A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at the 40 cm mark. A mas

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Question

A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at the 40 cm mark. A mass of 2 kg is suspended from the rod at the 20 cm mark, and another unknown mass 'm' is suspended from the rod at the 160 cm mark. Find the value of 'm' such that the rod is in equilibrium. (g = 10 m/s²)

Accepted Answer

Correct Answer: A — 1/12 kgStep-by-step Solution:Taking torques about the wedge at the 40 cm mark.Distances from wedge:2 kg mass (at 20 cm): 40 - 20 = 20 cm = 0.20 m to the left.Rod's centre of mass (at 100 cm): 100 - 40 = 60 cm = 0.60 m to the right.Unknown mass m (at 160 cm): 160 - 40 = 120 cm = 1.20 m to the right.Torque balance (anticlockwise = clockwise):Clockwise torque (2 kg on left): 2 × 10 × 0.20 = 4 N·m.Anticlockwise torques (rod weight + m on right): (0.5 × 10 × 0.60) + (m × 10 × 1.20) = 3 + 12m.Setting equal: 4 = 3 + 12m → 12m = 1 → m = 1/12 kg.Why other options are incorrect: Options B (0.5 kg), C (0.333 kg), and D (0.167 kg) do not satisfy the torque balance equation derived above.Key Concept: Condition for rotational equilibrium: Net torque about any point = 0. Torque = Force × perpendicular distance from pivot. Weight of uniform rod acts at its geometric centre.

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