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.
Two identical blocks of metal at different temperatures are brought into thermal contact in an isolated system. The process is:
Answer: C
Heat transfer between objects at different temperatures is irreversible. For an isolated system, ΔS_universe = ΔS_system > 0 (irreversible process), not equal to zero.
Q.43Medium
A gas expands from 1 L to 5 L against a constant external pressure of 2 atm. The work done by the gas is approximately:
The molar heat capacity of a diatomic ideal gas at constant pressure is (R = 8.314 J/mol·K):
Answer: B
For diatomic gas: Cv = (25)R, and Cp = Cv + R = (25)R + R = (27)R. This is at room temperature where vibration is not excited.
Q.45Medium
A cyclic process ABCA is shown on a P-V diagram where AB is isothermal expansion, BC is adiabatic compression, and CA is isochoric process. Which statement is correct?
Answer: A
In a complete cycle returning to initial state, ΔU = 0, so Q = W. For expansion-dominated processes in a typical cycle, W > 0 and Q > 0.
Q.46Medium
A heat engine absorbs 1000 J from a hot reservoir and rejects 600 J to a cold reservoir in one cycle. Its efficiency is:
A reversible heat engine operates between two thermal reservoirs. If the temperature of the cold reservoir decreases while hot reservoir temperature remains constant, the maximum efficiency:
Answer: A
Carnot efficiency η = 1 − (T_c/T_h). As T_c decreases while T_h remains constant, the ratio T_c/T_h decreases, so η increases.
Q.48Medium
Which process results in zero change of entropy for an ideal gas?
Answer: B
For a reversible adiabatic process, dQ = 0, so dS = dQ/T = 0, hence ΔS = 0. This is also called isentropic process.
Q.49Medium
In a throttling process (Joule-Thomson expansion), which of the following remains constant?
Answer: D
Throttling is an irreversible, adiabatic process where enthalpy remains constant (H_initial = H_final). Temperature and pressure both change, and internal energy remains nearly constant only for ideal gases.
Q.50Hard
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.51Hard
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.52Hard
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.53Easy
A heat pump delivers 5000 J of heat to a house while consuming 1500 J of work. Its coefficient of performance is:
Answer: A
For heat pump: COP = Q_h/W = 15005000 = 3.33. This indicates the pump delivers 3.33 J of heat for every 1 J of work input.
Q.54Easy
A diatomic ideal gas expands isothermally from volume V₁ to 2V₁. If the initial pressure is P₀, what is the work done by the gas?
Answer: A
For isothermal process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁) = P₀V₁ ln(2)
Q.55Easy
An ideal gas undergoes adiabatic compression from state (P₁, V₁, T₁) to (P₂, V₂, T₂). Which relation is correct?
Answer: A
For adiabatic process: TV^(γ-1) = constant, therefore T₁V₁^(γ-1) = T₂V₂^(γ-1)
Q.56Easy
The internal energy of an ideal gas depends on:
Answer: C
Internal energy U of ideal gas is U = nCᵥT, which depends only on temperature, not on pressure or volume individually
Q.57Easy
A Carnot engine operates between 400 K and 300 K. Its maximum efficiency is:
Two kilograms of water at 100°C is converted to steam at 100°C at 1 atm pressure. The change in entropy is (Latent heat of vaporization = 2.26 × 10⁶ J/kg):
