📌 Snapshot
- Establishes the physical distinction between heat (energy transferred by virtue of temperature difference) and temperature (a measure of hotness), and develops the three temperature scales (Celsius, Fahrenheit, Kelvin) with their conversions.
- Develops the ideal-gas equation PV = µRT and shows how a constant-volume gas thermometer leads to the absolute (Kelvin) scale with zero at −273.15 °C.
- Quantifies thermal expansion of solids, liquids and gases (αl, αv = 3αl, area expansion 2αl) and notes water's anomalous expansion with maximum density at 4 °C.
- Introduces specific heat capacity, molar specific heat capacity, latent heat of fusion and vaporisation, and the principle of calorimetry (heat lost = heat gained).
- Covers the three modes of heat transfer — conduction (H = KA(TC − TD)/L), convection, and radiation (Stefan-Boltzmann law H = AeσT⁴, Wien's displacement law λmT = constant) — and ends with Newton's law of cooling.
📖 Detailed Notes
2.1 Core concepts
- Temperature is a relative measure of the hotness or coldness of a body; heat is the form of energy transferred between two systems (or a system and its surroundings) by virtue of a temperature difference. SI unit of heat is the joule (J); SI unit of temperature is the kelvin (K), with degree Celsius (°C) as a common unit (NCERT §10.2, p. 203).
- Thermometers exploit a physical property that varies with temperature; common liquid-in-glass thermometers use mercury or alcohol whose volume varies linearly with temperature. Calibration uses two fixed points: the ice point and the steam point of water (NCERT §10.3, p. 203).
- On the Fahrenheit scale the ice and steam points are 32 °F and 212 °F (180 equal intervals); on Celsius they are 0 °C and 100 °C (100 intervals). The conversion is (tF − 32)/180 = tC/100 (NCERT §10.3, Eq. 10.1, p. 203).
- A constant-volume gas thermometer uses P ∝ T and gives the same reading regardless of which low-density gas is used; extrapolating P–T lines for different gases meets the T-axis at the same point, −273.15 °C, which is taken as absolute zero (0 K) on the Kelvin scale (NCERT §10.4, pp. 203–204).
- Boyle's law (PV = constant, T fixed) and Charles' law (V/T = constant, P fixed) combine into the ideal-gas equation PV = µRT, where µ is the number of moles and R = 8.31 J mol⁻¹ K⁻¹ is the universal gas constant (NCERT §10.4, Eq. 10.2, p. 204).
- Kelvin and Celsius scales have the same unit size but different origins: T = tC + 273.15 (NCERT §10.4, Eq. 10.3, p. 204).
- Most substances expand on heating and contract on cooling. For a long rod: ∆l/l = αl ∆T, where αl is the coefficient of linear expansion (a material constant) (NCERT §10.5, Eq. 10.4, p. 205).
- For a rectangular sheet, the coefficient of area expansion equals 2αl (the (αl ∆T)² term being negligible); for a cube, the coefficient of volume expansion αv = 3αl (NCERT §10.5, Eqs. 10.7–10.9, pp. 207–208).
- Liquids and gases are described by ∆V/V = αv ∆T. Gases expand much more than solids/liquids; for an ideal gas at constant pressure αv = 1/T, so at 0 °C, αv ≈ 3.7 × 10⁻³ K⁻¹ (NCERT §10.5, Eq. 10.6, p. 206).
- Water shows anomalous behaviour: between 0 °C and 4 °C it contracts on heating, so water has its maximum density at 4 °C. This makes lakes freeze at the top first, preserving aquatic life (NCERT §10.5, pp. 206–207).
- If thermal expansion of a rod is prevented by rigid supports, a compressive thermal stress is set up; for a steel rail with αl = 1.2 × 10⁻⁵ K⁻¹ and ∆T = 10 °C, the strain is 1.2 × 10⁻⁴ and thermal stress is 2.4 × 10⁷ N m⁻² (NCERT §10.5, p. 207).
- Heat capacity S = ∆Q/∆T; specific heat capacity s = (1/m)(∆Q/∆T) is the heat needed to change unit mass by one unit of temperature. SI unit: J kg⁻¹ K⁻¹ (NCERT §10.6, Eqs. 10.10–10.11, p. 208).
