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(B) The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$,which implies $R \propto \frac{l}{r^2}$.Since the volume of t

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$16$

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(B) The resistance of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$,which implies $R \propto \frac{l}{r^2}$.Since the volume of the wire remains constant during stretching,$V = A \cdot l = \pi r^2 l = 	ext{constant}$.Let the initial radius be $r$ and length be $l$. After stretching,the new radius is $r' = r/2$ and the new length is $l'$.Equating the volumes: $\pi r^2 l = \pi (r/2)^2 l'$.$\pi r^2 l = \pi (r^2/4) l' \implies l' = 4l$.The new resistance $R'$ is given by $R' = ho \frac{l'}{\pi (r')^2} = ho \frac{4l}{\pi (r/2)^2} = ho \frac{4l}{\pi r^2 / 4} = 16 \left( ho \frac{l}{\pi r^2} ight) = 16R$.Given $R' = xR$,we find $x = 16$.

$A$ wire of resistance $R$ and radius $r$ is stretched until its radius becomes $r / 2$. If the new resistance of the stretched wire is $x R$,then the value of $x$ is $\qquad$

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