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

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$0.2$

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(A) The resistance $R$ 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, we get $R = ho \frac{l^2}{V}$.Since $ho$ and $V$ are constants, $R \propto l^2$.Taking the derivative or using the approximation for small changes, we have $\frac{\Delta R}{R} \approx 2 \frac{\Delta l}{l}$.Given that the percentage increase in length $\frac{\Delta l}{l} 	imes 100 = 0.1\%$.Therefore, the percentage increase in resistance is $\frac{\Delta R}{R} 	imes 100 = 2 	imes (\frac{\Delta l}{l} 	imes 100) = 2 	imes 0.1\% = 0.2\%$.

If there is a $0.1\%$ increase in length due to stretching, the percentage increase in its resistance will be ............ $\%$

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