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 $$5 \text{ cm}$$. Another wire of the same material of length $$4L$$ and radius $$4r$$ is pulled by a force $$4F$$ under same conditions. The increase in length of this wire is ______ cm.

The elongation of a wire under a tensile force is given by:

$$\Delta L = \dfrac{FL}{\pi r^2 Y}$$

where $$F$$ is the applied force, $$L$$ is the original length, $$r$$ is the radius, and $$Y$$ is Young's modulus.

For the first wire: Length = $$L$$, radius = $$r$$, force = $$F$$

$$\Delta L_1 = \dfrac{FL}{\pi r^2 Y} = 5 \text{ cm}$$

For the second wire: Length = $$4L$$, radius = $$4r$$, force = $$4F$$ (same material, so same $$Y$$)

$$\Delta L_2 = \dfrac{4F \times 4L}{\pi (4r)^2 Y} = \dfrac{16FL}{\pi \times 16r^2 \times Y} = \dfrac{FL}{\pi r^2 Y}$$

$$\Delta L_2 = \Delta L_1 = 5 \text{ cm}$$

Therefore, the increase in length of the second wire is $$\boxed{5}$$ cm.