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.

Explanation:
Let us decode the phrasing and compute the resistance ratio step-by-step using the principles of conservation of volume:

1. Decoding the Phrase "Increased By":
There is a vital difference between increased "to" and increased "by":
- "Increased to 2 times" implies the final length is \( 2L_1 \).
- "Increased by 2 times" implies the extension added to the wire is \( 2L_1 \).
Therefore, the final length \( L_2 \) is:
\( L_2 = L_1 + 2L_1 = 3L_1 \)

2. Conservation of Volume:
When a wire is stretched, its total volume \( V \) remains constant. Let \( A_1 \) be the initial cross-sectional area and \( A_2 \) be the final cross-sectional area:
\( V = L_1 A_1 = L_2 A_2 \)
\( L_1 A_1 = (3L_1) A_2 \implies A_2 = \frac{A_1}{3} \)

3. Resistance Calculation:
The baseline resistance formula is \( R = \rho \frac{L}{A} \), where \( \rho \) is the material resistivity. Writing the ratio of final resistance \( R_2 \) to initial resistance \( R_1 \) gives:
\( \frac{R_2}{R_1} = \frac{\rho \frac{L_2}{A_2}}{\rho \frac{L_1}{A_1}} = \left(\frac{L_2}{L_1}\right) \times \left(\frac{A_1}{A_2}\right) \)

Substituting our derived values into this expression:
\( \frac{R_2}{R_1} = \left(\frac{3L_1}{L_1}\right) \times \left(\frac{A_1}{A_1 / 3}\right) = 3 \times 3 = 9 \)
Expressing this as a standard fraction ratio yields 9 : 1.

Hence, Option A is the correct choice.