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 resistance will be halved and the specific resistance will be doubled
D. The resistance and the specific resistance, will both remain unchanged

Answer

To solve the problem, we need to understand how the resistance of a wire changes with its dimensions. The resistance \( R \) of a wire is given by the formula:

\[ R = \rho \frac{L}{A} \]

where:
- \( R \) = resistance,
- \( \rho \) = resistivity of the material (which remains constant for the same material),
- \( L \) = length of the wire,
- \( A \) = cross-sectional area of the wire.

The cross-sectional area \( A \) of a wire with radius \( r \) is given by:

\[ A = \pi r^2 \]

### Step 1: Determine the initial resistance
Let the initial length of the wire be \( L \) and the initial radius be \( r \). The initial resistance \( R \) can be expressed as:

\[ R = \rho \frac{L}{\pi r^2} \]

### Step 2: Modify the dimensions
According to the problem, both the length and the radius of the wire are doubled. Therefore, the new length \( L' \) and new radius \( r' \) will be:

\[ L' = 2L \]
\[ r' = 2r \]

### Step 3: Calculate the new cross-sectional area
The new cross-sectional area \( A' \) can be calculated using the new radius:

\[ A' = \pi (r')^2 = \pi (2r)^2 = \pi \cdot 4r^2 = 4\pi r^2 \]

### Step 4: Calculate the new resistance
Now, we can find the new resistance \( R' \) using the modified length and area:

\[ R' = \rho \frac{L'}{A'} = \rho \frac{2L}{4\pi r^2} = \rho \frac{2L}{4A} = \frac{1}{2} \left( \rho \frac{L}{A} \right) = \frac{1}{2} R \]

### Conclusion
Thus, the new resistance \( R' \) is half of the original resistance \( R \):

\[ R' = \frac{R}{2} \]

### Final Answer
If the length and radius of the wire are both doubled, the new resistance will be \( \frac{R}{2} \).

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