A 100 m long wire having cross-sectional area $$6.25 \times 10^{-4}$$ m$$^2$$ and Young's modulus is $$10^{10}$$ N m$$^{-2}$$ is subjected to a load of 250 N, then the elongation in the wire will be:

Consider a wire of length $$L = 100$$ m, cross-sectional area $$A = 6.25 \times 10^{-4}$$ m², and Young’s modulus $$Y = 10^{10}$$ N/m², to which a force $$F = 250$$ N is applied.

Using the relation for elastic extension, the elongation $$\Delta L$$ is given by:

$$ \Delta L = \frac{F L}{A Y} = \frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}} = \frac{25000}{6.25 \times 10^6} = 4 \times 10^{-3} \text{ m} $$

Hence, the elongation of the wire is $$4 \times 10^{-3}$$ m.