Thermal Properties of Matter – Thermometry and Thermal Expansion: NEET MCQs with Answers

15 free questions on Thermal Properties of Matter – Thermometry and Thermal Expansion for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Thermometry and Thermal Expansion: Heat, temperature, thermal expansion

Practise all 156 Thermal Properties of Matter – Thermometry and Thermal Expansion questions free Take a free timed test

Question 1

Previous-year question
A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of the increase in temperature. The length of the aluminium rod is (α_Cu = 1.7 × 10⁻⁵ K⁻¹ and α_Al = 2.2 × 10⁻⁵ K⁻¹) [NEET 2019]
  1. 6.8 cm
  2. 113.9 cm
  3. 88 cm
  4. 68 cm
Show answer
Answer: (D) 68 cm
Explanation: The difference of lengths stays constant when ℓ_Cuα_Cu = ℓ_Alα_Al, so ℓ_Al = 88 × 1.7/2.2 = 68 cm.

Question 2

Previous-year question
An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by [NEET 2017]
  1. 3α/PK
  2. 3PKα
  3. P/3αK
  4. P/αK
Show answer
Answer: (C) P/3αK
Explanation: The compression gives ΔV/V = P/K, and heating restores it through ΔV/V = 3αΔT. Equating, ΔT = P/(3αK).

Question 3

Previous-year question
The coefficients of linear expansion of brass and steel rods are α₁ and α₂, and the lengths of the brass and steel rods are ℓ₁ and ℓ₂ respectively. If (ℓ₂ – ℓ₁) is maintained the same at all temperatures, which one of the following relations holds good? [NEET 2016]
  1. α₁ℓ₂² = α₂ℓ₁²
  2. α₁²ℓ₂ = α₂²ℓ₁
  3. α₁ℓ₂ = α₂ℓ₁
  4. α₁ℓ₁ = α₂ℓ₂
Show answer
Answer: (D) α₁ℓ₁ = α₂ℓ₂
Explanation: The difference stays fixed only if the two rods expand by the same amount: ℓ₁α₁ΔT = ℓ₂α₂ΔT, i.e. α₁ℓ₁ = α₂ℓ₂.

Question 4

Previous-year question
The value of the coefficient of volume expansion of glycerine is 5 × 10⁻⁴ K⁻¹. The fractional change in the density of glycerine for a rise of 40°C in its temperature is [AIPMT 2015]
  1. 0.020
  2. 0.025
  3. 0.010
  4. 0.015
Show answer
Answer: (A) 0.020
Explanation: Δρ/ρ = γΔT = 5 × 10⁻⁴ × 40 = 0.020.

Question 5

Previous-year question
A Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is then lowered until the Fahrenheit thermometer registers 140°. What is the fall in temperature as registered by the Centigrade thermometer? [AIPMT 1990]
  1. 80°
  2. 60°
  3. 40°
  4. 30°
Show answer
Answer: (C) 40°
Explanation: Convert 140°F to Celsius using C/100 = (F − 32)/180 : C = (140 − 32) × 100/180 = 108 × 5/9 = 60°C. The water started at the boiling point, 100°C, so the fall in temperature is 100 − 60 = 40°C. Note the question asks for the fall, not the final reading — 60° is the tempting wrong answer.  

Question 6

Previous-year question
A mercury thermometer can be used to measure temperature up to [AIPMT 1992]
  1. 260°C
  2. 100°C
  3. 360°C
  4. 500°C
Show answer
Answer: (C) 360°C
Explanation: A mercury thermometer relies on the liquid mercury expanding uniformly in its capillary, so it can only work while the mercury stays liquid. Mercury boils at about 357°C, so the practical upper limit of such a thermometer is close to 360°C. Beyond that the mercury vaporises and the reading is meaningless.

Question 7

Previous-year question
The temperature of inversion of a thermocouple is 620°C and its neutral temperature is 300°C. What is the temperature of the cold junction? [AIPMT 2005]
  1. 320°C
  2. 20°C
  3. −20°C
  4. 40°C
Show answer
Answer: (C) −20°C
Explanation: The neutral temperature is the mean of the cold-junction temperature and the temperature of inversion: T_n = (T_i + T_c)/2 300 = (620 + T_c)/2 600 = 620 + T_c T_c = −20°C.
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

Previous-year question
On a new linear scale of temperature 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? [AIPMT 2008]
  1. 78°W
  2. 117°W
  3. 200°W
  4. 139°W
Show answer
Answer: (B) 117°W
Explanation: For two linear scales the fraction of the fixed-point interval is the same: (39 − 0)/(100 − 0) = (t_W − 39)/(239 − 39). So 0.39 = (t_W − 39)/200, giving t_W − 39 = 78 and t_W = 117°W.

