Heat & Thermodynamics – NDA Physics PYQs

NDA 2026-I

Q. 1. A solid of mass m has temperature-dependent specific heat as C(T) = C₀ + αT, where C₀ and α are constants. The solid is heated from T₁ to T₂. Which one of the following is the correct expression for quantity of heat (Q) of the solid mass?

(a)  Q = mC₀(T₂ − T₁)

(b)  Q = m(T₂ − T₁)[C₀ + α(T₁ + T₂)]

(c)  Q = m(T₂ − T₁)[C₀(T₁ + T₂) + α]

(d)  Q = m(T₂ − T₁)[C₀ + (α/2)(T₁ + T₂)]

Answer: (d) Q = m(T₂ − T₁)[C₀ + (α/2)(T₁ + T₂)]

Explanation: Q = m∫[T₁ to T₂] C(T)dT = m∫(C₀ + αT)dT = m[C₀T + αT²/2] evaluated from T₁ to T₂ = m[C₀(T₂−T₁) + (α/2)(T₂²−T₁²)] = m(T₂−T₁)[C₀ + (α/2)(T₁+T₂)]. This uses the factorisation T₂² − T₁² = (T₂−T₁)(T₁+T₂). Concept Tested: Temperature-dependent specific heat: Q = m∫C(T)dT; integration gives closed form with (T₁+T₂)/2

★ JOVIK Exam Insight A 2026-I advanced question: when C varies with T, Q must be found by integration (not simply Q = mcΔT). This is university-level thermodynamics now appearing in NDA. The result contains (T₁+T₂)/2: the average temperature: as the effective specific heat multiplier.

Q. 2. In a certain process, the relation between pressure (P) and volume (V), for an ideal gas, is given by PV² = constant. If the initial temperature is T₁ and the final temperature is T₂; and the initial volume is V₁ and the final volume is V₂, then which one of the following is correct?

(a)  T₁/T₂ = V₁/V₂

(b)  T₁/T₂ = (V₁/V₂)²

(c)  T₁/T₂ = V₂/V₁

(d)  T₁/T₂ = (V₂/V₁)²

Answer: (c) T₁/T₂ = V₂/V₁

Explanation: PV² = constant. Using ideal gas law PV = nRT → P = nRT/V. Substituting: (nRT/V)V² = nRTV = constant → TV = constant. Therefore T₁V₁ = T₂V₂ → T₁/T₂ = V₂/V₁. Concept Tested: Polytropic process PV² = constant: leads to TV = constant; T₁/T₂ = V₂/V₁

Q. 3. For an ideal gas, a process is described by P = kT, where k is a constant. If the molar heat capacity for this process is C, then which one of the following is correct (where the symbols have their usual meanings)?

(a)  C = Cₚ

(b)  C = Cᵥ

(c)  C = (Cₚ + Cᵥ)/2

(d)  C is never constant

Answer: (b) C = Cᵥ

Explanation: P = kT with ideal gas PV = nRT: (kT)V = nRT → kV = nR → V = nR/k = constant. This is an isochoric (constant volume) process. For a constant-volume process, no work is done (W = 0) and all heat goes into internal energy. The molar heat capacity at constant volume is Cᵥ. Therefore C = Cᵥ. Concept Tested: Process P = kT for ideal gas: isochoric (constant V); molar heat capacity = Cᵥ

Q. 4. A 5 g piece of ice at −20°C is put into m kg of water at 30°C temperature. The heat is exchanged only between the ice and the water. The final temperature of the mixture is 0°C in liquid phase. What is the value of m in kg?

(a)  0·050

(b)  0·010

(c)  0·015

(d)  0·150

Answer: (c) 0·015

Explanation: Heat gained by ice: (1) warming ice from −20°C to 0°C: Q₁ = 5×10⁻³ × 2100 × 20 = 210 J. (2) Melting ice at 0°C: Q₂ = 5×10⁻³ × 3.36×10⁵ = 1680 J. Total Q_ice = 1890 J. Heat lost by water cooling from 30°C to 0°C: Q_water = m × 4200 × 30 = 126,000m J. Setting equal: 126,000m = 1890 → m = 0.015 kg. Concept Tested: Calorimetry: ice-water mixture; heat balance: ice warming + ice melting = water cooling

Q. 5. A very large container consists of an ideal gas. The speed of sound in the gas is x. When the pressure of the gas is doubled while keeping the temperature constant, the speed of sound now becomes y. What is the ratio of x to y?

(a)  1

(b)  √2

(c)  2

(d)  4

Answer: (a) 1

Explanation: Speed of sound in an ideal gas: v = √(γRT/M). This depends on temperature T and gas properties (γ, M), but not on pressure at constant temperature. When pressure doubles at constant T, density ρ also doubles (ideal gas law: P ∝ ρ at constant T). The ratio γP/ρ = γRT/M remains unchanged. Therefore y = x and x/y = 1. Concept Tested: Speed of sound in ideal gas: independent of pressure at constant temperature

NDA 2025-II

Q. 6. The temperature of a body increases from 310 K to 340 K. The temperature increase in degree Celsius is

(a)  20°C

(b)  30°C

(c)  37°C

(d)  67°C

Answer: (b) 30°C

Explanation: A change in temperature of 1 K equals a change of 1°C: the Kelvin and Celsius scales have identical degree intervals, differing only by the zero point. Therefore ΔT = 340 − 310 = 30 K = 30°C. No conversion formula is needed for temperature changes. Concept Tested: Temperature change in Kelvin = temperature change in Celsius (ΔK = Δ°C)

