A wire made of aluminum having resistivity `rho = 2.8 xx10^(-8) Omega - m ` with a circular cross - section and has a radius of `2xx10^(-3)` m. A current of 5 A flows through the wire . If the voltage difference between the ends is 1 V , the length of the wire in m is

Options

A. 50
B. 60
C. 90
D. 120

Answer

To find the length of the aluminum wire, we will use the relationship between resistance, resistivity, length, and cross-sectional area, as well as Ohm's law.

### Step-by-Step Solution:

1. **Identify the given values:**
- Resistivity of aluminum, \( \rho = 2.8 \times 10^{-8} \, \Omega \cdot m \)
- Radius of the wire, \( r = 2 \times 10^{-3} \, m \)
- Current flowing through the wire, \( I = 5 \, A \)
- Voltage difference across the wire, \( V = 1 \, V \)

2. **Calculate the resistance using Ohm's law:**
\[ V = I \cdot R \implies R = \frac{V}{I} \]
Substituting the values:
\[ R = \frac{1 \, V}{5 \, A} = 0.2 \, \Omega \]

3. **Calculate the cross-sectional area \( A \) of the wire:**
The area \( A \) of a circular cross-section is given by:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (2 \times 10^{-3})^2 = \pi (4 \times 10^{-6}) \approx 1.25664 \times 10^{-5} \, m^2 \]

4. **Use the formula for resistance:**
The resistance \( R \) can also be expressed as:
\[ R = \frac{\rho L}{A} \]
Rearranging for length \( L \):
\[ L = \frac{R \cdot A}{\rho} \]

5. **Substituting the known values into the length formula:**
\[ L = \frac{0.2 \, \Omega \cdot (1.25664 \times 10^{-5} \, m^2)}{2.8 \times 10^{-8} \, \Omega \cdot m} \]

6. **Calculating the length:**
\[ L = \frac{0.2 \cdot 1.25664 \times 10^{-5}}{2.8 \times 10^{-8}} = \frac{2.51328 \times 10^{-6}}{2.8 \times 10^{-8}} \approx 89.05 \, m \]

7. **Final result:**
Rounding to two decimal places, the length of the wire is approximately:
\[ L \approx 90 \, m \]

### Final Answer:
The length of the wire is approximately **90 meters**.