The resistance of a rectangular block of copper of dimension ` 2 mm xx 2 mm xx 5 m` between two square faces is `0.02 Omega` . What ist he resistivity of copper ?

Options

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

Answer

To find the resistivity of copper from the given dimensions and resistance of the rectangular block, we can follow these steps:

### Step 1: Identify the given values
- Dimensions of the rectangular block:
- Length (L) = 5 m
- Width (w) = 2 mm = 2 × 10^-3 m
- Height (h) = 2 mm = 2 × 10^-3 m
- Resistance (R) = 0.02 Ω = 2 × 10^-2 Ω

### Step 2: Calculate the cross-sectional area (A)
The cross-sectional area (A) of the rectangular block can be calculated using the formula:
\[ A = w \times h \]
Substituting the values:
\[ A = (2 \times 10^{-3} \, \text{m}) \times (2 \times 10^{-3} \, \text{m}) \]
\[ A = 4 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Use the formula for resistance
The resistance (R) of a conductor is given by the formula:
\[ R = \frac{\rho L}{A} \]
Where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity,
- \( L \) is the length,
- \( A \) is the cross-sectional area.

### Step 4: Rearrange the formula to solve for resistivity (ρ)
Rearranging the formula gives:
\[ \rho = \frac{R \times A}{L} \]

### Step 5: Substitute the known values into the formula
Substituting the values we have:
\[ \rho = \frac{(2 \times 10^{-2} \, \Omega) \times (4 \times 10^{-6} \, \text{m}^2)}{5 \, \text{m}} \]

### Step 6: Calculate the resistivity
Calculating the numerator:
\[ 2 \times 10^{-2} \times 4 \times 10^{-6} = 8 \times 10^{-8} \]
Now divide by the length:
\[ \rho = \frac{8 \times 10^{-8}}{5} = 1.6 \times 10^{-8} \, \Omega \cdot \text{m} \]

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

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