Thermal Properties of Matter NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Thermal Properties of Matter question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
- Chapter
- NCERT Class 11
- PYQs here
- 28
- Papers
- NEET & AIPMT 2005–2021
- In the app
- 499+ questions
Thermal Properties of Matter previous year questions
28 questions. Tap an option to check your answer.
Q1NEET 2021
A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :
Not quite — the answer is B.
Using the average form of Newton's law of cooling: 10/t = k(85 − 20) = 65k and 20/t' = k(70 − 20) = 50k. Dividing, t' = (20 × 65)/(10 × 50) t = 2.6t = (13/5)t. The inverse ratio (5/13)t would wrongly suggest the larger, cooler drop takes less time.
Q2NEET 2020 · Q174
The quantities of heat required to raise the temperature of two solid copper spheres of radii r₁ and r₂ (r₁ = 1.5 r₂) through 1 K are in the ratio :
Not quite — the answer is A.
Q = mcΔT with the same c and ΔT, and m ∝ volume ∝ r³, so Q₁/Q₂ = (1.5)³ = 27/8. Using surface area (r²) instead of volume would give 9/4.
Q3NEET 2019 · Q14
A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is: (α_Cu = 1.7 × 10⁻⁵ K⁻¹ and α_Al = 2.2 × 10⁻⁵ K⁻¹)
Not quite — the answer is D.
Equal expansions require L_Cu α_Cu = L_Al α_Al, so L_Al = 88 × 1.7/2.2 = 68 cm. 113.9 cm comes from inverting the ratio of expansion coefficients.
Q4NEET 2019 · Q20
The unit of thermal conductivity is:
Not quite — the answer is D.
From H = KA(ΔT)/L, K = HL/(AΔT), giving W·m/(m²·K) = W m⁻¹ K⁻¹. J m⁻¹ K⁻¹ misses the per-second (rate) part of heat flow.
Q5NEET 2018 · Q43
The power radiated by a black body is P and it radiates maximum energy at wavelength, λ₀. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength (3/4)λ₀, the power radiated by it becomes nP. The value of n is
Not quite — the answer is A.
Wien's law λ_mT = constant gives T′ = (4/3)T. By Stefan's law P ∝ T⁴, so n = (4/3)⁴ = 256/81. The inverse 81/256 would correspond to cooling, not heating.
Q6NEET 2017 · Q136
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:
Not quite — the answer is D.
By the Stefan–Boltzmann law P = σ(4πr²)T⁴, so P ∝ r²T⁴. Halving r multiplies P by 1/4 and doubling T multiplies it by 16, a net factor of 4: 450 × 4 = 1800 W. Keeping only the radius change would give the trap value of about 112 W, and assuming the effects cancel gives 450 W.
Q7NEET 2017 (Gujarati paper) · Q155
In a certain planetary system, it is observed that one of the celestial bodies having a surface temperature of 200 K, emits radiation of maximum intensity near the wavelength 12 μm. The surface temperature of a nearby star which emits light of maximum intensity at a wavelength λ = 4800 Å, is:
Not quite — the answer is A.
By Wien's law λ_max T is constant: 12 × 10⁻⁶ m × 200 K = 2.4 × 10⁻³ m K. For the star T = 2.4 × 10⁻³ / 4800 × 10⁻¹⁰ = 5000 K. Getting 10000 K means 4800 Å was taken as 2.4 × 10⁻⁷ m, a unit slip.
Q8NEET 2016 Phase 2 · Q15
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100 °C, while the other one is at 0 °C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is
Not quite — the answer is B.
The hot body has the larger heat capacity, so it gives out more heat per degree of cooling than the cold body absorbs per degree of warming. To balance heat lost and gained, the hot body cools through fewer degrees than the cold one warms, so the final temperature lies above 50 °C. Exactly 50 °C would need a constant heat capacity.
Q9NEET 2016 Phase 2 · Q16
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be
Not quite — the answer is B.
