NDA Properties of matter and heat previous year questions with solutions

Answer: (d) Pascal and N/⁠m² represent the same unit

Method. units and basic definitions

1 pascal is defined as 1 N/⁠m², so (d) is correct. (a) False: pressure is a scalar. (b) False: pressure P = ρgh increases with depth. (c) False: action and reaction act on different bodies (NCERT Class 11 Physics, Mechanical Properties of Fluids and Laws of Motion).

Answer: (b) 9ρ/⁠16

Method. average density = total mass / total volume

Solid sphere: volume (4/⁠3)π(R/⁠2)³ = πR³/⁠6, mass ρπR³/⁠6. Shell: volume (4/⁠3)πR³ − πR³/⁠6 = (7/⁠6)πR³, mass (ρ/⁠2)(7/⁠6)πR³ = 7ρπR³/⁠12. Total mass = (2 + 7)ρπR³/⁠12 = 3ρπR³/⁠4. Total volume = 4πR³/⁠3. Average density = (3/⁠4)/⁠(4/⁠3) ρ = 9ρ/⁠16.

Answer: (a) P_A > P_B > P_C

Method. pressure = weight / contact area

The weight W is the same in all three positions, so P = W/⁠A varies inversely with the contact area. A: 20×10 = 200 cm²; B: 10×40 = 400 cm²; C: 40×20 = 800 cm². The areas increase A < B < C, so P_A > P_B > P_C.

Answer: (b) 32ρ₀

Method. average density of a composite body

Let each part have volume V. Mass = ρ0·V + 2ρ0·V = 3ρ0V; volume = 2V. Average density = 3ρ0V / 2V = (3/⁠2)ρ0.

Answer: (d) 666.6 N/⁠m²

Method. pressure = force / area

Weight = mg = 2 × 10 = 20 N. Face resting on the table: 30 cm × 10 cm = 300 cm² = 0.03 m². P = 20 / 0.03 = 666.7 N/⁠m², nearest option (d) 666.6 (NCERT Class 11 Physics, Mechanical Properties of Fluids).

Answer: (b) The pressure depends on the height of liquid column

Method. hydrostatic pressure p = ρgh

Pressure at the base = ρgh depends on the liquid's density and the height of the column, and not on the base area or the shape of the vessel (NCERT Class 11 Physics, mechanical properties of fluids).

Answer: (a) 6, 2

Method. mixture density, equal volume and equal mass

Equal volumes: ρ_mix = (ρ₁ + ρ₂)/⁠2 = 4, so ρ₁ + ρ₂ = 8. Equal masses: ρ_mix = 2ρ₁ρ₂/⁠(ρ₁ + ρ₂) = 3, so ρ₁ρ₂ = 12. The roots of x² − 8x + 12 = 0 are 6 and 2. Check: (6+2)/⁠2 = 4 and 2(12)/⁠8 = 3. Option (a); (b) 3,5 gives 4 but 2.15/⁠8 ≠ 3, (c) and (d) have sums 16 and 12.

Shortcut. Only option (a) has sum 8.

Answer: (b) 2 and 3 only

Method. hydrostatic pressure in a fluid at rest

Pressure in a fluid at rest increases with depth (p = p0 + ρgh), so it is not the same at all points: statement 1 is false. The fluid presses on the walls of the vessel (2 true) and pressure exists at every point of the fluid (3 true). So 2 and 3 only (NCERT Class 11 Physics, mechanical properties of fluids).

Answer: (b) Barometer

Method. instrument for atmospheric pressure

A barometer measures atmospheric pressure. Ammeter measures current, potentiometer compares potential differences, lactometer measures density (purity) of milk.

Answer: (d) The centre of gravity of the body is below the metacentre.

Method. stability of floating bodies (metacentre)

A floating body is stable when its metacentre M lies above its centre of gravity G, i.e. G is below the metacentre. The relation of G to the centre of buoyancy B alone does not decide it (a ship has G above B and is still stable), so (a), (b) and (c) are not the condition.

