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Given: L0 = 50cm, ΔL = 0.03cm, αl = 1.2×10−5K−1, T1 = 20∘C. ΔL = L0αlΔT⇒0.03 = 50×1.2×10−5×ΔT. ΔT = 0.0350×1.2×10−5 = 500K. T2 = T1+ΔT = 20+500 = 520∘C.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Metals typically have a higher coefficient of linear expansion than glass, causing them to expand more for the same temperature increase. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Metal has a higher coefficient of expansion. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: ΔT = 120−20 = 100∘C, αl = 1.2×10−5K−1. αv = 3αl = 3×1.2×10−5 = 3.6×10−5K−1. ΔVV0 = αvΔT = 3.6×10−5×100 = 3.6×10−3. Percentage increase: ΔVV0×100 = 3.6×10−3×100 = 0.36%.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: r = 25cm, A0 = πr2 = π×252 = 625πcm2, ΔT = 130−30 = 100∘C, αl = 1.2×10−5K−1. ΔA = A0×2αlΔT = 625π×2×1.2×10−5×100. ΔA = 625π×2.4×10−3 = 1.5πcm2≈4.71cm2 (π≈3.14).
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: A0 = 225cm2 (one face), total surface area = 6×225 = 1350cm2, ΔT = 135−35 = 100∘C, αl = 1.8×10−5K−1. Area expansion: ΔA = A0×2αlΔT. For total surface area: ΔA = 1350×2×1.8×10−5×100 = 1350×3.6×10−3 = 4.86cm2.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
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