🔬 JEE Main PYQ Pattern (Critical)
Trend: Thermal Expansion (especially thermal stress + Young's modulus combination) appears most frequently. The JEE Main 2022–2024 trend shows increasing focus on multi-material slab conduction problems using thermal resistance analogy.
Repeated JEE Main Question Patterns (2015–2024)
🔥 Pattern 1: Thermal Stress Problem (Appears ~1/year)
A constrained rod/wire is heated. Find: (a) thermal stress OR (b) force on wall OR (c) change in length under combined thermal + mechanical load. Key formula: Stress = YαΔT.
🔥 Pattern 2: Series/Parallel Composite Slab Conduction
Two or more slabs in series or parallel. Find: steady-state heat current, interface temperature, or effective thermal conductivity. Method: thermal resistance analogy R = L/KA.
🔥 Pattern 3: Calorimetry with Phase Change
Mixture of ice + water OR steam + water. Always check if phase change is complete. Final answer can be 0°C (partial melt), 100°C (partial evaporation), or some T_f > 0°C.
🔥 Pattern 4: Wien's Law + Stefan's Law Combination
Given λ_max, find T (Wien's). Then find total power emitted (Stefan's). Or: compare power output of two stars given temperature ratio. T₁/T₂ = (λ₂/λ₁) from Wien's.
🔥 Pattern 5: Newton's Cooling Graph
A body cools from T₁ to T₂ in time t₁. Find time to cool from T₂ to T₃. Method: average temperature approximation: dT/dt ≈ k × (T_avg − T₀).
Representative JEE Main Questions
2023 JEE Main April Thermal Expansion
A body cools in 7 minutes from 60°C to 40°C. Temperature of the surroundings is 10°C. Find temperature at which it will cool to in next 7 minutes.
Approach: Newton's Cooling — average temp method. First 7 min: avg = 50°C, excess = 40°C. Second 7 min: starting from 40°C. Using dT/dt = k(T − T₀), set up ratio of rates. Answer: ≈ 28°C
2022 JEE Main Conduction
A compound slab consists of two slabs A and B of equal thickness. Thermal conductivity of A is twice that of B. The temperature at the interface is T_i. If T_hot = 200°C and T_cold = 0°C, find T_i.
Approach: Series slabs — same heat current. H = K_A × A × (200 − T_i)/d = K_B × A × (T_i − 0)/d. With K_A = 2K_B: 2(200 − T_i) = T_i → T_i = 400/3 ≈ 133.3°C. This type appeared in JEE Main 2022 June 29.
2021 JEE Main Radiation
Two bodies A and B at temperatures 700 K and 350 K. The ratio of rates of energy radiated by A and B is:
Approach: Stefan's Law: P ∝ T⁴. P_A/P_B = (700/350)⁴ = 2⁴ = 16. Answer: 16 : 1. Note: this is a direct formula question but JEE changes numbers each year.
🚀 JEE Advanced PYQ Reality
JEE Advanced rarely tests this chapter as a pure thermal problem. It combines thermal with: (1) Heat conduction in a rod that is also conducting electricity, (2) Stefan's law for a body undergoing temperature change (not just steady state), (3) Bimetallic strip questions with radius of curvature geometry, (4) First law of thermodynamics applied to solids/liquids.
Radiation + Joule heating
High
Newton's Cooling (ODE)
High
Calorimetry + First Law
Medium
Typical difficulty split for thermal questions in JEE Advanced:
🧠 Thinking Step
JEE Advanced thermal questions are almost always multi-concept. Expect: electric circuit + thermal radiation, or SHM + thermal expansion, or calorimetry + first law of thermodynamics.
Landmark JEE Advanced Questions (Thermal Properties)
2019 JEE Advanced Multi-concept
An electric heater heats a metal rod. The temperature of the rod T(t) varies as T(t) = T₀(1 + β·t^(1/4)). Find the rate of heat dissipation and power supplied.
