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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Rotational Mechanics

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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
  1. (A)300 g
  2. (B)190 g✓ Correct
  3. (C)200 g
  4. (D)290 g

Answer: (B)

Solution

For the rod to remain balanced, the net torque about the pivot at the 40 cm mark must vanish.

The rod's weight acts at its centre (the 50 cm mark), a distance of 10 cm from the pivot, on the same side as the 90 cm mark. The 400 g mass at the 10 cm mark is 30 cm from the pivot on the opposite side. The unknown mass at the 90 cm mark is 50 cm from the pivot, on the same side as the rod's weight.

Balancing torques about the pivot:



JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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