- Water has an unusually high specific heat capacity of 4186 J kg⁻¹ K⁻¹ — the highest in Table 10.3 — which is why water is used as a coolant in radiators and as a heater in hot-water bags, and why sea breezes have a cooling effect (NCERT §10.6, Table 10.3, p. 209).
- Molar specific heat capacity C = (1/µ)(∆Q/∆T); for gases one distinguishes Cp (constant pressure) and Cv (constant volume) (NCERT §10.6, Eq. 10.12, pp. 208–209).
- Calorimetry: in an isolated system, heat lost by the hotter body equals heat gained by the colder body. A calorimeter is a metallic vessel (copper/aluminium) kept inside an insulating jacket (NCERT §10.7, pp. 209–210).
- During a change of state (melting/fusion, freezing, vaporisation, condensation, sublimation) temperature remains constant; both phases coexist in thermal equilibrium. The temperature at which solid–liquid coexist is the melting point; liquid–vapour coexist at the boiling point (NCERT §10.8, pp. 210–212).
- Latent heat L = Q/m is the heat per unit mass required for a change of state at constant T and P. SI unit: J kg⁻¹. For water Lf = 3.33 × 10⁵ J kg⁻¹ and Lv = 22.6 × 10⁵ J kg⁻¹ — so steam at 100 °C carries 22.6 × 10⁵ J kg⁻¹ more energy than water at 100 °C, explaining why steam burns are more serious (NCERT §10.8.1, Eq. 10.13, Table 10.5, p. 213).
- Boiling point increases with pressure (pressure cooker) and decreases with reduced pressure (cooking is difficult on hills); regelation (refreezing under pressure) makes skating possible (NCERT §10.8, pp. 211–212).
- Three modes of heat transfer: conduction, convection, radiation (NCERT §10.9, p. 214).
- Conduction: in steady state, H = KA(TC − TD)/L, where K is thermal conductivity (SI unit J s⁻¹ m⁻¹ K⁻¹ = W m⁻¹ K⁻¹). Metals have large K (silver 406, copper 385); gases are poor conductors (air 0.024) (NCERT §10.9.1, Eq. 10.14, Table 10.6, pp. 214–215).
- Convection is heat transfer by bulk motion of matter and is possible only in fluids; it can be natural (gravity-driven, e.g., sea breeze, trade winds) or forced (pumps, heart, automobile cooling) (NCERT §10.9.2, pp. 217–218).
- Radiation requires no medium; energy is carried by electromagnetic waves at speed 3 × 10⁸ m s⁻¹. Stefan-Boltzmann law for a perfect radiator: H = AσT⁴ with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; for a real body H = AeσT⁴ where e is emissivity (e = 1 for perfect radiator). For a body at T with surroundings at Ts, net loss is H = eσA(T⁴ − Ts⁴) (NCERT §10.9.3, Eqs. 10.16–10.18, pp. 218–219).
- Blackbody radiation has a continuous spectrum. Wien's displacement law: λmT = constant = 2.9 × 10⁻³ m K. This explains why heated iron turns dull red → reddish yellow → white-hot, and lets us estimate Sun's surface temperature ≈ 6060 K from λm = 4753 Å (NCERT §10.9.4, Eq. 10.15, pp. 218–219).
- A Dewar flask reduces all three modes: silvered walls reflect radiation, evacuated space cuts conduction and convection (NCERT §10.9.3, p. 218).
- Newton's law of cooling: for small temperature differences, −dQ/dt = k(T2 − T1). Integrating gives loge(T2 − T1) = −Kt + c, so a plot of loge(T2 − T1) against t is a straight line with negative slope (NCERT §10.10, Eqs. 10.19–10.23, pp. 220–221).
- Triple point of water, the unique combination of pressure (4.58 mm Hg) and temperature (273.16 K) at which all three phases of water — ice, water and vapour — coexist in thermal equilibrium, is the modern SI defining point of the kelvin (NCERT §10.4, p. 204; Points to Ponder, p. 222).
- Anomalous expansion of water has profound ecological significance: as a lake cools, the densest (4 °C) water sinks while colder (less dense) water floats on top, eventually freezing into an insulating ice cap that protects fish and other aquatic life in the comparatively warmer (≈4 °C) water below (NCERT §10.5, p. 206–207).
