NEET PhysicsUpdated 19 Oct 2026

Thermal Properties of Matter: NEET PYQs and weightage

Thermal Properties of Matter carries 2.4 of 100 Physics marks in NEET by past-paper share: 18 of the 758 tagged NEET Physics PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 25 of 30 Physics chapters.

Take the 25-question Thermal Properties of Matter test Formula sheet ↓All NEET Physics chapters

2.4/100 Weight in Physics

Exam importance (2/5, by past-paper share)

18 NEET PYQs

360 Practice questions

334 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Physics PYQs in the bank (18 of 758), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Thermal Properties of Matter: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

Heat for a temperature change
$Q = m s Δ T$
Use it when: a body warms or cools with no change of state ( $s$ in $J k g^{- 1} K^{- 1}$ )
Latent heat
$Q = m L$
Use it when: a substance melts, freezes, boils or condenses at constant temperature (water: $L_{f} = 3.33 \times 1 0^{5} J/kg$ , $L_{v} = 22.6 \times 1 0^{5} J/kg$ )
Calorimetry
$heat lost by hot bodies = heat gained by cold bodies$
Use it when: things are mixed in an insulated vessel. Add one $m s Δ T$ or $m L$ term for each step, and include the calorimeter
Linear expansion
$Δ l = α_{l} l Δ T$
$L_{2} = L_{1} [1 + α_{l} (T_{2} - T_{1})]$
Use it when: a rod, wire, ring or hole changes length with temperature (a hole expands like the material around it)
Area and volume expansivity
$α_{A} = 2 α_{l}$
$α_{V} = 3 α_{l}$
Use it when: a sheet's area or a solid's volume changes and you are only given $α_{l}$
Thermal stress
$\frac{F}{A} = Y α_{l} Δ T$
Use it when: a rod or wire held between rigid supports is heated or cooled
Conduction
$H = \frac{K A ( T _{C} - T _{D} )}{L}$
Use it when: heat flows steadily through a slab or rod ( $K$ in $W m^{- 1} K^{- 1}$ , $L$ is the thickness along the flow)
Rods in series
$T_{0} = \frac{K _{1} T _{1} + K _{2} T _{2}}{K _{1} + K _{2}}$
$K_{e q} = \frac{2 K _{1} K _{2}}{K _{1} + K _{2}} (equal L and A)$
The same $H$ flows through each rod.
Use it when: two rods are joined end to end and you need the junction temperature or the equivalent $K$
Stefan–Boltzmann law
$H = e σ A T^{4}$
$σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$
Use it when: you need the power radiated by a surface ( $e = 1$ for a perfect blackbody)
Net radiation loss
$H = e σ A (T^{4} - T_{s}^{4})$
Use it when: a body at $T$ sits in surroundings at $T_{s}$
Wien's displacement law
$λ_{m} T = 2.9 \times 1 0^{- 3} m K$
Use it when: you have the peak wavelength and need the temperature, or the reverse (sun, stars, heated iron)
Newton's law of cooling
$- \frac{d Q}{d t} = k (T_{2} - T_{1})$
$T_{2} = T_{1} + C^{'} e^{- K t}$
$K = \frac{k}{m s}$
Use it when: a body cools with only a small excess temperature over its surroundings
Newton's law, average form
$\frac{fall in temperature}{time} = K \times (average temperature - surroundings)$
Use it when: a problem says "cools from $80^{\circ} C$ to $50^{\circ} C$ in 5 min…". Write it for both intervals and divide
Temperature scales
$t_{F} = \frac{9}{5} t_{C} + 32$
$T = t_{C} + 273.15$
Use it when: converting between scales (a change of $1^{\circ} C$ is the same as a change of $1 K$ )

Trap

putting °C into T⁴ or into λ_m T. Stefan's law and Wien's law need Kelvin, so convert first. ΔT in m s ΔT is the same in °C and K.

How does NEET ask Thermal Properties of Matter?

8 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1PhysicsNEET UG PYQThermal Expansion

Try right here · No sign-in

A copper rod of 88 cm and an aluminum rod of unknown length have their increase in length independent of increase in temperature. The length of aluminum rod is:

(α_{C u} = 1.7 \times 10^{- 5} K^{- 1} and α_{A l} = 2.2 \times 10^{- 5} K^{- 1})

Answer key (C) & worked solution

Correct answer: Option C

α_{C u} L_{C u} = α_{A l} L_{A l}

1.7 \times 1 0^{- 5} \times 88 c m = 2.2 \times 1 0^{- 5} \times L_{A l} L_{A l} = \frac{1.7 \times 88}{2.2} = 68 c m

Q2PhysicsNEET UG PYQHeat Transfer

A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

A1000

B1800

C225

D450

Answer key (B) & worked solution

Correct answer: Option B

Rate of power loss

P \propto R^{2} T^{4}

(where,

P

power loss,

R

is radius and

T

is temperature)

