The following four wires are made of the same material and are at the same temperature. Which one of them has highest electrical resistance

Options

A. Length = 50 cm, diameter = 0.5 mm
B. Length = 100 cm, diameter = 1 mm
C. Length = 200 cm, diameter = 2 mm
D. Length = 300 cm, diameter = 3 mm

Explanation

To determine which of the four wires has the highest electrical resistance, we can 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.

The cross-sectional area \( A \) can be calculated using the diameter \( D \) of the wire:

\[ A = \frac{\pi D^2}{4} \]

### Step-by-Step Solution:

1. **Identify the Length and Diameter of Each Wire:**
- Wire 1: Length = 50 cm, Diameter = 0.5 mm
- Wire 2: Length = 100 cm, Diameter = 1 mm
- Wire 3: Length = 200 cm, Diameter = 2 mm
- Wire 4: Length = 300 cm, Diameter = 3 mm

2. **Convert Measurements to Consistent Units:**
- Convert lengths to meters:
- Wire 1: \( L_1 = 0.50 \, m \)
- Wire 2: \( L_2 = 1.00 \, m \)
- Wire 3: \( L_3 = 2.00 \, m \)
- Wire 4: \( L_4 = 3.00 \, m \)
- Convert diameters to meters:
- Wire 1: \( D_1 = 0.5 \, mm = 0.0005 \, m \)
- Wire 2: \( D_2 = 1 \, mm = 0.001 \, m \)
- Wire 3: \( D_3 = 2 \, mm = 0.002 \, m \)
- Wire 4: \( D_4 = 3 \, mm = 0.003 \, m \)

3. **Calculate the Cross-Sectional Area for Each Wire:**
\[ A = \frac{\pi D^2}{4} \]
- Wire 1:
\[ A_1 = \frac{\pi (0.0005)^2}{4} = \frac{\pi (0.00000025)}{4} \approx 1.9635 \times 10^{-7} \, m^2 \]
- Wire 2:
\[ A_2 = \frac{\pi (0.001)^2}{4} = \frac{\pi (0.000001)}{4} \approx 7.854 \times 10^{-7} \, m^2 \]
- Wire 3:
\[ A_3 = \frac{\pi (0.002)^2}{4} = \frac{\pi (0.000004)}{4} \approx 3.142 \times 10^{-6} \, m^2 \]
- Wire 4:
\[ A_4 = \frac{\pi (0.003)^2}{4} = \frac{\pi (0.000009)}{4} \approx 7.0686 \times 10^{-6} \, m^2 \]

4. **Calculate the Resistance for Each Wire:**
Using the formula \( R = \frac{\rho L}{A} \) and noting that \( \rho \) is constant for all wires, we can compare the resistances directly:
- Wire 1:
\[ R_1 = \frac{\rho (0.50)}{1.9635 \times 10^{-7}} \approx \frac{0.50 \rho}{1.9635 \times 10^{-7}} \]
- Wire 2:
\[ R_2 = \frac{\rho (1.00)}{7.854 \times 10^{-7}} \approx \frac{1.00 \rho}{7.854 \times 10^{-7}} \]
- Wire 3:
\[ R_3 = \frac{\rho (2.00)}{3.142 \times 10^{-6}} \approx \frac{2.00 \rho}{3.142 \times 10^{-6}} \]
- Wire 4:
\[ R_4 = \frac{\rho (3.00)}{7.0686 \times 10^{-6}} \approx \frac{3.00 \rho}{7.0686 \times 10^{-6}} \]

5. **Compare the Resistances:**
- Since \( R \) is directly proportional to \( L \) and inversely proportional to \( A \), we can see that:
- Wire 1 has the smallest diameter and thus the largest resistance.
- Wire 4 has the largest diameter and thus the smallest resistance.

### Conclusion:
Wire 1 has the highest electrical resistance.