Question bank/NEET UG/Physics

NEET UGPhysicsModerateNEET 2025

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 g=10 m/s2g = 10\ \text{m/s}^2)

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Sol.
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For translational equilibrium

N1=MgN_1 = Mg
N2=fN_2 = f
For rotational equilibrium

Torque about A,
MgL2cos⁡θ=N2Lsin⁡θMg\frac{L}{2}\cos\theta = N_2L\sin\theta
Mg2cot⁡θ=N2=f\frac{Mg}{2}\cot\theta = N_2 = f
Mg2cot⁡30∘=f\frac{Mg}{2}\cot30^{\circ} = f
Mg23=N2\frac{Mg}{2}\sqrt{3} = N_2
1003=f100\sqrt{3} = f

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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