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(B) Since the volume $V$ of the wire remains constant during stretching,we have $V = A \ell = A^{\prime} \ell^{\prime}$.Given that the new length $\el

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$300$

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(B) Since the volume $V$ of the wire remains constant during stretching,we have $V = A \ell = A^{\prime} \ell^{\prime}$.Given that the new length $\ell^{\prime} = 2\ell$,we substitute this into the volume equation: $A \ell = A^{\prime} (2\ell)$,which gives $A^{\prime} = \frac{A}{2}$.The initial resistance is $R = ho \frac{\ell}{A}$.The new resistance is $R^{\prime} = ho \frac{\ell^{\prime}}{A^{\prime}} = ho \frac{2\ell}{A/2} = 4 ho \frac{\ell}{A} = 4R$.The percentage increase in resistance is given by $\frac{R^{\prime} - R}{R} 	imes 100\%$.Substituting $R^{\prime} = 4R$,we get $\frac{4R - R}{R} 	imes 100\% = 3 	imes 100\% = 300\%$.

The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be..... $\%$.

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