The resistance of a metallic wire becomes 8 times its original value when its
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → length is doubled and radius is halved
The resistance of a wire is \( R=\rho l/A \), with the cross-sectional area \( A=\pi r^{2} \). For the same metal at the same temperature, \( \rho \) does not change. So \( R \) is proportional to the length and inversely proportional to the square of the radius.
Option (1) length is doubled — Incorrect. With the area unchanged, doubling the length only doubles the resistance to \( 2R \).
Option (2) length is tripled — Incorrect. With the area unchanged, the resistance becomes \( 3R \).
Option (3) length is doubled and radius is halved — Correct. Halving the radius makes the area one quarter, so the resistance becomes four times. Doubling the length doubles it again, so the new resistance is \( 2\times4R=8R \).
Option (4) length is halved and radius is doubled — Incorrect. Doubling the radius makes the area four times, and halving the length halves the resistance, so the resistance falls to \( R/8 \).
NCERT (2026-27):
Ohm’s Law – “… In general, then resistance is proportional to length,
\( R\propto l \qquad (3.5) \)”
Ohm’s Law – “… In general, then the resistance R is inversely proportional to the cross-sectional area,
\( \displaystyle R\propto\frac{1}{A} \qquad (3.7) \)”