A wire is stretched to increase its length by 4 times. The resistance of the wire will:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → increase by 16 times
The resistance of a wire is \( R = \rho l/A \). When the wire is stretched, its volume stays constant, so \( Al = A'l' \): a longer wire is also thinner.
\( l' = 4l \Rightarrow A' = \displaystyle\frac{A}{4} \)
\( \displaystyle R' = \frac{\rho l'}{A'} = \frac{\rho\,(4l)}{A/4} = 16\,\frac{\rho l}{A} = 16R \)
In general, stretching to n times the length makes the resistance \( n^{2} \) times, so the resistance becomes 16 times the original value.