What was tested: Newton's Cooling applied as an ODE. dQ/dt = mc·dT/dt. Differentiate T(t) to get dT/dt = T₀β·(1/4)t^(−3/4). Then energy balance: P_in − P_cooling = mc·dT/dt. Multi-step calculus required.
2017 JEE Advanced Matching Column
Match: Bimetallic strip — thermostatic principle; Incandescent lamp — blackbody radiation; Electric fuse — Joule heating + melting; Steam engine — heat engine efficiency
Pattern: Device → Physical Principle matching. Tests breadth of thermal knowledge. Key: bimetallic = differential expansion; lamp = radiation (Wien's law gives color); fuse = Joule heating exceeds melting; steam engine = thermodynamics.
2015 JEE Advanced Conduction + Radiation
Two rods of different materials connected in series. Heat flows by conduction through them. The radiation from the junction is equal to the conduction loss. Set up the equation.
What was tested: At steady state at junction: Q_conduction_in = Q_conduction_out + Q_radiation. Set H₁ = H₂ + eσAT_j⁴. Requires combining Fourier's Law with Stefan's Law simultaneously.
🏥 NEET PYQ Pattern
NEET focuses on conceptual understanding and simple applications. ~1-2 questions per year from this chapter. Most are single-correct MCQ. Focus areas: Newton's Cooling (conceptual), anomalous expansion of water (application), calorimetry (calculation), and modes of heat transfer (conceptual).
Modes of Heat Transfer
Most frequent
Anomalous Expansion
Medium
- 1–2 questions guaranteed every year
- Focus: conceptual > numerical
- Most asked: convection vs radiation distinction
- Always: anomalous expansion application
- Newton's Cooling: know the graph shape
- Latent heat: distinguish fusion vs vaporization numerically
🎯 Strategy Tip
For NEET, this chapter is not high priority for time investment. Know the concepts, know 3–4 key formulas, and practice 15–20 NEET-style MCQs. Return time to investment is medium.
Frequently Repeated NEET Question Types
2022 NEET
In which of the following processes, convection does NOT take place primarily?
Answer: Radiation (sun heating earth). Conduction: iron rod heated at one end. Convection: sea/land breeze. Radiation: sun's energy reaching earth. This "mode identification" type repeats every 2–3 years.
2020 NEET
Woolen clothes keep us warm in winter because they: (a) are bad conductors of heat, (b) are good emitters, (c) trap air which is bad conductor, (d) reflect heat radiation
Answer: (c) trap air. Wool fibers trap air, and air is an excellent insulator (K = 0.024 W/mK). The wool itself matters less than the trapped air.
📚 CBSE Board Pattern
CBSE board asks 5–8 marks from this chapter across 1 and 2 mark questions. Derivations are important! Stefan's Law, Newton's Law of Cooling, and thermal expansion derivations are all asked explicitly.
- Derive: relationship between α, β, γ
- Explain: anomalous expansion of water
- State and apply: Newton's Law of Cooling
- Draw and label: heating curve of water
- Define: triple point, latent heat, specific heat
- Numericals: calorimetry with phase change
- Explain: why Cp > Cv for gases
CBSE Board Recent Trends (2019–2024)
2024 CBSE Board 5 marks
A piece of ice of mass 10 g at −10°C is mixed with 10 g water at 50°C. Find the final temperature. (c_ice = 0.5 cal/g°C, L = 80 cal/g, c_water = 1 cal/g°C)
Standard calorimetry problem. Heat available from water: 10×1×50 = 500 cal. Heat needed: warm ice = 10×0.5×10 = 50 cal; melt ice = 10×80 = 800 cal. Total needed = 850 cal > 500 cal. So NOT all ice melts. Final state: mixture at 0°C with some unmelted ice.
2023 CBSE Board 2 marks
A hole is drilled in a copper sheet. Does the hole expand or contract on heating? Give reason.
Answer: Expands. Reason: When copper expands, all dimensions increase including the boundary of the hole. Imagine filling the hole with copper — that phantom piece would also expand. Hence hole expands by the same coefficient β.