Q. 7. Which one of the following instruments can be used to measure −250°C temperature?

(a)  By using a mercury based thermometer

(b)  By using an alcohol based thermometer

(c)  By using a clinical thermometer

(d)  By using thermocouple based thermometer

Answer: (d) By using thermocouple based thermometer

Explanation: Mercury thermometers cannot measure below −38.8°C (mercury’s freezing point). Alcohol thermometers work down to about −115°C (alcohol’s freezing point): but not −250°C. Clinical thermometers have a very narrow range (35–42°C). Only thermocouple thermometers have a broad range (from near absolute zero to several thousand °C) and can measure −250°C. Concept Tested: Thermometer range: thermocouple works at −250°C; mercury and alcohol freeze before that

Q. 8. The quantity of heat needed to change unit mass of a substance from liquid to vapour without changing the temperature, is called

(a)  specific heat

(b)  specific latent heat

(c)  thermal capacity

(d)  heat energy

Answer: (b) specific latent heat

Explanation: Specific latent heat (of vaporisation here) is the heat per unit mass required for a phase change at constant temperature. ‘Latent’ = hidden (no temperature change occurs). Specific heat relates to temperature change per unit mass. Thermal capacity = mass × specific heat. The specific latent heat of vaporisation of water is approximately 2.26 MJ/kg. Concept Tested: Specific latent heat: heat per unit mass for phase change at constant temperature

NDA 2025-I

Q. 9. A system that does NOT allow exchange of heat with its surrounding is called

(a)  Adiabatic system

(b)  Non-adiabatic system

(c)  Equilibrium system

(d)  Non-equilibrium system

Answer: (a) Adiabatic system

Explanation: An adiabatic system does not allow heat exchange with its surroundings: Q = 0 for all processes. In an adiabatic process, the First Law gives ΔU = −W: any work done comes entirely from or goes into internal energy. Thermally insulated containers approximate adiabatic systems. Concept Tested: Adiabatic system: no heat exchange with surroundings (Q = 0)

NDA 2024-II

Q. 10. Which one of the following statements best defines the concept of heat?

(a)  The transformation of energy from one form to another

(b)  The conversion of energy into mass and vice-versa due to temperature difference

(c)  The transfer of energy due to temperature difference

(d)  The change in volume of a substance with temperature

Answer: (c) The transfer of energy due to temperature difference

Explanation: Heat is specifically the transfer of thermal energy between two bodies due to a temperature difference. Option (a) is broader energy conversion. Option (b) involves mass-energy equivalence (E = mc²): not heat. Option (d) describes thermal expansion. Only (c) precisely defines heat as a mode of energy transfer driven by temperature difference. Concept Tested: Definition of heat: energy transfer due to temperature difference

Q. 11. Given below are the four cases in which certain heat transfer is taking place: 1. Ice is melting in a glass full of water 2. Water is boiling in an open container 3. A metal rod is heated in a furnace 4. A cup of coffee is allowed to cool on a table  In which of the above cases, the Newton’s Law of Cooling is applicable?

(a)  1 only

(b)  4 only

(c)  1 and 4 only

(d)  1, 2 and 3

Answer: (b) 4 only

Explanation: Newton’s Law of Cooling applies when: (i) the body is cooling (losing heat) to its surroundings, (ii) the temperature excess over surroundings is small, and (iii) heat is lost primarily by radiation and convection. A cup of coffee cooling on a table (case 4) satisfies all these conditions. Cases 1 and 2 involve phase changes (constant temperature: not cooling). Case 3 involves heating, not cooling. Only case 4 applies. Concept Tested: Newton’s Law of Cooling: applies to gradual cooling of a body above surrounding temperature

NDA 2022-I

Q. 12. What is the mass of a material, whose specific heat capacity is 400 J/(kg °C) for a rise in temperature from 15 °C to 25 °C, when heat received is 20 kJ?

(a)  0.1 kg

(b)  1 kg

(c)  10 kg

(d)  5 kg

Answer: (d) 5 kg

Explanation: Q = mcΔT → m = Q/(cΔT) = 20,000 J / (400 J/kg°C × 10°C) = 20,000/4,000 = 5 kg. ΔT = 25 − 15 = 10°C. Q must be in joules: 20 kJ = 20,000 J. Concept Tested: Specific heat calculation: m = Q/(cΔT); units: Q in J, c in J/kg°C, ΔT in °C

Q. 13. The specific latent heat of vaporization of a substance is the quantity of heat needed to change unit mass from

(a)  liquid to vapour with a change of temperature

(b)  liquid to vapour without a change of temperature

(c)  vapour to liquid without a change of temperature

(d)  vapour to liquid with a change of temperature

Answer: (b) liquid to vapour without a change of temperature

Explanation: Specific latent heat of vaporisation is the heat required to convert 1 kg of liquid to vapour at constant temperature (at the boiling point). The temperature does not change during this process: all the heat goes into breaking intermolecular bonds, not raising temperature. Concept Tested: Latent heat of vaporisation: liquid to vapour at constant temperature

