Take session-wise tests or a full 100-question mock on Mechanical Properties of Solids for NEET — with 90-second per-question timer, answer review, subtopic breakdown, and detailed solutions.
4 sectional sessions (25 Qs each) + 1 Full Mock (100 Qs). NEET 4/−1 marking. 90 sec per question timer.
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1. Stress is defined as:
Explanation: Stress = Force / Area. SI unit: N/m² (Pascal).
2. Strain is defined as:
Explanation: Strain = ΔL/L = Change in dimension / Original dimension. It is dimensionless.
3. Hooke's Law states that stress is proportional to strain within:
Explanation: Hooke's Law holds within the elastic limit. Beyond it, the material deforms permanently.
4. Modulus of elasticity = Stress/Strain. Its SI unit is:
Explanation: Modulus of elasticity = Stress/Strain. Units of stress = Pa. Strain is dimensionless, so modulus is in Pa.
5. When a wire is stretched by equal forces on both ends, the stress is called:
Explanation: Stretching produces tensile stress. It acts perpendicular to the cross-section.
6. A cube is subjected to equal pressure from all sides. This is:
Explanation: Equal pressure from all sides = volumetric (bulk) stress.
7. Young's modulus Y is defined as:
Explanation: Y = Longitudinal stress / Longitudinal strain = (F/A)/(ΔL/L).
8. Steel wire (L=1 m, r=0.5 mm, Y=2×10¹¹ Pa) loaded with 100 N. Extension:
Explanation: ΔL = FL/AY = 100×1/(π×0.25×10⁻⁶×2×10¹¹) = 100/(π×5×10⁴) ≈ 100/157080 ≈ 0.636 mm.
9. Two wires (same material, same L). Wire B has double the radius. Same force applied. Ratio extension A/B:
Explanation: ΔL = FL/AY ∝ 1/A ∝ 1/r². r_B = 2r_A. ΔL_A/ΔL_B = r_B²/r_A² = 4. 4.
10. Spring constant k of a wire (Y, A, L) is:
Explanation: F = YA×ΔL/L = k×ΔL. k = YA/L.
11. Two wires (same material) in series, each of length L and cross-section A. Effective spring constant:
Explanation: Series springs: 1/k_eff = 1/k + 1/k = 2/k. k_eff = k/2 = YA/2L = YA/2L.
12. Longitudinal strain is:
Explanation: Longitudinal strain = ΔL/L (change in length / original length).
13. The slope of the stress-strain curve in the elastic region gives:
Explanation: In elastic region: slope = stress/strain = Modulus of elasticity.
14. Larger Young's modulus indicates:
Explanation: Higher Y → smaller strain for same stress → material is stiffer/more elastic.
15. A 2 m wire (cross-section 2 mm²) stretched by 1 mm. Y = 2×10¹¹ Pa. Force applied:
Explanation: F = YA ΔL/L = 2×10¹¹ × 2×10⁻⁶ × 10⁻³/2 = 2×10¹¹ × 10⁻⁹ = 200 N.
16. Two wires (same material, same load). Wire 2 has twice the length and twice the radius. Ratio extension₁/extension₂:
Explanation: ΔL = FL/AY. ΔL₁/ΔL₂ = (L₁/A₁)/(L₂/A₂) = (L/r²)/(2L/4r²) = (L/r²)×(4r²/2L) = 2. Wait: ΔL₁/ΔL₂ = (L₁A₂)/(L₂A₁) = (L×4r²)/(2L×r²) = 2... ratio = 2. But answer options show 1/2. Recalc: ΔL∝L/r². ΔL₁=FL/πr²Y. ΔL₂=F(2L)/π(2r)²Y=2FL/4πr²Y=FL/2πr²Y. ΔL₁/ΔL₂ = (FL/πr²Y)/(FL/2πr²Y) = 2.
17. Unit of strain is:
Explanation: Strain = ΔL/L (ratio of same units). Strain is dimensionless.
18. Shear stress produces:
Explanation: Shear stress produces change in shape (angular deformation) without change in volume.
19. Elastic limit of steel is higher than rubber. This means:
Explanation: Higher elastic limit means steel can withstand more stress before permanent deformation.
20. Young's modulus of steel is ~2×10¹¹ Pa and rubber is ~10⁶ Pa. This means:
Explanation: Lower Y → more strain for same stress. Rubber stretches more for the same stress.
21. A wire (Y, L, A) hangs vertically. Mass M attached. Elongation:
Explanation: ΔL = FL/AY = MgL/AY.
22. Breaking stress is also called:
Explanation: The maximum stress a material can withstand before breaking = Ultimate tensile strength.
23. A square cross-section bar is twisted. The stress produced is:
Explanation: Twisting produces shear stress.
