6 public questions tagged with this topic.
Given: L0 = 10cm, ΔL = −0.009cm, αl = 1.8×10−5K−1, T1 = 40∘C. ΔL = L0αlΔT⇒−0.009 = 10×1.8×10−5×ΔT. ΔT = −0.00910×1.8×10−5 = −0.0091.8×10−4 = −50K. T2 = 40−50 = −10∘C.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Given: L0 = 2m, ΔT = 125−25 = 100∘C, αl = 1.8×10−5K−1. ΔL = L0αlΔT = 2×1.8×10−5×100 = 3.6×10−3m = 3.6mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.6 mm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
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.
Shear modulus: G = (Shear stress) / (Shear strain). Shear stress: Shear stress = F / A = (4 × 104) / 0.05 = 8 × 105 N/m2. Shear strain: 2 × 10-5. Substitute: G = (8 × 105) / (2 × 10-5) = 4 × 1010 N/m2.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×5×10−4 = 4.5×107N/m2. Force: F = Stress×A = 4.5×107×3×10−6 = 135N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 135N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Young's modulus: Y = (F L) / (A ΔL). Substitute: ΔL = 0.25 × 10-3 m. Y = (150 × 1.5) / (2.5 × 10-6 × 0.25 × 10-3) = 225 / (6.25 × 10-10) = 3.6 × 1011 N/m2.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
