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Physics · Newton's Laws of Motion

JEE Main 2024 — 5 April, Shift 1 — Question 43

A wooden block of mass 5 kg rests on soft horizontal floor. When an iron cylinder of mass 25 kg is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of 0.1 ms−20.1 \mathrm{~ms}^{-2}. The action force of the system on the floor is equal to:

  1. Option A:

    297 N

  2. Option B:

    294 N

  3. Option C:

    291 N

    Correct
  4. Option D:

    196 N

Answer: C

Step-by-step solution

Taking g=9.8 m/s2\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}

294−N=30×0.1294-\mathrm{N}=30 \times 0.1

N=291\mathrm{N}=291

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 1
Subject
Physics
Chapter
Newton's Laws of Motion
Topic
Free Body Diagrams and Constraint Relations

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