1. Use the resistance formula

   For a wire, $R = \rho \frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

2. Volume remains constant on recasting

   Since the wire is recast, its material and volume remain the same.

   Initial volume: $V_1 = \pi r^2 \cdot 1$

   New radius: $r' = \frac{r}{2}$

   New area: $A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}$

   Let new length be $L'$. Using volume conservation: $\pi r^2 \cdot 1 = \frac{\pi r^2}{4} \cdot L'$ $L' = 4$

3. Compare old and new resistance

   Initial resistance: $R_1 = \rho \frac{1}{\pi r^2} = 100\,\Omega$

   New resistance: $R_2 = \rho \frac{4}{\pi r^2/4}$ $R_2 = \rho \frac{4 \cdot 4}{\pi r^2} = 16\rho \frac{1}{\pi r^2}$

   Therefore, $R_2 = 16 R_1 = 16 \times 100 = 1600\,\Omega$

4. Check options

   • A: $1600\,\Omega$ ✅
   • B: $400\,\Omega$ ❌
   • C: $200\,\Omega$ ❌
   • D: $100\,\Omega$ ❌

So, the correct answer is A.