With the average form: (3T − 2T)/10 = k(2.5T − T) gives k = 1/15 per minute; then (2T − θ)/10 = k((2T + θ)/2 − T) gives 6T − 3θ = θ, so θ = 3T/2. The exponential form agrees: the excess over the room halves from 2T to T, so it becomes T/2 and θ = 1.5T. Assuming equal fall per interval would wrongly give T.
Q10NEET 2016 Phase 1 · Q15
A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U₁, at wavelength 500 nm is U₂ and that at 1000 nm is U₃. Wien's constant, b = 2.88 × 10⁶ nmK. Which of the following is correct?
Not quite — the answer is D.
By Wien's law λmax = b/T = 2.88 × 10⁶/5760 = 500 nm, so emission peaks at 500 nm and U₂ is the largest; hence U₂ > U₁. A black body emits at all wavelengths, so neither U₁ nor U₃ is zero.
Q11NEET 2016 Phase 1 · Q16
Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ − l₁) is maintained same at all temperatures, which one of the following relations holds good?
Not quite — the answer is D.
For the difference to stay constant, both rods must expand by the same amount for any temperature change: l₁α₁ΔT = l₂α₂ΔT, so α₁l₁ = α₂l₂. Swapping the lengths (α₁l₂ = α₂l₁) is the common trap.
Q12NEET 2016 Phase 1 · Q35
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 × 10⁵ J/kg and g = 10 N/kg]
Not quite — the answer is C.
(1/4)mgh = mL, so h = 4L/g = 4 × 3.4 × 10⁵/10 = 1.36 × 10⁵ m = 136 km. Ignoring the one-quarter factor gives 34 km.
Q13AIPMT 2015
The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be:
Not quite — the answer is A.
The conduction rate is kAΔT/L and depends on the temperature difference, not the actual temperatures. ΔT is 10 °C in both cases, so the rate stays 4.0 J/s (ignoring change of k with temperature).
Q14AIPMT 2015
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:
Not quite — the answer is C.
By Wien's displacement law λ_max T = constant, so a shorter peak wavelength means a hotter star. λ_violet < λ_green < λ_red, hence T_P > T_R > T_Q.
Q15AIPMT 2014
Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5 minutes. The temperature of the surroundings is:
Not quite — the answer is D.
Using Newton's law with average temperatures: 10/5 = k(65 − T₀) and 6/5 = k(57 − T₀). Dividing, 5/3 = (65 − T₀)/(57 − T₀), so T₀ = 45°C.
Q16AIPMT 2014
Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be: [Take specific heat of water = 1 cal g⁻¹ °C⁻¹ and latent heat of steam = 540 cal g⁻¹]
Not quite — the answer is C.
Heat from steam condensing and cooling to 80°C: m(540 + 20) = 560m. Heat gained by water: 20 × 1 × 70 = 1400 cal, so m = 2.5 g. Total water = 20 + 2.5 = 22.5 g.
Q17NEET 2013
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 -
Not quite — the answer is B.
Wien's displacement law, λₘT = constant, says that the wavelength of maximum emission shifts to shorter values as temperature rises, so the colour changes from red towards yellow and white. Stefan's law gives only the total power radiated, not the colour.
Q18AIPMT 2012 Mains
A slab of stone of area 0.36 m² and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹)
Not quite — the answer is A.
Heat rate = mL/t = (4.8 × 3.36 × 10⁵)/3600 = 448 W. Setting KAΔT/d = 448 gives K × 0.36 × 100/0.1 = 360K = 448, so K ≈ 1.24 J m⁻¹ s⁻¹ °C⁻¹.
Q19AIPMT 2012 Prelims
If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q? (σ stands for Stefan's constant)
Not quite — the answer is A.
For a black body, power radiated = σT⁴ × area = 4πR²σT⁴. In steady state this equals the energy production rate Q, so T = (Q/(4πR²σ))^(1/4). The expression without the fourth root gives T⁴, not T.
Q20AIPMT 2010 Prelims · Q15
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?
Not quite — the answer is C.
Volume is fixed: halving the radius makes the area A/4, so the length becomes 4L. Heat flow ∝ A/L, so it becomes ¼ ÷ 4 = 1/16 of the original: Q/16. Taking only the area change gives Q/4.