Answer: (b) density of water is less than density of body

Method. sinking condition by density

A body sinks when its density exceeds that of water, i.e. the density of water is less than the density of the body. 'Heavy' alone (d) is not the condition: a large heavy ship floats (NCERT Class 11 Physics, Mechanical Properties of Fluids).

Answer: (b) 2 only

Method. Archimedes' principle

The upward buoyant force on an immersed body equals the weight of the fluid it displaces (not the mass of the body). So only statement 2 is correct (NCERT Class 11 Physics, Mechanical Properties of Fluids).

Answer: (d) ρ₁ < ρ_W < ρ₂

Method. floating condition for a massless container: weight of liquid = buoyancy

The cubes are massless, so the weight of the liquid inside equals the buoyant force: ρ·(filled fraction) = ρ_w·(submerged fraction). C1: about 3/⁠4 filled, about 2/⁠3 submerged, so ρ1 = (2/⁠3)/⁠(3/⁠4) ρ_w = 8ρ_w/⁠9 < ρ_w. C2: about 2/⁠3 filled, about 5/⁠6 submerged, so ρ2 = (5/⁠6)/⁠(2/⁠3) ρ_w = 5ρ_w/⁠4 > ρ_w. Hence ρ1 < ρ_w < ρ2.

Shortcut. Compare the filled fraction with the submerged fraction in the figure: filled > submerged means the liquid is lighter than water; filled < submerged means heavier.

Answer: (c) 1000 kg/⁠m³

Method. density = mass / volume

Mass = W/⁠g = 7.5/⁠10 = 0.75 kg. Volume = volume of water displaced = (3/⁠4) L = 0.75 × 10⁻³ m³. Density = 0.75 / 0.75×10⁻³ = 1000 kg/⁠m³.

Answer: (b) 2 and 3

Method. floating vs sinking by average density

The iron nail sinks because iron is denser than water (statement 2). The ship floats because its hollow hull makes its average density (steel plus enclosed air) less than that of water (statement 3). Statements 1 and 4 are false, so the code is 2 and 3 (NCERT Class 11 Physics, Mechanical Properties of Fluids).

Answer: (a) f₂ = f₁

Method. buoyancy in an accelerating frame (effective gravity)

In the ship's frame the effective gravity is a (pointing opposite to the acceleration), and the water also feels it. Floating: buoyant force ρ_w·(f2 V)·a = weight m·a, so f2 = m/⁠(ρ_w V) = f1. The effective gravity cancels, so the submerged fraction is unchanged: f2 = f1.

Answer: (a) The packet will float in both water and liquid B.

Method. floating condition by density comparison

Density of packet = 800 g / 1000 cm³ = 0.8 g/⁠cm³. This is less than 1 g/⁠cm³ (water) and less than 1.5 g/⁠cm³ (liquid B), so it floats in both (NCERT Class 11 Physics, Mechanical Properties of Fluids).

Answer: (c) an upward force

Method. definition of buoyancy

Buoyancy is the upward force a fluid exerts on a body immersed in it, caused by the pressure at the lower face exceeding that at the upper face. It is a force (unit N), not a pressure (NCERT Class 11 Physics, Mechanical Properties of Fluids, Archimedes' principle).

Answer: (d) upward force

Method. buoyancy

Buoyancy is the upward force a fluid exerts on a body immersed in it; it is a force, not a pressure (NCERT Class 11 Physics, mechanical properties of fluids).

Answer: (c) difference in the densities of the object and liquid

Method. floatation condition

A body floats if its density is less than the liquid's and sinks if greater; the deciding factor is the difference (comparison) of the two densities, not mass or shape alone. Mass alone can be large and still float (a ship) (NCERT Class 11 Physics, mechanical properties of fluids).

Answer: (b) Decrease

Method. Archimedes' principle

A submerged body feels an upward buoyant force equal to the weight of the water displaced, so its apparent weight = W − F_B is less than its weight in air (NCERT Class 11 Physics, mechanical properties of fluids).

Answer: (a) Surface tension

Method. surface tension and adhesion of water to a surface

Water poured slowly clings to the lip and flows down the side of the jug (the 'teapot effect'): the wetting film is held to the surface by adhesion and surface tension, which pulls the stream against the side (NCERT Class 11 Physics, Mechanical Properties of Fluids, surface tension). Vapour pressure and compressibility do not bear on this.

