A cylindrical wire of radius \(r\), having resistance \(R\) is stretched to \(n\) times its previous length \(l\). The new resistance of the wire will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(n^{2}R\)
When a wire is stretched, its volume stays the same, so the area of cross-section falls in the same ratio as the length rises.
\( \displaystyle R = \frac{\rho l}{A} \)
\( \displaystyle l' = nl, \quad A' = \frac{A}{n} \)
\( \displaystyle R' = \frac{\rho (nl)}{A/n} = n^{2}\frac{\rho l}{A} \)
\( \displaystyle R' = n^{2}R \)