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Given: L0 = 1m, T1 = 20∘C, T2 = 120∘C, αl = 1.2×10−5K−1. ΔT = 120−20 = 100∘C. Linear expansion: ΔL = L0αlΔT = 1×1.2×10−5×100 = 1.2×10−3m. New length: L = L0+ΔL = 1+0.0012 = 1.0012m.
Ref: University Physics, Chapter 17: Thermal Expansion.
Given: r = 8cm, V0 = 43π×83 = 20483πcm3, ΔT = 20−70 = −50∘C, αl = 1.9×10−5K−1. αv = 3αl = 3×1.9×10−5 = 5.7×10−5K−1. ΔV = V0αvΔT = 20483π×5.7×10−5×(−50). ΔV = 20483π×(−2.85×10−3)≈−6.12π≈−19.22cm3 (decrease of 19.22cm3).
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Given: A0 = 1.5m2, ΔT = 120−20 = 100∘C, αl = 1.9×10−5K−1. ΔA = A0×2αlΔT = 1.5×2×1.9×10−5×100 = 0.0057m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.0057 m². This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
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