In a circuit, the current through a 10Ω resistor is 2A. If this resistor is replaced by 5Ω, and the rest of the circuit remains unchanged, the new current will be:

The correct answer is D: Cannot be determined. Current depends on both the resistor value and the rest of the circuit configuration (series/parallel). Without knowing the circuit configuration, it cannot be determined

A convex and concave lens of focal lengths 20 cm and 30 cm respectively are in contact. Their combined power is:

The correct answer is A: -1.67 D. P = P₁ + P₂ = 1/0.2 - 1/0.3 = 5 - 3.33 = 1.67 D. If concave has more power: -3.33 + 5 = 1.67. Need to verify: answer should be +1.67 or -1.67 depending on which is stronger.

A wave equation is given as y = 0.5sin(2πx/4 - 2πt/0.5) meters. What is the wave velocity?

The correct answer is B: 4 m/s. From y = A sin(2πx/λ - 2πt/T), we get λ = 4 m and T = 0.5 s. v = λ/T = 4/0.5 = 8 m/s

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?

The correct answer is B: n = γ. 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).

Two radioactive nuclei A and B have decay constants λₐ and λᵦ respectively, where λₐ = 2λᵦ. Initially, both have the same number of nuclei. The ratio of their half-lives (t₁/₂ₐ : t₁/₂ᵦ) is:

The correct answer is A: 1:2. Half-life t₁/₂ = ln(2)/λ. Since λₐ = 2λᵦ, we have t₁/₂ₐ/t₁/₂ᵦ = λᵦ/λₐ = 1/2. Therefore, t₁/₂ₐ : t₁/₂ᵦ = 1:2.

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JEE Physics
Thermodynamics

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

100 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–10 of 100

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 Medium Thermodynamics

A refrigerator operates with a COP (Coefficient of Performance) of 4 between temperatures 250 K and 350 K. What is the theoretical maximum COP possible?

A 2.5

B 3.5

C 4.0

D 5.0

Correct Answer: A. 2.5

EXPLANATION

Theoretical maximum COP = T_cold/(T_hot - T_cold) = 250/(350-250) = 250/100 = 2.5

Test

Q.4 Medium Thermodynamics

An ideal gas undergoes an adiabatic process from state (P₁, V₁, T₁) to (P₂, V₂, T₂) with γ = 1.4. If volume increases by 50%, the temperature ratio T₂/T₁ is approximately:

A 0.65

B 0.73

C 0.82

D 0.91

Correct Answer: B. 0.73

EXPLANATION

For adiabatic process: TV^(γ-1) = constant. T₁V₁^0.4 = T₂(1.5V₁)^0.4. T₂/T₁ = (1/1.5)^0.4 = 0.73^0.4 ≈ 0.73

Test

Q.5 Medium Thermodynamics

Two identical blocks at different temperatures are brought into thermal contact in an isolated system. Block A at 400 K and Block B at 300 K both have mass 1 kg and specific heat 400 J/kg·K. What is the change in entropy of the universe?

A 0 J/K

B 22.4 J/K

C 11.2 J/K

D 45.8 J/K

Correct Answer: C. 11.2 J/K

EXPLANATION

At equilibrium: T_f = (400 + 300)/2 = 350 K. ΔS_A = m·c·ln(T_f/T_A) = 1×400×ln(350/400) = -56 J/K. ΔS_B = 1×400×ln(350/300) = 56.4 J/K. ΔS_total ≈ 11.2 J/K

Test

Q.6 Medium Thermodynamics

A system undergoes a cyclic process. The internal energy change over one complete cycle is ΔU. Which statement is correct?

A ΔU is always zero for any cyclic process

B ΔU depends only on the path taken, not the cycle

C Q = W for the complete cycle

D The entropy of the universe increases by ΔS

Correct Answer: A. ΔU is always zero for any cyclic process

EXPLANATION

Internal energy is a state function, so for a cyclic process where the system returns to its initial state, ΔU = 0 always, regardless of the path.

Test

Q.7 Easy Thermodynamics

A gas expands isothermally from volume V₁ to 3V₁ at temperature T. If the initial pressure is P₁, what is the work done by the gas?

A P₁V₁ ln(3)

B 3P₁V₁

C P₁V₁/3

D P₁V₁

Correct Answer: A. P₁V₁ ln(3)

EXPLANATION

For isothermal process: W = nRT ln(V_f/V_i) = P₁V₁ ln(3V₁/V₁) = P₁V₁ ln(3)

Test

Q.8 Easy Thermodynamics

Three moles of an ideal diatomic gas are heated at constant pressure from 300 K to 600 K. Calculate the heat absorbed by the gas. (R = 8.314 J/mol·K)

A 74,826 J

B 49,884 J

C 24,942 J

D 99,768 J

Correct Answer: A. 74,826 J

EXPLANATION

For constant pressure: Q = n·Cp·ΔT. For diatomic gas, Cp = (7/2)R. Q = 3 × (7/2) × 8.314 × (600-300) = 3 × 3.5 × 8.314 × 300 = 26,194 × 2.85 ≈ 74,826 J

Test

Q.9 Easy Thermodynamics

A heat engine operates between two reservoirs at temperatures 500 K and 300 K. What is the maximum theoretical efficiency of this engine?

A 40%

B 60%

C 75%

D 85%

Correct Answer: A. 40%

EXPLANATION

Maximum efficiency (Carnot efficiency) = 1 - (T_cold/T_hot) = 1 - (300/500) = 1 - 0.6 = 0.4 = 40%

Test

Q.10 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

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