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 3 Ω? (Given resistivities of iron and copper-nickel alloy wire are 12 μΩ cm and 51 μΩ cm respectively) (a) 82 m (b) 97 m (c) 110 m (d) 90 m

Answer: (b). Same l and A, in parallel: Req = l/A·(ρ₁ρ₂)/(ρ₁+ρ₂). (ρ₁ρ₂)/(ρ₁+ρ₂) = (12×51)/(63) = 9.71 μΩ cm = 9.71×10⁻⁸ Ω m A = π(10⁻³)² = 3.14×10⁻⁶ m² l = 3×3.14×10⁻⁶9.71×10⁻⁸ ≈ 97 m. The answer is (b).

What equal length of an iron wire and a copper-nickel alloy wire… | JEE Main 2021 (26 Aug, Shift 1)

Q36JEE Main 2021 · 26 AUG · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature

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 3 $\Omega$ ?

(Given resistivities of iron and copper-nickel alloy wire are 12 $\mu\Omega$ cm and 51 $\mu\Omega$ cm respectively)

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Answer: (b)

Same $l$ and $A$ , in parallel: $R_{eq} = \dfrac{l}{A}\cdot\dfrac{\rho_1\rho_2}{\rho_1+\rho_2}$ .
$\dfrac{\rho_1\rho_2}{\rho_1+\rho_2} = \dfrac{12\times51}{63} = 9.71\ \mu\Omega\,\text{cm} = 9.71\times10^{-8}\ \Omega\,\text{m}$
$A = \pi(10^{-3})^2 = 3.14\times10^{-6}\ \text{m}^2$
$l = \dfrac{3\times3.14\times10^{-6}}{9.71\times10^{-8}} \approx 97$ m. The answer is (b).

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