Viscosity also plays a part in the real teapot effect; surface tension is taken as the intended answer.

Answer: (b) It decreases when temperature increases

Method. temperature dependence of surface tension

Surface tension falls as temperature rises, because cohesive forces weaken with stronger molecular motion; it vanishes at the critical temperature (NCERT Class 11 Physics, Mechanical Properties of Fluids, surface tension). So (b).

Answer: (b) 1 cm

Method. weight depends on g; Hooke's law

Extension x = mg/⁠k, so x is proportional to g. On the Moon g' = g/⁠6, so x' = 6/⁠6 = 1 cm (NCERT Class 11 Physics, Mechanical Properties of Solids, Hooke's law).

Answer: (d) Surface tension

Method. surface tension and wetting

Soap lowers the surface tension of water, so the water wets and spreads into the fabric and lifts oily dirt away (NCERT Class 11 Physics, mechanical properties of fluids).

Answer: (a) Both Statement-I and Statement-II are correct and Statement-II explains Statement-I.

Method. anomalous expansion of water and survival of aquatic life

Water is densest at 4 °C, so as a lake cools the 4 °C water sinks and the colder, lighter water stays at the top and freezes. The bottom layer stays at about 4 °C(relatively warm compared with the ice above), and the ice sheet insulates it, so marine animals survive near the bottom (NCERT Class 11 Physics, Thermal Properties of Matter). Statement-II is correct and explains Statement-I.

Statement-II's 'remains warm' means about 4 °C, warm relative to the surface ice; read in that sense it is correct.

Answer: (b) 30°C

Method. change in temperature is the same in K and °C

A difference in kelvin equals the difference in degrees Celsius: 340 − 310 = 30 K = 30 °C. (Option 37 °C would be the temperature 310 K, not the increase.)

Answer: (d) By using thermocouple based thermometer

Method. range of thermometers

−⁠250 °C = 23 K. Mercury freezes at about −⁠39 °C and alcohol at about −⁠114 °C, so neither can read it; a clinical thermometer covers only about 35–⁠42 °C. A thermocouple works down to very low temperatures, so (d).

Answer: (b) 277 K

Method. anomalous expansion of water

Water has its maximum density at 4 °C = 273 + 4 = 277 K (NCERT Class 11 Physics, Thermal Properties of Matter, thermal expansion of water).

Answer: (a) Density of water at 4°C is 1000 kg/⁠m³.

Method. density of water

Water is densest at 4 °C, where its density is 1000 kg/⁠m³ (1 g/⁠cm³). At 0 °C it is slightly less (about 999.8 kg/⁠m³), so (b) is only approximately true, while (c) 100 and (d) 10 kg/⁠m³ are far off. (a) is the exact standard statement (NCERT Class 11 Physics, thermal properties of matter).

Answer: (d) – 273° Celsius

Method. absolute zero

The lowest possible temperature is absolute zero, 0 K = −⁠273.15 °C ≈ −⁠273 °C. Option (b) as printed ('– 073°') is −⁠73 °C, higher than −⁠173 and −⁠273.

Answer: (a) – 40°

Method. equal reading of the Celsius and Fahrenheit scales

F = 1.8C + 32; set F = C = x: x = 1.8x + 32, so −⁠0.8x = 32, x = −⁠40. At −⁠40° both read the same.

Answer: (a) 318 K

Method. Fahrenheit to Kelvin

C = (F − 32) × 5/⁠9 = 81 × 5/⁠9 = 45 °C. K = 45 + 273.15 = 318.15, about 318 K.

Answer: (a) 32

Method. Celsius-Fahrenheit conversion

F = 32 + 1.8 C(ice point 0 °C = 32 °F; steam point 100 °C = 32 + 180 = 212 °F), so X = 32 (NCERT Class 11 Physics, thermal properties of matter).

Answer: (c) −⁠273.15 °C

Method. Kelvin to Celsius

T(°C) = T(K) − 273.15, so 0 K = −⁠273.15 °C(NCERT Class 11 Physics, thermal properties of matter).

