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(C) The volume $V$ of the wire is given by $V = \pi r^2 l$,where $r$ is the radius and $l$ is the length.Since the volume remains constant during stre

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$0.4$

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(C) The volume $V$ of the wire is given by $V = \pi r^2 l$,where $r$ is the radius and $l$ is the length.Since the volume remains constant during stretching,$\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta l}{l} = 0$.This implies $\frac{\Delta l}{l} = -2\frac{\Delta r}{r} \dots (1)$.The resistance $R$ is given by $R = \frac{ho l}{\pi r^2}$.Taking the relative error,we get $\frac{\Delta R}{R} = \frac{\Delta l}{l} - 2\frac{\Delta r}{r}$.Substituting equation $(1)$ into this expression:$\frac{\Delta R}{R} = (-2\frac{\Delta r}{r}) - 2\frac{\Delta r}{r} = -4\frac{\Delta r}{r}$.Given the radius decreases by $0.1\%$,we have $\frac{\Delta r}{r} = -0.1\%$.Therefore,the percentage increase in resistance is $\frac{\Delta R}{R} = -4 	imes (-0.1\%) = 0.4\%$.

If a copper wire is stretched to make its radius decrease by $0.1\%$,then the percentage increase in resistance is approximately .......... $\%$

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