A certain wire has a resistance R. What is the resistance of a second wire, made of the same material, that is twice as long and has twice the diameter?

Answer

To find the resistance of the second wire, we start with the known properties of the first wire and apply the relevant formulas for resistance.

### Step-by-Step Solution:

1. **Identify the 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, and \( A \) is the cross-sectional area.

2. **Determine the area of the first wire**:
The area \( A \) of a wire with diameter \( D \) is:
\[ A = \frac{\pi D^2}{4} \]
Therefore, the resistance of the first wire can be expressed as:
\[ R = \frac{\rho L}{\frac{\pi D^2}{4}} = \frac{4 \rho L}{\pi D^2} \]

3. **Define the properties of the second wire**:
The second wire is twice the length and has twice the diameter of the first wire. Thus:
- Length of the second wire, \( L' = 2L \)
- Diameter of the second wire, \( D' = 2D \)

4. **Calculate the area of the second wire**:
The cross-sectional area \( A' \) of the second wire is:
\[ A' = \frac{\pi (D')^2}{4} = \frac{\pi (2D)^2}{4} = \frac{\pi \cdot 4D^2}{4} = \pi D^2 \]

5. **Calculate the resistance of the second wire**:
Using the resistance formula for the second wire:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{\pi D^2} = \frac{2\rho L}{\pi D^2} \]

6. **Relate the resistance of the second wire to the first wire**:
From the first wire, we have:
\[ R = \frac{4\rho L}{\pi D^2} \]
Therefore, we can express \( R' \) in terms of \( R \):
\[ R' = \frac{2\rho L}{\pi D^2} = \frac{2}{4} R = \frac{R}{2} \]

### Final Answer:
The resistance of the second wire is:
\[ R' = \frac{R}{2} \]