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(D) Let the initial length be $\ell$ and the initial resistance be $R = \rho \frac{\ell}{A}$.Since the length is increased by $100\,\%$,the new length

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$300$

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(D) Let the initial length be $\ell$ and the initial resistance be $R = ho \frac{\ell}{A}$.Since the length is increased by $100\,\%$,the new length $\ell' = \ell + 1.00\ell = 2\ell$.Since the volume $V = A 	imes \ell$ remains constant,$A' \ell' = A \ell \implies A' (2\ell) = A \ell \implies A' = \frac{A}{2}$.The new resistance $R' = ho \frac{\ell'}{A'} = ho \frac{2\ell}{A/2} = 4 \left( ho \frac{\ell}{A} ight) = 4R$.The percentage change in resistance is given by $\frac{R' - R}{R} 	imes 100\,\% = \frac{4R - R}{R} 	imes 100\,\% = 3 	imes 100\,\% = 300\,\%$.

The length of a given cylindrical wire is increased by $100\,\%$. Due to the consequent decrease in diameter,the percentage change in the resistance of the wire will be ................ $\%$.

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