Find the resistance of a hollow cylindrical conductor of length 1.0 mm and 2.0 mm respectively. The resistivity of the material is 2.0 × 10^-8 Ωm

2.1 × 10^-3 Ω

1.3 × 10^-4 Ω

3.2 × 10^-4 Ω

4.6 × 10^-2 Ω

Solution

2.1 × 10^-3 Ω

The area of cross-section of the conductor through which charges will be flow is

A = $π {(2.0 mm)}^{2} - π {(1.0 mm)}^{2}$

A = 3 $\times π {mm}^{2}$

The resistance of the wire is therefore

R = $ρ \frac{I}{A}$

= $\frac{(2.0 \times 10^{- 8}) \times (1.0)}{3.0 \times π \times 10^{- 6}}$

R = 2.1 × 10^-3 Ω

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