Question 9

Previous-year question
The density of water at 20°C is 998 kg/m³ and at 40°C it is 992 kg/m³. The coefficient of volume expansion of water is [NEET 2013]
  1. 3 × 10⁻⁴ /°C
  2. 2 × 10⁻⁴ /°C
  3. 6 × 10⁻⁴ /°C
  4. 10⁻⁴ /°C
Show answer
Answer: (A) 3 × 10⁻⁴ /°C
Explanation: Density falls with temperature as ρ₂ = ρ₁/(1 + γΔT), with ΔT = 40 − 20 = 20°C. 992 = 998/(1 + 20γ) 1 + 20γ = 998/992 = 1.00605 20γ = 6.05 × 10⁻³ γ = 3.02 × 10⁻⁴ ≈ 3 × 10⁻⁴ /°C.

Question 10

Previous-year question
If the cold junction of a thermocouple is kept at 0°C and the hot junction is kept at T°C, then the relation between the neutral temperature T_n and the temperature of inversion T_i is [AIPMT 2015]
  1. T_n = 2T_i
  2. T_n = T_i − T
  3. T_n = T_i + T
  4. T_n = T_i/2
Show answer
Answer: (D) T_n = T_i/2
Explanation: The neutral temperature always lies midway between the cold-junction temperature and the temperature of inversion: T_n = (T_i + T_c)/2. With the cold junction at 0°C, T_c = 0 : T_n = T_i/2.

Question 11

Previous-year question
The coefficient of volume expansion of glycerin is 5 × 10⁻⁴ K⁻¹. The fractional change in the density of glycerin for a rise of 40°C in its temperature is [AIPMT 2015]
  1. 0.015
  2. 0.020
  3. 0.025
  4. 0.010
Show answer
Answer: (B) 0.020
Explanation: On heating, ρ = ρ₀/(1 + γΔT) ≈ ρ₀(1 − γΔT) for small changes, so the fractional change in density is |Δρ|/ρ₀ = γΔT = 5 × 10⁻⁴ × 40 = 200 × 10⁻⁴ = 0.020. (A temperature rise in kelvin equals the rise in degrees Celsius, so the units are consistent.)

Question 12

Previous-year question
The coefficients of linear expansion of brass and steel rods are α₁ and α₂, and the lengths of the brass and steel rods are l₁ and l₂ respectively. If (l₂ − l₁) is to remain the same at all temperatures, which one of the following relations holds good? [NEET 2016, AIPMT 1999]
  1. α₁l₂² = α₂l₁²
  2. α₁²l₂ = α₂²l₁
  3. α₁l₁ = α₂l₂
  4. α₁l₂ = α₂l₁
Show answer
Answer: (C) α₁l₁ = α₂l₂
Explanation: After a temperature rise ΔT the lengths become l₁(1 + α₁ΔT) and l₂(1 + α₂ΔT), so their difference is (l₂ − l₁) + (l₂α₂ − l₁α₁)ΔT. For this to equal (l₂ − l₁) at every temperature, the coefficient of ΔT must vanish: l₂α₂ = l₁α₁, i.e. α₁l₁ = α₂l₂.

Question 13

Previous-year question
A metallic bar of Young's modulus 0.5 × 10¹¹ N m⁻² and coefficient of linear thermal expansion 10⁻⁵ °C⁻¹ has a length of 1 m and a cross-sectional area of 10⁻³ m². It is heated from 0°C to 100°C without being allowed to expand or bend. The compressive force developed in it is [NEET 2024]
  1. 5 × 10³ N
  2. 50 × 10³ N
  3. 100 × 10³ N
  4. 2 × 10³ N
Show answer
Answer: (B) 50 × 10³ N
Explanation: If the bar were free, it would expand by Δl = lαΔT = 1 × 10⁻⁵ × 100 = 10⁻³ m. Since expansion is prevented, this becomes a compressive strain Δl/l = 10⁻³. Thermal stress = Y × strain = 0.5 × 10¹¹ × 10⁻³ = 0.5 × 10⁸ N m⁻². F = stress × A = 0.5 × 10⁸ × 10⁻³ = 5 × 10⁴ = 50 × 10³ N.

Question 14

Previous-year question
The coefficient of volume expansion of glycerin is 5×10⁻⁴ K⁻¹. The fractional change in its density for a 40 °C rise in temperature is: [Re-AIPMT, 2015]
  1. 0.010
  2. 0.015
  3. 0.020
  4. 0.025
Show answer
Answer: (C) 0.020
Explanation: |Δρ/ρ| = γΔT = 5×10⁻⁴ × 40 = 0.020.

Question 15

Previous-year question
On a new linear temperature scale (W scale), the freezing and boiling points of water are 39 °W and 239 °W. What is the temperature on the new scale corresponding to 39 °C? [CBSE AIPMT, 2008]
  1. 78 °W
  2. 117 °W
  3. 200 °W
  4. 139 °W
Show answer
Answer: (B) 117 °W
Explanation: W = 39 + (C/100)(239 − 39) = 39 + (39/100)(200) = 39 + 78 = 117 °W.

Want more Thermal Properties of Matter – Thermometry and Thermal Expansion questions?

E-GYAN NIDHI has 156 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in PROPERTIES OF SOLIDS AND LIQUIDS

↑