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JEE Main 2003 (AIEEE) · Elastic Potential Energy

Elastic Potential Energy question from JEE Main 2003 (AIEEE)

Q16JEE Main 2003 · AIEEEEasy

Elastic Potential Energy

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N200\text{ N} to the lower end. The weight stretches the wire by 1 mm.1\text{ mm.} Then the elastic energy stored in the wire is

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Answer: (d)

  1. U=12Fl=12×200×10−3U = \dfrac{1}{2}Fl = \dfrac{1}{2} \times 200 \times 10^{-3}
  2. U=0.1 JU = 0.1\text{ J}
  3. So the answer is (d) 0.1 J.
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