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Laws of Motion question

2024 · 6 Apr · Shift 2 · Q63
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Laws of Motion question

2024 · 6 Apr · Shift 2 · Q63

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A body of weight 200 N200 \mathrm{~N}200 N is suspended from a tree branch through a chain of mass 10 kg10 \mathrm{~kg}10 kg. The branch pulls the chain by a force equal to (if g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2g=10 m/s2) :
  1. A
    300 N
  2. B
    100 N
  3. C
    150 N
  4. D
    200 N
View written solutionFree

Correct answer: A

  1. Given data

    • Weight of body =200 N= 200\,\text{N}=200N
    • Mass of chain =10 kg= 10\,\text{kg}=10kg
    • g=10 m/s2g = 10\,\text{m/s}^2g=10m/s2
  2. Find the weight of the chain Wchain=mg=10×10=100 NW_{\text{chain}} = mg = 10 \times 10 = 100\,\text{N}Wchain​=mg=10×10=100N

  3. Force exerted by the branch on the chain

    The branch supports:

    • the suspended body of weight 200 N200\,\text{N}200N
    • the chain itself of weight 100 N100\,\text{N}100N

    Therefore, the total downward force on the top of the chain is: F=200+100=300 NF = 200 + 100 = 300\,\text{N}F=200+100=300N

  4. Conclusion The branch pulls the chain upward with force 300 N\boxed{300\,\text{N}}300N​

  5. Option check

    • A: 300 N300\,\text{N}300N ✅
    • B: 100 N100\,\text{N}100N ❌
    • C: 150 N150\,\text{N}150N ❌
    • D: 200 N200\,\text{N}200N ❌

So, the correct option is A.

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