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(C) The resistance of a wire is given by $R = \rho \frac{l}{A} = 4 \,\Omega$ .... $(i)$When the wire is stretched to twice its original length,the new

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$16$

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(C) The resistance of a wire is given by $R = ho \frac{l}{A} = 4 \,\Omega$ .... $(i)$When the wire is stretched to twice its original length,the new length $l^{\prime} = 2l$.Since the volume of the wire remains constant during stretching,$V = lA = l^{\prime}A^{\prime}$.Substituting $l^{\prime} = 2l$,we get $lA = (2l)A^{\prime}$,which implies $A^{\prime} = \frac{A}{2}$.The new resistance $R^{\prime}$ is given by $R^{\prime} = ho \frac{l^{\prime}}{A^{\prime}}$.Substituting the values,$R^{\prime} = ho \frac{2l}{A/2} = 4 \left( ho \frac{l}{A} ight)$.Using equation $(i)$,$R^{\prime} = 4 	imes 4 \,\Omega = 16 \,\Omega$.

$A$ 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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