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A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is:
(Take

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NEET 2025 Official Paper (Held On: 04 May, 2025)
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  1. 100 N
  2. \(100\sqrt{3}~\text{N}\)

  3. 200 N

  4. \(200\sqrt{3}~\text{N}\)

Answer (Detailed Solution Below)

Option 2 : \(100\sqrt{3}~\text{N}\)
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Detailed Solution

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Calculation:

For translational equilibrium

N1 = Mg

N2 = f

1 (4)

For rotational equilibrium

Torque about A, MgL/2 cosθ = N2L sinθ

(Mg/2) cotθ = N2 = f

(Mg/2) cot 30° = f

(Mg/2) √3 = N2

100√3 = f

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