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A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is:

(1) 2.5%

(2) 0.5%

(3) 1.0%

(4) 2.0%
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Ans: (3) 1.0%

Sol: $R={\rho l\over A}$ and volume (V) = A$l$.

$R={\rho l^2\over V}$

$\implies{\Delta R\over R}={2\Delta l\over l}=1\%$

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