The resistance of a wire is \( 10\,\Omega \). Its length is increased by 10% by stretching. The new resistance of the wire will now be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 12.1\,\Omega \)
When a wire is stretched, its volume stays the same, so a longer wire is also thinner. The resistance is \( R = \rho L/A \), and with \( A = V/L \) this becomes
\( \displaystyle R = \frac{\rho L}{A} = \frac{\rho L^{2}}{V} \;\Rightarrow\; R \propto L^{2} \)
The new length is \( L' = 1.1\,L \).
\( \displaystyle \frac{R'}{R} = \left(\frac{L'}{L}\right)^{2} = (1.1)^{2} = 1.21 \)
\( \displaystyle R' = 1.21 \times 10\,\Omega = 12.1\,\Omega \)