14 public questions tagged with this topic.
Given: V0 = 1.5L = 1500cm3, ΔV = 0.0054L = 5.4cm3, αl = 1.2×10−5K−1, T1 = 55∘C. αv = 3αl = 3×1.2×10−5 = 3.6×10−5K−1. ΔV = V0αvΔT⇒5.4 = 1500×3.6×10−5×ΔT. ΔT = 5.41500×3.6×10−5 = 5.40.054 = 100K. T2 = 55+100 = 155∘C.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
The coefficient of volume expansion (αv) is three times the coefficient of linear expansion (αl), as αv = 3αl (Section 10.5). As per NCERT, applying relevant law/formula with correct units and sign convention leads to αv\=3αl. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Given: A0 = 0.8m2, ΔT = 10−60 = −50∘C, αl = 1.7×10−5K−1. ΔA = A0×2αlΔT = 0.8×2×1.7×10−5×(−50). ΔA = 0.8×3.4×10−5×(−50) = −0.00136m2 (decrease of 0.00136m2). As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.00136 m². This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Given: L0 = 31.4cm, ΔL = 0.0597cm, αl = 1.9×10−5K−1, T1 = 15∘C. ΔL = L0αlΔT⇒0.0597 = 31.4×1.9×10−5×ΔT. ΔT = 0.059731.4×1.9×10−5 = 0.05975.966×10−4≈100K. T2 = 15+100 = 115∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 115°C. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: L0 = 10cm, V0 = 103 = 1000cm3, ΔT = 130−30 = 100∘C, αl = 1.7×10−5K−1. Volume expansion: αv = 3αl = 3×1.7×10−5 = 5.1×10−5K−1. ΔV = V0αvΔT = 1000×5.1×10−5×100 = 5.1cm3. V = V0+ΔV = 1000+5.1 = 1005.1cm3.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Given: ΔT = 220−20 = 200∘C, ΔA = 0.018cm2, αl = 1.8×10−5K−1. Area expansion: ΔA = A0×2αlΔT. 0.018 = A0×2×1.8×10−5×200. 0.018 = A0×7.2×10−3⇒A0 = 0.0187.2×10−3 = 2.5cm2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 cm². This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: A0 = 1.2m2, ΔT = 5−45 = −40∘C, αl = 1.2×10−5K−1. Area expansion: ΔA = A0×2αlΔT = 1.2×2×1.2×10−5×(−40). ΔA = 1.2×2.4×10−5×(−40) = −0.001152m2 (decrease of 0.001152m2). As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.001152 m². This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: University Physics, Chapter 17: Thermal Expansion.
Given: V0 = 500cm3, ΔV = −0.85cm3, αl = 1.7×10−5K−1, T1 = 50∘C. αv = 3αl = 3×1.7×10−5 = 5.1×10−5K−1. ΔV = V0αvΔT⇒−0.85 = 500×5.1×10−5×ΔT. ΔT = −0.85500×5.1×10−5 = −0.850.0255≈−33.33K. T2 = 50−33.33≈16.67∘C≈16.7∘C.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Given: L0 = 5cm, L = 5.009cm, αl = 1.8×10−5K−1, T1 = 30∘C. ΔL = 5.009−5 = 0.009cm. ΔL = L0αlΔT⇒0.009 = 5×1.8×10−5×ΔT. ΔT = 0.0095×1.8×10−5 = 100K. T2 = 30+100 = 130∘C.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: L0 = 20cm, V0 = 203 = 8000cm3, ΔT = 125−25 = 100∘C, αl = 1.2×10−5K−1. αv = 3αl = 3×1.2×10−5 = 3.6×10−5K−1. ΔV = V0αvΔT = 8000×3.6×10−5×100 = 28.8cm3.
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.
