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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

Asked in AIEEE 2003 · Energy stored in a stretched wire

• (1) 0.2 J
• (2) 10 J
• (3) 20 J
• (4) 0.1 J

Step-by-step solution

Why the other options are wrong

• (1) comes from using U=Fl without the factor 1/2: 200×10⁻³=0.2 J; the tension builds from zero, so only half of that is stored.
• (2) comes from leaving the extension in millimetres and halving, 1/2×200×1=100, then losing a power of ten; either way it is the mm→m conversion that has gone missing.
• (3) comes from is the same mm slip without the factor 1/2, scaled down by ten: 200×1 mm read as 200×0.1.

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