A Copper (Cu) rod of length 25 cm and cross-sectional area 3 mm² is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B.
(Take Resistivity of Copper = 1.7 × 10⁻⁸ Ωm and Resistivity of Aluminium = 2.6 × 10⁻⁸ Ωm)

Solution

Physics sir · 4 steps

The rods share both ends A and B, so they are in parallel: find each resistance, then combine.

1. Read the figure. Both rods start at A and end at B, so they carry the current side by side. That is a parallel combination, not series.
2. Each rod alone. Use R = (ρ L)/A with L = 0.25 m and A = 3 mm² = 3 × 10⁻⁶ m². Copper gives R_(Cu) = (1.7 × 10⁻⁸ × 0.25)/(3 × 10⁻⁶) = 1.417 mΩ. Aluminium gives R_(Al) = (2.6 × 10⁻⁸ × 0.25)/(3 × 10⁻⁶) = 2.167 mΩ.

   R_(AB) = (R_(Cu) R_(Al))/(R_(Cu) + R_(Al)) = (1.417 × 2.167)/(1.417 + 2.167) mΩ

3. Put in numbers. The top is 3.071 and the bottom is 3.584, so R_(AB) ≈ 0.857 mΩ. The option rounds this to 0.858 mΩ.
4. So. The resistance between A and B is about 0.858 mΩ, option (c).

The idea to keep. First decide series or parallel from where the ends meet. In parallel the combined resistance is always smaller than the smallest single resistance, so use that to check your answer.

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