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JEE Main 2025 Jan 28 Shift 2, Physics Q17: Centre of Mass

JEE Main2025Jan 28, Shift 2Physics
Q.

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

    g

  2. B

    g

  3. C

    g

  4. D

    g

Solution

Take torques about the pivot at the 40 cm mark. The rod's weight (250 g) acts at its centre, the 50 cm mark, i.e. 10 cm to the right of the pivot. The 400 g mass at 10 cm is 30 cm to the left; the unknown at 90 cm is 50 cm to the right.

Anticlockwise torque (left of pivot) gcm.

Clockwise torque (right of pivot) .

Setting them equal: g.

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