Wires W₁ and W₂ are made of same material having the breaking stress of 1.25 × 10⁹ N/m². W₁ and W₂ have cross-sectional area of 8 × 10⁻⁷ m² and 4 × 10⁻⁷ m², respectively. Masses of 20 kg and 10 kg hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is ____________ kg. (Use g = 10 m/s²)

Solution

Physics sir · 3 steps

Each wire has its own tension limit. Find the pan mass each one allows and take the smaller one.

1. Maximum tension. A wire breaks when its stress reaches the breaking stress, so the largest tension it can take is Tₘₐₓ = σ_(break) × A. The two wires have different areas, so they have different limits.
2. Load on each wire. A wire holds up everything hanging below it. W₁ carries the 20 kg block, the 10 kg block and the pan mass m. W₂ carries only the 10 kg block and m. The pan's own mass is taken as negligible.

   Wire	Tₘₐₓ = σ A	Tension it carries	Limit on m
   W₁	1.25 × 10⁹ × 8 × 10⁻⁷ = 1000 N	(20 + 10 + m) × 10	m ≤ 70 kg
   W₂	1.25 × 10⁹ × 4 × 10⁻⁷ = 500 N	(10 + m) × 10	m ≤ 40 kg

3. So. Both wires must survive, so the stricter limit wins. W₂ snaps first, and the maximum mass in the pan is 40 kg.

The common slip: 70. Using only the top wire W₁, because it carries the most weight, gives 70. The thinner wire W₂ breaks first, at 40.

The idea to keep. In a chain of wires, every wire must survive. Find the limit each wire sets, using the full load hanging below it, and the smallest limit is your answer.

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