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JEE Main 2025 (28 Jan, Shift 2) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2025 (28 Jan, Shift 2)

Q10JEE Main 2025 · 28 JAN · SHIFT 2Medium

Torque, Couple & Equilibrium of Rigid Bodies

A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is

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Answer: (c)

  1. Take moments about the knife edge at 40 cm. The rod's weight acts at its centre, 50 cm.
  2. Anticlockwise: 400×(40−10)=12000400 \times (40-10) = 12000 g cm
  3. Clockwise: 250×(50−40)+m×(90−40)=2500+50m250 \times (50-40) + m \times (90-40) = 2500 + 50m
  4. 12000=2500+50m⇒m=19012000 = 2500 + 50m \Rightarrow m = 190 g
  5. So the answer is (c) 190 g.
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