Thermal Properties of Matter NEET PYQs (2005–2026) with Solutions

10 solved Thermal Properties of Matter PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyVolume expansion of solids

   The temperature of a metallic sphere of radius $R$ is increased by a small amount $\Delta T$. If the linear coefficient of thermal expansion of the metal is $\alpha$, the approximate increase in the volume of the sphere is:

   1. AOption A: $2\pi R^3\alpha\Delta T$
   2. BOption B: $3\pi R^3\alpha\Delta T$
   3. COption C: $4\pi R^3\alpha\Delta T$
   4. DOption D: $6\pi R^3\alpha\Delta T$

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   Correct answer: (C) $4\pi R^3\alpha\Delta T$

   Explanation

   The coefficient of volume expansion is $\alpha_V = 3\alpha$. $\Delta V = \alpha_V V\Delta T = 3\alpha\left(\frac{4}{3}\pi R^3\right)\Delta T = 4\pi R^3\alpha\Delta T$. (Equivalently, $\Delta R = R\alpha\Delta T$ and $\Delta V = 4\pi R^2\Delta R$.) Option (C).

2. Question 2 (NEET 2025)

   NEET 2025MediumHeat Transfer

   Three identical heat conducting rods are connected in series as shown in the figure. 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, the temperature at the left junction is $T_1$ and that at the right junction is $T_2$. The ratio $T_1$/$T_2$ is

   1. AOption A: 3/2
   2. BOption B: 4/3
   3. COption C: 5/3
   4. DOption D: 5/4

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   Correct answer: (C) 5/3

   Explanation

   In steady state the heat current through each rod is the same. Thermal resistance is inversely proportional to thermal conductivity. Thus the resistance ratio is 1:2:1. The total temperature drop is 2T, distributed as T/2, T and T/2. Therefore T₁ = 5T/2 and T₂ = 3T/2, giving T₁/T₂ = 5/3.

3. Question 3 (NEET 2021)

   NEET 2021MediumNewton's Law of Cooling

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

   1. AOption A: $\frac{5}{13}t$
   2. BOption B: $\frac{13}{10}t$
   3. COption C: $\frac{13}{5}t$
   4. DOption D: $\frac{10}{13}t$

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   Correct answer: (C) $\frac{13}{5}t$

   Explanation

   Using Newton's law of cooling: $$\frac{dT}{dt}=-k(T-T_s)$$ where $T_s=20^\circ C$is the surrounding temperature. For cooling from$90^\circ C$to$80^\circ C$:$$t=\frac{1}{k}\ln\left(\frac{90-20}{80-20}\right)=\frac{1}{k}\ln\left(\frac{70}{60}\right)=\frac{1}{k}\ln\left(\frac{7}{6}\right)$$For cooling from$80^\circ C$to$60^\circ C$:$$t'=\frac{1}{k}\ln\left(\frac{80-20}{60-20}\right)=\frac{1}{k}\ln\left(\frac{60}{40}\right)=\frac{1}{k}\ln\left(\frac{3}{2}\right)$$Thus,$$\frac{t'}{t}=\frac{\ln(3/2)}{\ln(7/6)}\approx \frac{13}{5}$$Therefore,$$t'=\frac{13}{5}t$$ Hence, option (C) is correct.

4. Question 4 (NEET 2020)

   NEET 2020MediumSpecific Heat & Calorimetry

   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 :

   1. AOption A: 27/8
   2. BOption B: 9/4
   3. COption C: 3/2
   4. DOption D: 5/3

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   Correct answer: (A) 27/8

   Explanation

   Heat required: Q = mcΔT For the same material and same temperature rise: Q ∝ mass ∝ volume ∝ r^3 Therefore: Q1/Q2 = (r1/r2)^3 = (1.5)^3 = (3/2)^3 = 27/8 Hence, option (A) is correct.

5. Question 5 (NEET 2019)

   NEET 2019MediumNewton's Law of Cooling

   An object kept in a large room having air temperature of 25°C takes 12 minutes to cool from 80°C to 70°C. The time taken to cool for the same object from 70°C to 60°C would be nearly:

   1. AOption A: 15 min
   2. BOption B: 10 min
   3. COption C: 12 min
   4. DOption D: 20 min

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   Correct answer: (A) 15 min

   Explanation

   Using Newton's law of cooling by average temperature method: (T₁ - T₂)/t = k[(T₁ + T₂)/2 - T₀] For cooling from 80°C to 70°C: (80 - 70)/12 = k(75 - 25) 10/12 = 50k For cooling from 70°C to 60°C: (70 - 60)/t = k(65 - 25) 10/t = 40k Dividing the two equations: (10/12)/(10/t) = 50/40 t = 15 min

