A wire 100cm long and 2.0mm diameter has a resistance of 0.7 ohm , the electrical resistivity of the material is

Options

A. `4.4 xx 10^(-6) ohm xx m`
B. `2.2 xx 10^(-6) ohm xx m`
C. `1.1 xx 10^(-6) ohm xx m`
D. `0.22 xx 10^(-6) ohm xx m`

Answer

To find the electrical resistivity of the material of the wire, we can use the formula for resistance:

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

Where:
- \( R \) is the resistance (in ohms),
- \( \rho \) is the resistivity (in ohm-meters),
- \( L \) is the length of the wire (in meters),
- \( A \) is the cross-sectional area of the wire (in square meters).

### Step 1: Convert Length and Diameter to Appropriate Units
The length of the wire is given as 100 cm, which we need to convert to meters:
\[ L = 100 \, \text{cm} = 1 \, \text{m} \]

The diameter of the wire is given as 2.0 mm, which we also need to convert to meters:
\[ \text{Diameter} = 2.0 \, \text{mm} = 2.0 \times 10^{-3} \, \text{m} \]

### Step 2: Calculate the Radius
The radius \( r \) is half of the diameter:
\[ r = \frac{\text{Diameter}}{2} = \frac{2.0 \times 10^{-3}}{2} = 1.0 \times 10^{-3} \, \text{m} \]

### Step 3: Calculate the Cross-Sectional Area
The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (1.0 \times 10^{-3})^2 = \pi (1.0 \times 10^{-6}) \]
Using \( \pi \approx 3.14 \):
\[ A \approx 3.14 \times 10^{-6} \, \text{m}^2 \]

### Step 4: Rearranging the Resistance Formula
We need to find the resistivity \( \rho \). Rearranging the formula gives us:
\[ \rho = \frac{R \cdot A}{L} \]

### Step 5: Substitute the Known Values
Now we can substitute the known values into the equation. The resistance \( R \) is given as 0.7 ohms:
\[ \rho = \frac{0.7 \cdot (3.14 \times 10^{-6})}{1} \]

### Step 6: Calculate the Resistivity
Calculating the above expression:
\[ \rho \approx 0.7 \times 3.14 \times 10^{-6} \]
\[ \rho \approx 2.198 \times 10^{-6} \, \text{ohm-meters} \]

### Final Answer
Thus, the electrical resistivity of the material is approximately:
\[ \rho \approx 2.2 \times 10^{-6} \, \text{ohm-meters} \]

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