\frac{P _{1}}{P _{2}} = \frac{R _{1}^{2} T _{1}^{4}}{R _{2}^{2} T _{2}^{4}}

\frac{P _{1}}{P _{2}} = 4 \times \frac{1}{16}

\frac{450}{P _{2}} = \frac{1}{4}

P_{2} = 1800 watt

Q3PhysicsNEET UG PYQThermal Expansion

The temperature of a metallic sphere of radius

R

is increased by a small amount

Δ T

. If the linear coefficient of thermal expansion of the metal is

α

, the approximate increase in the volume of the sphere is:

3 π R^{3} α Δ T

4 π R^{3} α Δ T

6 π R^{3} α Δ T

2 π R^{3} α Δ T

Answer key (B) & worked solution

Correct answer: Option B

For the metallic sphere,

\frac{Δ R}{R} = α Δ T

Volume of the sphere,

V = \frac{4}{3} π R^{3}

∴ \frac{Δ V}{V} = 3 \frac{Δ R}{R}

Increase in volume

= Δ V = 3 (\frac{Δ R}{R}) (\frac{4}{3} π R^{3}) = 4 π R^{3} α Δ T

Q4PhysicsNEET UG PYQHeat Transfer

On observing light from three different stars P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If T

_{P}

, T

_{Q}

and T

_{R}

are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observations that:

_{P}

< T

_{Q}

< T

_{R}

_{P}

> T

_{R}

> T

_{Q}

_{P}

> T

_{Q}

> T

_{R}

_{P}

< T

_{R}

< T

_{Q}

Answer key (B) & worked solution

Correct answer: Option B

Accordingly, to Wien’s law

λ \propto \frac{1}{T}

and

λ_{V} < λ_{G} < λ_{R}

∴ T

_{P}

> T

_{R}

> T

_{Q}

Q5PhysicsNEET UG PYQChange of State

A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [Latent heat of ice is 3.4 × 10

^{5}

J/kg and g = 10 m/s

^{2}

]

A34 km

B544 km

C136 km

D68 km

Answer key (C) & worked solution

Correct answer: Option C

\frac{m g h}{4} = m L \Rightarrow h = \frac{4 L}{g} = \frac{4 \times 3.4 \times 10 ^{5}}{10} = 136 k m

Q6PhysicsNEET UG PYQHeat Transfer

A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using

AStefan’s Law

BWien‘s displacement Law

CKirchhoff’s Law

DNewton’s Law of cooling

Answer key (B) & worked solution

Correct answer: Option B

It can be explained using the Wien’s displacement Law

λ_{m a x} = \frac{b}{T}

λ_{m a x} \propto \frac{1}{T}

λ_{m a x} T = C o n s t an t

Q7PhysicsNEET UG PYQHeat Transfer

The unit of thermal conductivity is:

J m^{- 1} K^{- 1}

W m K^{- 1}

W m^{- 1} K^{- 1}

J m K^{- 1}

Answer key (C) & worked solution

Correct answer: Option C

The heat current related to difference of temperature across the length

l

of a conductor of area A is

\frac{d H}{d t} = \frac{K A}{l} Δ T

(K = coefficient of thermal conductivity)

∴ K = \frac{l d H}{A d t Δ T}

Unit of K = Wm

^{- 1}

^{- 1}

Q8PhysicsNEET UG PYQChange of State

Steam at 100

^{o}

C is passed into 20 g of water at 10

^{o}

C. When water acquires a temperature of 80

^{o}

C, the mass of water present will be: [Take specific heat of water = 1cal g

^{- 1}

^{o}

^{-}

^{1}

and latent heat of steam = 540 cal g

^{- 1}

]

A42.5 g

B22.5 g

C24 g

D31.5 g

Answer key (B) & worked solution

Correct answer: Option B

Let m be the mass of steam than converted to the water at

8 0^{o} C

so the heat released by the m amount of water will be used to raise the temperature of

20 g

water

heat lost=heat gain

540 m + m (100 - 80) = 20 (80 - 10)

m = 2.5 g

∴ Total water at

8 0^{o} C

20 + 2.5 = 22.5 g

Which topics in Thermal Properties of Matter carry the most questions?

All 6 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
Heat Transfer	11	208
Thermal Expansion	3	63
Change of State	2	32
Specific Heat Capacity	2	20
Measurement of Temperature	0	20
Calorimetry	0	17

Why does Thermal Properties of Matter show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Thermal Properties of Matter, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

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What happens when you test Thermal Properties of Matter?

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Thermal Properties of Matter: NEET PYQs and weightage

Thermal Properties of Matter: formula sheet for NEET

Heat for a temperature change

Latent heat

Calorimetry

Linear expansion

Area and volume expansivity

Thermal stress

Conduction

Rods in series

Stefan–Boltzmann law

Net radiation loss

Wien's displacement law

Newton's law of cooling

Newton's law, average form

Temperature scales

How does NEET ask Thermal Properties of Matter?

Which topics in Thermal Properties of Matter carry the most questions?

Why does Thermal Properties of Matter show "–" until you answer 3 questions?

What happens when you test Thermal Properties of Matter?

Chapters either side of Thermal Properties of Matter by weight