If the length of a wire is increased by 0.1%, | vedclass

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

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$0.2$

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(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the volume $V = A 	imes l$ remains constant,we have $A = \frac{V}{l}$.Substituting this into the resistance formula,we get $R = ho \frac{l^2}{V}$.Since $ho$ and $V$ are constant,$R \propto l^2$.Taking the derivative,we get $\frac{\Delta R}{R} = 2 \frac{\Delta l}{l}$.Given $\frac{\Delta l}{l} = 0.1\%$,the percentage change in resistance is $\frac{\Delta R}{R} 	imes 100 = 2 	imes 0.1\% = 0.2\%$.

If the length of a wire is increased by $0.1\%$,what is the percentage increase in its resistance (in $\%$)?

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