1. Use the resistance formula

   For a wire, $R = \rho \frac{L}{A}$ where:

   • $\rho$ = resistivity (constant for same material)
   • $L$ = length
   • $A$ = cross-sectional area

2. Interpret the statement

   “Its length is increased by twice of its original length” means the new length becomes: $L' = L + 2L = 3L$

3. Use conservation of volume

   When a wire is drawn out, its volume remains constant: $AL = A'L'$

   Since $L' = 3L$, $A' = \frac{AL}{L'} = \frac{AL}{3L} = \frac{A}{3}$

4. Find the new resistance

   Original resistance: $R_1 = \rho \frac{L}{A}$

   New resistance: $R_2 = \rho \frac{L'}{A'} = \rho \frac{3L}{A/3} = \rho \frac{9L}{A} = 9R_1$

5. Compute the ratio

   $R_2 : R_1 = 9 : 1$

6. Check options

   • A: $9:1$ ✅
   • B: $1:9$ ❌
   • C: $4:1$ ❌
   • D: $3:1$ ❌

Therefore, the correct answer is A.