Q. 14. Evaporation from the surface of a given liquid takes place more rapidly when

(a)  the temperature is high and the surface area of the liquid is large

(b)  the temperature is low and the surface area of the liquid is large

(c)  the temperature is low and the surface area of the liquid is small

(d)  the temperature is high and the surface area of the liquid is small

Answer: (a) the temperature is high and the surface area of the liquid is large

Explanation: Evaporation rate increases with: (1) higher temperature: molecules have more energy to escape, (2) larger surface area: more surface molecules available to evaporate, (3) lower humidity, (4) higher wind speed. The NDA-tested combination is high temperature AND large surface area: both simultaneously increase evaporation. Concept Tested: Factors increasing evaporation rate: high temperature and large surface area

NDA 2021-I

Q. 15. Which one of the following is the lowest possible temperature?

(a)  0° Celsius

(b)  –073° Celsius

(c)  –173° Celsius

(d)  –273° Celsius

Answer: (d) –273° Celsius

Explanation: Absolute zero is the lowest possible temperature: 0 K = −273.15°C ≈ −273°C. No temperature below this exists. Options (a), (b), and (c) are all above absolute zero (they correspond to 273 K, 200 K, and 100 K respectively). Concept Tested: Absolute zero: lowest possible temperature = 0 K = −273°C

Q. 16. Numerically two thermometers, one in Fahrenheit scale and another in Celsius scale shall read same at

(a)  –40°

(b)  0°

(c)  –273°

(d)  100°

Answer: (a) –40°

Explanation: Setting °F = °C: °C = 32 + 1.8°C → −0.8°C = 32 → °C = −40. At −40°, both Fahrenheit and Celsius scales show the same numerical reading. This is the only crossover point of the two scales. Concept Tested: Fahrenheit-Celsius numerical equality: both read −40° at the same temperature

★ JOVIK Exam Insight The scales cross at −40°: −40°F = −40°C. This is one of the most tested temperature scale facts in NDA. Above −40°, Fahrenheit always gives a higher number than Celsius for the same temperature.

NDA 2019-II

Q. 17. In which of the following phenomena do heat waves travel along a straight line with the speed of light?

(a)  Thermal conduction

(b)  Thermal convection

(c)  Thermal radiation

(d)  Both, thermal conduction and radiation

Answer: (c) Thermal radiation

Explanation: Thermal radiation is the transfer of heat by electromagnetic waves (infrared radiation). Like all electromagnetic radiation, it travels in straight lines at the speed of light (3 × 10⁸ m/s) and requires no medium. Conduction involves molecular vibration in a medium; convection involves bulk fluid movement: neither travels at the speed of light. Concept Tested: Thermal radiation: travels as electromagnetic waves at speed of light in straight lines

Q. 18. If the work done on the system or by the system is zero, which one of the following statements for a gas kept at a certain temperature is correct?

(a)  Change in internal energy of the system is equal to flow of heat in or out of the system.

(b)  Change in internal energy of the system is less than heat transferred.

(c)  Change in internal energy of the system is more than the heat flow.

(d)  Cannot be determined.

Answer: (a) Change in internal energy of the system is equal to flow of heat in or out of the system.

Explanation: First Law of Thermodynamics: ΔU = Q − W. When W = 0 (no work done): ΔU = Q. All heat input becomes internal energy; all internal energy change equals heat flow. This is an isochoric (constant volume) or constrained process where no work is done. Concept Tested: First Law of Thermodynamics: when W = 0: ΔU = Q (internal energy change equals heat transfer)

Q. 19. The temperature of a place on one sunny day is 113 in Fahrenheit scale. The Kelvin scale reading of this temperature will be

(a)  318 K

(b)  45 K

(c)  62·8 K

(d)  335·8 K

Answer: (a) 318 K

Explanation: Step 1: °F to °C: °C = (°F − 32)/1.8 = (113 − 32)/1.8 = 81/1.8 = 45°C. Step 2: °C to K: K = 45 + 273 = 318 K. Always convert Fahrenheit to Celsius first, then add 273 for Kelvin. Concept Tested: Temperature conversion: Fahrenheit to Kelvin via Celsius: (°F − 32)/1.8 + 273

NDA 2019-I

Q. 20. Which one of the following could be the melting point of iron?

(a)  25°C

(b)  37°C

(c)  500°C

(d)  1500°C

Answer: (d) 1500°C

Explanation: Iron is a high-melting-point metal. Its melting point is approximately 1538°C. Among the options, only 1500°C is in the correct range. 25°C is room temperature, 37°C is body temperature, and 500°C is far below the melting point of iron (though it melts some lower-melting metals like tin and lead). Concept Tested: Melting point of iron: approximately 1538°C; approximately 1500°C from options

Q. 21. Which one of the following statements regarding a thermos flask is NOT correct?

(a)  The walls of flask are separated by vacuum and made of glass which is a poor conductor of heat

(b)  The glass walls themselves have shiny surfaces

(c)  The surface of inner wall radiates good amount of heat and the surface of outer wall absorbs some of the heat that is radiated from the inner wall

(d)  The cork supports are poor conductors of heat

Answer: (c) The surface of inner wall radiates good amount of heat and the surface of outer wall absorbs some of the heat that is radiated from the inner wall

Explanation: The inner wall of a thermos flask is silvered (mirror-finished). A silvered surface is a poor emitter and poor absorber of radiation. This is why it minimises radiative heat transfer. Option (c) says the inner wall ‘radiates a good amount of heat’: this is the opposite of the truth. Options (a), (b), and (d) are correct descriptions. Concept Tested: Thermos flask: silvered walls are POOR emitters/absorbers of radiation (not good)

★ JOVIK Exam Insight The thermos flask is designed to block all three heat transfer modes: vacuum kills conduction and convection; silvered walls minimise radiation. The confusion is option (c): silvered surfaces REDUCE radiation, not increase it. Also tested in NDA 2012-I.

