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Physics · Rotational Dynamics

JEE Main 2024 — 9 April, Shift 1 — Question 46

A heavy iron bar, of weight W is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle θ\theta with the horizontal. The weight experienced by the person is:

  1. Option A:

    W2\frac{W}{2}

    Correct
  2. Option B:

    W

  3. Option C:

    Wcos⁡θ\mathrm{W} \cos \theta

  4. Option D:

    Wsin⁡θ\mathrm{W} \sin \theta

Answer: A

Step-by-step solution

R=\mathrm{R}= net reaction force by shoulder

Balancing torque about pt of contact on ground: W(L2cos⁡θ)=R(Lcos⁡θ)\mathrm{W}\left(\frac{\mathrm{L}}{2} \cos \theta\right)=\mathrm{R}(\mathrm{L} \cos \theta)

⇒R=W2\Rightarrow \mathrm{R}=\frac{\mathrm{W}}{2}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 1
Subject
Physics
Chapter
Rotational Dynamics
Topic
Torque, Equation of Motion and Toppling

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