Elasticity
ICSE · Class 11 · Physics
Practice quiz for Elasticity — ICSE Class 11 Physics. MCQs and questions with answers to test your preparation.
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A steel wire of length 2 m and cross-sectional area 2 × 10⁻⁶ m² is stretched by a force of 400 N. If Young's modulus of steel is 2 × 10¹¹ N/m², what is the elastic potential energy stored per unit volume of the wire?
Two wires A and B of the same material have lengths L and 2L and radii r and 2r respectively. If the same force is applied to both wires, the ratio of the elongation of wire A to wire B (ΔL_A / ΔL_B) is:
A uniform rod of length L, cross-sectional area A, Young's modulus Y and density ρ is hung vertically from its upper end. The elongation of the rod due to its own weight is:
Two identical rods are joined end to end. The Young's modulus of the first rod is Y₁ and of the second is Y₂. When the combination behaves as a single wire under the same load, the equivalent Young's modulus Y_eq is:
Sample Questions
The Poisson's ratio of a material is 0.4. If a wire of this material is stretched so that its length increases by 0.2%, by what percentage does its radius decrease?
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0.08%
Step 1: Poisson's ratio σ = (lateral strain) / (longitudinal strain) = (-Δr/r) / (ΔL/L). Step 2: Longitudinal strain = ΔL/L = 0.2% = 0.002. Step 3: Lateral strain = σ × longitudinal strain = 0.4 × 0.002 = 0.0008. Step 4: Percentage decrease in radius = 0.0008 × 100% = 0.08%. Step 5: Common error — students directly multiply 0.4 × 0.2% = 0.08%, which happens to be correct, but they must understand the formula. Option C (0.4%) occurs when students confuse σ with the answer directly. Option D (0.8%) occurs when students forget to take σ as a ratio and just add.
A steel wire (Y = 2 × 10¹¹ N/m²) of length 5 m and diameter 10⁻³ m supports a sphere of mass 8π kg and radius 0.1 m. The wire hangs from a ceiling 5.22 m high. When the sphere is swung as a pendulum, what is the approximate velocity at the lowest point? (g = 9.8 m/s²)
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8.8 m/s
Step 1: Extension Δl = 5.22 - (l + 2r) = 5.22 - (5 + 0.2) = 0.02 m. Step 2: Tension T = YAΔl/l = 2×10¹¹ × π×(5×10⁻⁴)² × 0.02/5 = 2×10¹¹ × π×25×10⁻⁸ × 0.004 = 2×10¹¹ × π × 10⁻⁹ = 200π N ≈ 628 N. Step 3: At mean position of pendulum, T - Mg = Mv²/R, where R = 5.22 - 0.1 = 5.12 m and Mg = 8π × 9.8 ≈ 246.4 N. Step 4: Mv²/R = T - Mg = 200π - 8π×9.8 = π(200 - 78.4) = 121.6π. Step 5: v² = 121.6π × 5.12/(8π) = 121.6 × 5.12/8 = 622.6/8 ≈ 77.8; v ≈ 8.8 m/s.
The interatomic force constant k of a solid material is related to its Young's modulus Y and equilibrium interatomic distance r₀ by which of the following relations?
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k = Y × r₀
Step 1: Consider a wire of length L and cross-sectional area A. There are A/r₀² atomic chains in the cross-section. Step 2: Force on each chain = F/(A/r₀²) = Fr₀²/A. Step 3: Each chain has L/r₀ atoms, so strain = Δr/r₀ = (l/L) (where l is total elongation). Step 4: k = F'/Δr = (Fr₀²/A)/(r₀ × l/L) = (F/A)/(l/L) × r₀ = Y × r₀. Step 5: Option A (Y/r₀) is dimensionally wrong — Y has units N/m² and r₀ has units m, so Y×r₀ gives N/m (force per length = spring constant), which is correct for k. Y/r₀ gives N/m³ which is incorrect.
In the stress-strain graph of a ductile material, the region between yield point C and the point D of maximum stress is called:
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Plastic region
Step 1: In the stress-strain graph, OA is the proportional region (Hooke's law obeyed). Step 2: OB is the elastic region (up to elastic limit at B — body returns to original shape). Step 3: Beyond B, at point C (yield point), permanent deformation begins. Step 4: The region from C to D (maximum stress / ultimate stress) is the plastic region — here the material flows like a viscous fluid and permanent deformation occurs. Step 5: Beyond D, the wire thins (necking) and finally breaks at E. The breaking stress is at D, not E. The elastic region (option A) is from O to B, not C to D.
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