Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is 24/π × 10² Nm⁻². What is the minimum radius of the wire ? (Take g = 10 ms⁻²)

Solution

Physics sir · 4 steps

Find the tension in the wire first; the thinnest safe wire is the one whose stress equals the breaking stress.

1. Tension in wire. The heavier 5 kg block moves down and the 3 kg block moves up. For two masses on a smooth pulley, the tension is T = (2m₁ m₂ g)/(m₁ + m₂).

   T = (2 × 3 × 5 × 10)/(3 + 5) = 300/8 = 37.5 N

2. Stress condition. Stress is T/(π r²), and it must not go above 24/π × 10² N m⁻². At the minimum radius the two are equal: 37.5/(π r²) = 2400/π.
3. Solve for r. The π cancels, so r² = 37.5/2400 = 0.015625 m². Then r = 0.125 m.
4. So. r = 0.125 m = 12.5 cm, which is option (c).

The idea to keep. For a breaking-stress question, find the actual force in the wire first, not just the weight. Then the minimum area is force divided by breaking stress.

Answer from the NTA final answer key. Solution by PrepWiser.