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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 = 1.2m = 120cm, ΔL = −0.0228cm, αl = 1.9×10−5K−1, T1 = 25∘C. ΔL = L0αlΔT⇒−0.0228 = 120×1.9×10−5×ΔT. ΔT = −0.0228120×1.9×10−5 = −0.02282.28×10−3 = −10K. T2 = 25−10 = 15∘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.
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