A 1 m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.

We have a wire of total length 1 m broken into two parts: X of length $$l$$ and Y of length $$(1 - l)$$, both having the same cross-sectional area $$A$$ and resistivity $$\rho$$.

The resistance of Y is $$R_Y = \frac{\rho(1 - l)}{A}$$.

Now, wire X (length $$l$$, area $$A$$) is stretched to form wire W of length $$2l$$. Since the volume is conserved during stretching, $$l \cdot A = 2l \cdot A_W$$, which gives the new cross-sectional area $$A_W = \frac{A}{2}$$.

The resistance of W is $$R_W = \frac{\rho \cdot 2l}{A/2} = \frac{4\rho l}{A}$$.

We are given that $$R_W = 2R_Y$$, so: $$\frac{4\rho l}{A} = 2 \cdot \frac{\rho(1 - l)}{A}$$

Simplifying: $$4l = 2(1 - l) = 2 - 2l$$, which gives $$6l = 2$$, so $$l = \frac{1}{3}$$.

Therefore the length of X is $$\frac{1}{3}$$ m and the length of Y is $$\frac{2}{3}$$ m. The ratio $$\frac{X}{Y} = \frac{1/3}{2/3} = \frac{1}{2}$$.

Hence, the correct answer is Option 2.