Q21AIPMT 2010 Prelims · Q16
The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature T K is given by (where σ is Stefan's constant)
Not quite — the answer is B.
The star emits σT⁴ × 4πr² in total. At distance R this spreads over a sphere of area 4πR², so energy per unit area per second = σT⁴·4πr²/(4πR²) = σr²T⁴/R². Keeping the 4π in the numerator forgets to divide by the full sphere area.
Q22AIPMT 2009 Prelims · Q10
The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁ > T₂). The rate of heat transfer, dQ/dt, through the rod in a steady state is given by:
Not quite — the answer is C.
Fourier's law for steady conduction gives dQ/dt = kA(T₁ − T₂)/L: the heat current grows with area and temperature difference and falls with length. Forms with L in the numerator or A in the denominator have the wrong dependence.
Q23AIPMT 2009 Prelims · Q20
A black body at 227°C radiates heat at the rate of 7 cal/cm² s. At a temperature of 727°C, the rate of heat radiated in the same units will be:
Not quite — the answer is B.
Stefan's law: E ∝ T⁴ with absolute temperatures. 227°C = 500 K and 727°C = 1000 K, so E rises by (1000/500)⁴ = 16: 7 × 16 = 112 cal/cm² s. Using Celsius temperatures directly gives wrong ratios.
Q24AIPMT 2008 Prelims · Q15
On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39° W and 239° W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39° C on the Celsius scale?
Not quite — the answer is C.
For a linear scale, (W − 39)/(239 − 39) = (C − 0)/(100 − 0). With C = 39: W = 39 + 200 × 0.39 = 39 + 78 = 117° W. 78° W forgets to add the offset of the freezing point.
Q25AIPMT 2007 Prelims · Q40
A black body is at 727° C. It emits energy at a rate which is proportional to
Not quite — the answer is A.
By Stefan's law the emitted power ∝ T⁴ with T in kelvin: 727 + 273 = 1000 K, so rate ∝ (1000)⁴. Using 727 forgets to convert from Celsius.
Q26AIPMT 2007 Prelims · Q49
Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t°C, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the sun is (where σ is the Stefan's constant)
Not quite — the answer is C.
The sun radiates P = 4πr²σ(t + 273)⁴. At distance R this spreads over 4πR², giving r²σ(t + 273)⁴/R² per unit area. Options using t⁴ forget to convert to kelvin, and dividing by an extra 4π is wrong.
Q27AIPMT 2006 Prelims · Q7
A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be observed at :
Not quite — the answer is D.
By Wien's law λₘT is constant. T₁ = 1227 + 273 = 1500 K and T₂ = 2500 K, so λ₂ = 5000 × 1500/2500 = 3000 Å. Using Celsius temperatures (1227 and 2227) would give a wrong ratio; the peak must shift to a shorter wavelength as the body gets hotter.
Q28AIPMT 2005 Prelims · Q8
Which of the following circular rods, (given radius r and length l) each made of the same material and whose ends are maintained at the same temperature will conduct most heat ?
Not quite — the answer is B.
Heat current H = kAΔT/l ∝ r²/l. For r = 2r₀, l = l₀ this is 4r₀²/l₀, the largest; r = 2r₀, l = 2l₀ gives only 2r₀²/l₀.
How often Thermal Properties of Matter came up
Questions from this chapter in each paper on this site.
- NEET 2021: 1
- NEET 2020: 1
- NEET 2019: 2
- NEET 2018: 1
- NEET 2017: 1
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 2: 2
- NEET 2016 Phase 1: 3
- AIPMT 2015: 2
- AIPMT 2014: 2
- NEET 2013: 1
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 1
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 2
- AIPMT 2008 Prelims: 1
- AIPMT 2007 Prelims: 2
- AIPMT 2006 Prelims: 1
- AIPMT 2005 Prelims: 1
Keep going
PYQs show what NEET asks. For the full Thermal Properties of Matter bank (499+ questions with solutions), use the RankUp app.
