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Q.

An aluminium wire is stretched to make its length, 40 % larger. Then percentage change in resistance is:

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a

96%

b

40%

c

20%

d

80%

answer is D.

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Detailed Solution

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R=ρℓA=ρℓ2V

Here l becomes 1.4l

⇒R'=ρ(1.4l)2V=1.96ρlV

Percentage change is 

∆RR×100=1.96R-RR×100=96%

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