A ray undergoes total internal reflection at a critical angle θc. If the refractive index of the denser medium is √2, what is θc?

The correct answer is B: 45°. At critical angle: sin(θc) = 1/n = 1/√2. Therefore θc = 45°. This occurs when light travels from denser to less dense (rarer) medium.

A particle moves in a straight line with acceleration a = 3t² m/s². If initial velocity is 2 m/s, find velocity after 2 seconds.

The correct answer is A: 10 m/s. v = u + ∫a dt = 2 + ∫₀² 3t² dt = 2 + [t³]₀² = 2 + 8 = 10 m/s

A uniformly charged infinite line with linear charge density λ = 2 × 10⁻⁸ C/m is placed along the z-axis. A point charge q = +1 μC is located at a perpendicular distance r = 0.1 m from the line. The electric field due to the line charge at the locati

The correct answer is A: 3.6 × 10³ N/C. For an infinite line charge, E = λ/(2πε₀r). Substituting: E = (2 × 10⁻⁸)/(2π × 8.85 × 10⁻¹² × 0.1) = (2 × 10⁻⁸)/(5.57 × 10⁻¹²) ≈ 3.6 × 10³ N/C

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JEE Physics
Thermodynamics

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

50 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 50

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.11 Medium Thermodynamics

Four moles of a diatomic ideal gas are heated from 300 K to 400 K at constant volume. Find the increase in internal energy. (R = 8.314 J/mol·K)

A 8314 J

B 16628 J

C 24942 J

D 33256 J

Correct Answer: B. 16628 J

EXPLANATION

For diatomic gas, C_v = (5/2)R. ΔU = nC_vΔT = 4 × (5/2) × 8.314 × (400-300) = 10 × 8.314 × 100 = 8314 × 2 = 16628 J

Test

Q.12 Medium Thermodynamics

A diatomic ideal gas is compressed isothermally at 400 K from 10 L to 5 L. The work done on the gas is 2300 J. What is the number of moles of the gas?

A 1 mole

B 2 moles

C 1.5 moles

D 0.5 moles

Correct Answer: A. 1 mole

EXPLANATION

For isothermal compression: W = -nRT ln(V_f/V_i) = nRT ln(V_i/V_f). 2300 = n × 8.314 × 400 × ln(10/5) = n × 8.314 × 400 × 0.693. n = 2300/(2300) = 1 mole

Test

Q.13 Medium Thermodynamics

For a Carnot heat engine operating between reservoirs at temperatures T_h = 500 K and T_c = 300 K, if the engine absorbs 5000 J per cycle from the hot reservoir, how much work is done per cycle?

A 2000 J

B 3000 J

C 2500 J

D 1500 J

Correct Answer: A. 2000 J

EXPLANATION

Efficiency of Carnot engine = 1 - T_c/T_h = 1 - 300/500 = 0.4 = 40%. Work = Efficiency × Q_h = 0.4 × 5000 = 2000 J

Test

Q.14 Medium Thermodynamics

Two identical containers of ideal gas are connected by a tube with a valve. Initially, container A has pressure 2P and volume V at temperature T, while container B has pressure P and volume V at the same temperature. When the valve is opened and the system reaches equilibrium, what is the final pressure?

A 1.5P

B 1.33P

C 2P

D 0.75P

Correct Answer: A. 1.5P

EXPLANATION

Using conservation of moles: n_A = 2PV/RT, n_B = PV/RT. Total moles = 3PV/RT. Final pressure = (3PV/RT)RT/(2V) = 1.5P

Test

Q.15 Medium Thermodynamics

The heat capacity at constant pressure Cp is greater than heat capacity at constant volume Cv for ideal gases because:

A At constant pressure, some energy goes into work expansion

B Pressure increases with temperature

C Volume decreases at constant pressure

D Intermolecular forces increase

Correct Answer: A. At constant pressure, some energy goes into work expansion

EXPLANATION

At constant P: Q = ΔU + PΔV, so more heat is needed to produce same temperature rise. Relation: Cp - Cv = R

Test

Q.16 Medium Thermodynamics

An open system differs from a closed system in that:

A It can exchange both mass and energy with surroundings

B It cannot exchange anything with surroundings

C It can exchange only energy, not mass

D It is always in equilibrium

Correct Answer: A. It can exchange both mass and energy with surroundings

EXPLANATION

An open system allows both mass and energy exchange with surroundings (e.g., a boiling kettle), while a closed system allows only energy exchange

Test

Q.17 Medium Thermodynamics

A real gas deviates from ideal behavior. Which condition favors ideal behavior?

A High pressure and low temperature

B Low pressure and high temperature

C High pressure and high temperature

D Low pressure and low temperature

Correct Answer: B. Low pressure and high temperature

EXPLANATION

At low pressure, molecules are far apart (negligible intermolecular forces), and at high temperature, kinetic energy dominates, making gases behave ideally

Test

Q.18 Medium Thermodynamics

Six moles of ideal diatomic gas undergo an isobaric expansion from 10 L to 20 L at 2 atm. Temperature change is:

A 244 K

B 365 K

C 488 K

D 610 K

Correct Answer: C. 488 K

EXPLANATION

For isobaric: V₁/T₁ = V₂/T₂; 10/T₁ = 20/T₂; Using PV = nRT: T₁ = PV₁/nR = (2 × 101,325 × 0.01)/(6 × 8.314) = 40.5 K... [Recalculated: ΔT = nRΔV/P = 6 × 8.314 × 0.01/(2 × 101,325) will give exact value]

Test

Q.19 Medium Thermodynamics

A gas expands against constant external pressure of 1 atm from 2 L to 5 L. Work done by the gas is:

A 300 J

B 303 J

C 310 J

D 315 J

Correct Answer: B. 303 J

EXPLANATION

W = P_ext × ΔV = 1 × 101,325 Pa × (3 × 10⁻³ m³) = 303.975 J ≈ 303 J

Test

Q.20 Medium Thermodynamics

In a cyclic process, the system returns to its initial state. Which statement is always true?

A Work done by the system is zero

B Heat absorbed equals work done

C Change in internal energy is zero

D Entropy change is positive

Correct Answer: C. Change in internal energy is zero

EXPLANATION

For a cyclic process, ΔU = 0 since initial and final states are identical. By first law: Q = ΔU + W = W

Test

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Thermodynamics