- Mayer's relation Cp − Cv = R for an ideal gas follows from the first law plus the ideal-gas equation: the extra Cp − Cv accounts for the work done by the gas during a constant-pressure expansion. Cp and Cv are introduced individually here; their algebraic relation is fully developed in the next NCERT chapter (NCERT §10.6, p. 208–209).
- Black-body is an idealised body that absorbs (and re-emits) all incident radiation. Stefan-Boltzmann emission H = AσT⁴ refers to a perfect black-body (e = 1); for real bodies the emissivity e (0 < e < 1) multiplies the law. Wien's displacement λ_m T = b (b ≈ 2.9 × 10⁻³ m K) tracks the peak of the spectrum (NCERT §10.9.3–10.9.4, p. 218–219).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Heat | Form of energy transferred between systems (or system and surroundings) by virtue of temperature difference; SI unit J | 203 |
| Temperature | Relative measure of hotness/coldness; SI unit kelvin (K) | 203 |
| Absolute zero | The temperature −273.15 °C = 0 K, found by extrapolating P–T lines for low-density gases to P = 0 | 204 |
| Ideal-gas equation | PV = µRT, with R = 8.31 J mol⁻¹ K⁻¹ | 204 |
| Coefficient of linear expansion (αl) | ∆l/l = αl ∆T; characteristic of material | 205 |
| Coefficient of volume expansion (αv) | ∆V/V = αv ∆T; for solids αv = 3αl; for ideal gas αv = 1/T | 205–206 |
| Anomalous expansion of water | Water contracts on heating between 0 °C and 4 °C; maximum density at 4 °C | 206 |
| Thermal stress | Stress developed when expansion/contraction of a rigidly clamped rod is prevented | 207 |
| Specific heat capacity (s) | s = (1/m)(∆Q/∆T); SI unit J kg⁻¹ K⁻¹ | 208 |
| Molar specific heat capacity (C) | C = (1/µ)(∆Q/∆T); SI unit J mol⁻¹ K⁻¹ | 208–209 |
| Calorimetry principle | In an isolated system, heat lost by hotter body = heat gained by colder body | 209 |
| Melting point | Temperature at which solid and liquid phases coexist in thermal equilibrium | 210 |
| Boiling point | Temperature at which liquid and vapour phases coexist in thermal equilibrium | 211 |
| Latent heat (L) | L = Q/m; heat per unit mass required for a change of state at constant T, P | 213 |
| Latent heat of fusion (Lf) | Latent heat for solid → liquid change; Lf(water) = 3.33 × 10⁵ J kg⁻¹ | 213 |
| Latent heat of vaporisation (Lv) | Latent heat for liquid → vapour change; Lv(water) = 22.6 × 10⁵ J kg⁻¹ | 213 |
| Sublimation | Direct solid → vapour change (e.g., dry ice, iodine) | 212 |
| Conduction | Heat transfer between adjacent parts of a body due to temperature difference, without flow of matter | 214 |
| Thermal conductivity (K) | Constant in H = KA(TC − TD)/L; SI unit W m⁻¹ K⁻¹ | 215 |
| Convection | Heat transfer by bulk motion of matter; possible only in fluids | 217 |
| Radiation | Energy transfer by electromagnetic waves; needs no medium | 218 |
| Stefan-Boltzmann law | H = AeσT⁴ (e = emissivity); σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ | 219 |
| Wien's displacement law | λmT = 2.9 × 10⁻³ m K | 218 |
| Newton's law of cooling | −dQ/dt ∝ (T2 − T1) for small temperature differences | 220 |
| Triple point of water | 273.16 K, 4.58 mm Hg — the unique state where ice, water and vapour coexist | 204 |
| Heat capacity (S) | S = ∆Q/∆T; SI unit J K⁻¹ | 208 |
| Cp / Cv | Molar specific heat at constant pressure / volume; difference is R for ideal gas | 208–209 |
| Emissivity (e) | Fraction by which a real body radiates compared with a perfect black body | 219 |
| Regelation | Refreezing of water on release of high pressure; explains skating on ice | 212 |
| Coefficient of area expansion | ∆A/A = 2αl ∆T for isotropic solids | 208 |
| Universal gas constant (R) | R = 8.31 J mol⁻¹ K⁻¹ | 204 |
2.3 Diagrams / processes to remember
- Fig. 10.1 (p. 203): Straight-line plot of Fahrenheit tF versus Celsius tC — basis of conversion equation (tF − 32)/180 = tC/100.