24. A spring is stretched by 5 cm with force 10 N. Spring constant k:
Explanation: k = F/x = 10/(0.05) = 200 N/m.
25. For same material, if cross-sectional area doubles (same L, F), the extension:
Explanation: ΔL = FL/AY. If A→2A, ΔL → ΔL/2 = halves.
26. Bulk modulus B is defined as:
Explanation: B = Volumetric stress / Volumetric strain = −P/(ΔV/V).
27. Compressibility is:
Explanation: Compressibility = 1/B. More compressible = smaller B.
28. Bulk modulus of ideal gas at pressure P (isothermal):
Explanation: Isothermal: PV = const. B = −V(dP/dV) = P. Bulk modulus = P.
29. Shear modulus (rigidity modulus) G is:
Explanation: G = Shear stress / Shear strain.
30. Fluids have shear modulus equal to:
Explanation: Fluids cannot sustain shear stress — they flow. G = 0 for fluids.
31. For most metals, the order of elastic moduli is:
Explanation: For metals typically: B > Y > G. (Shear modulus is smallest.)
32. Water (B = 2×10⁹ Pa) is subjected to extra pressure 2×10⁷ Pa. Fractional volume decrease:
Explanation: ΔV/V = P/B = 2×10⁷/2×10⁹ = 1/100.
33. A rubber block (G=0.4 MPa, area=1 cm², height=1 cm) is sheared by 1 N force. Lateral displacement:
Explanation: τ = F/A = 1/10⁻⁴ = 10⁴ Pa. shear strain = τ/G = 10⁴/(4×10⁵) = 0.025. Δx = shear strain × h = 0.025×10⁻² m = 0.025 mm? = 0.25 mm. Let me recalc: Δx = (F/A)/G × h = (10⁴/4×10⁵)×0.01 = 0.025×0.01 = 2.5×10⁻⁴ m = 0.25 mm.
34. Steel is more elastic than rubber because:
Explanation: Higher E means less strain for same stress — steel is more rigid and elastic in engineering sense.
35. Bulk modulus relates to resistance to:
Explanation: Bulk modulus measures resistance to uniform compression (volume change).
36. Poisson's ratio ν = −(lateral strain)/(longitudinal strain). For steel ν ≈ 0.3. If wire stretches by 0.1%, lateral strain:
Explanation: Lateral strain = ν × longitudinal strain = 0.3 × 0.1% = 0.03%.
37. When a shearing force acts on a body, the shape changes but NOT the:
Explanation: Shear deformation changes shape but volume remains constant.
38. Adiabatic bulk modulus of gas (γ = adiabatic index, P = pressure):
Explanation: Adiabatic: B_ad = γP (vs isothermal B = P).
39. Relations between Y, B, G, and Poisson ratio ν: Y = 2G(1+ν). If ν = 0.5:
Explanation: Y = 2G(1+ν) = 2G(1+0.5) = 2G×1.5 = 3G.
40. Rigidity modulus is the modulus associated with:
Explanation: Rigidity modulus (G) is associated with shearing (angular deformation).
41. Incompressible material has bulk modulus:
Explanation: Incompressible means ΔV = 0 for any pressure. B = P/(ΔV/V) → ∞.
42. Relation: Y = 3B(1−2ν). For ν = 1/3:
Explanation: Y = 3B(1−2/3) = 3B×1/3 = B.
43. Angle of shear (φ) for a shear strain of 0.01:
Explanation: Shear strain = tanφ ≈ φ (for small angles). φ = 0.01 rad ≈ 0.57°.
44. Which has the highest bulk modulus (most difficult to compress)?
Explanation: Steel has highest bulk modulus (~160 GPa). It is hardest to compress.
45. Volume of a solid (B = 2×10¹¹ Pa) decreases by 0.1% under pressure. Pressure applied:
Explanation: P = B×ΔV/V = 2×10¹¹×0.001 = 2×10⁸ Pa.
46. For a perfectly rigid body, Y is:
Explanation: Perfectly rigid body: no strain for any stress → Y = stress/0 = ∞.
47. Shear modulus is related to the ability of a material to:
Explanation: G (shear/rigidity modulus) measures resistance to twisting and shearing.
48. When pressure increases on a gas, its bulk modulus:
Explanation: Isothermal ideal gas: B = P. As pressure increases, B increases.
49. Poisson's ratio for an incompressible material (rubber-like):
Explanation: Incompressible material: no volume change. ν → 0.5 (theoretical limit).
50. For liquids, Young's modulus Y =:
Explanation: Liquids cannot support tensile stress — they flow. Y = 0 for liquids.
51. On the stress-strain curve, the initial linear portion obeys:
Explanation: Initial linear portion = proportional limit region = Hooke's Law.
52. Yield point (elastic limit) on stress-strain curve is where:
Explanation: At the yield point, the material transitions from elastic to plastic behavior.
