Resistance of the conductor of length 5m and area of cross -section `4 mm^(2)` s `0.02 Omega `. Its resistivity is

Options

A. ` 1.6 xx 10^(-6) Omega ` m
B. `1.6 xx 10^(-5) Omega m `
C. ` 1.6 xx 10^(-8 ) Omega m `
D. `1 0^(-8) Omega m`

Answer

To find the resistivity of the conductor, we can use the formula for resistance:

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

Where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity,
- \( L \) is the length of the conductor,
- \( A \) is the area of cross-section.

### Step 1: Identify the given values
From the question, we have:
- Length \( L = 5 \, \text{m} \)
- Area \( A = 4 \, \text{mm}^2 \)
- Resistance \( R = 0.02 \, \Omega \)

### Step 2: Convert the area to square meters
Since the area is given in square millimeters, we need to convert it to square meters:
\[ A = 4 \, \text{mm}^2 = 4 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Rearrange the formula to solve for resistivity \( \rho \)
We can rearrange the formula to find resistivity:
\[ \rho = R \frac{A}{L} \]

### Step 4: Substitute the values into the formula
Now we can substitute the known values into the rearranged formula:
\[ \rho = 0.02 \, \Omega \times \frac{4 \times 10^{-6} \, \text{m}^2}{5 \, \text{m}} \]

### Step 5: Calculate the resistivity
Now, calculate the resistivity:
\[ \rho = 0.02 \times \frac{4 \times 10^{-6}}{5} \]
\[ = 0.02 \times 0.8 \times 10^{-6} \]
\[ = 1.6 \times 10^{-8} \, \Omega \cdot \text{m} \]

### Final Answer
Thus, the resistivity of the conductor is:
\[ \rho = 1.6 \times 10^{-8} \, \Omega \cdot \text{m} \]

---