A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm, then the elastic energy stored in the wire is

The correct option is 0.1J.

Step-by-step Explanation:
When a wire is suspended vertically and stretched by a load (force), work is done against the internal restoring elastic forces of the wire. This work done is stored in the wire as elastic potential energy.

The elastic potential energy ($U$) stored in a stretched wire is given by the formula:
$U=\frac{1}{2}\times \text{Stretching Force}\times \text{Extension}$

From the given information:
Stretching Force ($F$) = $200\text{N}$
Extension ($\mathrm{\Delta L}$) = $1\text{mm}=1\times {10}^{-3}\text{m}$

Substituting these values into the energy formula:
$U=\frac{1}{2}\times 200\times (1\times {10}^{-3})$
$U=100\times {10}^{-3}$
$U=0.1\text{J}$

Therefore, the elastic energy stored in the wire is $0.1\text{J}$.