Two wires with resistances \( R_1 \) and \( R_2 \) are stretched \( n \) times their original length. After stretching, the two wires are connected in parallel. The equivalent resistance of the combination will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \frac{n^{2} R_1 R_2}{R_1+R_2} \)
When a wire is stretched, its volume stays constant, so the area of cross-section falls as the length rises: \( A l = A' l' \).
\( \displaystyle l' = n l \;\Rightarrow\; A' = \frac{A}{n} \)
\( \displaystyle R' = \frac{\rho\,l'}{A'} = \frac{\rho\,(n l)}{A/n} = n^{2}\,\frac{\rho l}{A} = n^{2} R \)
So the stretched wires have resistances \( n^{2}R_1 \) and \( n^{2}R_2 \). In parallel:
\( \displaystyle R_{\mathrm{eq}} = \frac{(n^{2}R_1)(n^{2}R_2)}{n^{2}R_1 + n^{2}R_2} = \frac{n^{2}R_1 R_2}{R_1 + R_2} \)