Which of the following wires, all made of the same material, has the highest resistance?

Options

A. l= 100 cm, r = 0.1 cm
B. l = 200 cm, r = 0.2 cm
C. l= 300 cm, r = 0.3 cm
D. l = 400 cm, r = 0.4 cm

Explanation

To determine which of the wires has the highest resistance, we need to use the formula for resistance:

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

Where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material (which is the same for all wires),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

Since all wires are made of the same material, the resistivity \( \rho \) is constant for all options. Therefore, the resistance depends on the ratio \( \frac{L}{A} \). The wire with the highest value of \( \frac{L}{A} \) will have the highest resistance.

### Step-by-Step Solution:

1. **Identify Length and Radius for Each Wire**:
Let's assume we have four wires with the following lengths and radii:
- Wire 1: Length \( L_1 = 100 \) cm, Radius \( r_1 = 0.1 \) cm
- Wire 2: Length \( L_2 = 200 \) cm, Radius \( r_2 = 0.2 \) cm
- Wire 3: Length \( L_3 = 300 \) cm, Radius \( r_3 = 0.3 \) cm
- Wire 4: Length \( L_4 = 400 \) cm, Radius \( r_4 = 0.4 \) cm

2. **Calculate Cross-Sectional Area for Each Wire**:
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]
- For Wire 1: \( A_1 = \pi (0.1)^2 = \pi (0.01) = 0.01\pi \) cm²
- For Wire 2: \( A_2 = \pi (0.2)^2 = \pi (0.04) = 0.04\pi \) cm²
- For Wire 3: \( A_3 = \pi (0.3)^2 = \pi (0.09) = 0.09\pi \) cm²
- For Wire 4: \( A_4 = \pi (0.4)^2 = \pi (0.16) = 0.16\pi \) cm²

3. **Calculate the Ratio \( \frac{L}{A} \) for Each Wire**:
- For Wire 1:
\[ \frac{L_1}{A_1} = \frac{100}{0.01\pi} = \frac{10000}{\pi} \]
- For Wire 2:
\[ \frac{L_2}{A_2} = \frac{200}{0.04\pi} = \frac{5000}{\pi} \]
- For Wire 3:
\[ \frac{L_3}{A_3} = \frac{300}{0.09\pi} = \frac{3333.33}{\pi} \]
- For Wire 4:
\[ \frac{L_4}{A_4} = \frac{400}{0.16\pi} = \frac{2500}{\pi} \]

4. **Compare the Ratios**:
- \( \frac{L_1}{A_1} = \frac{10000}{\pi} \)
- \( \frac{L_2}{A_2} = \frac{5000}{\pi} \)
- \( \frac{L_3}{A_3} = \frac{3333.33}{\pi} \)
- \( \frac{L_4}{A_4} = \frac{2500}{\pi} \)

From these calculations, we can see that:
- Wire 1 has the highest ratio \( \frac{L}{A} \), followed by Wire 2, Wire 3, and Wire 4.

5. **Conclusion**:
Therefore, the wire with the highest resistance is **Wire 1**.