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 \( 1.7 \times 10^{-8}\,\Omega\,\mathrm{m} \), then the resistance of the tube will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 5.7 \times 10^{-5}\,\Omega \)
The current flows along the length of the tube, so \( R = \rho L / A \) with \( A \) the area of the ring-shaped cross-section of the wall.
\( R_{o} = 10/2 = 5\,\mathrm{cm} \)
\( R_{i} = 5 - 0.5 = 4.5\,\mathrm{cm} \)
\( \displaystyle A = \pi\,(R_{o}^{2} - R_{i}^{2}) = \pi\,(25 - 20.25)\times10^{-4}\,\mathrm{m^2} = 1.49\times10^{-3}\,\mathrm{m^2} \)
\( \displaystyle R = \frac{\rho L}{A} = \frac{1.7\times10^{-8} \times 5}{1.49\times10^{-3}} = 5.7\times10^{-5}\,\Omega \)
Note: A similar question was also asked in CUET (UG) 2025 on 23-05-2025 (Slot 2).