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(D) The resistance of a wire is given by $R = \frac{ho l}{A}$.Since volume $V = A 	imes l$ remains constant during stretching,we can write $A = \fr

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$4$

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(D) The resistance of a wire is given by $R = \frac{ho l}{A}$.Since volume $V = A 	imes l$ remains constant during stretching,we can write $A = \frac{V}{l}$.Substituting this into the resistance formula: $R = \frac{ho l^2}{V}$.Since $ho$ and $V$ are constant,$R \propto l^2$.Given the change in length $\frac{\Delta l}{l} = 2\% = 0.02$.Using the differential method for small changes: $\frac{\Delta R}{R} = 2 \frac{\Delta l}{l}$.$\frac{\Delta R}{R} = 2 	imes 2\% = 4\%$.Alternatively,using the ratio method:$R_1 \propto l^2$ and $R_2 \propto (1.02l)^2 = 1.0404l^2$.$\frac{R_2 - R_1}{R_1} 	imes 100\% = (1.0404 - 1) 	imes 100\% = 4.04\% \approx 4\%$.

$A$ wire is stretched to increase its length by $2\%$,then the percentage change in its resistance is . . . . . . . (in $\%$)

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