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NEET 2025
Exam
NEET
Year
2025
Subject
Physics
Chapter
Laws of Motion

NEET 2025 · Physics · Laws of Motion

A uniform rod of mass 20 kg and length 5 m leans against a smooth...

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60∘60^{\circ} with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 )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

Options

  1. A100N100 N
  2. B1003 N100 \sqrt{3} \mathrm{~N}
  3. C200N200 N
  4. D2003 N200 \sqrt{3} \mathrm{~N}

Correct answer

B. 1003 N100 \sqrt{3} \mathrm{~N}

Step-by-step solution

N2=mg=200 N\mathrm{N}_2=\mathrm{mg}=200 \mathrm{~N} 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
Taking torque about A : fs Lcos⁡60∘+mg L2sin⁡60∘=N2 Lsin⁡60∘fs(12)+200(12)32=20032fs=1003 N\begin{aligned} & \mathrm{f}_{\mathrm{s}} \mathrm{~L} \cos 60^{\circ}+\mathrm{mg} \frac{\mathrm{~L}}{2} \sin 60^{\circ}=\mathrm{N}_2 \mathrm{~L} \sin 60^{\circ} \\ & \mathrm{f}_{\mathrm{s}}\left(\frac{1}{2}\right)+200\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2}=200 \frac{\sqrt{3}}{2} \\ & \mathrm{f}_{\mathrm{s}}=100 \sqrt{3} \mathrm{~N} \end{aligned}
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