6. Question 6 (NEET 2017)

   NEET 2017MediumHeat Transfer

   Two rods $A$ and $B$ of different materials are welded together as shown in the figure. Their thermal conductivities are $K_1$ and $K_2$. The thermal conductivity of the composite rod will be:

   1. AOption A: $2(K_1+K_2)$
   2. BOption B: $\dfrac{K_1+K_2}{2}$
   3. COption C: $\dfrac{3(K_1+K_2)}{2}$
   4. DOption D: $K_1+K_2$

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   Correct answer: (B) $\dfrac{K_1+K_2}{2}$

   Explanation

   The rods are connected in parallel for heat flow. Total heat current: $$H = \frac{K_1A(T_1-T_2)}{d} + \frac{K_2A(T_1-T_2)}{d}$$ If equivalent conductivity is $K_{eq}$for total area$2A$:$$H = \frac{K_{eq}(2A)(T_1-T_2)}{d}$$Comparing:$$K_{eq} = \frac{K_1+K_2}{2}$$

7. Question 7 (NEET 2016)

   NEET 2016MediumSpecific Heat & Calorimetry

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

   1. AOption A: $50^\circ C$
   2. BOption B: More than $50^\circ C$
   3. COption C: Less than $50^\circ C$ but greater than $0^\circ C$
   4. DOption D: $0^\circ C$

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   Correct answer: (B) More than $50^\circ C$

   Explanation

   Since heat capacity increases with temperature, the hotter body has a larger heat capacity. Therefore, for the same temperature fall, the hot body loses more heat than the cold body gains for equal temperature rise. Thus the equilibrium temperature shifts towards the hotter body. Hence the final temperature is greater than $50^\circ C$. Hence option (B) is correct.

8. Question 8 (NEET 2016)

   NEET 2016HardNewton's Law of Cooling

   A body cools from a temperature $3T$ to $2T$ in 10 minutes. The room temperature is $T$. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of the next 10 minutes will be:

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (B) Option 2

   Explanation

   Using Newton’s law of cooling: $$\frac{d\theta}{dt} \propto (\theta - T)$$ Initially: Temperature excess decreases from: $$3T - T = 2T$$ to: $$2T - T = T$$ in 10 minutes. Thus the excess temperature halves in 10 minutes. After the next 10 minutes, excess temperature becomes: $$\frac{T}{2}$$ Therefore final temperature: $$T + \frac{T}{2} = \frac{3T}{2}$$ Hence option (B) is correct.

9. Question 9 (AIPMT 2015)

   AIPMT 2015MediumThermal Expansion

   The value of coefficient of volume expansion of glycerin is $5\times10^{-4}\,\text{K}^{-1}$. The fractional change in the density of glycerin for a rise of $40^{\circ}\text{C}$ in its temperature is:

   1. AOption A: 0.010
   2. BOption B: 0.015
   3. COption C: 0.020
   4. DOption D: 0.025

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   Correct answer: (C) 0.020

   Explanation

   For a liquid, $$V_f=V_i(1+\gamma\Delta T)$$ Since density is inversely proportional to volume, $$\rho_f=\frac{\rho_i}{1+\gamma\Delta T}$$ Fractional decrease in density: $$\frac{\rho_i-\rho_f}{\rho_i}=1-\frac{1}{1+\gamma\Delta T}$$ For small expansion, $$\frac{1}{1+x}\approx 1-x$$ Therefore, $$\frac{\Delta\rho}{\rho}\approx \gamma\Delta T$$ $$=(5\times10^{-4})(40)$$ $$=2.0\times10^{-2}=0.020$$ Hence option (C) is correct.

10. Question 10 (AIPMT 2014)

    AIPMT 2014MediumChange of State & Latent Heat

    Steam at $100^{\circ}\mathrm{C}$ is passed into $20\,\mathrm{g}$ of water at $10^{\circ}\mathrm{C}$. When the water acquires a temperature of $80^{\circ}\mathrm{C}$, the mass of water present will be: (Take specific heat of water $= 1\,\mathrm{cal\,g^{-1}\,^{\circ}C^{-1}}$ and latent heat of steam $= 540\,\mathrm{cal\,g^{-1}}$)

    1. AOption A: $24\,\mathrm{g}$
    2. BOption B: $31.5\,\mathrm{g}$
    3. COption C: $42.5\,\mathrm{g}$
    4. DOption D: $22.5\,\mathrm{g}$

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    Correct answer: (D) $22.5\,\mathrm{g}$

    Explanation

    Let $m$ grams of steam condense. Heat released by steam: $$mL + mc(100-80)$$ Heat gained by water: $$20 \times 1 \times (80-10)$$ Substituting values: $$m(540) + m(20) = 20 \times 70$$ $$560m = 1400$$ $$m = 2.5\,\mathrm{g}$$ Therefore, total mass of water present: $$20 + 2.5 = 22.5\,\mathrm{g}$$ Hence, the correct answer is option (D).