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Step-by-Step Solution
Step 1: Identify the Known Parameters
• Original wire has radius $r$ and resistance $R_1 = 100\,\Omega$.
• New wire has radius $r_2 = \dfrac{r}{2}$.
• We assume the material (and thus the resistivity) and the volume of the wire remain the same when it is recast.
Step 2: Write the General Formula for Resistance
For a wire of resistivity $\rho$, length $l$, and cross-sectional area $A$, the resistance $R$ is given by:
$R = \dfrac{\rho\,l}{A}$
Step 3: Relate Resistance to Radius When Volume is Constant
1. The volume $V$ of the wire remains constant during recasting.
2. The volume of a wire is $V = \text{(area of cross-section)} \times \text{(length)} = \pi r^2 l$.
3. When the radius changes, to keep the volume the same, the length also changes.
4. Combining these relationships shows that effectively:
$R \propto \dfrac{1}{r^4}$
(As derived often by expressing $R$ in terms of volume and radius, one obtains this fourth-power dependence on the radius.)
Step 4: Use the Ratio for Resistances
If $R_1$ corresponds to radius $r_1 = r$ and $R_2$ corresponds to radius $r_2 = \dfrac{r}{2}$, then:
$\dfrac{R_1}{R_2} = \left(\dfrac{r_2}{r_1}\right)^4$
Rearranging for $R_2$ gives:
$R_2 = R_1 \left(\dfrac{r_1}{r_2}\right)^4$
Step 5: Substitute the Values
Given:
$r_1 = r, \quad r_2 = \dfrac{r}{2}, \quad R_1 = 100\,\Omega.$
Hence,
$R_2 = 100\,\Omega \times \left(\dfrac{r}{\tfrac{r}{2}}\right)^4 = 100\,\Omega \times (2)^4 = 100\,\Omega \times 16 = 1600\,\Omega.$
Final Answer
$\boxed{1600\,\Omega}$
