A wire of resistance \( 5\,\Omega \) is drawn out so that its length is increased by thrice its original length. The new resistance would be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 45\,\Omega \)
The resistance of a wire is \( R = \rho l/A \). When the wire is drawn out its volume stays constant, so if the length becomes 3 times, the area becomes one-third.
\( l' = 3l,\qquad A' = \displaystyle\frac{A}{3} \)
\( \displaystyle R' = \rho\,\frac{3l}{A/3} = 9\,\rho\,\frac{l}{A} = 9R = 9\times5\,\Omega = 45\,\Omega \)
Shortcut: At constant volume \( R\propto l^{2} \). Length \( \times3 \) therefore gives resistance \( \times9 \), and \( 9\times5 = 45\,\Omega \) directly.
Note: The same question was also asked in CUET (UG) 2023 on 05-06-2023 (Slot 3), and a similar question in CUET (UG) 2026 on 14-05-2026 (Slot 2).