A refrigerator with COP = 4 requires 100 J of work per cycle. Heat removed from the cold reservoir is:

The correct answer is C: 400 J. COP = Q_c/W, where Q_c is heat removed from cold reservoir. Q_c = COP × W = 4 × 100 = 400 J.

Which of the following is NOT a state function?

The correct answer is C: Heat. Heat (Q) and work (W) are path functions, not state functions. They depend on the process, not just initial and final states.

In a cyclic process, the change in internal energy is:

The correct answer is C: Zero. In any cyclic process, system returns to initial state. Since U is a state function, ΔU = 0 for complete cycles.

For a reversible adiabatic process of an ideal diatomic gas (γ = 1.4), if pressure increases from 1 atm to 10 atm, the temperature ratio T_f/T_i is:

The correct answer is A: 10^0.286. For adiabatic process: T·P^(1-γ)/γ = constant, so T_f/T_i = (P_f/P_i)^((γ-1)/γ) = 10^(0.4/1.4) = 10^(2/7) ≈ 10^0.286

A particle moves along a circular path of radius 5 m with a constant speed of 20 m/s. What is the centripetal acceleration?

The correct answer is D: 80 m/s². Centripetal acceleration aᶜ = v²/r = (20)²/5 = 400/5 = 80 m/s²

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NEET Physics

Physics questions for NEET UG — Mechanics, Thermodynamics, Optics, Modern Physics.

72 Q 3 Topics Take Mock Test

Difficulty: All Easy Medium Hard 21–30 of 72

Topics in NEET Physics

All Mechanics & Laws of Motion 100 Thermodynamics 100 Optics 1

Q.21 Easy Thermodynamics

Which of the following is a path function?

A Internal energy

B Heat

C Enthalpy

D Gibbs free energy

Correct Answer: B. Heat

EXPLANATION

Heat and work are path functions because their values depend on the path followed during the process. Internal energy, enthalpy, and Gibbs free energy are state functions.

Test

Q.22 Easy Thermodynamics

A thermodynamic system undergoes a process where ΔU = -50 J and W = 30 J (work done by the system). Calculate the heat absorbed by the system.

A Q = -80 J

B Q = -20 J

C Q = 20 J

D Q = 80 J

Correct Answer: B. Q = -20 J

EXPLANATION

Using first law: ΔU = Q - W, so Q = ΔU + W = -50 + 30 = -20 J. Heat is released by the system.

Test

Q.23 Easy Thermodynamics

The maximum possible efficiency of a heat engine operating between 400 K and 500 K is:

A 11%

B 20%

C 25%

D 80%

Correct Answer: B. 20%

EXPLANATION

Maximum efficiency is Carnot efficiency: η_max = 1 - T_c/T_h = 1 - (400/500) = 1 - 0.8 = 0.2 = 20%.

Test

Q.24 Easy Thermodynamics

A heat engine absorbs 500 J of heat and does 200 J of work. The heat rejected is:

A 700 J

B 300 J

C 200 J

D 500 J

Correct Answer: B. 300 J

EXPLANATION

By energy conservation: Q_in = W + Q_out, so Q_out = Q_in - W = 500 - 200 = 300 J.

Test

Q.25 Easy Thermodynamics

For an isothermal process of an ideal gas, the relationship between pressure and volume is:

A PV = constant

B P/V = constant

C P + V = constant

D PV² = constant

Correct Answer: A. PV = constant

EXPLANATION

In isothermal process, T is constant. From ideal gas law PV = nRT, if T is constant, then PV = constant (Boyle's Law).

Test

Q.26 Easy Thermodynamics

Which process has the property that pressure remains constant?

A Isochoric

B Isobaric

C Isothermal

D Adiabatic

Correct Answer: B. Isobaric

EXPLANATION

Isobaric process is defined as a process at constant pressure. Prefix 'iso' means constant, and 'baric' refers to pressure.

Test

Q.27 Easy Thermodynamics

A Carnot engine operates between 300 K and 600 K. Its efficiency is:

A 25%

B 50%

C 75%

D 100%

Correct Answer: B. 50%

EXPLANATION

Carnot efficiency η = 1 - (T_cold/T_hot) = 1 - (300/600) = 1 - 0.5 = 0.5 = 50%.

Test

Q.28 Easy Thermodynamics

For a cyclic process, which of the following is always true?

A ΔU = 0 and Q = W

B ΔU ≠ 0 but Q = 0

C W = 0 but Q ≠ 0

D Both Q and W are zero

Correct Answer: A. ΔU = 0 and Q = W

EXPLANATION

In a cyclic process, the system returns to its initial state, so ΔU = 0. By first law: Q = ΔU + W = W.

Test

Q.29 Easy Thermodynamics

A gas undergoes an adiabatic compression. Which statement is correct?

A Temperature increases and internal energy decreases

B Temperature increases and internal energy increases

C Temperature decreases and internal energy increases

D Temperature and internal energy both decrease

Correct Answer: B. Temperature increases and internal energy increases

EXPLANATION

In adiabatic compression, Q=0. Work is done ON the gas (W<0), so ΔU = Q - W = -W > 0. Internal energy increases, causing temperature to rise.

Test

Q.30 Easy Thermodynamics

A gas expands from volume V to 2V at constant pressure P. The work done by the gas is:

A PV

B 2PV

C P(2V - V) = PV

D P/2V

Correct Answer: C. P(2V - V) = PV

EXPLANATION

Work in isobaric process: W = P(V_f - V_i) = P(2V - V) = PV

Test

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