NECO 2003 Physics · Question 6
A wire of length 100cm and cross-sectional area 2.0 × 10-3cm2 has a resistance of 0.10 ω. Calculate its electrical conductivity.
Options
- A.2.0 × 10-6 ω-1
- B.5.0 × 105 ω-1
- C.2.0 × 106 ω-1
- D.5.0 × 10-6 ω-1
- E.5.0 × 107 ω-1
Answer: B. 5.0 × 105 ω-1
Explanation written with AI and checked against the answer key
Analyze the question: We need the electrical conductivity (σ) of a wire, given its length, cross-sectional area and resistance.
Concept: Resistance R = ρL/A, where ρ is resistivity. Conductivity σ = 1/ρ, so σ = L/(RA).
Solution:
- L = 100 cm
- A = 2.0 × 10⁻³ cm²
- R = 0.10 Ω
ρ = RA/L = (0.10 × 2.0 × 10⁻³)/100 = 2.0 × 10⁻⁶ Ω cm
σ = 1/ρ = 1/(2.0 × 10⁻⁶) = 5.0 × 10⁵ Ω⁻¹ cm⁻¹
Answer: Option B, 5.0 × 10⁵ Ω⁻¹cm⁻¹.
Tip: Be careful not to stop at resistivity. The value 2.0 × 10⁻⁶ is the resistivity (this is the trap in option A). Conductivity is its reciprocal. Also, keep the units consistent: since the options are in cm, work in cm throughout.
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