A magnetic field B is applied perpendicular to a current-carrying wire of length L carrying current I. The magnetic force on the wire is F = BIL sin θ. What is the angle θ when the force is maximum?

The correct answer is A: 90°. Force F = BIL sin θ is maximum when sin θ = 1, which occurs at θ = 90°. This means the wire must be perpendicular to the magnetic field direction.

The work function of a metal is 3 eV. What is the threshold frequency for photoelectric emission?

The correct answer is B: 7.2×10¹⁴ Hz. Threshold frequency: f₀ = W/h = 3/(6.63×10⁻³⁴) = 4.5×10¹⁴ Hz. Wait, recalculating: f₀ = 3 eV/(4.14×10⁻¹⁵ eVs) ≈ 7.2×10¹⁴ Hz.

In pair production, a photon with energy 3 MeV converts near a nucleus into an electron-positron pair. The rest mass energy of electron/positron is 0.51 MeV. The excess energy appears as:

The correct answer is A: Kinetic energy of electron and positron. In pair production: E_photon = 2m_e c² + KE_total. Excess = 3 - 2(0.51) = 1.98 MeV becomes kinetic energy of the pair.

In FinFET technology (used in modern 7nm and 5nm nodes), multiple gates are used to:

The correct answer is B: Better control the channel and reduce leakage current. FinFETs employ multiple gates (typically 3 gates) wrapping around a thin silicon fin, providing superior electrostatic control of the channel, reducing short-channel effects and subthreshold swing.

A particle of mass m undergoes motion with velocity v(t) = 3t² + 2t (in m/s). What is the net force on the particle when t = 2 s, if m = 2 kg?

The correct answer is A: 28 N. Acceleration a = dv/dt = 6t + 2. At t=2s: a = 6(2) + 2 = 14 m/s². Force F = ma = 2 × 14 = 28 N

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JEE Physics
Thermodynamics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

23 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–10 of 23

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.1 Hard Thermodynamics

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?

A P₀(V₂ - V₁) - k(V₂² - V₁²)/2

B ∫P dV = RT ln(V₂/V₁)

C P₀V₂ - P₀V₁

D k(V₂ - V₁)

Correct Answer: A. P₀(V₂ - V₁) - k(V₂² - V₁²)/2

EXPLANATION

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

Test

Q.2 Hard Thermodynamics

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.

A All ice melts; final temperature ≈ 46°C

B Some ice remains; final temperature = 0°C

C All ice melts; final temperature ≈ 32°C

D All water freezes; final temperature < 0°C

Correct Answer: A. All ice melts; final temperature ≈ 46°C

EXPLANATION

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

Test

Q.3 Hard Thermodynamics

Which thermodynamic process results in maximum work extraction from an ideal gas expanding from the same initial to final states?

A Isobaric process

B Isothermal process

C Isochoric process

D Adiabatic process

Correct Answer: B. Isothermal process

EXPLANATION

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.

Test

Q.4 Hard Thermodynamics

A gas sample at 300 K has an entropy of 200 J/K. When heated at constant pressure to 600 K, its entropy becomes:

A 200 + R ln(2) J/K

B 200 + nC_p ln(2) J/K

C 200 ln(2) J/K

D 400 J/K

Correct Answer: B. 200 + nC_p ln(2) J/K

EXPLANATION

At constant pressure: ΔS = nC_p ln(T_f/T_i) = nC_p ln(600/300) = nC_p ln(2). Final entropy = 200 + nC_p ln(2) J/K

Test

Q.5 Hard Thermodynamics

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?

A n = 1

B n = γ

C n = 0

D n = ∞

Correct Answer: B. n = γ

EXPLANATION

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).

Test

Q.6 Hard Thermodynamics

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.

A 150 ln(2) J

B 300 ln(2) J

C 600 ln(2) J

D 900 ln(2) J

Correct Answer: C. 600 ln(2) J

EXPLANATION

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

Test

Q.7 Hard Thermodynamics

In a free expansion of ideal gas into vacuum, the entropy change of system is:

A Zero

B Positive

C Negative

D Depends on initial temperature

Correct Answer: B. Positive

EXPLANATION

Free expansion is irreversible with ΔU = 0 and W = 0, so Q = 0. Volume increases, so S = nR ln(V_f/V_i) > 0

Test

Q.8 Hard Thermodynamics

A reversible process has entropy change ΔS_sys = -100 J/K. The entropy change of universe is:

A Negative

B Positive

C Zero

D Cannot be determined

Correct Answer: C. Zero

EXPLANATION

For reversible process: ΔS_universe = ΔS_sys + ΔS_surr = 0. Since ΔS_sys = -100, ΔS_surr = +100, making total change zero

Test

Q.9 Hard Thermodynamics

For a van der Waals gas, the critical point is characterized by:

A dP/dV = 0 and d²P/dV² = 0

B dP/dV > 0

C Temperature at which gas liquefies

D Point where all intermolecular forces vanish

Correct Answer: A. dP/dV = 0 and d²P/dV² = 0

EXPLANATION

At critical point, both first and second derivatives of pressure with respect to volume are zero, marking the boundary of liquid-gas phase transition

Test

Q.10 Hard Thermodynamics

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:

A 2000 J

B 2500 J

C 3000 J

D 3500 J

Correct Answer: B. 2500 J

EXPLANATION

Carnot efficiency = 1 - 300/600 = 0.5; W_max = η × Q_h = 0.5 × 5000 = 2500 J

Test

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