Thermal Properties of Matter JEE Main previous year questions with solutions

5 solved JEE Main questions on Thermal Properties of Matter, free to read — no sign-in needed. The full chapter has 124 questions; sign in to attempt the remaining 119 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Calorimetry
   The heat extracted out of $x$ gram of water initially at $50°C$ to cool it down to $0°C$ is sufficient to evaporate $(1000 - x)$ gram of water also initially at $50°C$. The value of $x$ (closest integer) is _______. (Take latent heat of water $2256\text{ kJ/kg.K}$, specific heat capacity of water $4200\text{ J/kg.K}$)
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   Answer: 922

   Heat extracted to cool $x$ gram of water from $50^{\circ}\text{C}$ to $0^{\circ}\text{C}$ is given by: $Q_1 = m_1 c \Delta T_1$ $Q_1 = x \times 4.2 \times (50 - 0) = 210x \text{ J}$ Heat required to raise the temperature of $(1000 - x)$ gram of water from $50^{\circ}\text{C}$ to $100^{\circ}\text{C}$ and evaporate it is given by: $Q_2 = m_2 c \Delta T_2 + m_2 L$ $Q_2 = (1000 - x) \times [4.2 \times (100 - 50) + 2256]$ $Q_2 = (1000 - x) \times (210 + 2256) = 2466(1000 - x) \text{ J}$ Equating the heat extracted and the heat required: $Q_1 = Q_2$ $210x = 2466(1000 - x)$ $210x = 2466000 - 2466x$ $2676x = 2466000$ $x = \dfrac{2466000}{2676} \approx 921.52$ The closest integer value for $x$ is $922$. Answer: $922$
2. Q2JEE Main 2025 (07 Apr, Shift 1)Radiation
   A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is $1727^{\circ} \mathrm{C}$ and power radiated by the wire is 94.2 W. Its emissivity is $\frac{x}{8}$ where $x=$_______ (Given $\sigma=6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}, \pi=3.14$ and assume that the emissivity of wire material is same at all wavelength.)
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   Answer: 5

   $\mathrm{L}=10 \mathrm{~cm}, \mathrm{~d}=0.5 \mathrm{~mm}, \mathrm{~T}=1727^{\circ} \mathrm{C}=2000 \mathrm{~K}$ Power, $\mathrm{P}=94.2 \mathrm{~W}$ $\mathrm{P}=\varepsilon \sigma \mathrm{AT}^4$ $94.2=\varepsilon \times\left(6 \times 10^{-8}\right)(\pi \mathrm{dL})(2000)^4$ $94.2=\varepsilon \times\left(6 \times 10^{-8}\right)(3.14)(0.5)\left(10^{-3}\right)$ $\left(10 \times 10^{-2}\right)(2000)^4$ $\varepsilon=\frac{94.2}{(94.2)(16)}=\frac{5}{8}$
3. Q3JEE Main 2024 (31 Jan, Shift 1)Conduction
   Two conductors have the same resistances at $0^{\circ}C$ but their temperature coefficients of resistance are ${\alpha }_{1}$ and ${\alpha }_{2}$. The respective temperature coefficients for their series and parallel combinations are :
   1. A.${\alpha }_{1}+{\alpha }_{2},\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}$
   2. B.$\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2},\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}$
   3. C.${\alpha }_{1}+{\alpha }_{2},\dfrac{{\alpha }_{1}{\alpha }_{2}}{{\alpha }_{1}+{\alpha }_{2}}$
   4. D.$\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2},{\alpha }_{1}+{\alpha }_{2}$
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   Answer: (B)

