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The external diameter of a 5m long hollow tube is 10 cm and the thickness of it is 5 mm. The specific resistance of copper is $1 {.7×10}^{- 5} Ω m$ , then its resistance is

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5 {.7×10}^{- 5} Ω m

6 {.7×10}^{- 5} Ω m

0.057 Ω

0 {.67×10}^{- 5} Ω m

answer is D.

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Detailed Solution

R = \frac{ρ l}{π (R_{2}^{2} - R_{1}^{2})} = \frac{1.7 \times 10^{- 5} \times 5}{3.14 [{(0.05)}^{2} - {(0.045)}^{2}]} = 0.057 Ω

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