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

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$21$

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(C) The resistance of a wire is given by $R = ho \frac{l}{A}$. Since the volume $V = A 	imes l$ remains constant, $A = \frac{V}{l}$.Substituting this, we get $R = ho \frac{l^2}{V}$, which implies $R \propto l^2$.If the length increases by $10\%$, the new length $l_2 = l_1 + 0.10 l_1 = 1.1 l_1$.The ratio of resistances is $\frac{R_2}{R_1} = \left( \frac{l_2}{l_1} ight)^2 = (1.1)^2 = 1.21$.Thus, $R_2 = 1.21 R_1$.The percentage change in resistance is $\frac{R_2 - R_1}{R_1} 	imes 100 = (1.21 - 1) 	imes 100 = 21\%$.

The length of the resistance wire is increased by $10\%$. What is the corresponding change in the resistance of the wire? ................ $\%$

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