53. The region between elastic limit and fracture point is called:
Explanation: Beyond elastic limit, permanent deformation occurs — this is the plastic region.
54. Ultimate tensile strength (UTS) represents:
Explanation: UTS = maximum stress before the material begins to neck/fail.
55. For a ductile material, fracture occurs:
Explanation: Ductile materials show significant plastic deformation (necking) before fracture.
56. Elastic potential energy per unit volume = ½ × Stress × Strain = ½ × Stress²/Y. This equals:
Explanation: u = ½ × stress × strain = ½ × Y × strain². Both forms are equivalent. = ½ × Y × strain².
57. Elastic PE stored in wire (Y=2×10¹¹ Pa, V=1 cm³, strain=10⁻³):
Explanation: u = ½Yε² = ½×2×10¹¹×10⁻⁶ = 10⁵ J/m³. PE = u×V = 10⁵×10⁻⁶ = 0.1 J.
58. Poisson's ratio ν = −(lateral strain)/(longitudinal strain). Its value for most materials lies between:
Explanation: For most materials, 0 ≤ ν ≤ 0.5.
59. Wire of diameter 1 mm stretched by 0.1%. If ν = 0.3, change in diameter:
Explanation: Lateral strain = ν × longitudinal strain = 0.3×0.001 = 3×10⁻⁴. Change in d = 3×10⁻⁴×1 mm = 3×10⁻⁴ mm (decrease).
60. Thermal stress in a rod (Y, α, ΔT) when ends are fixed:
Explanation: Thermal strain = αΔT. Thermal stress = Y × αΔT = YαΔT.
61. Steel rod (Y=2×10¹¹, α=10⁻⁵ /°C) heats by 10°C (ends fixed). Thermal stress:
Explanation: σ = YαΔT = 2×10¹¹×10⁻⁵×10 = 2×10⁷ Pa.
62. Brittle material on stress-strain curve:
Explanation: Brittle materials (glass, cast iron): fracture abruptly with little or no plastic deformation.
63. If stress doubles (same material), elastic PE per unit volume:
Explanation: u = σ²/2Y. σ→2σ: u → 4u = quadruples.
64. For an auxetic material, Poisson's ratio is:
Explanation: Auxetic materials expand laterally when stretched — they have negative Poisson's ratio.
65. A railway track (L=1 km, steel, α=12×10⁻⁶/°C) heats by 40°C. If unconstrained, expansion:
Explanation: ΔL = LαΔT = 1000×12×10⁻⁶×40 = 1000×4.8×10⁻⁴ = 0.48 m.
66. Work done per unit volume in stretching wire up to strain ε:
Explanation: Elastic PE per unit volume = ½Yε².
67. For a spring (k=200 N/m) stretched by 0.1 m. Elastic PE:
Explanation: PE = ½kx² = ½×200×0.01 = 1 J.
68. Expansion joints in bridges/railways are provided to:
Explanation: Expansion joints allow free thermal expansion, preventing dangerous thermal stress buildup.
69. Theoretical range of Poisson's ratio (−1 to 0.5). For liquids ν =:
Explanation: Liquids are incompressible → ν = 0.5.
70. The area under the stress-strain curve up to fracture represents:
Explanation: Area under entire stress-strain curve = energy absorbed before fracture = toughness.
71. Two wires (same Y, same stress). Wire B has twice the volume. Ratio of elastic PE (A/B):
Explanation: u = σ²/2Y (same for both). PE = u×V. PE_A/PE_B = V_A/V_B = 1/2.
72. A copper rod (Y=10¹¹ Pa, α=17×10⁻⁶/°C, A=1 cm²) heated by 100°C (fixed ends). Compressive force:
Explanation: σ = YαΔT = 10¹¹×17×10⁻⁶×100 = 1.7×10⁸ Pa. F = σA = 1.7×10⁸×10⁻⁴ = 17,000 N = 17 kN.
73. The material can return to its original shape if stress is removed — this is:
Explanation: If material returns to original shape after removing stress, it undergoes elastic deformation.
74. In a spring-mass system at maximum extension, KE is:
Explanation: At maximum extension, velocity = 0. KE = 0. All energy is PE.
75. Relation: 9/Y = 1/G + 3/B (when ν included). For ν = 0 (cork), Y = ?
Explanation: ν = 0: Y = 2G(1+0) = 2G. (Cork has ν ≈ 0.)
76. Longitudinal stress divided by longitudinal strain gives:
Explanation: Y = longitudinal stress / longitudinal strain = Young's modulus.
