The electric resistance of a certain wire of iron is R . If its length and radius are both doubled, 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 resistancce will be halved and the specific resistance will be doubled.
D. both the resistance and the specific resistance, will remain unchanged.

Answer

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

### Step-by-Step Solution:

1. **Understanding the Formula for Resistance**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance
- \( \rho \) = resistivity of the material
- \( L \) = length of the wire
- \( A \) = cross-sectional area of the wire

2. **Cross-Sectional Area**:
The area \( A \) of a circular wire can be expressed as:
\[ A = \pi r^2 \]
where \( r \) is the radius of the wire.

3. **Initial Conditions**:
Let the initial length of the wire be \( L \) and the initial radius be \( r \). Thus, the initial resistance \( R \) can be expressed as:
\[ R = \frac{\rho L}{\pi r^2} \]

4. **New Conditions**:
According to the problem, both the length and radius are doubled:
- New length \( L' = 2L \)
- New radius \( r' = 2r \)

5. **Calculating New Area**:
The new cross-sectional area \( A' \) becomes:
\[ A' = \pi (r')^2 = \pi (2r)^2 = \pi (4r^2) = 4\pi r^2 \]

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

7. **Relating New Resistance to Old Resistance**:
We can relate the new resistance \( R' \) to the original resistance \( R \):
\[ R' = \frac{1}{2} R \]
This shows that the new resistance \( R' \) is half of the original resistance \( R \).

8. **Conclusion**:
Therefore, if the length and radius of the wire are both doubled, the new resistance is:
\[ R' = \frac{R}{2} \]
The specific resistance \( \rho \) remains unchanged as it is a property of the material.

### Final Answer:
The resistance of the wire becomes half of its original value, and the specific resistance remains unchanged.