A composite Carnot engine and refrigerator system operates between three reservoirs. If the engine operates between 500 K and 300 K, and the refrigerator between 300 K and 200 K, what is the theoretical relationship for reversible operation?

The correct answer is A: Q_h,engine/T_h,engine = Q_c,ref/T_c,ref. For reversible processes on the middle reservoir (300K), entropy change must be zero: Q_h,engine/500 = (Q_h,engine - W_engine)/300 and similar relations apply to refrigerator at 300K.

Two reservoirs are at temperatures 400 K and 300 K. A reversible Carnot engine operates between them. If work output is 100 J per cycle, what is the heat input from the hot reservoir?

The correct answer is C: 400 J. For Carnot engine: η = 1 - Tc/Th = 1 - 300/400 = 0.25. W = ηQh, so 100 = 0.25 × Qh, Qh = 400 J

A body of mass 2 kg experiences forces of 3 N and 4 N at right angles. The net acceleration is:

The correct answer is C: 2.5 m/s². Net force = √(3² + 4²) = 5 N. Using F = ma, a = 5/2 = 2.5 m/s²

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

The correct answer is B: 405 J. W = P_ext × ΔV = 1 atm × (5-1) L = 4 L·atm = 4 × 101.325 = 405 J (positive, work done by gas).

A block of mass m slides down a smooth inclined plane of angle θ. If the height of the plane is h and its length is L, what is the acceleration along the incline?

The correct answer is A: g sin(θ). Component of gravity along smooth incline = g sin(θ). Since there is no friction, a = g sin(θ)

Home

Home › Subjects › NEET Physics

NEET Physics

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

97 Q 3 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 97

Topics in NEET Physics

All Mechanics & Laws of Motion 100 Thermodynamics 100 Optics 1

Q.11 Medium Thermodynamics

Two identical gases at the same temperature and pressure occupy different volumes. Which thermodynamic property must be the same for both?

A Internal energy

B Entropy

C Heat capacity

D Chemical potential per molecule

Correct Answer: D. Chemical potential per molecule

EXPLANATION

At same T and P, the chemical potential per molecule is identical. Internal energy and entropy depend on amount (volume), heat capacity depends on mass.

Test

Q.12 Medium Thermodynamics

For a reaction where ΔH = -200 kJ/mol and ΔS = 50 J/mol·K at 300 K, calculate ΔG:

A -185 kJ/mol

B -215 kJ/mol

C 15 kJ/mol

D -200 kJ/mol

Correct Answer: A. -185 kJ/mol

EXPLANATION

ΔG = ΔH - TΔS = -200 - (300)(50/1000) = -200 - 15 = -185 kJ/mol

Test

Q.13 Medium Thermodynamics

A diatomic gas undergoes free expansion into a vacuum. The entropy of the universe:

A Decreases

B Remains unchanged

C Increases

D First increases then decreases

Correct Answer: C. Increases

EXPLANATION

Free expansion is an irreversible process. For the isolated system, ΔS_universe > 0. The gas entropy increases as volume increases.

Test

Q.14 Medium Thermodynamics

When a substance changes phase from solid to liquid at its melting point at constant temperature and pressure, which is true?

A ΔG is positive

B ΔG is zero

C ΔG is negative

D ΔS decreases

Correct Answer: B. ΔG is zero

EXPLANATION

At equilibrium phase transition (melting point), ΔG = 0. This is the condition for phase equilibrium at constant T and P.

Test

Q.15 Medium Thermodynamics

For a monatomic ideal gas, the ratio Cp/Cv is:

A 1.2

B 1.33

C 1.4

D 1.67

Correct Answer: D. 1.67

EXPLANATION

For monatomic gas: Cv = (3/2)R, Cp = (5/2)R. Therefore γ = Cp/Cv = (5/2)/(3/2) = 5/3 ≈ 1.67

Test

Q.16 Medium Thermodynamics

The coefficient of performance (COP) of a refrigerator is 5. If 100 J of work is done on it, how much heat is removed from the cold reservoir?

A 20 J

B 500 J

C 600 J

D 400 J

Correct Answer: B. 500 J

EXPLANATION

COP = Q_cold/W. Therefore, Q_cold = COP × W = 5 × 100 = 500 J

Test

Q.17 Medium Thermodynamics

A gas undergoes compression where both pressure and temperature increase. This process is most likely:

A Isobaric

B Isochoric

C Isothermal

D Adiabatic

Correct Answer: D. Adiabatic

EXPLANATION

In adiabatic compression (q=0), work is done on gas, increasing both P and T without heat exchange. Other processes would have different P-T relationships.

Test

Q.18 Medium Thermodynamics

In an isothermal process of an ideal gas, the work done by the gas is:

A Zero

B nRT ln(V_f/V_i)

C Depends only on temperature change

D Equal to heat absorbed

Correct Answer: B. nRT ln(V_f/V_i)

EXPLANATION

For isothermal process: W = nRT ln(V_f/V_i) = nRT ln(P_i/P_f). Since ΔU = 0, q = W.

Test

Q.19 Medium Thermodynamics

Which statement about entropy is correct for an isolated system?

A Entropy always decreases

B Entropy always increases or remains constant

C Entropy remains constant for all processes

D Entropy has no defined value

Correct Answer: B. Entropy always increases or remains constant

EXPLANATION

Second law of thermodynamics: For an isolated system, ΔS ≥ 0 (entropy increases for irreversible processes, constant for reversible).

Test

Q.20 Medium Thermodynamics

The molar heat capacity of a gas at constant pressure (Cp) is greater than at constant volume (Cv) because:

A Pressure is constant at higher values

B Additional energy is needed to do work against external pressure

C Volume is always larger than pressure

D Temperature changes are larger at constant pressure

Correct Answer: B. Additional energy is needed to do work against external pressure

EXPLANATION

At constant pressure, supplied heat does two things: increases internal energy and does expansion work. Hence Cp > Cv by amount R.

Test

1 2 3 4…10

All Subjects & Topics

Govt. Exams

Entrance Exams

Teacher Exams

Subjects

Quick Links

NEET Physics