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

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$2$

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(D) The resistance of a wire is given by $R = ho \frac{l}{A}$. Since the volume $V = A 	imes l$ remains constant during stretching, we can write $A = \frac{V}{l}$.Substituting this into the resistance formula: $R = ho \frac{l}{V/l} = ho \frac{l^2}{V}$.Since $ho$ and $V$ are constant, $R \propto l^2$.Taking the derivative or using the approximation for small changes: $\frac{\Delta R}{R} \approx 2 \frac{\Delta l}{l}$.Given $\frac{\Delta R}{R} = 4 \%$, we have $4 \% = 2 	imes \frac{\Delta l}{l} 	imes 100 \%$.Therefore, $\frac{\Delta l}{l} 	imes 100 \% = \frac{4 \%}{2} = 2 \%$.

$A$ cylindrical metallic wire is stretched to increase its length. If the resistance of the wire is increased by $4 \%$, then the percentage increase in its length is (in $\%$)

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