A uniform wire of resistance \( 8\,\Omega \) is folded at its midpoint and the two halves are twisted together. The resistance of the twisted wire is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\,\Omega \)
Folding the wire at its midpoint gives two equal halves, each with half the length and so half the resistance of the whole wire. Twisted together, the two halves have their ends common, so they are in parallel.
The resistance of each half is:
\( \begin{aligned} R_1 &= \frac{8\,\Omega}{2} \\ &= 4\,\Omega \end{aligned} \)
For the two halves in parallel:
\( \begin{aligned} R' &= \frac{R_1R_1}{R_1+R_1} \\ &= \frac{4\,\Omega\times4\,\Omega}{4\,\Omega+4\,\Omega} \\ &= 2\,\Omega \end{aligned} \)
The same value follows from \( R=\rho l/A \). The twisted wire has half the length and twice the area of cross-section, so its resistance is one-fourth of \( 8\,\Omega \).