🟢 Easy Level — Boards / NCERT
Direct concepts and definitions
The coefficient of volume expansion γ is related to the coefficient of linear expansion α by:
A γ = α
B γ = 2α
C γ = 3α
D γ = α/3
✓ For isotropic solids: α:β:γ = 1:2:3, so γ = 3α. This is derived from the cube expansion: (1+αΔT)³ ≈ 1 + 3αΔT for small α. Derivation: ΔV/V₀ = γΔT = 3αΔT.
Heat transfer through vacuum is possible only by:
A Conduction
B Convection
C Radiation
D Both conduction and convection
✓ Radiation requires no medium — it travels as electromagnetic waves. Conduction needs molecular contact; convection needs fluid bulk motion. Hence only radiation works in vacuum. Example: Sun's energy reaching Earth.
Which of the following is correct about the boiling point of water at high altitudes?
A Decreases because atmospheric pressure decreases
B Increases because temperature is lower
C Remains same at 100°C
D Decreases because air is thinner
✓ Boiling point depends on external pressure. Higher altitude = lower atmospheric pressure = lower boiling point. Water boils at ~70°C at 8000 m altitude. This is why cooking takes longer at high altitudes.
Water has maximum density at:
A 0°C
B 4°C
C 37°C
D 100°C
✓ Water exhibits anomalous expansion from 0°C to 4°C — it CONTRACTS (becomes denser). Maximum density at 4°C. Above 4°C, normal expansion resumes. This property allows aquatic life to survive in winter.
The triple point of water is at:
A 0 K and 1 atm
B 273.16 K and 611.6 Pa
C 273.15 K and 1 atm
D 273 K and 0.006 bar
✓ Triple point is 273.16 K (0.01°C) at 611.6 Pa (≈ 0.006 atm). At this unique point, all three phases coexist. Used as a calibration point for thermometers.
🟡 Medium — NEET / JEE Main Level
Application and numerical problems
Two spheres of the same material have radii 1m and 4m. They are heated to the same temperature and allowed to cool in the same surroundings. The ratio of their initial rates of cooling will be:
A 1:1
B 4:1
C 4:1 (surface area) but rate of cooling per unit mass goes as 1/r
D 1:16
✓ Rate of cooling dT/dt = eσA(T⁴−T₀⁴)/(mc) = eσ(4πr²)(T⁴−T₀⁴)/[(4/3)πr³ρc] = 3eσ(T⁴−T₀⁴)/(rρc). Rate ∝ 1/r. Smaller sphere (r=1) cools 4 times faster than larger (r=4). Ratio = 4:1.
Two rods of the same length and cross-section area — rod A (K = 2K₀) and rod B (K = K₀) — are connected in series. The effective thermal conductivity is:
A 3K₀/2
B 4K₀/3
C 2K₀/3
D K₀
✓ For series with equal lengths: K_eff = 2K₁K₂/(K₁+K₂) = 2(2K₀)(K₀)/(2K₀+K₀) = 4K₀²/3K₀ = 4K₀/3. This is the harmonic mean formula for series combination.
100 g of water at 80°C is mixed with 100 g of water at 20°C. The final temperature is:
A 40°C
B 80°C
C 50°C
D 60°C
✓ Principle of calorimetry: m·c·(80−T) = m·c·(T−20). Since same m and c: 80−T = T−20 → 100 = 2T → T = 50°C. When mixing equal masses of same liquid, T_f is simply the average.
The wavelength of maximum intensity radiation from a star is 400 nm. If Wien's constant b = 2.9 × 10⁻³ m·K, the temperature of the star is approximately:
A 3625 K
B 7250 K
C 14500 K
D 1800 K
✓ T = b/λ_max = 2.9×10⁻³ / (400×10⁻⁹) = 2.9×10⁻³ / 4×10⁻⁷ = 7250 K. Note: 400 nm ≈ violet light, so this star appears blue-white (hotter than our Sun at ~5800 K).
A steel rail of length 5 m and cross-section 40 cm² is rigidly fixed at both ends. If temperature rises by 10°C, what is the force on the walls? (Y = 2×10¹¹ Pa, α = 12×10⁻⁶ K⁻¹)
A 9.6 × 10⁴ N
B 9.6 × 10⁵ N
C 9.6 × 10³ N
D 9.6 × 10⁶ N
✓ F = Y·α·ΔT·A = 2×10¹¹ × 12×10⁻⁶ × 10 × 40×10⁻⁴ = 2×10¹¹ × 12×10⁻⁵ × 40×10⁻⁴ = 9.6×10⁵ N. Note: length of rail doesn't matter! Thermal stress = YαΔT is independent of length.
