A wire of resistance $$R_1$$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

A wire of resistance $$R_1$$ is drawn out so that its length increases by twice its original length. We need to find the ratio of new resistance to original resistance.

The length is increased by twice the original length, meaning the new length is:

$$L' = L + 2L = 3L$$

When a wire is drawn out, its volume remains constant:

$$A \cdot L = A' \cdot L'$$

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

Resistance $$R = \frac{\rho L}{A}$$

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

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

$$\frac{R_2}{R_1} = \frac{9}{1}$$

Hence, the correct answer is Option A: 9 : 1.