alleryA wire of resistance 10 Ω is stretched to double its length without changing volume. Its new resistance:
A5 Ω
B40 Ω
C80 Ω
D20 Ω
Answer & Solution
Correct answer: B. 40 Ω
Volume V = lA = constant.
When l doubles, A halves.
R_new = ρ(2l)/(A/2) = 4ρl/A = 4R = 4 × 10 = 40 Ω.
**So stretching doubles l → resistance quadruples.**
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