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.
During an isobaric expansion of an ideal gas, the work done by the gas is 400 J. If pressure is constant at 2 atm, what is the change in volume? (1 atm = 101325 Pa)
Answer: B
W = PΔV. Here W = 400 J, P = 2 × 101325 = 202650 Pa. So ΔV = W/P = 202650400 ≈ 0.00197 m³
Q.25Medium
For a diatomic ideal gas undergoing an isothermal process, which quantity remains constant?
Answer: C
In an isothermal process, temperature is constant. For an ideal gas, internal energy depends only on temperature, so ΔU = 0. Pressure and volume change according to PV = constant.
Q.26Medium
Two samples of the same ideal gas at the same temperature have volumes V and 2V respectively. The ratio of their internal energies is:
Answer: D
Internal energy U = nCvT. Without knowing the number of moles in each sample, the ratio cannot be determined. Same temperature doesn't mean same internal energy.
Q.27Medium
A heat engine absorbs 1000 J of heat and rejects 600 J to the cold reservoir in one cycle. What is its efficiency?
The entropy change for a reversible isothermal process is given by:
Answer: C
For reversible isothermal process: ΔS = Q/T = nR ln(Vf/Vi) since Q = nRT ln(Vf/Vi) for ideal gas isothermal expansion.
Q.29Medium
Which process would result in the maximum work output from an ideal gas expansion between two fixed pressures?
Answer: A
For expansion between fixed initial and final states, isothermal process yields maximum work because the gas maintains maximum pressure throughout the expansion compared to adiabatic or other processes.
Q.30Medium
A cylinder with a movable piston contains 2 moles of ideal gas at 300 K and 1 atm. When heated at constant pressure to 600 K, what is the work done by the gas? (R = 8.314 J/mol·K)
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.34Hard
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.35Hard
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.36Hard
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.37Hard
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.38Easy
A system absorbs 500 J of heat and performs 200 J of work on the surroundings. What is the change in internal energy of the system?
Answer: A
By first law of thermodynamics: ΔU = Q − W = 500 − 200 = 300 J
Q.39Easy
An ideal gas undergoes an adiabatic process. If the gas is compressed, which of the following is true?
Answer: B
In adiabatic compression (Q = 0), work is done on the gas (W < 0). By first law: ΔU = Q − W = 0 − W > 0. Since ΔU ∝ ΔT for ideal gas, temperature increases.
Q.40Easy
What is the efficiency of a Carnot engine operating between 400 K and 300 K?
