Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1 : 4, the ratio of their diameters is:

Solution

Physics sir · 4 steps

Same load, same steel: energy per volume goes as stress squared, so it falls as 1/d⁴.

1. Energy per volume. For a stretched wire, the energy stored per unit volume is 1/2 × stress × strain. Strain is stress divided by Y, so this becomes a formula in stress alone.

   u = (stress)²/2Y = 1/2Y(F/(π d²/4))² ∝ 1/d⁴

2. What stays fixed. Both wires are steel, so Y is the same. They carry the same load, so F is the same. The length does not appear in u, so only the diameter matters.
3. Take the ratio. u₁/u₂ = (d₂/d₁)⁴ = 1/4. Taking the fourth root gives d₂/d₁ = 1/√2.
4. So. d₁ : d₂ = √2 : 1. The wire storing less energy per volume is the thicker one, because its stress is lower. That is option (d).

The idea to keep. For the same load and material, energy per unit volume goes as 1/A², which is 1/d⁴. Square the stress, turn area into diameter, and check that the thicker wire stores less.

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