Two wires each of radius 0.2 cm and negligible mass, one made of steel and the other made of brass are loaded as shown in the figure. The elongation of the steel wire is __________ × 10⁻⁶ m. [Young's modulus for steel = 2 × 10¹¹ Nm⁻² and g = 10 ms⁻² ]

Solution

Physics sir · 4 steps

The steel wire holds up everything below it, so its tension comes from both masses: 2 kg plus 1.14 kg.

1. Tension in steel. The brass wire is massless, so the steel wire supports the 2 kg block and also the 1.14 kg block hanging below it. Total mass = 3.14 kg, so T = 3.14 × 10 = 31.4 N.
2. Cross-section area. With r = 0.2 cm = 2 × 10⁻³ m, A = π r² = 3.14 × 4 × 10⁻⁶ m². Keep the 3.14 as it is, because it cancels against the 3.14 in the tension.

   Δ L = (T L)/(A Y) = (3.14 × 10)(1.6)/(3.14 × 4 × 10⁻⁶)(2 × 10¹¹)

3. Simplify. After cancelling 3.14: Δ L = 16/(8 × 10⁵) = 2 × 10⁻⁵ m.
4. So. Δ L = 20 × 10⁻⁶ m, so the answer is 20.

The common slip: 12.7. Taking the tension in steel as only the 2 kg weight (20 N) gives about 12.7, not 20. The 1.14 kg block also pulls on the steel wire through the brass wire.

The idea to keep. In a hanging chain of wires and blocks, the tension in any wire is the weight of everything below it. Find that tension first, then use Δ L = TL/AY.

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