistance \( R \) of a conductor is given by the formula:
\[ R = \frac{\rho L}{A} \]
For the original conductor:
\[ R = \frac{\rho L}{\pi r^2} \]
For the new conductor:
\[ R' = \frac{\rho (2L)}{\frac{1}{4} \pi r^2} = \frac{8\rho L}{\pi r^2} \]

5. **Determine the Specific Resistance**:
The specific resistance \( \rho \) remains unchanged because it is a property of the material. Therefore, regardless of the changes made to the dimensions of the conductor (doubling the length and halving the radius), the specific resistance \( \rho \) does not change.

### Conclusion:
The specific resistance of the conductor remains unchanged.

### Answer:
**Unchanged**