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Work Power and Energy question

2003 · Shift 0 · Q181
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Work Power and Energy question

2003 · Shift 0 · Q181

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N200N200N to the lower end. The weight stretches the wire by 1mm.1mm.1mm. Then the elastic energy stored in the wire is
  1. A
    0.2J0.2J0.2J
  2. B
    10J10J10J
  3. C
    20J20J20J
  4. D
    0.1J0.1J0.1J
View written solutionFree

Correct answer: D

  1. Given data

    • Load applied: F=200 NF = 200\,\text{N}F=200N
    • Extension produced: x=1 mm=10−3 mx = 1\,\text{mm} = 10^{-3}\,\text{m}x=1mm=10−3m

  2. Elastic potential energy stored

    For a wire obeying Hooke’s law, the force increases linearly from 000 to FFF as extension increases from 000 to xxx.

    Hence, elastic energy stored is U=12FxU = \frac{1}{2}FxU=21​Fx

  3. Substitute the values U=12×200×10−3U = \frac{1}{2} \times 200 \times 10^{-3}U=21​×200×10−3

    U=100×10−3=0.1 JU = 100 \times 10^{-3} = 0.1\,\text{J}U=100×10−3=0.1J

  4. Match with options 0.1 J0.1\,\text{J}0.1J corresponds to Option D.

  5. Comparison with stored correct answer Stored correct answer is D, which matches our derived answer.

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