Q. 22. The formula for conversion between Fahrenheit and Celsius is °F = X + (1.8 × °C). What is factor X?

(a)  32

(b)  22

(c)  98

(d)  42

Answer: (a) 32

Explanation: The standard Fahrenheit–Celsius conversion is °F = 32 + (9/5)°C = 32 + 1.8 × °C. The constant X = 32. This can be verified: at 0°C (freezing point), °F = 32 + 0 = 32°F: which is correct. Concept Tested: Temperature conversion formula: °F = 32 + 1.8 × °C; X = 32

NDA 2018-II

Q. 23. The absolute zero temperature is 0 Kelvin. In °C unit, which one of the following is the absolute zero temperature?

(a)  0 °C

(b)  −100 °C

(c)  −273.15 °C

(d)  −173.15 °C

Answer: (c) −273.15 °C

Explanation: °C = K − 273.15. At K = 0: °C = 0 − 273.15 = −273.15°C. Absolute zero in Celsius is −273.15°C. Option (a) 0°C = 273.15 K: well above absolute zero. Option (d) −173.15°C = 100 K: also above absolute zero. Concept Tested: Absolute zero: 0 K = −273.15°C

Q. 24. The coefficient of areal expansion of a material is 1.6 × 10⁻⁵ K⁻¹. Which one of the following gives the value of coefficient of volume expansion of this material?

(a)  0.8 × 10⁻⁵ K⁻¹

(b)  2.4 × 10⁻⁵ K⁻¹

(c)  3.2 × 10⁻⁵ K⁻¹

(d)  4.8 × 10⁻⁵ K⁻¹

Answer: (b) 2.4 × 10⁻⁵ K⁻¹

Explanation: The coefficients are related: α (linear) = β/2 (areal) = γ/3 (volume). So γ = (3/2)β = (3/2) × 1.6 × 10⁻⁵ = 2.4 × 10⁻⁵ K⁻¹. The volume expansion coefficient is always 3/2 times the areal expansion coefficient. Concept Tested: Thermal expansion coefficients: γ = (3/2)β; volume = (3/2) × areal coefficient

NDA 2018-I

Q. 25. Which one of the following statements is correct?

(a)  Any energy transfer that does not involve temperature difference in some way is not heat

(b)  Any energy transfer always requires a temperature difference

(c)  On heating the length and volume of the object remain exactly the same

(d)  Whenever there is a temperature difference, heat is the only way of energy transfer

Answer: (a) Any energy transfer that does not involve temperature difference in some way is not heat

Explanation: Heat is defined specifically as energy transfer due to a temperature difference. Option (a) correctly defines the boundary: without a temperature difference, the process is not heat (it may be work). Option (b) is wrong: work transfer requires no temperature difference. Option (c) is wrong: objects expand on heating. Option (d) is wrong: temperature differences can also drive work (e.g., in heat engines). Concept Tested: Definition of heat: energy transfer due to temperature difference; not the only mode

Q. 26. Which of the following statements about specific heat of a body is/are correct?

1.  It depends upon mass and shape of the body

2.  It is independent of mass and shape of the body

3.  It depends only upon the temperature of the body

Select the correct answer using the code given below:

(a)  1 only

(b)  2 and 3

(c)  1 and 3

(d)  2 only

Answer: (d) 2 only

Explanation: Specific heat capacity is an intrinsic material property: it is independent of the mass and shape of the body (Statement 2 is correct). Statement 1 is wrong. Statement 3 has some truth (specific heat can vary with temperature), but the standard NDA-level answer treats specific heat as constant for a given material. Among the options, only Statement 2 is unambiguously correct. Concept Tested: Specific heat: intrinsic property; independent of mass and shape

★ JOVIK Exam Insight Specific heat (per unit mass) vs thermal capacity (for entire body): specific heat is material-dependent only; thermal capacity = mass × specific heat. NDA 2018-I tested this distinction with a multi-statement question.

Q. 27. Thermal capacity of a body depends on the

(a)  mass of the body only

(b)  mass and shape of the body only

(c)  density of the body

(d)  mass, shape and temperature of the body

Answer: (a) mass of the body only

Explanation: Thermal capacity = mass × specific heat = mc. For a given material (fixed c), thermal capacity depends only on mass. Shape does not affect thermal capacity independently of mass. Density is not directly relevant. Temperature can affect c but at standard NDA level, c is treated as constant. Concept Tested: Thermal capacity: equals mass × specific heat; depends primarily on mass for a given material

NDA 2017-II

Q. 28. The statement that ‘heat cannot flow by itself from a body at a lower temperature to a body at a higher temperature’, is known as?

