NEET Thermal Properties of Matter PYQs & Practice Questions

10 free NEET Physics questions on Thermal Properties of Matter, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Heat Conduction in Series · Temperature Scales · Thermal Expansion · Specific Heat Capacity · Calorimetry · Latent Heat · Heat Transfer – Conduction · Heat Transfer – Radiation · Newton's Law of Cooling · Anomalous Expansion of Water · Thermal Expansion - Angular Momentum · Thermal Expansion - Pendulum

1. Q1NEET 2026Thermal Expansion - Volume Expansion of a Sphere

   The temperature of a metallic sphere of radius R is increased by a small amount ΔT. If the linear coefficient of thermal expansion of the metal is α, the approximate increase in the volume of the sphere is:

   • A. 3πR³αΔT
   • B. 4πR³αΔT
   • C. 6πR³αΔT
   • D. 2πR³αΔT
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   Answer: A — 3πR³αΔT

   Volume V = (4/3)πR³. Differentiating: dV = 4πR² dR. Using linear expansion dR = RαΔT, we get dV = 4πR²(RαΔT) = 4πR³αΔT.

2. Q2NEET 2025Heat Conduction in Series

   Three identical heat conducting rods are connected in series. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is T1 and that at the right junction is T2. The ratio T1/T2 is

   • A. 3/2
   • B. 4/3
   • C. 5/3
   • D. 5/4
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   Answer: B — 4/3

   Equal heat current through each rod: (3T−T1)/(L/2kA) = (T1−T2)/(L/kA) = (T2−T)/(L/2kA). This gives 3T−T1 = T2−T ⟹ 4T = T1+T2, and 2(3T−T1) = T1−T2 ⟹ 3T1−T2 = 6T. Solving: T1 = (5/2)T, T2 = (3/2)T, so T1/T2 = 5/3.

3. Q3NEET 2021Newton's law of cooling time ratio

   A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken for a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is:

   • A. (13/10)t
   • B. (13/5)t
   • C. (10/13)t
   • D. (5/13)t
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   Answer: A — (13/10)t

   Using Newton's law of cooling approximation for average temperatures, solving gives time=(13/5)t for the 80→60°C interval.

4. Q4NEET 2020Heat required to raise temperature of spheres of different radii

   The quantities of heat required to raise the temperature of two solid copper spheres of radii r1 and r2 (r1=1.5 r2) through 1 K are in the ratio:

   • A. 27/8
   • B. 9/4
   • C. 3/2
   • D. 5/3
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   Answer: —

   Heat required Q = mcΔT ∝ mass ∝ volume ∝ r³ (for same material, same ΔT). Ratio = (r1/r2)³ = (1.5)³ = 3.375 = 27/8.

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5. Q5NEET 2019Thermal Conductivity

   The unit of thermal conductivity is:

   • A. W m K^-1
   • B. W m^-1 K^-1
   • C. J m K^-1
   • D. J m^-1 K^-1
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   Answer: A — W m K^-1

   Using the thermal conductivity formula k = (dQ/dt x l)/(A x dT), the SI unit of thermal conductivity is W m^-1 K^-1.

6. Q6NEET 2019Linear Thermal Expansion

   A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of the aluminium rod is: (alpha_Cu = 1.7x10^-5 K^-1 and alpha_Al = 2.2x10^-5 K^-1)

   • A. 88 cm
   • B. 68 cm
   • C. 6.8 cm
   • D. 113.9 cm
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   Answer: A — 88 cm

   For the difference in length to be independent of temperature, alpha_Cu x L_Cu = alpha_Al x L_Al. So (1.7x10^-5)(88) = (2.2x10^-5)(L_Al), giving L_Al = 1.7(88)/2.2 = 68 cm.

7. Q7NEET 2018Stefan-Boltzmann law: power radiated at a different wavelength of maximum emission

   The power radiated by a black body is P and it radiates maximum energy at wavelength, λ0. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength (3/4)λ0, the power radiated by it becomes nP. The value of n is

   • A. 3/4
   • B. 81/256
   • C. 256/81
   • D. 4/3
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   Answer: B — 81/256

   By Wien's law, λmT = constant, so T ∝ 1/λm. New temperature T' = T×(4/3) (since λ' = (3/4)λ0). By Stefan's law, P ∝ T⁴, so n = (T'/T)⁴ = (4/3)⁴ = 256/81.

8. Q8NEET 2017Thermal conductivity of composite rod

   Two rods A and B of different materials are welded together side by side (same length d, combined cross-section). Their thermal conductivities are K₁ and K₂. The thermal conductivity of the composite rod will be:

   • A. 3(K₁+K₂)/2
   • B. K₁+K₂
   • C. 2(K₁+K₂)
   • D. (K₁+K₂)/2
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   Answer: C — 2(K₁+K₂)

   For two identical rods of equal cross-sectional area placed side by side (in parallel for heat flow) with the same length, the effective thermal conductivity of the composite is the average: K_eff = (K₁+K₂)/2.

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9. Q9NEET 2017Stefan-Boltzmann law

   A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be:

   • A. 450
   • B. 1000
   • C. 1800
   • D. 225
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   Answer: B — 1000

   Power ∝ R²T⁴. New power = 450×(1/2)²×(2)⁴ = 450×(1/4)×16 = 450×4 = 1800 W.

10. Q10NEET 2016Heat Capacity and Common Temperature

    Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100°C, while the other one is at 0°C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is:

    • A. 0°C
    • B. 50°C
    • C. More than 50°C
    • D. Less than 50°C but greater than 0°C
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    Answer: B — 50°C

    Since heat capacity increases with temperature, the hotter body (100°C) has a higher heat capacity, so it loses less temperature change for the same heat transferred compared to the colder body gaining; the equilibrium temperature shifts closer to the hotter body's value, making it more than 50°C.