A wire of resistance 5 Ω is drawn out so that its length becomes three times its original length. The new resistance would be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 45 Ω
Drawing a wire does not change its volume. If the length becomes n times, the area of cross-section becomes \( 1/n \) times, and the resistance \( R = \rho l/A \) becomes \( n^{2} \) times.
Here the length becomes three times the original, \( l' = 3l \).
\( Al = A'l' \Rightarrow A' = \frac{A}{3} \)
\( \displaystyle R' = \frac{\rho l'}{A'} = \frac{\rho\,(3l)}{A/3} = 9\,\frac{\rho l}{A} = 9R \)
\( R' = 9\times5\,\Omega = 45\,\Omega \)
Note: The same question was also asked in CUET (UG) 2023 on 13-06-2023 (Slot 2), and a similar question in CUET (UG) 2026 on 14-05-2026 (Slot 2).