- Fig. 10.2 and 10.3 (p. 204): P–T lines for low-density gases at constant volume; all extrapolations meet at absolute zero (−273.15 °C).
- Fig. 10.4 (p. 204): Side-by-side comparison of Kelvin, Celsius and Fahrenheit scales.
- Fig. 10.5 (p. 205): Linear, area and volume expansion sketches with the relations ∆l/l = αl ∆T, ∆A/A = 2αl ∆T, ∆V/V = 3αl ∆T.
- Fig. 10.7 (p. 206): Volume and density of water versus temperature, showing the dip in volume (peak in density) at 4 °C — anomalous expansion of water.
- Fig. 10.9 (p. 211): Temperature vs time when ice is heated steadily — flat plateaus at 0 °C (melting) and 100 °C (boiling) where latent heat is absorbed.
- Fig. 10.12 (p. 213): Temperature vs heat for water at 1 atm, showing slopes (specific heats of ice, water, steam) and flat segments of length Lf and Lv.
- Fig. 10.14 (p. 215): Steady-state conduction along an insulated bar between reservoirs at TC and TD.
- Fig. 10.17 (p. 217): Convection cycles — day-time sea breeze (warm air rises over land) and reversed night-time circulation.
- Fig. 10.18 (p. 218): Blackbody emission curves at different temperatures — λm shifts to shorter wavelengths as T rises (Wien's law).
- Fig. 10.19 (p. 220): Cooling curve T2 − T1 versus t — exponential decay (Newton's law of cooling).
- Fig. 10.20 (p. 220): Verification of Newton's law of cooling — straight-line plot of loge(T2 − T1) versus t with negative slope.
2.4 Common confusions / NTA trap points
- Tc = T − 273.15, not −273.16. The triple point of water is defined as 273.16 K, but the offset between Celsius and Kelvin is 273.15 (NCERT §10.4, p. 204).
- αv = 3αl applies only to isotropic solids (and to liquids by definition of αv). For gases αv = 1/T at constant pressure and depends on T (NCERT §10.5, Eq. 10.6, p. 206).
- Anomalous expansion of water is between 0 °C and 4 °C (not "below 0 °C" or "below 4 °C for all temperatures") — maximum density is at 4 °C (NCERT §10.5, p. 206).
- Steam at 100 °C carries 22.6 × 10⁵ J kg⁻¹ more heat than water at 100 °C; that is why steam burns are more serious than burns from boiling water — a frequent assertion–reason trap (NCERT §10.8.1, p. 213).
- Stefan-Boltzmann law uses absolute (Kelvin) temperature, not Celsius, and net loss has (T⁴ − Ts⁴), not (T − Ts)⁴ (NCERT §10.9.3, Eq. 10.18, p. 219).
- Newton's law of cooling holds only for small differences in temperature between body and surroundings (NCERT §10.10, p. 220).
- Convection requires a fluid (and gravity for natural convection); conduction does not need bulk motion; radiation does not need any medium. Mixing these up is a classic NTA distractor (NCERT §10.9, pp. 214–218).
- Stress = Y × αl × ∆T for a clamped rod whose expansion is prevented — independent of the rod's length. Many students wrongly include L in the formula.
- Latent heat is absorbed at constant T, so heating curves for water show two horizontal plateaus (at 0 °C and 100 °C). Sloped segments give specific heat; plateau lengths give latent heat.
- Boiling vs evaporation: boiling is bulk vaporisation at the boiling point; evaporation occurs at any temperature from the surface, and is what drives sweat-based cooling.
- K (thermal conductivity) ≠ k (Newton's-law cooling constant); they have different units (W m⁻¹ K⁻¹ vs s⁻¹) and arise from unrelated phenomena. The same letter "k" causes confusion in problems.
