If a wire of resistance 2Ω gets stretched to thrice its original length :
Calculate the new resistance of the wire.
Topic: Heat and calorimetry
Answer
Given: original resistance R = 2 Ω; the wire is stretched to three times its original length.
Stretching does not change the VOLUME of the wire, so if the length becomes 3l, the area of cross-section must become A/3: volume = l × A = 3l × A′, so A′ = A ÷ 3
Resistance, R = ρ l ÷ A, so for the new wire: R′ = ρ × (3l) ÷ (A/3) = ρ × 3l × 3 ÷ A = 9 × (ρ l ÷ A) = 9R
R′ = 9 × 2 = 18 Ω
New resistance of the wire = 18 Ω
The trap is answering 6 Ω by thinking only about the length. Two things change when a wire is stretched, and they both push the resistance UP: it gets longer (×3) and it gets thinner (area ÷3). Resistance is proportional to length and inversely proportional to area, so the two effects multiply: 3 × 3 = 9.
The general rule is worth memorising: stretch a wire to n times its length and its resistance becomes n² times as great. Stretch it to twice the length and the resistance quadruples.
What makes it work is that the volume of metal is conserved — no material is added or removed, it is just redistributed. Say that sentence in your answer; it is where the reasoning mark is.