What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3Ω ?
(Given resistivities of iron and copper-nickel alloy wire are 12 μΩ and 51 μΩ cm respectively)

Solution

Physics sir · 4 steps

Same length, same thickness, so only the resistivities differ. Combine them first, then solve for L once.

1. Convert the units. 1 μΩ cm = 10⁻⁶ Ω × 10⁻² m = 10⁻⁸ Ω m. So ρ_(Fe) = 12 × 10⁻⁸ Ω m and ρ_(alloy) = 51 × 10⁻⁸ Ω m. The radius is 1 mm, so A = π r² = π × 10⁻⁶ m².
2. Parallel combination. Each wire has R = ρ L / A. Both wires share the same L/A, so in R₁ R₂ / (R₁ + R₂) the factor L/A comes out once.

   R_(eq) = L/A·(ρ₁ ρ₂)/(ρ₁ + ρ₂) = L/A·(12 × 51)/63 × 10⁻⁸ = L/A × 9.71 × 10⁻⁸ Ω m

3. Solve for L. L = (3 × π × 10⁻⁶)/(9.71 × 10⁻⁸) = (9.42 × 10⁻⁶)/(9.71 × 10⁻⁸) ≈ 97 m. As a check, at 97 m the iron wire has 3.71 Ω and the alloy wire has 15.7 Ω. In parallel these give 3.0 Ω.
4. So. Each wire must be about 97 m long, which is option (b).

The idea to keep. When parallel wires share the same length and area, combine the resistivities like resistors, ρ₁ρ₂/(ρ₁+ρ₂), and then use R = ρ L/A once. Always convert μΩ cm to Ω m first.

Answer from the NTA final answer key. Solution by PrepWiser.