If the length of a wire is increased by 10% b | vedclass

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

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$21$

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(C) The resistance of a wire is given by $R = ho \frac{\ell}{A}$.Since the volume $V = A \ell$ remains constant when the wire is stretched,$A = \frac{V}{\ell}$.Substituting this into the resistance formula,we get $R = ho \frac{\ell^2}{V}$,which implies $R \propto \ell^2$.Given that the length increases by $10\%$,the new length $\ell_2 = 1.1 \ell_1$.Therefore,the ratio of the new resistance to the original resistance is $\frac{R_2}{R_1} = \left(\frac{\ell_2}{\ell_1}ight)^2 = (1.1)^2 = 1.21$.This means $R_2 = 1.21 R_1$.The percentage increase in resistance is $\Delta R \% = \frac{R_2 - R_1}{R_1} 	imes 100\% = (1.21 - 1) 	imes 100\% = 21\%$.

If the length of a wire is increased by $10\%$ by stretching it,the percentage increase in its resistance is ..............$\%$

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