Heat Transfer - JEE Main Previous Year Questions with Solutions

eSaral › JEE › JEE Main ›Heat Transfer - JEE Main Previous Year Questions with Solutions

🚀 Checkout eSaral Courses

What Is Heat Transfer and Why Does It Matter for JEE Main? 

Heat Transfer is a chapter from Class 11 Physics (Chapter 11 — Thermal Properties of Matter) that consistently delivers 1–2 questions in JEE Main every year. According to NTA's official syllabus, the chapter is part of the Thermodynamics and Thermal Physics unit, which carries roughly 3–5% of the total paper weightage.

What makes this chapter valuable for JEE Main preparation is the predictability of its questions. NTA has returned repeatedly to four themes: temperature distribution along conducting rods, Newton's law of cooling graphs, Stefan–Boltzmann radiation problems, and composite-rod (series/parallel) conduction setups. A student who has worked through every previous year question from AIEEE 2009 onwards will recognise the template of almost every new question that appears.

Heat Transfer JEE Main Previous Year Questions & Solutions

Download eSaral app for free study material and video tutorials.   Simulator Previous Years AIEEE/JEE Mains Questions [esquestion] A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $\theta$ along the length x of the bar from its hot end is best described by which of the following figures ? #tag# [AIEEE - 2009] #sol# (4) The heat flow rate is given by $\frac{d Q}{d t}=\frac{k A\left(\theta_{1}-\theta\right)}{x}$ $\Rightarrow \theta=\theta_{1}-\frac{x}{k A} \frac{d Q}{d t}$ where $\theta_{1}$ is the temp of hot end and $\theta$ is temp at a distance x from hot end. [/esquestion] [esquestion] A liquid in a beaker has temperature $\theta(\mathrm{t})$ at time t and $\theta_{0}$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $\log _{\mathrm{e}}\left(\theta-\theta_{0}\right)$ and t is :- #tag# [AIEEE 2012] #sol# (2) Newtons law of cooling $\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)$ $\Rightarrow \frac{d \theta}{\left(\theta-\theta_{0}\right)}=-k d t$ Integrating $\ln \left(\theta-\theta_{0}\right)=-k t+C$ [/esquestion] [esquestion] If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature $\theta_{0}$ the graph between the temperature T of the metal and time t will be closed to: #tag# [JEE-Main- 2013] #sol# (3) According to Newtons law of cooling, The temp goes on decreasing with time non-linearly. [/esquestion] [esquestion] Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = $4 \mathrm{cm}^{2}$. End of copper rod is maintained at $100^{\circ} \mathrm{C}$ where as ends of brass and steel are kept at $0^{\circ} \mathrm{C}$. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is : (1) 4.8 cal/s (2) 6.0 cal/s (3) 1.2 cal/s (4) 2.4 cal/s #tag# [JEE-Main-2014] #sol# (1) Heat flow per unit time through copper rod $=\frac{(100-40)}{\ell_{\mathrm{c}}}\left(\mathrm{K}_{\mathrm{c}} \mathrm{A}_{\mathrm{c}}\right)$ $=\frac{60}{46} \times 0.92 \times 4$ $=4.8 \mathrm{cal} / \mathrm{s}$ [/esquestion]

For Heat Transfer and other Physics topics, the JEE Mains PYQ chapter wise page provides chapter-wise questions for focused practice.

For a true exam simulation, pair this Heat Transfer practice with our JEE Main previous year question paper, which has complete papers from past sessions.