Home › Subjects › JEE Physics

JEE Physics

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

434 Q 9 Topics Take Test

Difficulty: All Easy Medium Hard 341–350 of 434

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.341 Medium Thermodynamics

For a Carnot refrigerator operating between 250 K (cold reservoir) and 300 K (hot reservoir), the coefficient of performance (COP) is:

A 5

B 6

C 1.2

D 0.2

Correct Answer: A. 5

EXPLANATION

For Carnot refrigerator: COP = T_c/(T_h - T_c) = 250/(300-250) = 250/50 = 5

Test

Q.342 Medium Thermodynamics

A thermally insulated container holds 2 moles of oxygen gas at 300 K. A heating element supplies 5000 J of heat to the gas. What is the final temperature? (C_v for O₂ = 5R/2, R = 8.314 J/mol·K)

A 360 K

B 420 K

C 380 K

D 400 K

Correct Answer: C. 380 K

EXPLANATION

Q = nC_vΔT. 5000 = 2 × (5/2) × 8.314 × ΔT. ΔT = 5000/(5 × 8.314) = 5000/41.57 ≈ 120.2 K. Final T = 300 + 120.2 ≈ 380 K

Test

Q.343 Medium Thermodynamics

A gas undergoes a process where PV² = constant. If the gas expands from volume V to 2V, what is the ratio of final to initial pressure?

A 1/4

B 1/2

C 1/8

D 2

Correct Answer: A. 1/4

EXPLANATION

From PV² = constant: P₁V₁² = P₂V₂². P₂/P₁ = (V₁/V₂)² = (V/2V)² = 1/4

Test

Q.344 Medium Thermodynamics

For a reversible isothermal process of an ideal gas, the entropy change of the universe is:

A Positive

B Negative

C Zero

D Equal to entropy change of surroundings

Correct Answer: C. Zero

EXPLANATION

In a reversible process, total entropy change = 0. ΔS_system + ΔS_surroundings = 0. This is the definition of a reversible process.

Test

Q.345 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.346 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.347 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.348 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.349 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.350 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

1…33 34 35 36 37…44

All Subjects & Topics

Govt. Exams

Entrance Exams

Teacher Exams

Subjects

Quick Links

JEE Physics