#thermal expansion

PHYSICS

A steel rod has a length of 1 m at 20° C . What will be its length at 120° C if the coefficient of linear expansion of

Given: A steel rod has a length of 1 m at 20° C . What will be its length at 120° C if the coefficient of linear expansion of steel is 1.2 × 10⁻⁵ K^{-1 ? These values define the system as per NCERT data. Formula: Given: L_0 = 1 m, T_1 = 20° C, T_2 = 120° C, α_l = 1.2 × 10⁻⁵ K^{-1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Δ T = 120 - 20 = 100° C . Linear expansion: Δ L = L_0 α_l Δ T = 1 × 1.2 × 10⁻⁵ × 100 = 1.2 × 10⁻³ m . New length: L = L_0 + Δ L = 1 + 0.0012 = 1.0012 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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Ideal Gas Laws and Thermal Properties

A steel rod of length 50cm at 20∘C is heated until its length increases by 0.03cm. What is the final temperature? (αl\=1

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.

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Ideal Gas Laws and Thermal Properties

Why does a metallic lid expand more than a glass jar when heated together?

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.

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Ideal Gas Laws and Thermal Properties

A steel sphere of radius 5cm at 20∘C is heated to 120∘C. What is the percentage increase in its volume? (αl\=1.2×10−5K−1

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.

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Ideal Gas Laws and Thermal Properties

A steel cylinder has a volume of 1.5L at 55∘C. What temperature must it be heated to for the volume to increase by 0.005

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.

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Fluid Pressure, Pascal's Law and Manometers

Which of the following is true about the coefficient of volume expansion compared to the coefficient of linear expansion

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.

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Latent Heat and Phase Transitions

A copper plate has an area of 0.8m2 at 60∘C. What is the decrease in area when cooled to 10∘C? (αl\=1.7×10−5K−1)

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.

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Latent Heat and Phase Transitions

A silver ring has an inner circumference of 31.4cm at 15∘C. What temperature must it be heated to for the circumference

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.

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Latent Heat and Phase Transitions

A brass ring has an inner diameter of 10cm at 40∘C. What temperature must it be cooled to for the diameter to decrease b

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.

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Latent Heat and Phase Transitions

Why do gases expand more than solids for the same temperature increase?

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.

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Latent Heat and Phase Transitions

A silver wire of length 1.2m at 25∘C is cooled until its length decreases by 0.0228cm. What is the final temperature? (α

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.

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Latent Heat and Phase Transitions

A brass rod of length 2m at 25∘C is heated to 125∘C. Calculate the increase in length. (Coefficient of linear expansion

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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