A wire of length $$L$$ and radius $$r$$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $$f$$, its length increases by $$l$$. Another wire of same material of length $$2L$$ and radius $$2r$$ is pulled by a force $$2f$$. Then the increase in its length will be:

Using Young's modulus formula: $$Y = \frac{FL}{Al} = \frac{FL}{\pi r^2 l}$$

For the first wire: $$Y = \frac{fL}{\pi r^2 l}$$

For the second wire: $$Y = \frac{2f \times 2L}{\pi (2r)^2 \times l'}$$

Since both wires are of the same material (same $$Y$$):

$$\frac{fL}{\pi r^2 l} = \frac{4fL}{\pi \times 4r^2 \times l'}$$

$$\frac{fL}{r^2 l} = \frac{4fL}{4r^2 l'}$$

$$\frac{1}{l} = \frac{1}{l'}$$

$$l' = l$$

The increase in length is $$\mathbf{l}$$.