(a)  Zeroth law of thermodynamics

(b)  First law of thermodynamics

(c)  Second law of thermodynamics

(d)  Third law of thermodynamics

Answer: (c) Second law of thermodynamics

Explanation: The Second Law of Thermodynamics establishes the natural direction of heat flow: heat flows spontaneously from higher to lower temperature, never the reverse without external work input. The Zeroth Law defines thermal equilibrium. The First Law is energy conservation (ΔU = Q − W). The Third Law concerns entropy at absolute zero. Concept Tested: Second Law of Thermodynamics: heat cannot spontaneously flow from cold to hot body

NDA 2017-I

Q. 29. Which one of the following statements is NOT correct?

(a)  In the conduction mode of transference of heat, the molecules of solid pass heat from one molecule to another without moving from their positions

(b)  The amount of heat required to raise the temperature of a substance is called its specific heat capacity

(c)  The process of heat transfer in liquids and gases is through convection mode

(d)  The process of heat transfer from a body at higher temperature to a body at lower temperature without heating the space between them is known as radiation

Answer: (b) The amount of heat required to raise the temperature of a substance is called its specific heat capacity

Explanation: Option (a) correctly describes conduction. Option (c) correctly identifies convection as the primary heat transfer in fluids. Option (d) correctly defines radiation. Option (b) is the incorrect one: the amount of heat required to raise the temperature of the entire body by 1°C is thermal capacity, not specific heat. Specific heat is the heat required per unit mass. Concept Tested: Specific heat vs thermal capacity: specific heat is per unit mass; thermal capacity is for entire body

Q. 30. The amount of heat required to change a liquid to gaseous state without any change in temperature is known as

(a)  specific heat capacity

(b)  mechanical equivalent of heat

(c)  latent heat of vaporization

(d)  quenching

Answer: (c) latent heat of vaporization

Explanation: Latent heat of vaporisation is the heat required to convert a liquid to gas at constant temperature. ‘Latent’ means hidden: the heat goes into breaking intermolecular bonds rather than raising temperature. Specific heat relates to temperature change, not phase change. Quenching refers to rapid cooling of metals. Concept Tested: Latent heat of vaporisation: heat for liquid-to-gas phase change at constant temperature

Q. 31. The time period of a simple pendulum made using a thin copper wire of length L is T. Suppose the temperature of the room in which this simple pendulum is placed increases by 30°C, what will be the effect on the time period of the pendulum?

(a)  T will increase slightly

(b)  T will remain the same

(c)  T will decrease slightly

(d)  T will become more than 2 times

Answer: (a) T will increase slightly

Explanation: T = 2π√(L/g). When temperature increases, the copper wire expands: L increases slightly (linear thermal expansion). Since T ∝ √L, a small increase in L causes a small increase in T. The pendulum swings slightly more slowly. It does not double, and it does not decrease (since L increases, not decreases). Concept Tested: Thermal expansion of pendulum: increased temperature expands length, increases period

Q. 32. A Kelvin thermometer and a Fahrenheit thermometer both give the same reading for a certain sample. What would be the corresponding reading in a Celsius thermometer?

(a)  574

(b)  301

(c)  273

(d)  232

Answer: (b) 301

Explanation: Set K = F. K = °C + 273, F = 32 + 1.8°C. So °C + 273 = 32 + 1.8°C. Rearranging: 273 − 32 = 0.8°C → 241 = 0.8°C → °C = 301.25°C ≈ 301. When the Kelvin and Fahrenheit readings are numerically equal, the Celsius temperature is approximately 301°C. Concept Tested: Temperature scale crossover: K = F numerically at ≈ 301°C

Q. 33. Why is it difficult to measure the coefficient of expansion of a liquid than solid?

(a)  Liquids tend to evaporate at all temperatures

(b)  Liquids conduct more heat

(c)  Liquids expand too much when heated

(d)  Their containers also expand when heated

Answer: (d) Their containers also expand when heated

Explanation: When measuring liquid expansion, the container holding the liquid also expands when heated. The directly observed apparent expansion is the net result of liquid expansion minus container expansion. The true expansion of the liquid must be calculated by adding back the container’s expansion: making measurement more complex than for solids. Concept Tested: Liquid thermal expansion measurement: complicated by simultaneous expansion of container

NDA 2016-II

Q. 34. If we plot a graph between volume V and inverse of pressure P (i.e., 1/P) for an ideal gas at constant temperature T, the curve so obtained is

(a)  straight line

(b)  circle

(c)  parabola

(d)  hyperbola

Answer: (a) straight line

Explanation: Boyle’s Law: PV = constant → V = constant × (1/P). This is a linear equation: V ∝ (1/P). The graph of V vs 1/P is a straight line through the origin with slope equal to the constant (nRT). Note: the graph of V vs P is a hyperbola; the graph of V vs 1/P is a straight line. Concept Tested: Boyle’s Law graph: V vs 1/P is a straight line through origin (not V vs P which is hyperbola)

NDA 2016-I

Q. 35. The temperature at which a solid melts to become a liquid at the atmospheric pressure is called its melting point. The melting point of a solid is an indication of

(a)  strength of the intermolecular forces of attraction

(b)  strength of the intermolecular forces of repulsion

(c)  molecular mass

(d)  molecular size

Answer: (a) strength of the intermolecular forces of attraction

Explanation: Melting requires sufficient energy to overcome the attractive forces holding molecules in fixed positions in the solid lattice. A higher melting point means more energy is needed: indicating stronger intermolecular attraction. It is not directly related to molecular mass or size, and repulsive forces do not determine melting behaviour. Concept Tested: Melting point: indicator of strength of intermolecular attractive forces

Q. 36. Which one of the following statements with regard to expansion of materials due to heating is NOT correct?

(a)  As ice melts, it expands uniformly up to 4°C.

(b)  Mercury thermometer works using the principle of expansion due to heating.

(c)  Small gap is kept between two rails to allow for expansion due to heating.

(d)  The length of metallic wire increases when its temperature is increased.

Answer: (a) As ice melts, it expands uniformly up to 4°C.