2.5 Key formulas table
| Quantity | Symbol / Formula | NCERT reference |
|---|---|---|
| Celsius–Fahrenheit conversion | (tF − 32)/180 = tC/100 | §10.3, Eq. 10.1, p. 203 |
| Kelvin–Celsius conversion | T = tC + 273.15 | §10.4, Eq. 10.3, p. 204 |
| Ideal gas equation | PV = µRT | §10.4, Eq. 10.2, p. 204 |
| Linear expansion | ∆l/l = αl ∆T | §10.5, Eq. 10.4, p. 205 |
| Area expansion | ∆A/A = 2αl ∆T | §10.5, p. 208 |
| Volume expansion (solid) | ∆V/V = 3αl ∆T | §10.5, Eq. 10.9, p. 208 |
| Volume expansion (ideal gas) | αv = 1/T | §10.5, Eq. 10.6, p. 206 |
| Thermal stress | σ = Y αl ∆T | §10.5, p. 207 |
| Specific heat capacity | s = (1/m) ∆Q/∆T | §10.6, Eq. 10.11, p. 208 |
| Molar specific heat | C = (1/µ) ∆Q/∆T | §10.6, Eq. 10.12, p. 209 |
| Calorimetry | Heat lost = heat gained | §10.7, p. 209 |
| Latent heat | Q = m L | §10.8.1, Eq. 10.13, p. 213 |
| L_f (water) | 3.33 × 10⁵ J kg⁻¹ | §10.8.1, p. 213 |
| L_v (water) | 22.6 × 10⁵ J kg⁻¹ | §10.8.1, p. 213 |
| Conduction (steady) | H = KA(TC − TD)/L | §10.9.1, Eq. 10.14, p. 215 |
| Stefan-Boltzmann | H = A e σ T⁴ | §10.9.3, Eq. 10.16, p. 219 |
| Stefan-Boltzmann (net) | H_net = e σ A (T⁴ − Ts⁴) | §10.9.3, Eq. 10.18, p. 219 |
| Wien's displacement | λ_m T = 2.9 × 10⁻³ m K | §10.9.4, Eq. 10.15, p. 218 |
| Newton's law of cooling | −dQ/dt = k (T₂ − T₁) | §10.10, Eq. 10.19, p. 220 |
| Cooling-curve integrated form | ln(T₂ − T₁) = −Kt + c | §10.10, Eq. 10.23, p. 221 |
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. The temperature at which a Fahrenheit thermometer reads the same numerical value as a Celsius thermometer is
▸ Show answer & explanation
Answer: B
Setting tF = tC = x in (x − 32)/180 = x/100 gives 100x − 3200 = 180x, so x = −40°. Both scales coincide at −40°. Option (C) is the obvious wrong-sign trap.
Q2. The constant volume gas thermometers using different low-density gases all give the same absolute zero because
▸ Show answer & explanation
Answer: B
Experimentally, P–T lines for all low-density gases at constant V extrapolate to P = 0 at the same temperature, −273.15 °C, which is taken as absolute zero. Option (A) is wrong because Boyle's law holds at constant T, not at every temperature.
Q3. A steel rod of length 1.00 m at 27 °C is heated to 127 °C. If the coefficient of linear expansion of steel is 1.2 × 10⁻⁵ K⁻¹, the increase in length of the rod is
▸ Show answer & explanation
Answer: B
∆l = αl l ∆T = 1.2 × 10⁻⁵ × 1.00 × 100 = 1.2 × 10⁻³ m = 1.2 mm.
🔒 10 more practice MCQs
Create a free account to unlock every MCQ in this chapter — answers and explanations included. No payment needed.
Already registered? Just log in and they'll all appear here.
Q4. Water has its maximum density at
▸ Show answer & explanation
Answer: B
Between 0 °C and 4 °C, water contracts on heating; below 4 °C its volume increases on cooling. Hence the maximum density is at 4 °C. This is why lakes freeze from the top.
Q5. Assertion (A): Steam at 100 °C produces more severe burns than boiling water at 100 °C. Reason (R): Each kilogram of steam carries 22.6 × 10⁵ J more heat than the same mass of water at 100 °C, which is released on condensation.
▸ Show answer & explanation
Answer: A
Lv(water) = 22.6 × 10⁵ J kg⁻¹, and this extra heat is released when steam condenses on the skin, making steam burns more serious.
