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(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the volume $V = A 	imes l$ remains constant when the wire is stretched,we can w

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$16$

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(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the volume $V = A 	imes l$ remains constant when the wire is stretched,we can write $A = \frac{V}{l}$.Substituting this into the resistance formula,we get $R = ho \frac{l}{V/l} = ho \frac{l^2}{V}$.Since $ho$ and $V$ are constant,$R \propto l^2$.If the length is doubled $(l' = 2l)$,the new resistance $R'$ becomes $R' = R 	imes (2)^2 = 4R$.Given the initial resistance $R = 4\,\Omega$,the new resistance is $R' = 4 	imes 4\,\Omega = 16\,\Omega$.

$A$ wire of resistance $4\,\Omega$ is stretched to twice its original length. The resistance of the stretched wire would be ................ $\Omega$.

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