A wire has a resistance 10 Ω. It is stretched by one-tenth of its original length. Then its resistance will be

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10 Ω

12 Ω

9 Ω

11 Ω

answer is B.

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Detailed Solution

Here volume remains constant. Thusπr12l=πr22(11l/10)or πr22=1011πr12{ ∴When wire is stretched by 1/10 of its original length, the new length of wire becomes (11 / / 10)}Let the new resistance be R2. ThenR2=ρ⋅1110lπr22=(11/10)ρl(10/11)πr12=(11/10)10/11)ρlπr12=(11/10)(10/11)×10=12110=12⋅1Ω

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