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(C) The resistance $R$ of a metallic wire is given by the formula $R = \rho \frac{\ell}{A} = \rho \frac{\ell}{\pi r^2}$,where $\rho$ is resistivity,$\

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length is doubled and radius is halved

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(C) The resistance $R$ of a metallic wire is given by the formula $R = ho \frac{\ell}{A} = ho \frac{\ell}{\pi r^2}$,where $ho$ is resistivity,$\ell$ is length,and $r$ is the radius.From this,we see that $R \propto \frac{\ell}{r^2}$.Let the initial resistance be $R = k \frac{\ell}{r^2}$.For option $C$,the new length $\ell' = 2\ell$ and the new radius $r' = \frac{r}{2}$.The new resistance $R'$ is given by $R' = k \frac{\ell'}{(r')^2} = k \frac{2\ell}{(r/2)^2} = k \frac{2\ell}{r^2/4} = 8 	imes k \frac{\ell}{r^2} = 8R$.Therefore,the resistance becomes $8$ times when the length is doubled and the radius is halved.

The resistance of a metallic wire becomes $8$ times when :

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