A wire of 1 ,Omega has a length of 1, m. It i | vedclass

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(A) Initial resistance $R_{0} = 1\, \Omega$ and initial length $\ell_{0} = 1\, m$.When the wire is stretched,its volume remains constant. Let the new

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$56$

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(A) Initial resistance $R_{0} = 1\, \Omega$ and initial length $\ell_{0} = 1\, m$.When the wire is stretched,its volume remains constant. Let the new length be $\ell_{1}$.Given that the length increases by $25\, \%$,so $\ell_{1} = \ell_{0} + 0.25\ell_{0} = 1.25\ell_{0} = 1.25\, m$.Since volume $V = A \ell$ is constant,$A_{0}\ell_{0} = A_{1}\ell_{1}$,which implies $A_{1} = A_{0}(\ell_{0} / \ell_{1})$.The resistance $R$ is given by $R = ho \frac{\ell}{A}$.Therefore,the new resistance $R_{1} = ho \frac{\ell_{1}}{A_{1}} = ho \frac{\ell_{1}}{A_{0}(\ell_{0} / \ell_{1})} = ho \frac{\ell_{1}^{2}}{A_{0}\ell_{0}} = R_{0} \left( \frac{\ell_{1}}{\ell_{0}} ight)^{2}$.Substituting the values: $R_{1} = 1 	imes (1.25)^{2} = 1.5625\, \Omega$.The percentage change in resistance is $\frac{R_{1} - R_{0}}{R_{0}} 	imes 100\, \% = \frac{1.5625 - 1}{1} 	imes 100\, \% = 56.25\, \%$.Rounding to the nearest integer,the percentage change is $56\, \%$.

$A$ wire of $1 \,\Omega$ has a length of $1\, m$. It is stretched until its length increases by $25\, \%$. The percentage change in resistance to the nearest integer is .....$\%$

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