Focuses on stress, strain, and Hooke's Law with problems on elastic deformation and material properties. Helps students apply these concepts in physics and engineering contexts.
Fluids lack rigidity and cannot resist shear stress without flowing. Any attempt to apply shear stress causes deformation (flow), as fluids have no fixed shape, unlike solids, which resist shear due to strong intermolecular forces.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Gases have weaker intermolecular forces and larger intermolecular spaces compared to liquids, allowing them to compress significantly under pressure, while liquids’ strong forces resist volume changes. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Weaker intermolecular forces. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.4×0.3 = 0.12m2, L = 0.1m. Substitute: Δx = 3×104×0.10.12×3.6×1010 = 30004.32×109≈6.94×10−7m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.94×10−7m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Shear modulus: G = (Shear stress) / (Shear strain). Shear stress: Shear stress = F / A, A = 0.2 × 0.1 = 0.02 m2. Shear stress = (1 × 104) / 0.02 = 5 × 105 N/m2. Shear strain: Shear strain = Shear stress / G = (5 × 105) / (8.4 × 1010) ≈ 5.95 × 10-6.
Ref: University Physics, Chapter 17: Thermal Expansion.
Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −4×1062.2×109≈−1.82×10−3. Volume: V = 3litres = 3×10−3m3. Change in volume: ΔV = ΔVV×V = −1.82×10−3×3×10−3≈−5.45×10−6m3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.45×10−6m3. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Shear modulus: G = Shear stressShear strain. Shear stress: Shear stress = FA, A = 0.5×0.2 = 0.1m2. Shear stress: 2×1040.1 = 2×105N/m2. Shear strain: Shear strain = Shear stressG = 2×1053.6×1010≈5.56×10−6.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 7×1010×2×10−4 = 1.4×107N/m2. Force: F = Stress×A = 1.4×107×2×10−6 = 28N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 28N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Shear modulus: G = (Shear stress) / (Shear strain). Shear stress: Shear stress = F / A, A = 0.5 × 0.2 = 0.1 m2. So, Shear stress = (6 × 104) / 0.1 = 6 × 105 N/m2. Shear strain: Shear strain = Shear stress / G = (6 × 105) / (2.5 × 1010) = 2.4 × 10-5.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.52×10−3m. F = 2×1011×2×10−6×0.52×10−32.6 = 2082.6 = 80N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 80N. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: University Physics, Chapter 17: Thermal Expansion.
Shear modulus: G = Shear stressShear strain. Shear stress: Shear stress = FA, A = 0.5×0.2 = 0.1m2. Shear stress: 5×1040.1 = 5×105N/m2. Shear strain: Shear strain = Shear stressG = 5×1053.6×1010≈1.39×10−5.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 160×2.82×10−6×1.1×1011 = 4482.2×105≈2.04×10−3m = 2.04mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.04mm. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Stress: Stress = FA = 2502.5×10−6 = 1×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 1×1082×1011 = 5×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5×10−4. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: University Physics, Chapter 17: Thermal Expansion.
