A steel rod has a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is $$2.0 \times 10^{11}$$ N m$$^{-2}$$. The longitudinal strain produced in the wire is ______ $$\times 10^{-5}$$.

We have a rod of radius $$r = 20\;\text{mm} = 0.02\;\text{m}$$, length $$L = 2.0\;\text{m}$$, subjected to a force $$F = 62.8\;\text{kN} = 62800\;\text{N}$$, with Young's modulus $$Y = 2.0 \times 10^{11}\;\text{N/m}^2$$.

The cross-sectional area of the rod is

$$A = \pi r^2 = \pi (0.02)^2 = \pi \times 4 \times 10^{-4} = 1.2566 \times 10^{-3}\;\text{m}^2$$

Now, the stress is

$$\sigma = \frac{F}{A} = \frac{62800}{1.2566 \times 10^{-3}} = 4.998 \times 10^{7}\;\text{N/m}^2$$

So the longitudinal strain is

$$\text{strain} = \frac{\sigma}{Y} = \frac{4.998 \times 10^{7}}{2.0 \times 10^{11}} = 2.5 \times 10^{-4} = 25 \times 10^{-5}$$

Hence, the longitudinal strain produced is $$25 \times 10^{-5}$$. So, the answer is $$25$$.