Answer: (b) 2.4 × 10^-5 K^-1

Method. β = 3α, areal coefficient = 2α

Areal coefficient = 2α, volume coefficient = 3α, so γ = (3/⁠2)(areal) = 1.5 × 1.6 × 10^-5 = 2.4 × 10^-5 K^-1.

Answer: (b) 301

Method. equate the Kelvin and Fahrenheit scales

K = C + 273.15 and F = 1.8C + 32. Setting K = F: C + 273.15 = 1.8C + 32, so 0.8C = 241.15 and C ≈ 301.4, i.e. about 301 °C. Option (b) is the nearest (with 273 in place of 273.15 the value is 301.25, same option).

Answer: (d) Their containers also expand when heated.

Method. apparent expansion of a liquid

A liquid must be held in a vessel, and the vessel expands too when heated, so only the apparent expansion is seen: γ_real = γ_apparent + γ_vessel. This is why it is hard to measure a liquid's true coefficient (NCERT Class 11 Physics, thermal properties of matter).

Answer: (c) 550 Cal/⁠g

Method. calorimetry: heat lost by condensing steam = heat gained by water

Water gains: 250 × 1 × (25 − 20) = 1250 cal. Steam condensed = 252 − 250 = 2 g. It loses 2L on condensing plus 2 × 1 × (100 − 25) = 150 cal on cooling to 25 °C. So 2L + 150 = 1250 gives L = 550 cal/⁠g (container's heat capacity neglected).

Answer: (d) Q = m(T₂ − T₁)[C₀ + α2(T₁ + T₂)]

Method. heat from temperature-dependent specific heat: Q = m∫C dT

Q = m∫(C0 + αT) dT from T1 to T2 = m[C0(T2 − T1) + (α/⁠2)(T2² − T1²)]. Since T2² − T1² = (T2 − T1)(T1 + T2), Q = m(T2 − T1)[C0 + (α/⁠2)(T1 + T2)].

Answer: (c) 0.015

Method. calorimetry: heat lost by water = heat gained by ice

Ice gains: warming to 0 °C: 5 × 0.5 × 20 = 50 cal; melting: 5 × 80 = 400 cal; total 450 cal. Water loses m × 1 × (30 − 0) = 30m cal (m in g). 30m = 450 gives m = 15 g = 0.015 kg (c_ice = 0.5 cal g⁻¹ °C⁻¹, L = 80 cal/⁠g; NCERT Class 11 Physics, Thermal Properties of Matter, calorimetry).

Answer: (b) specific latent heat

Method. definition of specific latent heat of vaporisation

The heat per unit mass needed to change a substance from liquid to vapour at constant temperature is the specific latent heat of vaporisation (NCERT Class 11 Physics, Thermal Properties of Matter, change of state). Specific heat and thermal capacity refer to a temperature change.

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

Method. definition of heat

Heat is energy in transit from one body to another because of a temperature difference (NCERT Class 11 Physics, Thermal Properties of Matter). (a) describes energy conversion in general; (d) is thermal expansion.

Answer: (d) 5 kg

Method. Q = m c ΔT

ΔT = 25 − 15 = 10 °C. m = Q/⁠(c ΔT) = 20000/⁠(400 × 10) = 5 kg.

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

Method. latent heat of vaporization definition

Specific latent heat of vaporization is the heat needed to change unit mass of liquid to vapour at constant temperature (the boiling point), with no change in temperature. (c) and (d) describe the reverse change.

Answer: (c) 1050 cal

Method. calorimetry: heat for the whole path

Ice −⁠10 to 0 °C: 10 × 0.5 × 10 = 50 cal. Melting 10 g: 10 × 80 = 800 cal. Then 20 g of water 0 to 10 °C: 20 × 1 × 10 = 200 cal. Total 50 + 800 + 200 = 1050 cal (c_ice = 0.5 cal/⁠g°C, L = 80 cal/⁠g; checked in Python).

