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(A) Given: Initial resistance $R_1 = 100\, \Omega$,initial radius $r_1 = r$,final radius $r_2 = r/2$.The resistance of a wire is given by $R = \frac{\

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$1600$

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(A) Given: Initial resistance $R_1 = 100\, \Omega$,initial radius $r_1 = r$,final radius $r_2 = r/2$.The resistance of a wire is given by $R = \frac{ho l}{A}$.Since the volume $V$ of the wire remains constant during recasting,$V = A \cdot l = 	ext{constant}$.Substituting $l = V/A$ into the resistance formula,we get $R = \frac{ho V}{A^2}$.Since $ho$ and $V$ are constant,$R \propto \frac{1}{A^2}$.Given $A = \pi r^2$,we have $R \propto \frac{1}{(\pi r^2)^2} \propto \frac{1}{r^4}$.Therefore,$\frac{R_2}{R_1} = \left( \frac{r_1}{r_2} ight)^4$.Substituting the values: $\frac{R_2}{100} = \left( \frac{r}{r/2} ight)^4 = (2)^4 = 16$.$R_2 = 16 	imes 100 = 1600\, \Omega$.

$A$ uniform wire of length $l$ and radius $r$ has a resistance of $100\, \Omega$. It is recast into a wire of radius $\frac{r}{2}$. The resistance of the new wire will be ............... $\Omega$.

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