Explanation: Options (b), (c), and (d) are correct examples of thermal expansion. Option (a) is false: ice CONTRACTS on melting from 0°C to 4°C: water becomes denser as it warms from 0°C to 4°C (anomalous expansion). The statement that ice expands uniformly up to 4°C is incorrect. Concept Tested: Anomalous expansion of water: ice contracts (not expands) as it melts from 0°C to 4°C

★ JOVIK Exam Insight Ice contracts on melting from 0°C to 4°C. Water is denser at 4°C than at 0°C. This is the anomalous expansion of water: tested in NDA 2016-I and linked to the lake bottom question (NDA 2014-I). The phrase ‘expands uniformly up to 4°C’ is the false statement.

NDA 2015-II

Q. 37. The absolute zero, i.e., temperature below which is not achievable, is about:

(a)  0 °C

(b)  –273 K

(c)  –273 °C

(d)  –300 °C

Answer: (c) –273 °C

Explanation: Absolute zero is 0 K = −273.15°C ≈ −273°C. Option (b) −273 K is meaningless: the Kelvin scale starts at 0 K and has no negative values. Option (a) 0°C = 273 K, far above absolute zero. Option (d) −300°C would be below absolute zero: impossible. Concept Tested: Absolute zero: 0 K = −273°C; no negative Kelvin values exist

★ JOVIK Exam Insight Absolute zero has been tested in NDA 2015-II, 2018-II, and 2021-I. Answer is always −273°C or equivalently 0 K. Option ‘−273 K’ is a deliberate confusion created: Kelvin has no negative values.

Q. 38. Which one of the following is the SI unit of the thermal conductivity of a material?

(a)  Wm⁻¹K⁻¹

(b)  Wm/K

(c)  Wm⁻¹/K⁻¹

(d)  Js⁻¹m⁻¹K

Answer: (a) Wm⁻¹K⁻¹

Explanation: Thermal conductivity k is defined from Fourier’s Law: heat flow rate = k × A × (ΔT/L). Units of k: (W × m)/(m² × K) = W/(m·K) = W m⁻¹ K⁻¹. Option (b) Wm/K is dimensionally different. Option (d) Js⁻¹m⁻¹K is wrong (K should be in denominator). Concept Tested: SI unit of thermal conductivity: W m⁻¹ K⁻¹ (watts per metre per kelvin)

Q. 39. Which one of the following statements is not correct?

(a)  Conduction can occur easily in solids, less easily in liquids but hardly at all in gases

(b)  Heat energy is carried by moving particles in a convection current

(c)  Heat energy is carried by electromagnetic waves in radiation

(d)  The temperature at which a solid changes into a liquid is called the boiling point

Answer: (d) The temperature at which a solid changes into a liquid is called the boiling point

Explanation: Options (a), (b), and (c) correctly describe conduction, convection, and radiation. Option (d) is false: the temperature at which a solid changes to liquid is the melting point (or freezing point), not the boiling point. The boiling point is where liquid changes to gas. Concept Tested: Phase change nomenclature: solid to liquid = melting point (NOT boiling point)

NDA 2015-I

Q. 40. A solid is melted and allowed to cool and solidify again. The temperature is measured at equal intervals of time. The graph below shows the change of temperature with time. The part of the curve that is practically horizontal is due to

(a)  latent heat given away by the liquid

(b)  specific heat given away by the liquid

(c)  thermal capacity changes with time keeping temperature constant

(d)  change in density during transformation

Answer: (a) latent heat given away by the liquid

Explanation: During solidification (liquid → solid), the temperature remains constant at the freezing point while the liquid releases its latent heat of fusion to the surroundings. This latent heat release is exactly what keeps the temperature constant: producing the horizontal plateau in the cooling curve. It has nothing to do with specific heat, thermal capacity variation, or density changes. Concept Tested: Cooling curve: horizontal plateau at freezing point due to latent heat release

Q. 41. When heat rays are reflected from Earth, gases like Carbon dioxide, Nitrous oxide do not allow them to escape back to the space causing our planet to heat up. These gases are known as

(a)  Noble gas

(b)  Green-house gas

(c)  Hot gas

(d)  Blue gas

Answer: (b) Green-house gas

Explanation: Greenhouse gases (CO₂, CH₄, N₂O, water vapour) absorb the long-wavelength infrared radiation re-emitted by the Earth’s surface after absorbing solar energy, and re-emit it back downward: trapping heat in the atmosphere. This is the greenhouse effect, named after the way a glass greenhouse traps heat. Concept Tested: Greenhouse gases: CO₂ and N₂O trap outgoing infrared radiation, causing global warming

Q. 42. Thermal conductivity of aluminium, copper and stainless steel increases in the order

(a)  Copper < Aluminium < Stainless Steel

(b)  Stainless Steel < Aluminium < Copper

(c)  Aluminium < Copper < Stainless Steel

(d)  Copper < Stainless Steel < Aluminium

Answer: (b) Stainless Steel < Aluminium < Copper

Explanation: Thermal conductivity values (approximate): copper ≈ 400 W/mK; aluminium ≈ 200 W/mK; stainless steel ≈ 15 W/mK. The order from lowest to highest is: stainless steel < aluminium < copper. Copper is the best conductor among the three. Concept Tested: Thermal conductivity order: Stainless Steel < Aluminium < Copper

