1. Which of the following unit of Young's modulus is in MKS system:[IOM 2011]
(a) Young's modulus = Stress/Strain = (N/m²)/(unitless) = N/m² (Pa in MKS). 2. The elasticity of highly elastic body is:[IOM 2009]
(c) Perfectly elastic bodies have elasticity = 1 (complete recovery after deformation). 3. Breaking stress (in N/m²) of a wire of radius 3 mm is F. The breaking stress of the same material of radius 6 mm will be:[MOE 2013, 2012]
(c) Breaking stress is intrinsic to material, independent of dimensions. 4. The stress-strain graph for copper and rubber shows slopes tanθ₁ (rubber) and tanθ₂ (copper). Then:[MOE 2010]
- tanθ₁ < tanθ₂
- tanθ₁ ≥ tanθ₂
- tanθ₁ = tanθ₂
- tanθ₁ > tanθ₂
(a) Copper (steeper slope) has higher Young's modulus than rubber. 5. The energy density of a wire of strain S and Young's modulus Y is:[MOE 2014]
(d) Energy density = ½ × Stress × Strain = ½ × (Y×S) × S = S²Y/2. 6. When length and load are doubled for a wire, the ratio of stress to strain:[KU 2014, 2011]
- Doubles
- Remains unchanged
- Becomes 4×
- Becomes 16×
(b) Stress/strain = Young's modulus (material property, unchanged). 7. Which statement is correct about elasticity?[KU 2014]
- Stress ∝ strain (elastic limit)
- Plasticity: strain occurs up to a limit
- Elasticity only in solids
- Perfectly plastic body has ∞ elasticity
(a) Hooke's Law: Stress ∝ strain within elastic limit. 8. Modulus of rigidity is:[KU 2013]
- Bulk stress/bulk strain
- Shear stress/shear strain
- Shear stress/bulk strain
- Normal stress × normal strain
(b) Modulus of rigidity (G) = Shear stress/Shear strain. 9. Energy stored per unit volume under pressure P (Young's modulus Y):[IE 2013]
(b) Energy density = ½ × Stress²/Y = P²/2Y. 10. For Poisson's ratio 0.20 and longitudinal strain 2×10⁻³, % volume change is:[IE 2013]
(a) ΔV/V = (1-2σ)×longitudinal strain = 0.6×2×10⁻³ = +0.12%. 11. Energy stored per unit volume (stress S, Young's modulus Y):[BP 2010]
(b) Energy density = ½ × Stress²/Y = S²/2Y. 12. Breaking strength of nylon rope (1.5cm dia) if 3cm rope breaks at 1.5×10⁵ N:[BP 2009]
- 0.75×10⁵ N
- 0.375×10⁵ N
- 1.5×10⁵ N
- 6×10⁵ N
(b) Breaking strength ∝ area ∝ r² → (1.5/3)² × 1.5×10⁵ = 0.375×10⁵ N. 13. Energy stored in spring B (K_A=2K_B) when spring A stores E:[BP 2012]
(b) Energy ∝ 1/K → E_B/E_A = K_A/K_B = 2 → E_B = 2E. 14. Young's modulus for wire (300cm, 0.003cm², ΔL=0.5cm, F=10⁷ dyne):[KU 2012]
- 4×10¹¹ dyne/cm²
- 4×10¹¹ N/m²
- 2×10¹¹ dyne/cm²
- 10¹² dyne/cm²
(c) Y = (FL)/(AΔL) = (10⁷×300)/(0.003×0.5) = 2×10¹¹ dyne/cm². 15. Force to double length of material (Y=2×10¹⁰ N/m², A=100 m²):[KU 2010]
- 4×10¹² N
- 2×10¹² N
- 2×10¹⁰ N
- 10¹⁸ N
(b) F = YAΔL/L = 2×10¹⁰ × 100 × 1 = 2×10¹² N. 16. Steel vs. rubber elasticity means for same stress:[Bangladesh 2009]
- More strain in steel
- Less strain in steel
- Equal strain
- Variable strain
(b) Higher Y (steel) → less strain for same stress. 17. Stress required to double wire length:[IOM 1996]
(b) Strain = 1 → Stress = Y×1 = Y (theoretical, beyond elastic limit). 18. Energy stored per unit volume (Y=2×10¹¹ N/m², strain=0.05):[MOE 2065]
- 2.5×10⁸ J/m³
- 5×10⁸ J/m³
- 3.5×10⁸ J/m³
- 4×10⁸ J/m³
(a) Energy density = ½Y×(strain)² = ½×2×10¹¹×(0.05)² = 2.5×10⁸ J/m³. 19. Extension of steel wire (L=2.5m, A=0.8×10⁻⁶ m², Y=2×10¹¹ Pa, F=8N):[MOE 2062]
- 0.125 mm
- 0.250 mm
- 0.500 mm
- 0.750 mm
(a) ΔL = FL/AY = (8×2.5)/(0.8×10⁻⁶×2×10¹¹) = 1.25×10⁻⁴ m = 0.125 mm. 20. Elastic energy per unit volume (stress S, Y):[MOE 2010]
(b) Energy density = ½ × Stress × Strain = ½ × S × (S/Y) = S²/(2Y). 21. Work done stretching wire (L, A, Y) by x:[IE-08]
- Yx²A/L
- Yx²A/2L
- 2Yx²A/L
- Yx²/2AL
(b) Work = ½kx² = ½(YA/L)x² = Yx²A/2L. 22. PE when string stretched from 2cm (U) to 10cm:[IE-08]
