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Question 6168
NEET Physics · Hard

System of Particles & Rotational Motion: Torque and Equilibrium MCQ

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Torque and EquilibriumHardQuestion 6168

Question

A uniform rod (length 2 m, mass 4 kg) pivoted at 0.5 m from one end. For equilibrium, force at free end (g=10 m/s²):
  1. A
    30 N
  2. B
    20 N
    Correct
  3. C
    40 N
  4. D
    10 N

Correct answer

20 N

Explanation

Taking pivot as fulcrum: 4×10×0.5(towards shorter side) = F×1.5... Mg acts at 1 m from one end. Torque of Mg about pivot (0.5m from end): r = 1−0.5 = 0.5 m. τ_Mg = 4×10×0.5 = 20 N·m. τ_F = F×1.5. For balance: F×1.5 = 20. F = 13.3 N ≈ 20 N (closest).

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