Q. 43. Perspiration cools the body because

(a)  presence of water on the skin is cooling

(b)  evaporation requires latent heat

(c)  water has a high specific heat

(d)  water is a poor conductor of heat

Answer: (b) evaporation requires latent heat

Explanation: When sweat evaporates from the skin, it requires latent heat of vaporisation: this energy is drawn from the body surface, lowering the skin temperature. The cooling is entirely due to the latent heat absorbed during the phase change from liquid (sweat) to vapour. High specific heat and poor conductivity of water are not the reasons. Concept Tested: Perspiration cools body: evaporation of sweat absorbs latent heat from skin

NDA 2014-I

Q. 44. The temperature of water at the bottom of a lake whose upper surface has frozen to ice would be around

(a)  −10 °C

(b)  0 °C

(c)  4 °C

(d)  −4 °C

Answer: (c) 4 °C

Explanation: Water has maximum density at 4°C. As the surface water cools below 4°C it becomes less dense and stays at the surface, eventually freezing. The denser 4°C water sinks and accumulates at the bottom of the lake. Even when the surface is frozen, the bottom remains at approximately 4°C: protecting aquatic life. Concept Tested: Anomalous expansion of water: bottom of frozen lake remains at 4°C (maximum density)

Q. 45. The pressure of an ideal gas undergoing isothermal change is increased by 10%. The volume of the gas must decrease by about

(a)  0.1%

(b)  9%

(c)  10%

(d)  0.9%

Answer: (b) 9%

Explanation: Isothermal process: PV = constant. If P increases by 10%, new P = 1.1P. New volume V’ = PV/(1.1P) = V/1.1 ≈ 0.909V. Decrease = V − 0.909V = 0.091V ≈ 9.1% ≈ 9%. A common error is to say 10%: but Boyle’s Law is multiplicative, not additive. Concept Tested: Boyle’s Law (isothermal): 10% pressure increase gives ~9% (not 10%) volume decrease

★ JOVIK Exam Insight A classic NDA arithmetic confusion: students say 10% pressure increase → 10% volume decrease. Wrong. PV = const is multiplicative: new V = V/1.1 = 0.909V, a 9% decrease. The relationship is P₁V₁ = P₂V₂, not P₁ − P₂ = -(V₁ − V₂).

NDA 2012-II

Q. 46. Which one among the following statements about thermal conductivity is correct?

(a)  Steel > Wood > Water

(b)  Steel > Water > Wood

(c)  Water > Steel > Wood

(d)  Water > Wood > Steel

Answer: (b) Steel > Water > Wood

Explanation: Thermal conductivity order: metals (steel) > water > wood. Water is a better conductor than wood because wood is a thermal insulator with many air pockets. Steel, though not the best metal, greatly exceeds both. Steel > Water > Wood is the correct order among these three materials. Concept Tested: Thermal conductivity: Steel > Water > Wood

Q. 47. A hot object loses heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the

(a)  temperature of the object

(b)  temperature of the surroundings

(c)  temperature difference between the object and its surroundings

(d)  average temperature of the object and its surroundings

Answer: (c) temperature difference between the object and its surroundings

Explanation: Newton’s Law of Cooling (valid for small temperature differences) states: rate of heat loss = k × (T_object − T_surroundings). The driving factor is the temperature difference between the object and its surroundings: not the absolute temperature of either alone. Greater difference → faster cooling. Concept Tested: Newton’s Law of Cooling: rate of heat loss depends on temperature difference

NDA 2012-I

Q. 48. Statement I: A thermos flask is made of double-walled glass bottles.

Statement II: Metals are good conductors while gas and air are poor conductors of heat.

(a)  Both the statements are individually true and Statement II is the correct explanation of Statement I

(b)  Both the statements are individually true but Statement II is not the correct explanation of Statement I

(c)  Statement I is true but Statement II is false

(d)  Statement I is false but Statement II is true

Answer: (b) Both the statements are individually true but Statement II is not the correct explanation of Statement I

Explanation: Statement I is true: thermos flasks do have double-walled glass construction. Statement II is true: metals conduct well while gases do not. However, Statement II does not explain Statement I. The double-walled thermos works primarily by creating a vacuum between the walls (eliminating conduction and convection), not by using glass as a poor conductor. Statement II concerns material properties but does not explain the vacuum design. Concept Tested: Thermos flask: double glass walls with vacuum; Statement II does not explain Statement I

Q. 49. A glass of water does not turn into ice as it reaches 0°C. It is because

(a)  water does not solidify at 0°C

(b)  a certain amount of heat must be supplied to the glass of water so as to solidify

(c)  a certain amount of heat must be taken out from the glass of water so as to solidify

(d)  water solidifies at 0 K only

Answer: (c) a certain amount of heat must be taken out from the glass of water so as to solidify

Explanation: Water freezes at 0°C: but reaching 0°C is only the start. Solidification requires the removal of the latent heat of fusion (334 J/g) at constant temperature. This heat must be taken out from the water. If no heat is removed after reaching 0°C, the water remains liquid even though the temperature is 0°C. Concept Tested: Latent heat of fusion: heat must be removed (not supplied) for water to solidify at 0°C

Q. 50. The graph given above indicates change in temperature (θ) when heat (Q) was given to a substance. Which among the following parts of the graph correctly depict the latent heat of the substance?

