Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
Three identical conducting rods are arranged in series between two heat reservoirs at 100°C and 0°C. At steady state, what is the temperature at the junction between the second and third rod?
Answer: A
In series arrangement with identical rods, temperature difference is equally distributed. ΔT_total = 100°C, so ΔT per rod = 3100 = 33.3°C. Second junction = 100 - 2(33.3) = 33.3°C
Q.2Hard
A gas undergoes a cyclic process ABCA where AB is isothermal, BC is adiabatic, and CA is isochoric. If work is done on the gas in the cycle, what can be concluded?
Answer: A
If W_net < 0 (work done on gas), then from first law: ΔU_cycle = 0 = Q - W, so Q = W < 0, meaning net heat flows out
Q.3Hard
Two bodies at temperatures T₁ = 400 K and T₂ = 300 K are brought into thermal contact. If entropy change of universe is 0.575 J/K and heat capacity of both bodies is 1000 J/K, what is the final equilibrium temperature? (Assume no heat loss to surroundings)
Answer: B
Heat lost by body 1: Q = C(T₁ - T_f) = 1000(400 - T_f). Heat gained by body 2: Q = 1000(T_f - 300). ΔS_univ = C ln(T_f/T₁) + C ln(T_f/T₂) = 1000[ln(T_f/400) + ln(T_f/300)] = 0.575. Solving: T_f = 350 K
Q.4Hard
If a reversible process occurs in an isolated system, the entropy of the system:
Answer: C
For a reversible process in an isolated system, ΔS_total = ΔS_sys + ΔS_surr = 0 (no heat exchange with surroundings). Therefore entropy remains constant at its initial value.
Q.5Hard
An ideal gas undergoes a process where PV^n = constant. If n = 1, this process is _____ and if n = γ, this process is _____
Answer: A
When n = 1: PV = constant (isothermal). When n = γ = Cp/Cv: PV^γ = constant (adiabatic process).
Q.6Hard
Consider a system where entropy decreases by 50 J/K. Which statement must be true?
Answer: B
By second law, ΔS_total ≥ 0. If ΔS_sys = -50 J/K, then ΔS_surr ≥ +50 J/K to maintain ΔS_total ≥ 0.
Q.7Hard
A polytropic process with polytropic index n = 1.3 is performed on 1 mole of air (diatomic). The work done when volume changes from 1 m³ to 0.5 m³ at initial pressure 100 kPa is:
Answer: A
For polytropic process: W = [P₁V₁ - P₂V₂]/(n-1). Using PVⁿ = const: P₂ = 100 × (01.5)^1.3 ≈ 245.7 kPa. W = [100×1 - 245.7×0.5]/0.3 ≈ 81.2 kJ
Q.8Hard
For a van der Waals gas with equation (P + a/V²m)(Vm - b) = RT, at the critical point, which relationship is valid?
Answer: A
At the critical point, both first and second derivatives of pressure with respect to volume (at constant T) are zero: (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0. This defines the critical point.
Q.9Hard
Two moles of an ideal diatomic gas undergo a process where temperature changes from 400 K to 800 K. If the process is such that PV^(1.4) = constant, then the work done by the gas is approximately (R = 8.314 J/mol·K):
Answer: C
For polytropic process: W = nR(T_i − T_f)/(γ − n) = 2 × 8.314 × (400 − 800)/(1.4 − 1.4). Since n = γ, this is adiabatic: W = nCvΔT = 2 × (25) × 8.314 × (400 − 800) = 5 × 8.314 × (−400) ≈ −16,628 J. But if heating occurs, work done is positive. Actual calculation needs clarification on process direction.
Q.10Hard
A system undergoes a process where both pressure and volume increase. Which thermodynamic quantity must increase?
Answer: A
If both P and V increase for an ideal gas (PV = nRT), then T must increase, hence internal energy U ∝ T increases. Work done by system depends on process path; entropy and heat depend on specific process.
Q.11Hard
Four moles of an ideal gas are heated from 250 K to 350 K at constant volume. Simultaneously, it is allowed to expand at constant temperature. Which process involves greater entropy change?
Answer: B
Isochoric: ΔS = nCv ln(T_f/T_i) = 4 × Cv × ln(250350). Isothermal: ΔS = nR ln(V_f/V_i). For large expansions, isothermal entropy change is typically larger.
Q.12Hard
Three moles of ideal gas undergo polytropic process with n = 1.5. If temperature increases from 300 K to 450 K, the work done by gas is:
Answer: B
W = nR(T₂-T₁)/(1-n) = 3 × 8.314 × 150/(1-1.5) = 3,741/(-0.5) = -3,741 J (compression), |W| ≈ 3,372 J accounting for polytropic work formula
Q.13Hard
A heat engine operates between 600 K and 300 K reservoirs. It absorbs 5000 J from hot reservoir. For a Carnot engine operating between same temperatures, maximum work output would be:
For a van der Waals gas, the critical point is characterized by:
Answer: A
At critical point, both first and second derivatives of pressure with respect to volume are zero, marking the boundary of liquid-gas phase transition
Q.15Hard
A reversible process has entropy change ΔS_sys = -100 J/K. The entropy change of universe is:
Answer: C
For reversible process: ΔS_universe = ΔS_sys + ΔS_surr = 0. Since ΔS_sys = -100, ΔS_surr = +100, making total change zero
Q.16Hard
In a free expansion of ideal gas into vacuum, the entropy change of system is:
Answer: B
Free expansion is irreversible with ΔU = 0 and W = 0, so Q = 0. Volume increases, so S = nR ln(V_f/V_i) > 0
Q.17Hard
A monatomic ideal gas undergoes a cyclic process ABCA where: A→B is isothermal expansion, B→C is isochoric process, C→A is adiabatic compression. If at point A, P = 1 atm, V = 1 L, and T = 300 K, and the volume doubles from A to B, find the heat absorbed during the isothermal process.
Answer: C
For isothermal process of ideal gas: Q = nRT ln(V_f/V_i) = W. n = PV/RT = (101325 × 0.001)/(8.314 × 300) ≈ 0.0405 mol. Q = nRT ln(2) = 0.0405 × 8.314 × 300 × ln(2) ≈ 600 ln(2) J
Q.18Hard
For an ideal gas undergoing a polytropic process (PV^n = constant), the heat capacity is C = C_v + R/(1-n). For which value of n does the polytropic process become adiabatic?
Answer: B
For adiabatic process, Q = 0, so C = 0. This occurs when 1-n approaches infinity, which happens when n = γ. At n = γ, PV^γ = constant (adiabatic relation).
Q.19Hard
A gas sample at 300 K has an entropy of 200 J/K. When heated at constant pressure to 600 K, its entropy becomes:
Which thermodynamic process results in maximum work extraction from an ideal gas expanding from the same initial to final states?
Answer: B
For expansion between the same P-V states, isothermal process produces maximum work because W = nRT ln(V_f/V_i) is maximum when temperature is highest throughout the process.
