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 is $1.7×10^{-8}Ω m$, then the resistance of the tube will be:

$5.7 × 10^{-5}Ω$

The correct answer is Option (3) → $5.7 × 10^{-5}Ω$

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

External diameter: $D_\text{ext} = 10\ \text{cm} = 0.1\ \text{m}$

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

Internal diameter: $D_\text{int} = D_\text{ext} - 2t = 0.1 - 2 \cdot 0.005 = 0.09\ \text{m}$

Cross-sectional area of the hollow tube:

$A = \frac{\pi}{4} (D_\text{ext}^2 - D_\text{int}^2) = \frac{\pi}{4} (0.1^2 - 0.09^2) = \frac{\pi}{4} (0.01 - 0.0081) = \frac{\pi}{4} \cdot 0.0019 \approx 1.492 \times 10^{-3}\ \text{m}^2$

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

Resistance of the tube: $R = \rho \frac{L}{A} = 1.7 \times 10^{-8} \cdot \frac{5}{1.492 \times 10^{-3}} \approx 5.7 \times 10^{-5}\ \Omega$

Final Answer: $R \approx 5.7 \times 10^{-5}\ \Omega$