Solution:
Resistance of a stretched wire of constant volume is proportional to the square of its length: \( R \propto l^2 \). Thus, \( \frac{\Delta R}{R}\text{ (%)} = 2 \frac{\Delta l}{l}\text{ (%)} = 2 \times 0.1\% = 0.2\% \).
Resistance of a stretched wire of constant volume is proportional to the square of its length: \( R \propto l^2 \). Thus, \( \frac{\Delta R}{R}\text{ (%)} = 2 \frac{\Delta l}{l}\text{ (%)} = 2 \times 0.1\% = 0.2\% \).
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