Question

The resistance of 2 km of a subway track with cross-sectional area 20 $cm^2$ is (in ohm) as a multiple of the specific resistance of steel $\rho$: ________.

• A. $10^{2} \rho$
• B. $10^{3} \rho$
• C. $10^{4} \rho$
• D. $10^{5} \rho$
• E. $10^{6} \rho$

Hint:

Always convert all values to SI units (meters and $m^2$) before calculating.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
The resistance of a conductor is related to its material properties (specific resistance or resistivity, \(\rho\)), its length (L), and its cross-sectional area (A). We need to use this relationship and ensure all units are consistent (SI units).
Step 2: Key Formula or Approach
The formula for resistance (R) is:
\[ R = \rho \frac{L}{A} \] We are asked to find R as a multiple of \(\rho\), which means we need to calculate the value of the geometric factor \(\frac{L}{A}\).
Step 3: Detailed Explanation
1. List the given values and convert to SI units.
- Specific resistance = \(\rho\) (units of \(\Omega \cdot m\))
- Length, \(L = 2 \text{ km} = 2 \times 10^3 \text{ m}\).
- Cross-sectional area, \(A = 20 \text{ cm}^2\). We need to convert this to m\(^2\).
Since \(1 \text{ m} = 100 \text{ cm}\), then \(1 \text{ m}^2 = (100 \text{ cm})^2 = 10000 \text{ cm}^2 = 10^4 \text{ cm}^2\).
Therefore, \(1 \text{ cm}^2 = 10^{-4} \text{ m}^2\).
\(A = 20 \times 10^{-4} \text{ m}^2 = 2 \times 10^1 \times 10^{-4} \text{ m}^2 = 2 \times 10^{-3} \text{ m}^2\).
2. Calculate the geometric factor \(\frac{L}{A}\).
\[ \frac{L}{A} = \frac{2 \times 10^3 \text{ m}}{2 \times 10^{-3} \text{ m}^2} \] \[ \frac{L}{A} = \frac{2}{2} \times \frac{10^3}{10^{-3}} = 1 \times 10^{3 - (-3)} = 10^6 \text{ m}^{-1} \] Wait, let's recheck the area conversion. It's the most likely source of error. Area = 20 cm\(^2\). \(1 \text{cm} = 10^{-2} \text{m}\). \(1 \text{cm}^2 = (10^{-2} \text{m})^2 = 10^{-4} \text{m}^2\). \(A = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{m}^2\). This is correct. Let's recheck the division. \(\frac{L}{A} = \frac{2 \times 10^3}{20 \times 10^{-4}} = \frac{2}{20} \times 10^{3 - (-4)} = \frac{1}{10} \times 10^7 = 10^{-1} \times 10^7 = 10^6\). This calculation seems correct. This means \(R = 10^6 \rho\). This matches option (E). Okay, my initial thought was wrong. Let's proceed. Final Calculation:
1. Convert units to SI: - \(L = 2 \text{ km} = 2000 \text{ m} = 2 \times 10^3 \text{ m}\) - \(A = 20 \text{ cm}^2 = 20 \times (10^{-2} \text{ m})^2 = 20 \times 10^{-4} \text{ m}^2 = 2 \times 10^{-3} \text{ m}^2\) 2. Apply the resistance formula: \[ R = \rho \frac{L}{A} \] \[ R = \rho \frac{2 \times 10^3}{2 \times 10^{-3}} \] \[ R = \rho \times 10^{3 - (-3)} = \rho \times 10^6 \] This result matches option (E). Step 4: Final Answer
The resistance of the track is \(10^6 \rho\).