A body cools from 80°C to 64°C in 5 minutes and from 64°C to 52°C in the next 5 minutes. The temperature of the surroundings is:
A 10°C
B 24°C
C 16°C
D 20°C
✓ Using average temperature method: Rate₁ ∝ (72−T₀), Rate₂ ∝ (58−T₀). Rate₁/Rate₂ = (80−64)/(64−52) = 16/12 = 4/3. So (72−T₀)/(58−T₀) = 4/3 → 3(72−T₀) = 4(58−T₀) → 216−3T₀ = 232−4T₀ → T₀ = 16... wait: let me recalculate: 3×72−3T₀ = 4×58−4T₀ → T₀ = 232−216 = 16. Check: (72−16)/(58−16) = 56/42 = 4/3 ✓. Actually T₀ = 16°C. But standard answer often given as 24 — let's verify: at T₀=24: (72−24)/(58−24)=48/34 ≠ 4/3. Correct T₀ = 16°C.
🔴 Hard — JEE Advanced Level
Multi-concept and derivation-based problems
🔬 Before Attempting
These questions require combining 2+ concepts. Approach: identify all physical mechanisms, write energy balance equations, then solve. Don't rush — think first.
A body radiates energy at a rate of P when its temperature is T K. If the temperature is doubled to 2T K, the new rate of energy radiation (assuming same emissivity and area) will be:
A 2P
B 4P
C 8P
D 16P
✓ Stefan's Law: P ∝ T⁴. If T→2T: P_new = eσA(2T)⁴ = eσA·16T⁴ = 16P. The T⁴ dependence is the key. Doubling temperature gives 2⁴ = 16 times the power. This is a classic JEE trap — many students say 2P or 4P.
A pendulum clock keeps perfect time at 20°C. If temperature rises to 30°C (α_steel = 12×10⁻⁶ K⁻¹), how many seconds does the clock gain or lose per day?
A Gains 5.18 seconds
B Loses 5.18 seconds
C Loses 10.36 seconds
D Gains 10.36 seconds
✓ ΔT/T = (1/2)αΔθ = (1/2)(12×10⁻⁶)(10) = 6×10⁻⁵. Loss per day = 6×10⁻⁵ × 86400 s ≈ 5.18 s. Clock LOSES time because longer pendulum → longer period → fewer oscillations per day. T = 2π√(L/g), longer L → longer T.
5 g of steam at 100°C is passed into 6 g of ice at 0°C. Find the resulting temperature. (L_steam = 540 cal/g, L_ice = 80 cal/g, c_water = 1 cal/g°C)
A 100°C (all steam)
B 100°C (some steam condenses)
C 40°C
D 0°C (not all ice melts)
✓ Phase check: Heat available from steam → condense to water: 5×540 = 2700 cal. Heat needed: melt 6g ice = 6×80 = 480 cal; warm 6g water from 0→100: 600 cal. Total needed = 1080 cal << 2700 cal. So all ice melts AND water reaches 100°C with some steam remaining. Final state: water at 100°C (with some steam still present).
Three slabs (K₁=K, K₂=2K, K₃=K) of same area A and equal thickness L are in series. If T_hot = 400K and T_cold = 0K, the interface temperature between slab 2 and 3 is:
A 100 K
B 100 K
C 200 K
D 150 K
✓ R₁ = L/KA, R₂ = L/2KA, R₃ = L/KA. R_total = L/KA(1 + 0.5 + 1) = 2.5L/KA. H = ΔT_total/R_total = 400/(2.5L/KA) = 160KA/L. T after slab 3 from hot side: T₂₃ = T_hot − H(R₁+R₂) = 400 − (160KA/L)(L/KA)(1.5) = 400−240 = 160 K? Let me recalculate: T₂₃ = 0 + H×R₃ = (160KA/L)×(L/KA) = 160 K. Wait: from cold side: T₂₃ = T_cold + H×R₃ = 0 + 160 = 160 K. So actually 160 K, but answer options show 100K — re-examining with correct values gives 160K from cold interface.
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Time management, attempt order, and exam-specific tactics.
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