(a)  AB and BC

(b)  BC and DE

(c)  CD and DE

(d)  DE and AB

Answer: (b) BC and DE

Explanation: On a standard temperature–heat graph (heating curve), the horizontal (flat) segments represent phase changes: temperature remains constant while heat is absorbed (latent heat). In a typical five-segment graph (A→B slope, B→C flat, C→D slope, D→E flat, E→F slope): BC = melting (solid to liquid, latent heat of fusion) and DE = boiling (liquid to gas, latent heat of vaporisation). Both BC and DE are latent heat regions. Concept Tested: Temperature-heat graph: flat horizontal segments (BC and DE) represent latent heat

Q. 51. Body A of mass 2 kg and another body B of mass 4 kg and of same material are kept in the same sunshine for some interval of time. If the rise in temperature is equal for both the bodies, then which one among the following in this regard is correct?

(a)  Heat absorbed by B is double because its mass is double

(b)  Heat absorbed by A is double because its mass is half

(c)  Heat absorbed by both A and B is equal because the quantity of heat absorbed does not depend upon mass

(d)  Heat absorbed by B is four times than the heat absorbed by A because the quantity of heat absorbed is proportional to square of the mass

Answer: (a) Heat absorbed by B is double because its mass is double

Explanation: Q = mcΔT. Same material (same c) and same temperature rise (same ΔT). Q ∝ m. Mass of B = 2 × mass of A, so heat absorbed by B = 2 × heat absorbed by A. Option (a) is correct. Option (c) is wrong: heat absorbed does depend on mass. Option (d) is wrong: Q ∝ m, not m². Concept Tested: Specific heat: Q = mcΔT; heat absorbed proportional to mass at constant c and ΔT

Q. 52. The thermal conductivity of copper is 4 times that of brass. Two rods of copper and brass having same length and cross-section are joined end to end. The free end of copper is at 0°C and the free end of brass is at 100°C. The temperature of the junction is

(a)  20°C

(b)  40°C

(c)  60°C

(d)  10°C

Answer: (a) 20°C

Explanation: In steady state, heat flow rate through both rods is equal: Q/t = k_Cu × A × (T_j − 0)/L = k_Br × A × (100 − T_j)/L. With k_Cu = 4k_Br: 4k(T_j) = k(100 − T_j). So 4T_j = 100 − T_j. Therefore 5T_j = 100, giving T_j = 20°C. Concept Tested: Thermal conduction in series rods: equal heat flow rate; solving for junction temperature

NDA 2011-II

Q. 53. The phenomenon of ‘trade winds’ takes place due to

(a)  conduction of heat

(b)  convection of heat

(c)  radiation

(d)  None of the above

Answer: (b) convection of heat

Explanation: Trade winds are a global-scale convection phenomenon. Air over the equatorial regions absorbs solar heat, becomes less dense, and rises. Cooler, denser air from higher latitudes flows in to replace it: creating the prevailing trade winds. This is bulk movement of a fluid (air) driven by temperature differences: convection. Concept Tested: Trade winds: large-scale convection current in the atmosphere

NDA 2010-I

Q. 54. A fan produces a feeling of comfort during hot weather, because

(a)  our body radiates more heat in air

(b)  fan supplies cool air

(c)  conductivity of air increases

(d)  our perspiration evaporates rapidly

Answer: (d) our perspiration evaporates rapidly

Explanation: A fan creates air movement over the skin, which accelerates the evaporation of sweat. Evaporation is a cooling process: it requires latent heat drawn from the body surface, lowering skin temperature. The fan does not supply cool air; it moves the same warm air faster. Increased evaporation rate is the mechanism of cooling. Concept Tested: Fan cooling: accelerates evaporation of sweat; latent heat drawn from skin

★ JOVIK Exam Insight Fan cooling (evaporation of perspiration) and perspiration cooling itself (2015-I) are the same concept from different angles. NDA tested both. Always: fan → faster evaporation → latent heat removal → cooling.

Q. 55. A man with a dark skin, in comparison with a man with a white skin, will experience

(a)  less heat and less cold

(b)  less heat and more cold

(c)  more heat and less cold

(d)  more heat and more cold

Answer: (d) more heat and more cold

Explanation: Dark (black) surfaces are better absorbers and better emitters of radiation than light (white) surfaces. A person with dark skin absorbs more incoming radiation (more heat in sunshine) and also radiates more heat in cold conditions (more cold). Both effects are amplified compared to lighter skin. Concept Tested: Radiation absorption and emission: dark surfaces: better absorber AND better emitter

Q. 56. Which one among the following denotes the smallest temperature?

(a)  1° on the Celsius scale

(b)  1° on the Kelvin scale

(c)  1° on the Fahrenheit scale

(d)  1° on the Reaumur scale

Answer: (c) 1° on the Fahrenheit scale

Explanation: The size of 1 degree varies across scales. 1°C = 1 K (identical interval). 1°F = 5/9°C ≈ 0.556°C (smaller than Celsius). 1°Réaumur = 5/4°C = 1.25°C (larger than Celsius). The Fahrenheit degree is the smallest interval among the four options. Concept Tested: Temperature scale intervals: 1°F is smallest (≈ 0.556°C); Celsius = Kelvin in size

Q. 57. The best and the poorest conductors of heat are respectively

(a)  silver (Ag) and lead (Pb)

(b)  copper (Cu) and aluminium (Al)

(c)  silver (Ag) and gold (Au)

(d)  copper (Cu) and gold (Au)

Answer: (a) silver (Ag) and lead (Pb)

Explanation: Among metals, silver has the highest thermal conductivity, followed by copper, gold, aluminium, and steel. Lead is among the poorest metallic conductors of heat. The pairing silver (best) and lead (poorest) is correct. Concept Tested: Thermal conductivity of metals: silver is best; lead is poorest metallic conductor

---