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

Method. definition of heat as energy in transit due to temperature difference

Heat is energy transferred because of a temperature difference; transfer without such a difference (e.g. work) is not heat, so (a) is correct. (b) is false since work transfers energy with no temperature difference, (c) is false because bodies expand on heating, (d) is false because work can also transfer energy when a temperature difference exists (NCERT Class 11 Physics, thermodynamics).

Answer: (d) 1 and 4

Method. latent heat is a property of the phase change at a given temperature

Latent heat of a given substance is fixed at the given transition temperature (statement 1), and it varies with that temperature (for example, L of vaporization of water changes with the boiling point under different pressure) but does not depend on the volume (statement 4). Statements 2 and 3 are false: 2 includes a volume dependence, 3 denies any temperature dependence. So 1 and 4 (NCERT Class 11 Physics, thermal properties of matter).

Answer: (d) 2 only

Method. specific heat is a material property

Specific heat c = Q/⁠(m ΔT) is a property of the material (intensive): it does not depend on the mass or shape of the body, so 2 is correct and 1 is false. Statement 3 is false because it says c depends ONLY on temperature; c is chiefly fixed by the material (it varies with temperature only slightly). So 2 only (NCERT Class 11 Physics, thermal properties of matter).

Answer: (a) mass of the body only

Method. thermal capacity = m c

Thermal capacity = m c, so for a given body it is proportional to its mass; shape plays no part, which rules out (b) and (d). Density alone does not fix it (c). Of the offered options only (a) is acceptable, though it ignores the material.

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

Method. definition of specific heat capacity

Specific heat capacity is the heat needed to raise the temperature of UNIT MASS of a substance by 1 degree (c = Q/⁠(m ΔT)); statement (b) omits unit mass and unit rise, so it describes heat capacity of a sample, not specific heat. (a) conduction passes energy between neighbouring molecules that stay in place, (c) convection is the mode in liquids and gases, (d) radiation needs no heating of the intervening space: all correct (NCERT Class 11 Physics, thermal properties of matter).

Answer: (c) latent heat of vaporization

Method. latent heat of vaporization

Heat absorbed in the liquid to vapour change at constant temperature is the latent heat of vaporization (Q = mL). Specific heat capacity involves a temperature change; the mechanical equivalent of heat is the J-to-calorie factor; quenching is rapid cooling of hot metal (NCERT Class 11 Physics, thermal properties of matter).

Answer: (b) Boiling point

Method. definition of boiling point

The boiling point is the temperature at which the vapour pressure of the liquid equals the external (atmospheric) pressure (NCERT Class 11 Physics, Thermal Properties of Matter, change of state).

No correct option. This question is defective as printed in UPSC's paper.

Method. boiling point definition

A liquid boils when its vapour pressure equals the external pressure. Options (a) and (d) both state exactly this (atmospheric pressure = vapour pressure; vapour pressure = air pressure), and are logically identical; (b) and (c) are wrong.

Options (a) and (d) state the same condition (the vapour pressure equal to the atmospheric or air pressure), which is the boiling condition, so two options are correct and no single key exists. (b) and (c) are wrong.

Answer: (a) increasing the boiling point of water

Method. effect of pressure on boiling point

The lid of a pressure cooker traps steam, so the pressure inside rises above 1 atm. Boiling point rises with pressure, so water boils at about 120 °C, and food cooks faster at the higher temperature (NCERT Class 11 Physics, Thermal Properties of Matter, change of state).

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

Method. factors governing evaporation

Evaporation is a surface process: it speeds up with a higher temperature (more molecules with enough kinetic energy to escape) and a larger exposed surface area. Both conditions are met only in (a) (NCERT Class 9 Science, matter around us; standard result).

Answer: (d) Boiling is a bulk phenomenon, but evaporation is a surface phenomenon.

Method. boiling versus evaporation

Boiling takes place throughout the liquid (bubbles form in the bulk) at a fixed temperature; evaporation takes place only at the surface at any temperature. So boiling is bulk, evaporation is surface.

Answer: (d) 1500°C

Method. melting point of iron

Iron melts at about 1538 °C, so of the options only 1500 °C is possible; 25, 37 and 500 °C are far too low (iron is solid in all of these).

