A cylindrical wire is stretched to increase its length by 10%. The percentage increase in resistance will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 21%
When a wire is stretched its volume stays constant, so the area of cross-section falls as the length rises, \( A = V/L \).
\( \displaystyle R = \frac{\rho L}{A} = \frac{\rho L^{2}}{V} \;\Rightarrow\; R \propto L^{2} \)
\( \displaystyle L' = 1.1\,L \)
\( \displaystyle \frac{R'}{R} = \left(\frac{L'}{L}\right)^{2} = (1.1)^{2} = 1.21 \)
\( \displaystyle \text{Percentage increase} = (1.21 - 1) \times 100\% = 21\% \)