   Change in resistance due to temperature is given by, ${R}^{'}=R\left(1+\alpha ∆T\right)$ For Series Combination: ${R}_{eq}={R}_{1}+{R}_{2}$ $\Rightarrow 2R\left(1+{\alpha }_{eq}∆T\right)=R\left(1+{\alpha }_{1}∆T\right)+R\left(1+{\alpha }_{2}∆T\right)$ $\Rightarrow 2R\left(1+{\alpha }_{eq}∆T\right)=2R+\left({\alpha }_{1}+{\alpha }_{2}\right)R∆T$ $\Rightarrow {\alpha }_{eq}=\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}$ For Parallel Combination: $\dfrac{1}{{R}_{eq}}=\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}$ $\Rightarrow \dfrac{1}{\dfrac{R}{2}\left(1+{\alpha }_{eq}∆T\right)}=\dfrac{1}{R\left(1+{\alpha }_{1}∆T\right)}+\dfrac{1}{R\left(1+{\alpha }_{2}∆T\right)}$ $\Rightarrow \dfrac{2}{1+{\alpha }_{eq}∆T}=\dfrac{1}{1+{\alpha }_{1}∆T}+\dfrac{1}{1+{\alpha }_{2}∆T}$ $\Rightarrow \dfrac{2}{1+{\alpha }_{eq}∆T}=\dfrac{1+{\alpha }_{2}∆T+1+{\alpha }_{1}∆T}{\left(1+{\alpha }_{1}∆T\right)\left(1+{\alpha }_{2}∆T\right)}$ $\Rightarrow 2\left[\left(1+{\alpha }_{1}∆T\right)\left(1+{\alpha }_{2}∆T\right)\right]=\left[2+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T\right]\left[1+{\alpha }_{eq}∆T\right]$ $\Rightarrow 2\left[1+{\alpha }_{1}∆T+{\alpha }_{2}∆T+{\alpha }_{1}{\alpha }_{2}∆T\right]=2+2{\alpha }_{eq}∆T+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T+{\alpha }_{eq}\left({\alpha }_{1}+{\alpha }_{2}\right)∆{T}^{2}$ Neglecting small terms $2+2\left({\alpha }_{1}+{\alpha }_{2}\right)∆T=2+2{\alpha }_{eq}∆T+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T$ $\Rightarrow \left({\alpha }_{1}+{\alpha }_{2}\right)∆T=2{\alpha }_{eq}∆T$ $\Rightarrow {\alpha }_{eq}=\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}$
4. Q4JEE Main 2023 (08 Apr, Shift 2)Thermal Expansion
   A steel rod of length $1m$ and cross-sectional area ${10}^{-4}{m}^{2}$ is heated from $0^{\circ}C$ to $200^{\circ}C$ without being allowed to extend or bend. The compressive tension produced in the rod is _____$\times {10}^{4}N$. (Given Young's modulus of steel $=2\times {10}^{11}N{m}^{-2}$, coefficient of linear expansion $={10}^{-5}{K}^{-1}$ )
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   Answer: 4

   The formula to calculate the Young's modulus of the material of the wire is given by $Y=\dfrac{FL}{A∆L}...\left(1\right)$ The increase in length of the wire due to increase in temperature is given by $∆L=L\alpha \left({T}_{2}-{T}_{1}\right)...\left(2\right)$ Substitute the expression for the extension in length from equation (2) into equation (1) and simplify to obtain the required compressive tension. $\begin{matrix}Y & = & \dfrac{FL}{AL\alpha \left({T}_{2}-{T}_{1}\right)} \\ & \Rightarrow & F=A\alpha Y\left({T}_{2}-{T}_{1}\right)...\left(3\right)\end{matrix}$ Substitute the values of the known parameters into equation (3) to calculate the required compressive tension in the wire. $\begin{matrix}F & = & {10}^{-4}{m}^{2}\times {10}^{-5}{K}^{-1}\times 2\times {10}^{11}{Nm}^{-2}\times \left(200^{\circ}C-0^{\circ}C\right) \\ & = & 4\times {10}^{4}N\end{matrix}$
5. Q5JEE Main 2013 (09 Apr)Thermometry
   On a linear temperature scale $\mathrm{Y}$, water freezes at $-160^{\circ} \mathrm{Y}$ and boils at $-50^{\circ} \mathrm{Y}$. On this $\mathrm{Y}$ scale, a temperature of $340 \mathrm{~K}$ would be read as : (water freezes at $273 \mathrm{~K}$ and boils at $373 \mathrm{~K}$ )
   1. A.$-73.7^{\circ} \mathrm{Y}$
   2. B.$-233.7^{\circ} \mathrm{Y}$
   3. C.$-86.3^{\circ} \mathrm{Y}$
   4. D.$-106.3^{\circ} \mathrm{Y}$
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   Answer: (C)

   Reading on any scale - LFP UFP - LFP $=$ constant for all scales $\frac{340-273}{373-273}=\frac{{ }^{\circ} \mathrm{y}-(-160)}{-50-(-160)}$ $\Rightarrow \frac{67}{100}=\frac{\mathrm{y}+160}{110}$ $\therefore \quad y=-86.3^{\circ} \mathrm{y}$