1. Use Hooke’s law idea for gradual stretching

When a wire is stretched elastically from zero load to a final force $F$, the extension increases from $0$ to $l$.

So the force is not constant during the stretching; it rises linearly from $0$ to $F$.

2. Work done = area under the force-extension graph

For an elastic wire, $F \propto x$ where $x$ is the extension.

Thus the $F$ vs $x$ graph is a straight line from $(0,0)$ to $(l,F)$.

Hence, the work done is the triangular area under the graph: $W = \frac{1}{2} \times l \times F$

So, $W = \frac{Fl}{2}$

3. Check options

• A: $2Fl$ ❌
• B: $Fl$ ❌
• C: $\dfrac{F}{2l}$ ❌
• D: $\dfrac{Fl}{2}$ ✅

Therefore, the correct answer is: $\boxed{\frac{Fl}{2}}$