A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively

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1.2 times, 1.1 times

1.2 L times, same

both remain the same

1.1 times, 1.1 times

answer is B.

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Detailed Solution

Specific resistance ρ will remain same as it is the intrinsic property of material.Original resistance of the wire, R= ρlAor R= lA or R ∝ l2V as V= Aland R' ∝ l+10% l2VTherefore, R'R= l+10100 l2l2or R'R= 11l210l2 = 121100or R' =1.21 R

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