Answer: (d) water vapour

Method. boiling is a change of phase

Bubbles in boiling water are formed by water changing to steam inside the liquid, so they are water vapour. Dissolved air is driven out well before vigorous boiling, and water is not decomposed into H₂ and O₂ by boiling.

Answer: (a) the air pressure is less

Method. boiling point depends on external pressure

A liquid boils when its vapour pressure equals the external pressure. At high altitude the air pressure is lower, so the vapour pressure reaches it at a lower temperature (NCERT Class 11 Physics, thermal properties of matter, effect of pressure on boiling point).

Answer: (c) T₁T₂ = V₂V₁

Method. ideal-gas law combined with the process equation

PV = nRT, so P = nRT/⁠V. Then PV² = (nRT/⁠V)V² = nRTV = constant, so TV = constant. Hence T1V1 = T2V2, i.e. T1/⁠T2 = V2/⁠V1 (NCERT Class 11 Physics, Thermodynamics, polytropic process).

Answer: (b) C = C_V

Method. molar heat capacity of the process P = kT

For an ideal gas V = nRT/⁠P = nR/⁠k, a constant. The process is isochoric, so no work is done and the heat is dQ = n C_V dT. Hence C = C_V.

Answer: (a) Adiabatic system

Method. definition of adiabatic system

A system that cannot exchange heat with its surroundings is called an adiabatic system (NCERT Class 11 Physics, Thermodynamics).

Answer: (b) 4 only

Method. Newton's law of cooling

Newton's law of cooling describes a body at a temperature above its surroundings losing heat at a rate proportional to the temperature difference, for a small difference. Only case 4 (coffee cooling on the table) is of this kind. Melting ice and boiling water are at constant temperature (latent heat); heating a rod in a furnace is a warming process (NCERT Class 11 Physics, Thermal Properties of Matter, Newton's law of cooling).

Answer: (a) Both the statements are individually true and Statement-2 is the correct explanation of Statement-1.

Method. infrared radiation and its absorption by water

Infrared waves are called heat waves because they make molecules vibrate and so heat the bodies that absorb them (Statement-1 true). Water molecules absorb infrared strongly, which is why IR heats moist objects and skin (Statement-2 true). Statement-2 gives the reason for Statement-1, so (a) (NCERT Class 12 Physics, Electromagnetic Waves, infrared waves).

Reading of 'correct explanation' is judgement: the heating effect of IR comes from absorption by molecules, of which water is the common example.

Answer: (c) Thermal radiation

Method. thermal radiation

Radiation (heat waves, infrared electromagnetic waves) travels in straight lines at the speed of light and needs no medium. Conduction and convection need a material medium and are slow.

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

Method. first law, W = 0

ΔU = Q − W. With W = 0, ΔU = Q: the change in internal energy equals the heat flowing in or out (NCERT Class 11 Physics, thermodynamics).

Answer: (c) light and short wavelength infrared.

Method. solar spectrum

Sunlight is mostly visible light plus near (short wavelength) infrared, with a smaller ultraviolet part. The long wavelength (far) infrared is absorbed by the atmosphere or is a minor part of the Sun's output. So light and short wavelength infrared.

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

Method. construction of a thermos flask

The walls are silvered, shiny surfaces which are POOR radiators and good reflectors, so the inner wall radiates little and the outer wall reflects back what reaches it. Statement (c) says the inner wall radiates a good amount, which is wrong. (a), (b), (d) are correct: vacuum between double glass walls, silvered walls, cork supports poor conductors (NCERT Class 11 Physics, thermal properties of matter).

Answer: (c) P′ = P/⁠2

Method. ideal gas law at fixed T and V

At fixed T and V, P = nkT/⁠V is proportional to the number of molecules, independent of the kind of gas. n becomes n/⁠2, so P′ = P/⁠2.

Answer: (c) Second law of thermodynamics

Method. Clausius statement of the second law

Heat does not flow by itself from a colder to a hotter body: this is the Clausius form of the second law of thermodynamics. Zeroth law concerns thermal equilibrium and temperature, first law is energy conservation, third law is the unattainability of absolute zero (NCERT Class 11 Physics, thermodynamics).