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 \( 1.7 \times 10^{-8}\,\Omega\text{-}\mathrm{m} \), then its resistance would be approximately
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 6.0 \times 10^{-5}\,\Omega \)
The resistance of a conductor is \( R = \rho l/A \), where \( \rho \) is the specific resistance (resistivity). Current flows only through the copper wall, so \( A \) is the area of the ring-shaped cross-section between the outer and inner radii.
\( r_2 = 10\,\mathrm{cm}/2 = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)
\( t = 5\,\mathrm{mm} = 0.5\,\mathrm{cm} \)
\( r_1 = 5 - 0.5 = 4.5\,\mathrm{cm} = 0.045\,\mathrm{m} \)
\( \displaystyle A = \pi\left(r_2^{2} - r_1^{2}\right) = \pi\left(0.05^{2} - 0.045^{2}\right) = \pi \times 4.75\times10^{-4} = 1.49\times10^{-3}\,\mathrm{m^{2}} \)
\( \displaystyle R = \frac{\rho\,l}{A} = \frac{1.7\times10^{-8} \times 5}{1.49\times10^{-3}} = 5.7\times10^{-5}\,\Omega \)
This is approximately \( 6.0\times10^{-5}\,\Omega \), the nearest option.
Note: A similar question was also asked in CUET (UG) 2025 on 02-06-2025 (Slot 2).