A wire suspended vertically from one of its ends stretched by attaching weight of 200 N to the lower end. The weight stretches the wire by 1 mm . Then - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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A wire suspended vertically from one of its ends stretched by attaching 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:

Solution & Step-by-step Explanation

Elastic potential energy stored U = \frac{1}{2} \times Force \times Extension . U = \frac{1}{2} \times 200 \times (1 \times 1 0^{- 3}) m U = 100 \times 1 0^{- 3} = 0.1 J

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A wire suspended vertically from one of its ends stretched by attaching 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:

0.2 J

10 J

20 J

0.1 J

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