Elastic Properties of solids

The wire follows a dotted path parallel to OA back to zero stress but retains a permanent set (residual strain).

Step 1: Beyond the elastic limit, the wire behaves plastically — permanent deformation occurs in the crystal structure. Step 2: When load is removed, the wire does recover partially (elastic recovery), following a line parallel to the initial OA region of the curve. Step 3: However, it does NOT return to zero strain; a residual strain called 'permanent set' remains. Step 4: Option A is wrong because complete recovery is only possible within the elastic limit. Option C is wrong because the wire breaks only at the fracture point F, which is well beyond B. Option D incorrectly suggests higher str

5.17 km

Step 1: The weight of a wire of cross-sectional area A and length ℓ is W = Aℓρg. Step 2: Stress at the top due to its own weight = W/A = ℓρg. Step 3: At maximum length, this equals breaking stress: ℓ_max = Breaking stress / (ρg). Step 4: ℓ_max = (4.0 × 10⁸) / (7.9 × 10³ × 9.8) = (4.0 × 10⁸) / (7.742 × 10⁴) ≈ 5.17 × 10³ m ≈ 5.17 km. Option B results from dividing by 2ρg (factor of 2 error); option C from using ρ = 3.9 × 10³; option D from using g = 16 m/s². The key insight is that cross-sectional area cancels out, so maximum length is independent of wire thickness.

4 : 1

Step 1: Elastic potential energy U = ½kx², so U ∝ x². Step 2: In case 1: x₁ = 4 cm = 0.04 m; U₁ = ½ × 500 × (0.04)² = ½ × 500 × 0.0016 = 0.4 J. Step 3: In case 2: x₂ = 8 cm = 0.08 m; U₂ = ½ × 500 × (0.08)² = ½ × 500 × 0.0064 = 1.6 J. Step 4: Ratio U₂/U₁ = 1.6/0.4 = 4:1. This can also be seen directly: since x₂ = 2x₁, U₂/U₁ = (2x₁)²/x₁² = 4. Option A (2:1) is wrong — it assumes U ∝ x (linear), not quadratic. Option C (8:1) is wrong — it assumes U ∝ x³.

Rubber wire stretches more for the same force; hence more stress is needed in steel for the same strain, so steel is more elastic.

Step 1: Elasticity is defined as the ability to resist deformation — a body requiring more stress to produce the same strain is more elastic. Step 2: When equal forces are applied, rubber stretches far more than steel (lower Young's modulus for rubber). Step 3: To produce the SAME strain in steel as in rubber, a much larger stress is required — this means steel's internal restoring force is much stronger. Step 4: Hence steel is more elastic. Option A is self-contradictory. Option C is factually wrong — steel stretches far less. Option D confuses 'stretchability' (large strain capability) with