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(C) The resistance of a wire is given by $R = \frac{\rho \ell}{A}$.Since the volume $V = \ell A$ remains constant during stretching,we have $A = \frac

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$0.8$

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(C) The resistance of a wire is given by $R = \frac{ho \ell}{A}$.Since the volume $V = \ell A$ remains constant during stretching,we have $A = \frac{V}{\ell}$.Substituting this into the resistance formula,we get $R = \frac{ho \ell^2}{V}$.For small changes,the relative change in resistance is given by $\frac{\Delta R}{R} = 2 \frac{\Delta \ell}{\ell}$.Given that the percentage change in length is $\frac{\Delta \ell}{\ell} 	imes 100 = 0.4 \%$.Therefore,the percentage change in resistance is $\frac{\Delta R}{R} 	imes 100 = 2 	imes 0.4 \% = 0.8 \%$.

An aluminium wire is stretched to make its length $0.4 \%$ larger. Then the percentage change in resistance is $.....\,\%$.

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