Q6. 0.15 kg of ice at 0 °C is mixed with 0.30 kg of water at 50 °C. Using Lf(ice) = 3.34 × 10⁵ J kg⁻¹ and s(water) = 4186 J kg⁻¹ K⁻¹, the final equilibrium temperature is approximately
▸ Show answer & explanation
Answer: B
Heat lost by water = msw(50 − Tf); heat gained = 0.15 × Lf + 0.15 × sw × Tf. Setting them equal gives Tf ≈ 6.7 °C, the exact answer worked out in Example 10.4.
Q7. In the steady state, the rate of heat flow through a rod of length L and uniform cross-section A, with its ends at temperatures TC and TD (TC > TD), is
▸ Show answer & explanation
Answer: C
The heat current H is directly proportional to A and to (TC − TD), and inversely proportional to L, giving H = KA(TC − TD)/L. Options (A), (B), (D) misplace one of these dependencies.
Q8. Wien's displacement law states that the wavelength λm at which a blackbody radiates maximum energy is related to its absolute temperature T by λm T = 2.9 × 10⁻³ m K. If solar radiation has its maximum at λm = 4753 Å, the estimated surface temperature of the Sun is closest to
▸ Show answer & explanation
Answer: C
T = (2.9 × 10⁻³)/(4753 × 10⁻¹⁰) ≈ 6060 K — the figure quoted by NCERT for the Sun's surface temperature.
Q9. A perfect black-body (e = 1) of surface area 0.3 cm² is maintained at a temperature of 3000 K. Taking σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, the power radiated is approximately (for the same surface a tungsten lamp with e = 0.4 radiates 60 W)
▸ Show answer & explanation
Answer: D
For a tungsten lamp H = AeσT⁴ = 0.3 × 10⁻⁴ × 0.4 × 5.67 × 10⁻⁸ × (3000)⁴ = 60 W (NCERT figure). For a perfect black-body with e = 1, the rate is 60/0.4 = 150 W.
Q10. Match List-I with List-II and choose the correct answer: | List-I (Mode/Law) | List-II (Key feature) | |---|---| | (i) Conduction | (P) Requires no medium; carried by EM waves | | (ii) Convection | (Q) Heat current proportional to T⁴ | | (iii) Radiation (Stefan-Boltzmann) | (R) Possible only in fluids; involves bulk motion | | (iv) Newton's law of cooling | (S) Valid only for small temperature differences | | | (T) Heat current proportional to (TC − TD)/L in a bar |
▸ Show answer & explanation
Answer: A
Conduction in a bar gives H = KA(TC − TD)/L (T); convection involves bulk motion in fluids (R); Stefan-Boltzmann radiation gives H ∝ T⁴ (Q); Newton's law of cooling holds only for small temperature differences (S). Option (P) "needs no medium" applies to radiation, not conduction — eliminating (B).
Q11. The coefficient of volume expansion of an ideal gas at constant pressure at 273 K is approximately
▸ Show answer & explanation
Answer: B
At 273 K, αv = 1/273 ≈ 3.66 × 10⁻³ K⁻¹. Compared with αv of solids (~10⁻⁵ K⁻¹), gases expand much more on heating.
Q12. A black body at 6000 K emits radiation with peak wavelength λm = 4830 Å. If the same body is cooled to 3000 K, the new peak wavelength is approximately
▸ Show answer & explanation
Answer: C
λm ∝ 1/T. Halving T from 6000 → 3000 K doubles λm from 4830 → 9660 Å — i.e. the peak shifts into the near-infrared, consistent with a cooler radiator glowing red rather than yellow-white.
Q13. Assertion (A): Two large containers of water at 100 °C — one boiling, the other condensing steam onto the same area of skin — would produce very different burns even though both deliver heat at 100 °C. Reason (R): The latent heat of vaporisation of water (22.6 × 10⁵ J kg⁻¹) is released when the steam condenses onto the skin, in addition to the heat given up as the resulting water cools.
▸ Show answer & explanation
Answer: A
Steam at 100 °C delivers an extra 22.6 × 10⁵ J per kg upon condensation that boiling water at 100 °C cannot, hence the more serious burns from steam. R correctly explains A.
📊 Previous-Year Questions
Practise with real CUET Physics previous-year papers — every question solved, with the correct answer and a step-by-step explanation.
View solved CUET PYQ papers →Ready to drill Physics?
Unlock all MCQs, chapter tests, mocks & PYQs for ₹199/year.
Get UniDrill Pro