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Question

# A wire of resistance R is stretched uniformly so that its length becomes n times the original value. The new resistance of the wire is:

A
n2R
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B
Rn2
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C
nR
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D
Rn
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Solution

## The correct option is A n2RLet l be the original length of A.Original area of cross-section, thenOriginal resistance R=ρlANow, length of the wire =l′=nlIf A′ be the new cross-sectional area, then l′A′=lA(∵ volume of metal is a constant).A′=lAl′=lAnl=AnNew resistance of the wire isR′=ρl′Al=ρ(nl)An=ρnll×nA=n2(ρlA)=n2R

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