A copper rod of length 20 cm and cross-sectional area 2 mm is joined with a similar aluminium rod as shown below

The resistance of pair of rods is (ρ_Al =2.6 x 10^-8 Ω-m and ρ_Cu =1.7 x 10^-8 Ω-m)

1.0 mΩ

2.0 mΩ

3.0 mΩ

None of these

Solution

1.0 mΩ

Given

Cross-sectional area a = 2 mm²

a = 2 × 10^-6 m²

Length of rod l = 20 cm

= 20 × 10^-2 m

The resistance of the Copper rod

R = $ρ \frac{l}{a}$

R₁ = $\frac{1.7 \times 10^{- 8} \times 20 \times 10^{- 2}}{20 \times 10^{- 6}}$

R₁ = 1.7 × 10^-3 Ω

The net resistance of the Aluminium rod

R₂ = $\frac{2.6 \times 10^{- 8} \times 20 \times 10^{- 2}}{2.0 \times 10^{- 6}}$

R₂ = 2.6 × 10^-3 Ω

The equivalent resistance

$\frac{1}{R} = \frac{1}{1.7 \times 10^{- 3}} + \frac{1}{2.6 \times 10^{- 3}}$

⇒ R = $\frac{1.7 \times 2.6}{4.42}$ × 10^-3

⇒ R = 1.0 × 10^-3Ω

⇒ R = 1.0 mΩ

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