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 new wire will be-

We recall the basic relation between the resistance $$R$$ of a uniform wire, its length $$l$$, cross-sectional area $$A$$ and the material’s resistivity $$\rho$$. The formula is

$$R=\dfrac{\rho\,l}{A}.$$

The wire is melted and redrawn, so the material remains the same; therefore $$\rho$$ is unchanged and the total volume of the wire is conserved. We first write the volume of the original wire:

$$V_{\text{old}}=\pi r^{2}l.$$

After recasting, the new radius becomes $$\dfrac{r}{2}$$ and we denote the new length by $$l_2$$. The new volume must equal the old volume, so

$$V_{\text{new}}=\pi\left(\dfrac{r}{2}\right)^{2}l_2=\pi r^{2}l.$$

We simplify the left side:

$$\pi\left(\dfrac{r}{2}\right)^{2}l_2=\pi\left(\dfrac{r^{2}}{4}\right)l_2=\dfrac{\pi r^{2}l_2}{4}.$$

Equating the two volumes we have

$$\dfrac{\pi r^{2}l_2}{4}=\pi r^{2}l.$$

Dividing by $$\pi r^{2}$$ gives

$$\dfrac{l_2}{4}=l,$$

so

$$l_2=4l.$$

Now we calculate the resistance of the new wire. Its cross-sectional area is

$$A_2=\pi\left(\dfrac{r}{2}\right)^{2}=\dfrac{\pi r^{2}}{4}.$$

Applying the resistance formula to the new wire, we write

$$R_2=\dfrac{\rho\,l_2}{A_2}.$$

Substituting $$l_2=4l$$ and $$A_2=\dfrac{\pi r^{2}}{4}$$, we get

$$R_2=\dfrac{\rho\,(4l)}{\dfrac{\pi r^{2}}{4}}.$$

Dividing by a fraction is the same as multiplying by its reciprocal, so

$$R_2= \rho\,(4l)\left(\dfrac{4}{\pi r^{2}}\right)=16\,\dfrac{\rho\,l}{\pi r^{2}}.$$

But the original resistance is

$$R_1=\dfrac{\rho\,l}{\pi r^{2}}=100\;\Omega.$$

Therefore

$$R_2 = 16R_1 = 16 \times 100\;\Omega = 1600\;\Omega.$$

Hence, the correct answer is Option A.