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Thermal Properties of Matter is a collection of small, separate ideas, and JEE Main picks one of them at random each year. Conduction through joined rods, a calorimetry mixture, thermal expansion, a latent heat sum. None of them is deep, but all of them have to be there when you need them. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 1 to 2 questions every year, which is 4 to 8 marks.
- Conduction through combined rods and calorimetry mixtures are the regular ones.
- Thermal expansion and latent heat fill the rest.
What the syllabus says
In the syllabus (Unit 7, Properties of Solids and Liquids): Heat, temperature, thermal expansion, specific heat capacity, calorimetry, change of state, latent heat, and heat transfer by conduction, convection and radiation.
Taken out in 2024: Newton’s law of cooling.
Never in the JEE Main syllabus: Stefan’s law, Wien’s displacement law and black-body radiation. They are in NCERT, so they still turn up in old question banks.
Taken from the NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| Linear expansion | ΔL = LαΔT, with β = 2α and γ = 3α |
| Heat gained or lost | Q = mcΔT |
| Latent heat | Q = mL, ice 80 cal/g and water 540 cal/g |
| Conduction | Q/t = kA(T1 − T2)/L |
| Two rods in series | k = 2k1k2/(k1 + k2) |
| Two rods in parallel | k = (k1 + k2)/2 |
| Thermal resistance | R = L / kA |
| Principle of calorimetry | heat lost by the hot body = heat gained by the cold body |
| Latent heat values | ice 80 cal/g, water 540 cal/g |
Notice that thermal conductivities in series and in parallel behave exactly like electrical resistances and conductances. Heat flow down a temperature difference is the same mathematics as current flowing down a potential difference.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. Two rods of the same length and cross-section, of conductivities k1 and k2, are joined end to end. The conductivity of the combination is
- (a) k1 + k2
- (b) (k1 + k2)/2
- (c) 2k1k2/(k1 + k2)
- (d) √(k1k2)
Q2. A metal rod of length 1 m and coefficient of linear expansion 2 × 10⁻⁵ /°C is heated through 50 °C. Its increase in length is
- (a) 10⁻³ m
- (b) 10⁻⁴ m
- (c) 10⁻² m
- (d) 10⁻⁵ m
Q3. 100 g of ice at 0°C is mixed with 100 g of water at 80°C. Taking the latent heat of ice as 80 cal/g, the final temperature of the mixture is
- (a) 0°C
- (b) 40°C
- (c) 20°C
- (d) 10°C
Q4. A bimetallic strip bends when it is heated because the two metals differ in their
- (a) densities
- (b) specific heats
- (c) coefficients of linear expansion
- (d) thermal conductivities
Q5. Steam at 100 °C is passed into 100 g of water at 20 °C until the water reaches 100 °C. Taking the latent heat of steam as 540 cal/g, the mass of steam condensed is about
- (a) 14.8 g
- (b) 8 g
- (c) 54 g
- (d) 1.5 g
Q6. Water is used as a coolant in radiators mainly because it has a high
- (a) density
- (b) specific heat capacity
- (c) thermal conductivity
- (d) boiling point
Q7. A metal plate with a circular hole in it is heated. The hole becomes
- (a) larger
- (b) smaller
- (c) exactly the same size
- (d) oval in shape
Q8. Two bodies of different masses are at the same temperature and are placed in contact. Then
- (a) heat flows from the heavier to the lighter
- (b) no net heat flows between them
- (c) heat flows from the larger to the smaller
- (d) both of them cool down
Q9. Two rods of the same length and area, of thermal conductivities k and 3k, are joined end to end. The equivalent thermal conductivity is
- (a) 2k
- (b) 3k/2
- (c) 4k
- (d) k/2
Q10. A body cools from 80°C to 70°C in 5 minutes in surroundings at 30°C. The time it takes to cool from 70°C to 60°C is about
- (a) 5 minutes
- (b) 6.4 minutes
- (c) 4 minutes
- (d) 10 minutes
Answer key, with a hint for each
| Q1 | (c) | In series the thermal resistances add, exactly as electrical resistances do. |
| Q2 | (a) | ΔL = LαΔT = 1 × 2 × 10⁻⁵ × 50 = 10⁻³ m, that is one millimetre. |
| Q3 | (a) | The hot water can give out 8000 cal, which is exactly enough to melt all 100 g of ice and nothing more. |
| Q4 | (c) | One side grows more than the other, so the strip curls towards the metal that expands less. |
| Q5 | (a) | Heat needed = 100 × 1 × 80 = 8000 cal. Mass of steam = 8000 ÷ 540 ≈ 14.8 g. |
| Q6 | (b) | A high specific heat means it can absorb a lot of heat for a small rise in its own temperature. |
| Q7 | (a) | The plate expands in every direction, and the hole expands along with it, as if it were made of the same metal. |
| Q8 | (b) | Heat flows because of a temperature difference, not a mass difference. At the same temperature there is no flow. |
| Q9 | (b) | End to end means the thermal resistances add: L/kA + L/3kA = 4L/3kA = 2L/kₑA, so kₑ = 3k/2. |
| Q10 | (b) | From 10/5 = k(75 − 30), k = 2/45. Then 10/t = (2/45)(65 − 30), giving t = 450/70. |
Now mark yourself honestly
JEE Main marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 25 Physics questions are out of 100.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in January.
The full chapter, on a clock
Ten questions tell you a little. Twenty-five tell you the truth.
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