A uniform metallic wire is elongated by 0.04 m when subjected to a linear force $$F$$. The elongation, if its length and diameter is doubled and subjected to the same force will be ______ cm.

The elongation of a wire under a linear force $$F$$ is given by $$\Delta L = \frac{FL}{AY}$$, where $$L$$ is the length of the wire, $$A$$ is the cross-sectional area, and $$Y$$ is the Young's modulus of the material.

For the original wire with length $$L$$, diameter $$d$$, and area $$A = \frac{\pi d^2}{4}$$, the elongation is $$\Delta L_1 = \frac{FL}{AY} = 0.04$$ m.

When both the length and diameter are doubled, the new length is $$2L$$ and the new diameter is $$2d$$, giving a new cross-sectional area of $$A' = \frac{\pi (2d)^2}{4} = 4A$$.

The new elongation under the same force $$F$$ is $$\Delta L_2 = \frac{F(2L)}{(4A)Y} = \frac{2FL}{4AY} = \frac{1}{2} \cdot \frac{FL}{AY} = \frac{\Delta L_1}{2}$$.

Substituting the given value: $$\Delta L_2 = \frac{0.04}{2} = 0.02$$ m $$= 2$$ cm.

Therefore, the elongation is $$2$$ cm.