The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then :

Options

A. the resistance and the specific resistance will both remain unchanged
B. the resistance will be doubled, and the specific resistance will be halved
C. the resistance will be halved, and the specific resistance will remain unchanged
D. the resistance will be halved, and the specific resistance will be doubled

Answer

To solve the problem, we need to analyze how the resistance of a wire changes when both its length and radius are doubled.

### 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 \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire.

2. **Determine the Original Parameters**:
Let's denote the original length of the wire as \( L \) and the original radius as \( r \). The cross-sectional area \( A \) of the wire, which is circular, can be calculated as:
\[ A = \pi r^2 \]

3. **Calculate the Original Resistance**:
Substituting the values into the resistance formula, we get:
\[ R = \frac{\rho L}{\pi r^2} \]

4. **Change the Length and Radius**:
If the length \( L \) is doubled, the new length \( L' \) becomes:
\[ L' = 2L \]
If the radius \( r \) is also doubled, the new radius \( r' \) becomes:
\[ r' = 2r \]

5. **Calculate the New Cross-Sectional Area**:
The new cross-sectional area \( A' \) with the new radius \( r' \) is:
\[ A' = \pi (r')^2 = \pi (2r)^2 = \pi \cdot 4r^2 = 4\pi r^2 \]

6. **Calculate the New Resistance**:
Now, substituting the new length and new area into the resistance formula, we get the new resistance \( R' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{4\pi r^2} \]
Simplifying this gives:
\[ R' = \frac{2\rho L}{4\pi r^2} = \frac{\rho L}{2\pi r^2} \]

7. **Relate the New Resistance to the Original Resistance**:
We can express the new resistance \( R' \) in terms of the original resistance \( R \):
\[ R' = \frac{1}{2} \left( \frac{\rho L}{\pi r^2} \right) = \frac{R}{2} \]

8. **Conclusion About Specific Resistance**:
The specific resistance \( \rho \) is a property of the material and does not change with the dimensions of the wire. Therefore, the specific resistance remains unchanged.

### Final Answer:
The new resistance \( R' \) is half of the original resistance \( R \), and the specific resistance remains unchanged.

### Summary of Results:
- The resistance \( R' = \frac{R}{2} \)
- The specific resistance \( \rho \) remains unchanged.