Question
Two wires made of the same material have lengths in the
ratio 1:2 and diameters in the ratio 2:1. If the same force is applied to both, what is the ratio of work done (elastic potential energy) stored in the shorter to longer wire?Solution
The work done (or elastic potential energy ) stored in a stretched wire is given by: 
Where:
- F is the force applied,
- L is the length of the wire,
- A is the cross-sectional area,
- Y is Young’s modulus (same material ⇒ same Y).
- Lengths in the ratio: L1:L2 = 1:2
- Diameters in the ratio: d1:d2 = 2:1
So, energy stored in the shorter wire is 1/8 of that in the longer wire.
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