The external diameter of a 5 m long hollow copper tube is 10 cm and the thickness of its wall is 5 mm. If the specific resistance of copper be $1.7 × 10^{-8} Ω-m$, then its resistance would be approximately

$6.0 × 10^{-5}Ω$

The correct answer is Option (2) → $6.0 × 10^{-5}Ω$

Given:

Length of tube: $L = 5\ \text{m}$

External diameter: $D_{\text{ext}} = 10\ \text{cm} = 0.10\ \text{m}$

Wall thickness: $t = 5\ \text{mm} = 0.005\ \text{m}$

Resistivity of copper: $\rho = 1.7 \times 10^{-8}\ \Omega\text{-m}$

Internal diameter:

$D_{\text{int}} = D_{\text{ext}} - 2t = 0.10 - 2(0.005) = 0.09\ \text{m}$

Cross-sectional area of hollow tube:

$A = \frac{\pi}{4} (D_{\text{ext}}^2 - D_{\text{int}}^2) = \frac{\pi}{4} (0.10^2 - 0.09^2)$

$A = \frac{\pi}{4} (0.01 - 0.0081) = \frac{\pi}{4} (0.0019) \approx 0.001492\ \text{m}^2$

Resistance:

$R = \rho \frac{L}{A} = \frac{1.7 \times 10^{-8} \cdot 5}{0.001492} \approx 5.7 \times 10^{-5}\ \Omega$

Approximate resistance: $5.7 \times 10^{-5}\ \Omega$