A rectabgykar carbon block has dimensions `1.0 cmxx 1.0 xx50 cm`.
(i) What is the resistance measured between the two square ends ?
(ii) Between two opposing rectangular faces ?
Resistivity of carbon at `20^(@)C" is "3.5xx10^(-5)Omegam`.

Answer

To solve the problem, we will calculate the resistance of a rectangular carbon block in two scenarios: (i) between the two square ends and (ii) between the two opposing rectangular faces.

### Given Data:
- Dimensions of the carbon block:
- Width (w) = 1.0 cm = 0.01 m
- Height (h) = 1.0 cm = 0.01 m
- Length (l) = 50.0 cm = 0.50 m
- Resistivity of carbon (ρ) = \(3.5 \times 10^{-5} \, \Omega \cdot m\)

### (i) Resistance between the two square ends:
1. **Identify the length and area**:
- Length (l) = 50 cm = 0.50 m
- Area (A) = width × height = \(1.0 \, cm \times 1.0 \, cm = 0.01 \, m \times 0.01 \, m = 0.0001 \, m^2\)

2. **Use the resistance formula**:
\[ R = \frac{\rho \cdot l}{A} \]

3. **Substitute the values**:
\[ R = \frac{3.5 \times 10^{-5} \, \Omega \cdot m \times 0.50 \, m}{0.0001 \, m^2} \]

4. **Calculate the resistance**:
\[ R = \frac{1.75 \times 10^{-5} \, \Omega \cdot m^2}{0.0001 \, m^2} = 0.175 \, \Omega \]

### (ii) Resistance between the two opposing rectangular faces:
1. **Identify the length and area**:
- Length (l) = 1 cm = 0.01 m (the distance between the two opposing rectangular faces)
- Area (A) = length × height = \(50.0 \, cm \times 1.0 \, cm = 0.50 \, m \times 0.01 \, m = 0.005 \, m^2\)

2. **Use the resistance formula**:
\[ R = \frac{\rho \cdot l}{A} \]

3. **Substitute the values**:
\[ R = \frac{3.5 \times 10^{-5} \, \Omega \cdot m \times 0.01 \, m}{0.005 \, m^2} \]

4. **Calculate the resistance**:
\[ R = \frac{3.5 \times 10^{-7} \, \Omega \cdot m^2}{0.005 \, m^2} = 7 \times 10^{-5} \, \Omega \]

### Summary of Results:
- (i) Resistance between the two square ends: \(0.175 \, \Omega\)
- (ii) Resistance between the two opposing rectangular faces: \(7 \times 10^{-5} \, \Omega\)