29 public questions tagged with this topic.
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: 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, 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.
Gases have a much higher coefficient of volume expansion than solids due to weaker intermolecular forces, allowing greater volume changes with temperature. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Gases have a higher expansion coefficient. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
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.
Metals have a higher coefficient of linear expansion (αl) than glass (Section 10.5, Table 10.1). For the same ΔT, ΔL = L0αlΔT is greater for metals due to their larger αl.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
