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:

The correct answer is B: 0.73. 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

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?

The correct answer is A: 1 mole. 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

A force of 10 N acts on a mass for 5 seconds, changing its momentum by:

The correct answer is C: 50 kg·m/s. By impulse-momentum theorem, Δp = F·Δt = 10 × 5 = 50 kg·m/s

Two coherent sources vibrate in phase and produce waves of wavelength 2 cm. If they are separated by 6 cm, what is the phase difference between the waves at a point equidistant from both sources?

The correct answer is A: 0°. At equidistant points, path difference = 0, hence phase difference = 0°. This leads to constructive interference

The resistance of a wire at 0°C is R₀. If its temperature coefficient is α, its resistance at temperature T is:

The correct answer is A: R₀(1 + αT). R_T = R₀(1 + αT), where T is temperature rise from reference point (0°C in this case)

Home

Home › Subjects › JEE Physics › Thermodynamics

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 21–23 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.21 Hard Thermodynamics

Two bodies at temperatures T₁ = 400 K and T₂ = 300 K are brought into thermal contact. If entropy change of universe is 0.575 J/K and heat capacity of both bodies is 1000 J/K, what is the final equilibrium temperature? (Assume no heat loss to surroundings)

A T_f = 347.5 K

B T_f = 350 K

C T_f = 352.5 K

D T_f = 345 K

Correct Answer: B. T_f = 350 K

EXPLANATION

Heat lost by body 1: Q = C(T₁ - T_f) = 1000(400 - T_f). Heat gained by body 2: Q = 1000(T_f - 300). ΔS_univ = C ln(T_f/T₁) + C ln(T_f/T₂) = 1000[ln(T_f/400) + ln(T_f/300)] = 0.575. Solving: T_f = 350 K

Test

Q.22 Hard Thermodynamics

A gas undergoes a cyclic process ABCA where AB is isothermal, BC is adiabatic, and CA is isochoric. If work is done on the gas in the cycle, what can be concluded?

A Net heat flows out of the gas

B Internal energy increases

C The process is impossible

D Entropy of the gas decreases

Correct Answer: A. Net heat flows out of the gas

EXPLANATION

If W_net < 0 (work done on gas), then from first law: ΔU_cycle = 0 = Q - W, so Q = W < 0, meaning net heat flows out

Test

Q.23 Hard Thermodynamics

Three identical conducting rods are arranged in series between two heat reservoirs at 100°C and 0°C. At steady state, what is the temperature at the junction between the second and third rod?

A 33.3°C

B 50°C

C 66.7°C

D 75°C

Correct Answer: A. 33.3°C

EXPLANATION

In series arrangement with identical rods, temperature difference is equally distributed. ΔT_total = 100°C, so ΔT per rod = 100/3 = 33.3°C. Second junction = 100 - 2(33.3) = 33.3°C

Test

1 2 3

All Subjects & Topics

Govt. Exams

Entrance Exams

Teacher Exams

Subjects

Quick Links

JEE Physics Thermodynamics

JEE Physics
Thermodynamics