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

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Initial Resistance Formula**:
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,
- \( A \) is the cross-sectional area of the wire.

2. **Cross-Sectional Area Calculation**:
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]

3. **New Dimensions**:
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:
\[ r' = 2r \]

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

5. **New Resistance Calculation**:
Now, substituting the new values into the resistance formula, we have:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{4A} \]
Simplifying this gives:
\[ R' = \frac{2\rho L}{4A} = \frac{\rho L}{2A} = \frac{R}{2} \]

6. **Conclusion**:
Therefore, the new resistance \( R' \) is half of the original resistance:
\[ R' = \frac{R}{2} \]

### Final Answer:
The resistance of the wire when both the length and radius are doubled is \( \frac{R}{2} \).