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 ........ $J$
AIEEE 2003, Easy
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(a) $U = \frac{1}{2} \times F \times l = \frac{1}{2} \times 200 \times {10^{ - 3}} = 0.1\;J$
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