A copper wire is stretched so that its length is increased by 0.2%. What is the percentage change in the resistance?

Correct option is A

​$$\text{The resistance of a wire is given by:} \\[6pt]R = \rho \frac{L}{A} \\[10pt]\text{When a wire is stretched, its:} \\[6pt]\bullet \ \text{Length increases} \\[2pt]\bullet \ \text{Cross-sectional area decreases} \\[2pt]\bullet \ \text{Volume remains constant (elastic deformation)} \\[12pt]\textbf{Given} \\[6pt]\text{Percentage increase in length:} \\[6pt]\frac{\Delta L}{L} = 0.2\% = 0.002 \\[10pt]\text{Since volume is constant:} \\[6pt]\frac{\Delta A}{A} = -\frac{\Delta L}{L} = -0.002 \\[12pt]\textbf{Percentage change in resistance} \\[8pt]\frac{\Delta R}{R} = \frac{\Delta L}{L} - \frac{\Delta A}{A} \\[8pt]\frac{\Delta R}{R} = 0.002 - (-0.002) = 0.004 \\[10pt]\Rightarrow \ \text{Percentage increase in resistance} = 0.4\%$$​​