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.
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.22Hard
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.23Hard
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.24Hard
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.26Hard
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.27Hard
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.28Hard
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.29Hard
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.30Hard
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.
Q.32Hard
A 2 kg mass of ice at 0°C is mixed with 5 kg of water at 80°C in a thermally insulated container. If latent heat of fusion = 3.36 × 10⁵ J/kg and specific heat of water = 4200 J/kg·K, determine the final state of the system.
Answer: A
Heat available from water cooling from 80°C to 0°C: Q = 5 × 4200 × 80 = 1.68 × 10⁶ J. Heat needed to melt ice: Q = 2 × 3.36 × 10⁵ = 6.72 × 10⁵ J. Since 1.68 × 10⁶ > 6.72 × 10⁵, all ice melts. Remaining heat: 1.68 × 10⁶ - 6.72 × 10⁵ = 1.008 × 10⁶ J raises temperature of 7 kg water: ΔT = 1.008 × 10⁶/(7 × 4200) ≈ 34.3°C → final temp ≈ 34°C
Q.33Hard
A gas sample undergoes a process where pressure decreases linearly with volume: P = P₀ - kV, where k is a constant. For 1 mole of ideal gas at constant temperature, what is the work done when volume changes from V₁ to V₂?
Answer: A
Work done: W = ∫P dV = ∫(P₀ - kV) dV from V₁ to V₂ = [P₀V - kV²/2] from V₁ to V₂ = P₀(V₂ - V₁) - k(V₂² - V₁²)/2
Q.34Hard
A charged soap bubble of radius R has surface charge density σ. The excess pressure inside the bubble due to electrostatic force is:
Answer: A
Electrostatic pressure = ε₀E²/2 at surface. E = σ/ε₀ just outside. Excess pressure p = σ²/(2ε₀)
Q.35Hard
A point charge q is placed at distance r from an infinite grounded conducting plane. The force on the charge is:
Answer: A
By method of images, image charge -q is at distance r behind plane. Total distance = 2r. F = kq²/(2r)² = q²/(16πε₀r²)
Q.36Hard
The self-energy of a uniformly charged sphere of radius R and total charge Q is:
Answer: A
Self-energy of uniformly charged sphere: U = 3Q²/(20πε₀R) = 3kQ²/(5R)
Q.37Hard
If potential varies as V = 3x² + 4y in a region, the electric field at point (1,2) is:
Answer: A
E = -∇V = -(∂V/∂x i + ∂V/∂y j) = -(6x i + 4j) = -6i - 4j at (1,2)
Q.38Hard
A charged rod of length L with linear charge density λ is placed along the x-axis. The electric field at a point on the perpendicular bisector at distance y from the center is:
Answer: A
By symmetry, perpendicular components cancel. Axial component: E = λ/(2πε₀y) × L/√(L²/4 + y²) = λL/(2πε₀y√(L²/4 + y²))
Q.39Hard
Two point charges Q and -Q are at distance 2d apart. The potential difference between two points on the perpendicular bisector at distances x and 2x from the midpoint is proportional to:
Answer: C
Using potential superposition and the dipole configuration, ΔV = 2kQd(1/x - 21x) for points on the perpendicular bisector.
Q.40Hard
An insulating rod of length L is uniformly charged with total charge Q. The electric potential at a point on the axis at distance x from one end is:
Answer: A
Integrating potential contributions from small elements: V = (kQ/L)ln[(x+L)/x].
