If a copper wire is stretched to make it 0.1 % longer. The percentage change in its resistance is

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increase by 0.2%

decrease by 0.2%

decrease by 0.05%

increase by 0.05%

answer is A.

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Detailed Solution

The resistance will increase by 0.2%
Let the length of the wire be 'L' and area of cross-section be 'A' and resistivity of the material be ρ
Therefore R =

\frac{ρ L}{A}

On stretching a wire its length increases but its area of cross section decreases, and volume remains constant.
V = AL
R =

\frac{ρ L^{2}}{V}

R ∝ L2
Hence

\frac{ΔR}{R}

= 2

\frac{ΔL}{L}

Hence, for

\frac{ΔL}{L}

= 0.1%

\frac{ΔR}{R}

= 0.2%

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