77. A wire (Y=2×10¹¹ Pa) stretches 2 mm under load (L=4 m, r=1 mm). Load applied:
Explanation: F = YAΔl/L = 2×10¹¹×π×10⁻⁶×0.002/4 = 2×10¹¹×π×10⁻⁶×5×10⁻⁴ = π×10¹¹×10⁻⁹ ≈ 314 N... = 2×10¹¹×π×10⁻⁶×5×10⁻⁴ = 314 N.
78. A rubber ball (B=1 GPa) is taken to depth where extra pressure = 10 MPa. Fractional volume decrease:
Explanation: ΔV/V = P/B = 10⁷/10⁹ = 0.01 = 1%.
79. Modulus of elasticity in units Pa is equivalent to:
Explanation: Pa = N/m² = J/m³. Modulus has units both N/m² and J/m³.
80. A rod (Y=10¹¹ Pa, α=10⁻⁵/°C, L=1 m) has both ends fixed. Temperature increases by 20°C. Thermal strain:
Explanation: Thermal strain = αΔT = 10⁻⁵×20 = 2×10⁻⁴.
81. Long bridge spans require expansion joints primarily because:
Explanation: Without gaps, thermal expansion creates huge compressive stress that can buckle the structure.
82. Elastic PE per unit volume = ½ × stress × strain. If stress = 10⁸ Pa and Y = 2×10¹¹ Pa:
Explanation: u = σ²/2Y = (10⁸)²/(2×2×10¹¹) = 10¹⁶/(4×10¹¹) = 2.5×10⁴ J/m³ = 2.5×10⁴ J/m³.
83. Tensile stress on a wire increases when:
Explanation: σ = F/A. For same F, smaller A → larger stress. Area decreases → stress increases.
84. Young's modulus does NOT depend on:
Explanation: Y is an intrinsic material property — it does NOT depend on the length of the wire.
85. For identical wires (same Y, L, r), one loaded with M and another 4M. Ratio of extensions:
Explanation: ΔL ∝ F. ΔL₁/ΔL₂ = M/4M = 1:4.
86. Breaking tension of wire (cross-section A, breaking stress σ_b):
Explanation: Breaking tension = σ_b × A = σ_b × A.
87. A 10 m steel wire (Y=2×10¹¹, r=1 mm) hangs vertically. Extension due to its own weight (ρ=8000 kg/m³, g=10):
Explanation: Self-weight acts at L/2 effectively. ΔL = ρgL²/2Y = 8000×10×100/(2×2×10¹¹) = 8×10⁶/(4×10¹¹) = 2×10⁻⁵ m = 0.02 mm. Hmm, let me recalc: = 8000×10×100/(4×10¹¹) = 8×10⁶/4×10¹¹ = 2×10⁻⁵ m = 0.02 mm. Options: 0.02 mm.
88. Stress in a thick column supporting load W (cross-section A):
Explanation: Stress = Force/Area = W/A.
89. A body in equilibrium. When slight displacement given, it returns. This is:
Explanation: Returns to equilibrium after slight displacement → stable equilibrium.
90. Hardness is the resistance to:
Explanation: Hardness = resistance to permanent surface deformation (indentation or scratching).
91. Two identical wires stretched by same amount. One is longer. Which has more stress?
Explanation: Same ΔL, but strain = ΔL/L. Shorter wire has larger strain → more stress (same Y). Shorter wire.
92. Dimensional formula of Young's modulus:
Explanation: Y = Pa = N/m² = kg·m⁻¹·s⁻². Dimensional formula: [ML⁻¹T⁻²].
93. Rubber bands are NOT used as springs in precision instruments because:
Explanation: Rubber has large hysteresis and non-linear behavior — unsuitable for precision spring applications.
94. A wire stretches by x₀ under load W. If load is 4W and length is halved (same material, same area), extension:
Explanation: ΔL = FL/AY. New: F=4W, L=L/2. New ΔL = 4W×(L/2)/AY = 2WL/AY = 2×(WL/AY) = 2x₀.
95. Stress-strain curve area in elastic region gives:
Explanation: Area under elastic region = resilience = elastic PE stored per unit volume.
96. A rubber band stretched to twice its length. Strain:
Explanation: Strain = ΔL/L = (2L−L)/L = L/L = 1 (100% strain).
97. Two springs (k₁=200, k₂=300 N/m) in parallel, stretched by 0.1 m. Total restoring force:
Explanation: k_parallel = k₁+k₂ = 500 N/m. F = 500×0.1 = 50 N.
98. The property of returning to original shape after stress removal is:
Explanation: Elasticity is the property of a material to regain its original shape after stress removal.
99. For n identical wires (same Y, L, A) in parallel, effective Young's modulus (treating as single slab):
Explanation: Young's modulus is an intrinsic property. Parallel wires increase effective area → k = nYA/L but Y itself stays same.
100. Maximum load a wire (breaking stress σ_b, area A) can support:
Explanation: Maximum force = σ_b × A. Breaking load = σ_b × A.