(c) PE ∝ x² → (10/2)² = 25 → PE = 25U. 23. Change in seawater density (ρ₀, B) at depth h:[BPKIHS-09]
- Bρ₀²gh
- ρ₀gh/B
- ρ₀²gh/B
- ρ₀gh/B²
(c) Δρ/ρ₀ = ΔP/B = ρ₀gh/B → Δρ = ρ₀²gh/B. 24. Substance with highest elasticity:[BPKIHS-05]
(c) Steel has highest Young's modulus (stiffest). 25. Length of wire (breaking stress=10⁷ N/m², density=3×10³ kg/m³) breaking under own weight:
(b) L = Breaking stress/(ρg) = 10⁷/(3×10³×9.8) ≈ 34 m. 26. Wire elongating most under same load:
- L=1m, d=1mm
- L=3m, d=3mm
- L=2m, d=2mm
- L=4m, d=0.5mm
(d) ΔL ∝ L/d² → highest for L=4m, d=0.5mm. 27. Angle of shear at surface for twisted wire (30° twist, L=1m, r=4mm):[BPKIHS]
(c) Surface shear angle = rθ/L = (4×10⁻³×30°)/1 = 0.12°. 28. Original spring length if elongation changes from 1cm to 5cm when ω doubled:
(b) F = kx = mω²(L+x) → solve for L = 15 cm. 29. Work done stretching wire by l under weight Mg:
(a) Work = ½kx² = ½ × (Mg/l) × l² = ½Mgl. 30. Elastic energy per unit volume in water (B, ρ, depth h):
- B(hρg)
- (hρg)²/2B
- ½B(hρg)²
- (hρg)/B
(b) Energy density = ½ × Stress²/B = (hρg)²/2B. 31. Relation between R_B (brass) and R_S (steel) for same ΔL under same F (Y_steel=2Y_brass):
- R_S = √2 R_B
- R_S = R_B/√2
- R_S = 4R_B
- R_S = R_B/4
(a) ΔL = FL/AY → A ∝ 1/Y → R_S/R_B = √(Y_B/Y_S) = √(1/2). 32. Spring constant K for metal wire (L, A, Y):
- K = YA
- K = YA/L
- K = YL/A
- K = Y/AL
(b) F = (YA/L)ΔL → K = YA/L. 33. Depression at center of loaded beam is proportional to:
(b) Depression δ ∝ 1/Y (for same load and geometry). 34. % volume change for longitudinal strain 10⁻³ (Poisson's ratio=0.2):
(c) ΔV/V = (1-2σ)×strain = 0.6×10⁻³ = 0.06%. 35. Depth of lake if bubble radius increases n times (atm pressure = H mmHg):
- h = (n³-1)H×13.6
- h = (n³-1)H
- h = (n-1)H×13.6
- h = (n-1)H
(a) P₁V₁ = P₂V₂ → hρg + 13.6Hg = 13.6Hg × n³ → h = (n³-1)H×13.6. 36. Depth for 0.2% density increase in rubber ball (B=10⁹ N/m², ρ=10³ kg/m³):
(b) Δρ/ρ = ΔP/B → h = (0.002×10⁹)/(10³×9.8) ≈ 200 m. 37. Decrease in 1L water volume under 2×10⁷ N/m² (compressibility=5×10⁻¹⁰ m²/N):
(b) ΔV = βVΔP = 5×10⁻¹⁰ × 1000 × 2×10⁷ = 10 cc. 38. Interatomic force constant for iron (Y=2×10¹¹ N/m², spacing=3×10⁻¹⁰ m):
- 60 N/m
- 120 N/m
- 3 N/m
- 180 N/m
(a) k = Y × r₀ = 2×10¹¹ × 3×10⁻¹⁰ = 60 N/m. 39. Hooke's Law: ratio of stress to strain when stress increases:
- Increases
- Decreases
- Remains constant
- None
(c) Within elastic limit, stress/strain = Y (constant). 40. Breaking load for wire divided into two identical parts:
(b) Breaking load depends on material strength, not length (each part still holds 80 kg). 41. Breaking force for wire radius 2r vs. radius r:
(d) Breaking force ∝ area ∝ r² → (2r)²/r² × F = 4F. 42. Ratio of twist angles for rods A and B (radii r₁ and r₂) under same torque:
(b) Twist angle ∝ 1/r⁴ → θ_A/θ_B = r₂⁴/r₁⁴. 43. Force developed in bar heated 0°C→100°C while constrained:
(a) F = YAαΔT → F ∝ α (thermal expansion coefficient). 44. Length of spring when tension is 9N (4N→a, 5N→b):
(b) Natural length L = (5a-4b); at 9N: L + (9/k) = 5b-4a. 45. Stress that changes object shape:[IOM 2015]
- Bulk stress
- Shear stress
- Tensile stress
- Longitudinal stress
(b) Shear stress causes shape deformation without volume change. 46. Young's modulus is:[IOM 2015]
- Normal stress/lateral strain
- Normal stress/longitudinal strain
- Normal stress × longitudinal strain
- Shear stress/lateral strain
(b) Y = Longitudinal stress/Longitudinal strain. 47. Effect of temperature on Young's modulus:[IOM 2016]
- Increases
- Decreases
- Remains constant
- First increases then decreases
(b) Y decreases with temperature (atomic bonds weaken). 48. Strain ratio for wires (length ratio 1:2, same radius/material, same force):[KU 2017]
(c) Strain = Stress/Y = (F/A)/Y → same for both (same material and stress). 