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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 during melting and recasting,we hav

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$R/4$

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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 during melting and recasting,we have $A = \frac{V}{l}$.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,$R \propto l^2$.Therefore,$\frac{R_1}{R_2} = \left( \frac{l_1}{l_2} ight)^2$.Given $l_2 = \frac{l_1}{2}$,we have $\frac{R}{R_2} = \left( \frac{l_1}{l_1/2} ight)^2 = (2)^2 = 4$.Thus,the new resistance $R_2 = \frac{R}{4}$.

If a wire of resistance $R$ is melted and recasted to half of its length,then the new resistance of the wire will be

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